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CHAPTER – 14 Electric Current and Its Effects | CLASS 7TH | NCERT SCIENCE IMPORTANT QUESTIONS & MCQS | EDUGROWN

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Get Chapter Wise MCQ Questions for Class 7 Science with Answers prepared here according to the latest CBSE syllabus and NCERT curriculum. Students can practice CBSE Class 7 Science MCQs Multiple Choice Questions with Answers to score good marks in the examination. Students can also visit the most accurate and elaborate NCERT Solutions for Class 7Science. Every question of the textbook has been answered here.

Chapter - 14 Electric Current and Its Effects

Question 1.
In making a battery
(a) positive terminal of one cell is connected to the negative terminal of the next cell
(b) positive terminal of one cell is connected to the positive terminal of the next cell
(c) negative terminal of one cell is connected to the negative terminal of the next cell
(d) none of the above

Answer

Answer: (a) positive terminal of one cell is connected to the negative terminal of the next cell

Question 2.
Where can the key or switch be placed in the circuit?
(a) Left side of the battery
(b) Right side of the battery
(c) Can be placed anywhere in the circuit
(d) Near the positive terminal of the bulb

Answer

Answer: (c) Can be placed anywhere in the circuit

Question 3.
Which one of the following is based on the heating effect of current?
(a) Geyser
(b) Hair dryer
(c) Immersion rod
(d) All of these

Answer

Answer: (d) All of these

Question 4.
The coil of wire contained in an electric heater is known as
(a) component
(b) element
(c) circuit
(d) spring

Answer

Answer: (b) element

Question 5.
The amount of heat produced in a wire depends on
(a) material
(b) length
(c) thickness
(d) all of these

Answer

Answer: (d) all of these

Question 6.
Which mark is necessary on electric appliances?
(a) AGMARK
(b) ISI
(c) FICCI
(d) KSK

Answer

Answer: (b) ISI

Question 7.
When an electric current flows through a copper wire AB as shown in Figure, the wire

(a) deflects a magnetic needle placed near it
(b) becomes red hot
(c) gives electric shock
(d) behaves like a fuse

Answer

Answer: (a) deflects a magnetic needle placed near it

Question 8.
Choose the statement which is not correct in the case of an electric fuse.
(a) Fuses are inserted in electric circuits of all buildings.

(b) There is a maximum limit on the current which can safely flow through the electric circuits.

(c) There is a minimum limit on the current which can safely flow in the electric circuits.
(d) If a proper fuse is inserted in a circuit it will blow off if current exceeds the safe limit.

Answer

Answer: (c) There is a minimum limit on the current which can safely flow in the electric circuits.

Question 9.
When a switch is in OFF position.
(i) circuit starting from the positive terminal of the cell stops at the switch.
(ii) circuit is open.
(iii) no current flows through it.
(iv) current flows after some time.
Choose the combination of correct answer from the following:
(a) all are correct
(b) (ii) and (iii) are correct
(c) only (iv) is correct
(d) only (i) and (ii) are correct

Answer

Answer: (b) (ii) and (iii) are correct

Question 10.
Which of the following precautions need not be taken while using electric gadgets / appliances/circuit?
(a) We should never touch a lighted electric bulb connected to the mains.

(b) We should never experiment with the electric supply from the mains or a generator or an inverter.
(c) We should never use just any wire or strip of metal in place of a fuse.
(d) We should never turn the switch in ON position.

Answer

Answer: (d) We should never turn the switch in ON position.

Match the following:

Column A	Column B
(i) Switch	(a) Coil of wire which heats up when electricity current is supplied
(ii) Battery	(b) Blows off, if the current exceeds safe limit
(iii) Element	(c) Consumes less energy than a bulb
(iv) Filament	(d) Mark that ensures that the electric appliance is safe to handle
(v) Fuse	(e) Supplies current to the circuit
(vi) MCBs	(f) Turns the circuit ON and OFF
(vii) CFL	(g) Turn OFF if current exceeds safe limit
(viii) ISI	(h) Wire in the bulb which glows

Answer

Answer:

Column A	Column B
(i) Switch	(f) Turns the circuit ON and OFF
(ii) Battery	(e) Supplies current to the circuit
(iii) Element	(a) Coil of wire which heats up when electricity current is supplied
(iv) Filament	(h) Wire in the bulb which glows
(v) Fuse	(b) Blows off, if the current exceeds safe limit
(vi) MCBs	(g) Turn OFF if current exceeds safe limit
(vii) CFL	(c) Consumes less energy than a bulb
(viii) ISI	(d) Mark that ensures that the electric appliance is safe to handle

Fill in the blanks:

1. Our body is a ………………….. of electricity.

Answer

Answer: conductor

2. An electric cell produces electricity from the ………………….. in it.

Answer

Answer: chemicals stored

3. In an electric circuit a fuse is a ………………….. to prevent possible fire.

Answer

Answer: safety device

4. A combination of two or more cells is called a ………………….. .

Answer

Answer: battery

5. The coil of wire in an electric heater is called an ………………….. .

Answer

Answer: element

6. A ………………….. is a safety device which prevents damages to electrical circuits and possible fires.

Answer

Answer: fuse

7. The wire gets ………………….. when an electric current passes through it.

Answer

Answer: hot

8. We must look for ………………….. mark on electrical appliances.

Answer

Answer: ISI

9. When electric current passes through a wire, it behaves like a magnet. It is the ………………….. effect of current.

Answer

Answer: magnetic

10. Crane has a strong ………………….. attached to it.

Answer

Answer: electromagnet

Choose the true and false statements from the following:

1. It is convenient to represent electric components by symbols.

Answer

Answer: True

2. A connecting wire is symbolized by a zig-zag line in the circuit diagram.

Answer

Answer: False

3. When an electric current flows through a wire, the wire gets heated.

Answer

Answer: True

4. The key or switch can be placed anywhere in the circuit.

Answer

Answer: True

5. The amount of heat produced in a wire depends on its material, length and thickness.

Answer

Answer: True

6. CFLs consume more electricity than ordinary bulbs.

Answer

Answer: False

7. For different requirements, the wires of different materials, different lengths and thicknesses are used.

Answer

Answer: True

8. A fuse is used to save energy in electrical circuits.

Answer

Answer: False

9. MCBs are the switches which automatically turn off when current in a circuit exceeds the safe limit.

Answer

Answer: True

10. When an electric current flows through a wire, it behaves like a magnet.

Answer

Answer: True

Question 1.
Mention the name of the two devices that work on the basis of magnetic effects of current.
Answer:
The devices that work on the basis of magnetic effects of current are loudspeaker and electric bell.

Question 2.
Name the device used these days in place of eiectric fuses in electrical circuits.
Answer:
The device used in these days in place of electric fuse is MCB (Miniature Circuit Breaker).

Question 3.
State the property of a conducting wire is utilised in making electric fuse. INCERT Exemplar]
Answer:
Electric fuse wire is made up of special material which has low melting point. As if high amount of current is passed, it melts to disconnect the electric circuit and prevent us from causing any damage.

Question 4.
Explain why are CFLs (Compact Fluorescent Lamps) preferred over electric bulbs. [NCERT Exemplar; HOTS]
Answer:
Compact fluorescent lamps are preferred over electric bulbs because electric bulbs use more power of electricity and it also losses electrical energy in the form of heat but it is not so in compact fluorescent lamps.

Question 5.
Name the type of mark for which we should look at an electrical appliance before buying, [HOTS]
Answer:
ISI mark is a mark for which we should look at an electrical appliance before buying.

Question 6.
Briefly mention which part of the symbol of battery shows positive and negative terminals?
Answer:

The positive terminal is represented by longer vertical line while negative terminal is represented by shorter vertical line.

Question 7.
Briefly state the effects of electricity.
Answer:
The three effects of electricity are

Chemical effect
Heating effect
Magnetic effect

Question 8.
When does a circuit is said to be overloaded?
Answer:
It is overloaded by connecting too many devices to it. When too many devices get connected, then a circuit is said to be overloaded.

Question 9.
Why is an electric fuse required in all electrical appliance? [NCERT Exemplar]
Answer:
Electric fuse is required in all electrical appliances to prevent damage from excessive current flow and during short circuit.

Question 10.
Give the name any two electrical appliances in which electromagnets are used.
Answer:
Electric fan and electrical motor are the electrical appliances in which electromagnets are used.

Question 11.
Paheli does not have a night lamp in her room. She covered the bulb of her room with a towel in the night to get dim light. Has she taken the right step? Give one reason to justify your answer. [NCERT Exemplar; HOTS]
Answer:
No, she has not taken the right step. Because due to excessive heat of bulb, the towel may burn and it also results in the wastage of electrical energy.

Question 12.
The nails attract the pins. Comment.
Answer:
When the current is passed through the wire wound on the nail, it behave as electromagnet. Due to this reason, the nails attract the pins.

Question 13.
Name the scientist who discovered the magnetic effect of electric current.
Answer:
The scientist who discovered the magnetic effect of electric current is Hans Christian Oersted.

Question 14.
If the filament of the bulb is broken, would the circuit be complete? Would the bulb still glow?
Answer:
If the filament of the bulb is broken, the circuit will not be complete.
So, the bulb will not glow.

Question 15.
Name some electric appliances where the heating effect of the electric current is used.
Answer:
Some of the electric appliances where the heating effect of the electric current is used are electric heater, geyser, micro-oven, room heater, boiler, etc.

Question 16.
If the current through the coil stops flowing. Will the coil remain an electromagnet?
Answer:
When the current through the coil stops flowing, the coil does not remain an electromagnet.

Question 17.
State whether the bulb glows when the circuit is opened.
Answer:
No, the bulb does not glow when the circuit is opened. The bulb glows only when the circuit is closed.

Question 18.
What happens to the current in a circuit at the time of short circuit?
Answer:
The value of the current in a circuit increases heavily at the time of short circuit.

Question 19.
We should not touch the lighted bulb. Explain, why.
Answer:
The lighted bulb connected to main supply should not be touched by our hand directly, as a lighted bulb can be very hot and can burn our hand as well.

Electric Current and Its Effects Class 7 Science Extra Questions Short Answer Type Questions

Question 1.
Name two electric devices for each where
(a) heating effect of current is used and
(b) magnetic effect of current is used. [NCERT Exemplar]
Answer:
(a) Heating effect of current is used in electric heater and geyser.
(b) Magnetic effect of current is used in electric bell and cranes to lift heavy magnetic materials from one place to other.

Question 2.
Why do we cover plug pinholes which are within the reach of children with cellotape or a plastic cover when not in use? [NCERT Exemplar; HOTS]
Answer:
We do cover plug pinholes which are within the reach of children with cellotape or plastic cover to avoid electric shocks. If unconsciously, a child puts his finger in the electric socket, the shock may be fatal.

Question 3.
Can we use the same fuse in a geyser and a television set? Explain. [NCERT Exemplar; HOTS]
Answer:
No, we cannot use same fuse in a geyser and in a television set because the fuse used in every appliances has some limit to withstand the current flows through it. So, different appliances have different fuses.

Question 4.
If cells are placed side by side. Then, how are the terminals of the cells connected?
Answer:
If cells are placed side by side, then with the help of some connecting wires, the positive terminal of one cell is connected to the negative terminal of other to produce a combined power of all cells which can be called a battery.

Question 5.
Explain how a battery can be constructed.
Answer:
As we know that a battery is a combination of two or more cells and it can be constructed by placing cells property on cell holder in such a way that the positive terminal of one cell is connected to the negative terminal of other.
A piece of wire is connected to each of the two metal clips on the cells holder.

Question 6.
Distinguish between an open circuit and a closed circuit.
Answer:
Difference between an open circuit and a closed circuit

Open (Electric) circuit	Closed (Electric) circuit
An open electric circuit is or electric path which begins from the positive terminal of a battery or cell, gets broken at some point.	A closed circuit is an electric path which begins from the positive terminal of a cell or battery and terminates at its negative terminal without any break.

Question 7.
Boojho made an electromagnet by winding 50 turns of wire over an iron screw. Paheli also made an electromagnet by winding 100 turns over a similar iron screw. Which electromagnet will attract more pins? Give reason. [NCERT Exemplar; HOTS]
Answer:
Since the magnetic effect directly depends on the number of turns of the coil. As, Paheli’s coil has more number of turns than Boojho. So, her electromagnet is stronger than Boojho.
So, electromagnet of Paheli attracts more pins as compared to Boojho.

Question 8.
Does the electric current have other effect except heating? Name it.
Answer:
Yes, electric current have other effect except heating, i.e. magnetic effect of current.
When electric current is passed through a coil, there is a magnetic field developed around the coil or wire, if magnetic compass is placed near by, it deflects the magnetic needle.

Question 9.
Explain the following.
(a) Copper and aluminium wires are usually employed for electricity transmission. Explain the reason.
(b) Explain how does the resistance of a wire vary with its length.
(c) The tungsten is used almost exclusively for filament of electric lamp. Comment.
Answer:
(a) Due to the low resistance and strength of aluminium and coppers, both these metals are usually employed for the transmission of electricity.
(b) There is always an increase in the length of the wire with the increase in its resistance.
(c) Since tungsten has high melting point and high resistance. So, that is why, the tungsten is used exclusively for filament of electric lamp.

Question 10.
If we connect more cells in the circuit, then what will happen?
Answer:
If we connect more of cells in the circuit, then the nail will attract more pins. It is due to the reason that the current flowing through the wire wound on the nail will get increase which in turn will increase the strength of the electromagnet.

Question 11.
Electromagnets are better than permanent magnets. Explain why.
Answer:
An electromagnet can be switched OFF or switched ON as desired and this is not possible in the permanent magnets. That is why the electromagnets are better than the permanent magnets.

Question 12.
If the current flows through wire, does the wire behave like a magnet?
Answer:
When the current flows through any wire, a magnetic field is developed around that wire or coil and it behaves like magnet. It can be analysed by placing a magnetic compass around the wire, it will show deflection of the needle.

Question 13.
If current is passed through a coil, does the pins cling to the coil?
Answer:
When an electric current is passed through a coil, it gets magnetised due to phenomena of magnetic effect of current. When magnetic materials such as pins are placed near to it. It gets attracted by the coil or we can say that pins cling to the coil.

Question 14.
Batteries used in tractors, trucks and inverters are also made from cells. Then why it is called a battery? [HOTS]
Answer:
The cell is the unit of battery when more than one cell are combined together, it forms a battery. In trucks, tractors and inverters, cells are internally arranged and we need not to connect it externally, so we called it as batteries.

Electric Current and Its Effects Class 7 Science Extra Questions Long Answer Type Questions

Question 1.
One day, Pinki was ironing the clothes in her room. After half an iron of ironing, the light went off and Pinki went outside to the lobby of her house to check it there was any problem in the household circuit. At the same time, she listened the voice of her 4 years old daughter from the same room where she was ironing the clothes. Her daughter was about to touch the hot electric iron but at the same moment, Pinki entered in the room and pushed her daughter back from that place.
(a) On which effect of electric current, does the electric iron works?
(b) Mention the values showed by Pinki here. [Value Based Question]
Answer:
(a) The electric iron works on the basis of heating effect of electric current.
(b) The values showed by Pinki here is a great concern and love towards her own 4 years old daughter.

Question 2.
Explain with the help of a diagram, how does the magnetic effect of electric current help in the working of an electric bell.
Answer:
In order to ring the bell first of all we press the push button switch in order to ring the bell. So, when we press the switch, then the electric circuit of the bell is completed and a current passes through the coil of the electromagnet and it gets magnetised. The electromagnet attracts the iron armature towards itself.

So, as the armature moves towards the poles of the electromagnet, the clapper attached to it strikes the gong and produces a ringing sound. It implies that the bell rings.

When the armature moves towards the magnet, its contact with the contact screw is broken. Due to this, the electric circuit breaks and no current flows in the electromagnet coil. The electromagnet loses its magnetism for a moment and the armature is no longer attracted by it. The flat spring brings back the iron armature to its original position and the clapper also moves away from the gong.

As soon as the armature comes back and touches the contact screw the circuit is completed and current starts flowing in the electromagnet coil again. The electromagnet attracts the iron armature once again and the clapper strikes the gong again producing a ringing sound.

So, this process of ‘make and break’ of the electric circuit continues as long as we are pressing the switch. Due to this, the armature vibrates forwards and backwards rapidly each time making the clapper strike the gong. Thus, the clapper strikes the gong rapidly producing almost continuous sound.

Question 3.
State one measure to avoid overloading in an electrical circuit. Also mention the name given to a situation in which the live and the neutral wires accidently come in contact. Describe the role of a safety device in this situation.
Answer:
Overloading can be avoided if too many appliances are not connected to a single socket. Short circuiting is a name given to a situation in which the live and the neutral wires accidently come in contact.

Electric fuse is a safety device. When a short circuit takes place or when overloading takes place, then the current becomes large and heats the fuse wire too much and thus, it gets melted and circuit breaks which prevents the damage of the wiring and electrical appliances.

Question 4.
Paheli took a wire of length 10 cm. Boojho took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.
(a) Will the heat produced in both the cases be equal? Explain.
(b) Will the heat produced be the same, if the wires taken by them are of equal lengths but of different thickness? Explain.

Answer:
(a) No, the amount of heat produced in both the wires will be different because amount of heat produced in a wire on passing electric current depends on the length of wire and here length is different for both the wires.
(b) No, the amount of heat produced in the wire of same length but different thickness cannot be same because amount of heat produced in a wire also depends on the thickness of the wire.

Question 5.
Last Sunday, Pulpit was playing videotape in his room. While playing, the electricity of his house went off due to which he could not able to see anything around him. Then anyhow, he managed to get his mobile in his hand and with the help of its light, he went outside of his room to check if there was any problem in the electric circuit board of his house.
At the same time, his father suggested him to maintain the distance from the circuit board and decided to call some electrician to check the problem.
(a) Explain, what happens when live wire and neutral wire touches each other directly?
(b) State the values here showed by Pulkit’s father. [Value Bated Question]
Answer:
(a) When live and neutral wire touches each other directly then it leads to short circuit in which the large amount of current flows through the household wiring and this large current may heat the wires to a dangerously high temperature and a fire may be started.
(b) Pulkit’s father seems very sensible in taking the decisions and he showed a very great concern towards his son

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CHAPTER – 13 Motion and Time | CLASS 7TH | NCERT SCIENCE IMPORTANT QUESTIONS & MCQS | EDUGROWN

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 13 Motion and Time

Question 1.
A bus travels 54 km in 90 minutes. The speed of the bus is
(a) 0.6 m/s
(b) 10 m/s
(c) 5.4 m/s
(d) 3.6 m/s

Answer

Answer: (b) 10 m/s

Question 2.
If we denote speed by S, distance by D and time by T, the relationship between these quantities is
(a) S = D T
(b) T = SD
(c) S = 1T × D
(d) S = TD

Answer

Answer: (c) S = 1T × D

Question 3.
Observe the figure given below:
MCQ Questions for Class 7 Science Chapter 13 Motion and Time with Answers 1
The time period of a simple pendulum is the time taken by it to travel from
(a) A to B and back to A
(b) O to A, A to B and B to A
(c) B to A, A to B and B to O
(d) A to B

Answer

Answer: (a) A to B and back to A

Question 4.
Nearly all the clocks make use of
(a) straight line motion
(b) periodic motion
(c) random motion
(d) circular motion

Answer

Answer: (b) periodic motion

Question 5.
A simple pendulum takes 42 sec. to complete 20 oscillations. What is its time period?
(a) 2.1 s
(b) 4.2 s
(c) 21 s
(d) 8.40 s

Answer

Answer: (a) 2.1 s

Question 6.
Time period of a simple pendulum depends upon its
(a) weight of bob
(b) length
(c) both (a) and (b)
(d) None of these

Answer

Answer: (b) length

Question 7.
Which of the following cannot be used for measurement of time?
(a) A leaking tap
(b) Simple pendulum
(c) Shadow of an object during the day
(d) Blinking of eyes

Answer

Answer: (d) Blinking of eyes

Question 8.
On which axis is dependent variable represented?
(a) x-axis
(b) y-axis
(c) On any axis
(d) Depends on the data

Answer

Answer: (b) y-axis

Question 9.
The correct symbol to represent the speed of an object is:
(a) 5 m/s
(b) 5 mp
(c) 5 m/s^-1
(d) 5 s/m

Answer

Answer: (d) 5 s/m

Question 10.
Boojho walks to his school which is at a distance of 3 km from his home in 30 minutes. On reaching he finds that the school is closed and comes back by a bicycle with his friend and reaches home in 20 minutes. His average speed in km/h is
(a) 8.3
(b) 7.2
(c) 5
(d) 3.6

Answer

Answer: (b) 7.2

Match the following:

Column A	Column B
(i) Years	(a) Rotation period of earth on its axis
(ii) Hour	(b) Time to complete 100 m race
(iii) Minute	(c) Gestation period in humans
(iv) Days	(d) Age of a person
(v) Second	(e) In scientific research
(vi) Month	(f) Age of stars
(vii) Microseconds	(g) Time to reach your school
	(h) Train reaches from one city to other

Answer

Answer:

Column A	Column B
(i) Years	(d) Age of a person
(ii) Hour	(h) Train reaches from one city to other
(iii) Minute	(g) Time to reach your school
(iv) Days	(a) Rotation period of earth on its axis
(v) Second	(b) Time to complete 100 m race
(vi) Month	(c) Gestation period in humans
(vii) Microseconds	(e) In scientific research

Fill in the blanks:

1. The distance covered by an object in a ………………… is called its speed.

Answer

Answer: unit time

2. Time between one sunrise and the next sunrise was called a …………………

Answer

Answer: day

3. In a simple pendulum, the metallic ball suspended by thread is called its …………………

Answer

Answer: bob

4. Time period of a pendulum depends on its …………………

Answer

Answer: length

5. The symbols of all units are written in …………………

Answer

Answer: singular

6. A ………………… is one billionth of a second.

Answer

Answer: nanosecond

7. Motion of objects can be presented in pictorial form by their ………………… graph.

Answer

Answer: distance time

8. The distance-time graph for the motion of an object moving with a constant speed is a …………………

Answer

Answer: straight line

Choose the true and false statements from the following:

1. Faster vehicle has a higher speed.

Answer

Answer: True

2. All the clocks make use of some periodic motion.

Answer

Answer: True

3. The basic unit of speed is km/h.

Answer

Answer: False

4. The symbols of all units are written in singular.

Answer

Answer: True

5. The basic unit of time is second.

Answer

Answer: True

6. Age of a person is expressed in days.

Answer

Answer: False

7. The distance-time graph of standing vehicle is a straight line parallel to x-axis.

Answer

8. Periodic events are used for the measurement of time.

Answer

Question 1.
Give the basic unit of speed.
Answer:
Metre/second is the basic unit of speed.

Question 2.
Mention which is the most common thing in almost all the clocks.
Answer:
The most common thing in almost all the clocks is that all of them shows periodic motion.

Question 3.
Paheli and Boojho have to cover different distances to reach their school but they take the same time to reach the school. What can you say about their speed? [NCERT Exemplar; HOTS]
Answer:
They do not have equal speed because they cover unequal distance in equal intervals of time. One of them has higher speed whom has to cover larger distance with respect to other.

Question 4.
If Boojho covers a certain distance in one hour and Paheli covers the same distance in two hours, who travels with a higher speed? [NCERT Exemplar; HOTS]
Answer:
Boojho travels with a higher speed as he has covered same distance in lesser time with respect to Paheli.

Question 5.
A simple pendulum is oscillating between two points A and B as shown in figure. Is the motion of the bob uniform or non-uniform?
Motion and Time Class 7 Extra Questions Science Chapter 13 1
Answer:
The motion of bob is non-uniform as it does not cover equal distance in equal intervals of time.

Question 6.
State the characteristic of distance-time graph for the motion of an object moving with a constant speed.
Answer:
The distance-time graph for the motion of an object moving with a constant speed is a straight line inclined on x-axis at some angles.

Question 7.
Name the axis that represents dependent variable.
Answer:
y-axis represents the dependent variable.

Question 8.
Briefly explain how will decide which object is moving fast and which one is moving slow.
Answer:
If an object covers more distance in equal intervals of time with respect to other, then we can say that object is moving faster with respect to the other object.

Question 9.
If you did not have a clock, how would you decide what time of the day is? [HOTS]
Answer:
We can decide time of the day without clock by seeing shadow formed by the sun, e.g. at noon, shadow formed by the sun is shorter than at evening.

Question 10.
State how do we measure time interval of a month.
Answer:
Time interval of a month is measured by one new moon to the next.

Question 11.
Determine the number of seconds there in a day.
Answer:
In a day, we have 24 h
We know that 1 h = 3600 s
So, 24 h = 3600 × 24 = 86400s

Question 12.
Estimate that how many hours are there in a year.
Answer:
In a year, we have 365 days and 1 day = 24 h
So, for 365 days, we have 24 × 365 h = 8760 h

Question 13.
Explain how time was measured when pendulum clocks were not available.
Answer:
Before the discovery of pendulum clocks, sundials, water clocks and sand clocks were used.

Question 14.
A spaceship travels 36000 km in one hour.Express its speed in km/s.
Answer:
Motion and Time Class 7 Extra Questions Science Chapter 13 2

Question 15.
Out of the two given clocks, a pendulum clock or a quartz clock. Mention which of the clock between the given ones requires an electric cell for its working?
Answer:
A quartz clock requires an electric cell for its working.

Question 16.
Name the most ancient unit of time in which it was earlier measured.
Answer:
Solar day is the most ancient unit of time in which it was earlier measured.

Question 17.
State the condition under which the distance-time graph of an object is a horizontal line parallel to the axis of time.
Answer:
When the object is at rest, then the distance-time graph of an object is a horizontal line parallel to the axis of time.

Question 18.
State the factors on which the time period of a simple pendulum depend.
Answer:
The factors on which the time period of a simple pendulum depend are

Length of the pendulum
Acceleration due to gravity.

Question 19.
Name the different types of graphs.
Answer:
Different types of graphs are

Line graph
Pie chart
Bar graph

Question 20.
State the way through which we can prepare a pendulum by our own.
Answer:
By suspending a metal ball with the help of a cotton thread and the other end of the thread can be tied to some different supports, so in this way, we can prepare a pendulum by our own.

Motion and Time Class 7 Science Extra Questions Short Answer Type Questions

Question 1.
Last Sunday, Priyanka decided to go outside with her family for the dinner in some particular restaurant.
Then, she thought to go through her own car. So, while moving she increased the speed of the car from 36 km/h to 54 km/h. All of sudden, a rickshaw came in front of her car but due to applying of brakes at the right time, both of them got safe from meeting with an accident. [Value Based Question]
(a) Calculate the increase in the speed of the car in the terms of m/s.
(b) Mention a positive action performed by Priyanka here.
Answer:
Motion and Time Class 7 Extra Questions Science Chapter 13 3
∴ Increase in speed = Final speed – Initial speed =(15-10) m/s = 5 m/s
(b) Priyanka applied the brakes at a right time while driving the car and prevented her family and the person on rickshaw from meeting with an accident.

Question 2.
Differentiate between a uniform and non-uniform motion.
Answer:
When a body travels equal distance in an equal interval of time, then the body is said to be in a uniform motion.
When a body travels unequal distance in an equal interval of time, then the body is said to be in non-uniform motion.

Question 3.
Briefly mention how many types of motion are there. Also, name all of them.
Answer:
Different types of motion are there and they can be named as follows

Periodic motion
Rolling motion
Rotatory motion
Oscillatory and vibratory motions
Translatory motion

Question 4.
The average age of children of class VII is 12 years and 3 months. Express this age in second. [NCERT Exemplar]
Answer:
Given, the average age of children 12 years and 3 months
1 year = 365 days = 365 × 24 h [1 day = 24 h]
= 365 × 24 × 3600 s [1 h = 3600 s]
12 years = 31536000 s × 12 = 378432000 s
3 months = 30x3days = 30 × 3 × 24h = 30 × 3 × 24 × 3600s = 7776000s
So, total age in second = 378432000 + 7776000 = 386208000s

Question 5.
Plot a distance-time graph of the tip of the second hand of a clock by selecting 4 points on X-axis and /-axis, respectively. The circumference of the circle traced by the second hand is 64 cm. [NCERT Exemplar]
Answer:

Time (s)	X	15	30	45	60
Distance (cm)	Y	16	32	48	64

As, here equal distance is covered in equal intervals of time.
So, graph will be straight line.
Motion and Time Class 7 Extra Questions Science Chapter 13 4

Question 6.
With the help of the data given in the following table, show graphically the relationship between the diameter and circumference of a circle.

Diameter of the circle (cm)	Circumference of the circle (cm)
1	3.2
2	6.3
3	9.4
4	12.6
5	15.7
6	18.8

Answer:
With the help of a given data in a tabular form, the graph can be obtained by taking diameter of the circle on the x-axis and the circumference on y-axis. On x-axis, 1 cm diameter is shown by 1 cm and on the y-axis, 3 cm of circumference is shown by 1 cm.
Motion and Time Class 7 Extra Questions Science Chapter 13 5

Question 7.
Starting from A, Paheli moves along a rectangular path ABCD as shown in figure. She takes 2 min to travel each side. Plot a distance-time graph and explain whether the motion is uniform or non-uniform.
Motion and Time Class 7 Extra Questions Science Chapter 13 6
Answer:
Since, the distance covered per unit time for the entire distance covered is not the same, the motion is a non-uniform motion.

Question 8.
One day, a five years old boy, Chintoo was playing with his toys in his room. After few minutes, he saw a wristwatch on his study table, so after seeing it, he got crazy and so much in eager to play with that watch. Then, for 2 to 3 min, he was trying to analyse it that whether its a toy or something else, then suddenly, he was just about to throw it but at the same his mother entered into the room and take the watch back immediately from his hand.
(a) State the benefit of a wristwatch.
(b) Briefly mention the values showed by Chintoo’s mother here. [Value Based Question]
Answer:
(a) The wristwatch can be handled very easily to check the time in comparison to the wall clock.
(b) The values showed by Chintoo’s mother here is the concern towards her four years old child and also the concern towards that wristwatch while considering the importance of money.

Question 9.
The following distance-time graph of three objects (D, E and F) are given (see figure given alongside). What can you say about the motion of the objects?
Motion and Time Class 7 Extra Questions Science Chapter 13 8
Answer:
Since, the distance-time graph of an object D is a straight line therefore, its motion is uniform. Also here, we can see that the distance-time graph of object E is parallel to x-axis (time-axis), therefore object E is at rest.
As, the distance-time graph of object F is a curved line, therefore the object C is moving with a non-uniform speed and its motion is non-uniform.

Question 10.
The following figure shows the distance-time graph for the motion of two objects D and E. Which one of them is moving slower?
Motion and Time Class 7 Extra Questions Science Chapter 13 9
Answer:
After studying the figure given alongside, D or E is moving slower as the slope of distance-time graph of E is less than that of graph D.

Question 11.
The following figure shows the distance-time graph for the motion of two vehicles (C and D). At what time and distance (from the starting point), they will cross each other.
Answer:
Firstly, draw perpendicular lines from point E on x-axis and y-axis.
Motion and Time Class 7 Extra Questions Science Chapter 13 10
The two vehicles (C and D) will meet (or cross) each other at 5:00 PM and after travelling a distance of 4 km.

Motion and Time Class 7 Science Extra Questions Long Answer Type Questions

Question 1.
Given below as a figure is the distance-time graph of the motion of
(a) What will be the position of the object at 20 s?
(b) What will be the distance travelled by the object in 12 s?
(c) What is the average speed ofthe object? [NCERT Exemplar]
Motion and Time Class 7 Extra Questions Science Chapter 13 11
Answer:
(a) From the graph, it is clear that the distance at 20 s is 8 m.
(b) Distance travelled by the body object in 12 s is 6 m.
(c) As, average speed = Total distance Total time =820=0.4m/s

Question 2.
Boojho goes to the football ground to play football. The distance-time graph of his journey from his home to the ground is given as figure. [NCERT Exemplar]
(a) What does the graph between points B and C indicate about the motion of Boojho?
(b) Is the motion between 0 to 4 min uniform or non-uniform?
(c) What is his speed between 8 to 12 min of his journey?
Answer:
(a) Since, the graph between B and C is parallel to time-axis, so it indicates that Boojho is at rest.
(b) Since, the graph is not straight line, so it is a non-uniform motion.
(c) As, speed = Distance Time =225−15012−8=754=18.75m/min

Question 3.
Explain by giving example that how can we choose a suitable scale.
Answer:
While choosing the most suitable scale for drawing a graph, we should keep the following points in mind
(i) The difference between the highest and lowest values of each quantity.
(ii) The intermediate values of each quantity so that with the scale chosen, it is easy to mark the values on the graph.
(iii) To utilise the maximum part of the paper on which the graph is to be drawn, assume that we have a graph paper of size 25 cm x 25 cm and we have to accommodate following data:

Time (AM)	Odometer reading	Distance from the starting point
8:00 AM	36540 km	0 km
8:30 AM	36560 km	20 km
9:00 AM	36580 km	40 km
9:30 AM	36600 km	60 km
10:00 AM	36620 km	80 km

Since, one of the scales will be
Distance 5 km = 1 cm and Time 6 min = 1 cm
Motion and Time Class 7 Extra Questions Science Chapter 13 12

Question 4.
Distance between Bholu’s and Golu’s house is 9 km. Bholu has to attend Golu’s birthday party at 7 O’ clock. He started his journey from his home at 6 O’ clock on his bicycle and covered a distance of 6 km in 40 min. At that point, he met Chintu and he spoke to him for 5 min and reached Golu’s birthday party at 7 O’ clock. With what speed, did he cover the second part of the journey? Calculate-his average speed for the entire journey. [NCERT Exemplar]
Answer:
From the question, Bholu covers 3 km distance in 15 min.
Motion and Time Class 7 Extra Questions Science Chapter 13 13

Question 5.
The given data regarding the motion of two different objects C and D is in a tabular form. Check them carefully and state whether the motion of the objects is uniform or non-uniform.

Time	Distance travelled by C (in m)	Distance travelled by D (in m)
9:30 AM	10	12
9:45 AM	20	19
10:00 AM	30	23
10:15 AM	40	35
10:30 AM	50	37
10:45 AM	60	41
11:00 AM	70	44

Answer:
We can check the motion of the objects C and D by plotting the distance-time graph for the two objects.
Motion and Time Class 7 Extra Questions Science Chapter 13 14
Graph the object C on x-axis
Scale 1 cm = 15 min and on y-axis 1 cm = 10 m
Since, the distance-time graph of the object C is a straight line. Therefore, its motion is uniform.
Graph of object D
Scale On x-axis, 1 cm = 15 min, On y-axis 1 cm=5m
Motion and Time Class 7 Extra Questions Science Chapter 13 15
Since, the distance-time graph of the object D is not a straight line, therefore, its motion is non-uniform.

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CHAPTER – 12 Reproduction in Plants | CLASS 7TH | NCERT SCIENCE IMPORTANT QUESTIONS & MCQS | EDUGROWN

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 12 Reproduction in Plants

MCQs

Question 1.
Vegetative propagation in potato takes place by
(a) leaves
(b) stem
(c) root
(d) seed

Answer

Answer: (b) stem

Question 2.
In which of the following plants buds are present on the margins of leaves?
(a) Bryophyllum
(b) Touch me not
(c) Chandan
(d) Coriander

Answer

Answer: (a) Bryophyllum

Question 3.
In yeasts reproduction occurs by
(a) fragmentation
(b) binary fission
(c) budding
(d) spore formation

Answer

Answer: (c) budding

Question 4.
Which of the following parts of a plant take part in sexual reproduction?
(i) Flower
(ii) Seed
(iii) Fruit
(iv) Branch
Choose the correct answer from below:
(a) (i) and (ii)
(b) (i), (ii) and (iii)
(c) (iii) and (iv)
(d) (ii), (iii) and (iv)

Answer

Answer: (b) (i), (ii) and (iii)

Question 5.
Lila observed that a pond with clear water was covered up with a green algae within a week. By which method of reproduction did the algae spread so rapidly ?
(a) Budding
(b) Sexual reproduction
(c) Fragmentation
(d) Pollination

Answer

Answer: (c) Fragmentation

Question 6.
Seeds of drumstick and maple are carried to long distances by wind because they possess
(a) winged seeds
(b) large and hairy seeds
(c) long and ridged fruits
(d) spiny seeds

Answer

Answer: (a) winged seeds

Question 7.
The ‘eye of the potato plant is what
(a) the root is to any plant
(b) the bud is to a flower
(c) the bud is to Bryophyllum leaf
(d) the anther is to stamen

Answer

Answer: (c) the bud is to Bryophyllum leaf

Question 8.
The ovaries of different flowers may contain
(a) only one ovule
(b) many ovules
(c) one to many ovules
(d) only two ovules

Answer

Answer: (c) one to many ovules

Question 9.
Which of the following statements is/are true for sexual reproduction in plants?
(i) Plants are obtained from seeds
(ii) Two plants are always essential
(iii) Fertilisation can occur only after pollination
(iv) Only insects are agents of pollination
Choose from the options given below:
(a) (i) and (ii)
(b) (i) only
(c) (i) and (ii)
(d) (i) and (iv)

Answer

Answer: (a) (i) and (ii)

Question 10.
Pollination refers to the:
(a) transfer of pollen from anther to ovary
(b) transfer of male gametes from anther to stigma
(c) transfer of pollen from anther to stigma
(d) transfer of pollen from anther to ovule

Answer

Answer: (c) transfer of pollen from anther to stigma

Match the following:

Column A	Column B
(i) Bread mould	(a) Cutting
(ii) Yeast	(b) Leaves
(iii) Potato	(c) Fragmentation
(iv) Bose	(d) Detached body part
(v) Sweet potato	(e) Spores
(vi) Bryophyllum	(f) Eye
(vii) Cactus	(g) Roots
(viii) Spirogyra	(h) Budding

Answer

Answer:

Column A	Column B
(i) Bread mould	(e) Spores
(ii) Yeast	(h) Budding
(iii) Potato	(f) Eye
(iv) Bose	(a) Cutting
(v) Sweet potato	(g) Roots
(vi) Bryophyllum	(b) Leaves
(vii) Cactus	(d) Detached body part
(viii) Spirogyra	(c) Fragmentation

Fill in the blanks:

1. In …………………. reproduction, one individual can produce many individuals from its body parts.

Answer

Answer: asexual

2. Both stamen and carpel are present in …………………. flowers.

Answer

Answer: bisexual

3. Budding is a type of …………………. reproduction.

Answer

Answer: asexual

4. The process of …………………. ensures continuity of life on the earth.

Answer

Answer: reproduction

5. …………………. are the reproductive parts of a plant.

Answer

Answer: Flowers

6. Buds in potato are also called ………………….

Answer

Answer: eyes

7. Plants produce seeds as a result of …………………. reproduction.

Answer

Answer: sexual

8. The small bulb like projection coming out from the yeast cell is called a ………………….

Answer

Answer: bud

9. The …………………. develops into an embryo.

Answer

Answer: zygote

10. The fruits are ripened ………………….

Answer

Answer: ovary

Choose the true and false statements from the following:

1. Two individuals are needed for a sexual reproduction.

Answer

Answer: True

2. Seed is the only structure which develops into new plant.

Answer

Answer: False

3. Plants such as cacti produce new plants when their parts get detached from the main plant body.

Answer

Answer: True

4. Plants produced by vegetative propagation take less time to grow and bear flower and fruit.

Answer

Answer: True

5. The spores are asexual reproductive bodies.

Answer

Answer: True

6. Anther contains female gametes called eggs.

Answer

Answer: False

7. The fruit is ripened ovary.

Answer

Answer: True

8. Seed dispersal in coconut is aided by winds.

Answer

Answer: False

9. Fusion of male and female gametes is called pollination.

Answer

Answer: False

10. A bisexual flower has both male and female reproductive parts.

Answer

Answer: True

Question 1.
Why is the process of reproduction necessary?
Answer:
The process of reproduction is necessary for the perpetuation and preservation of species and to increase the number of members of species.

Question 2.
If the filament of Spirogyra is broken into fragments. What will you observe?
Answer:
If the filament of Spirogyra are broken into fragments, then each fragment will develop into a new plant.

Question 3.
Fungus, moss and fern reproduce by a common method of asexual reproduction. Name the method. [NCERT Exemplar]
Answer:
Fungus, moss and fern reproduce by the common method of spore formation which is a type of asexual reproduction.

Question 4.
A flower consists of different parts. Name these parts of a flower.
Answer:
The four main parts of a flower are
(i) Sepals
(ii) Petals
(iii) Stamen
(iv) Pistil

Question 5.
Mention the mode of reproduction in the following plants
(a) Spirogyra
(b) Yeast
(c) Money plants
Answer:
(a) Spirogyra – fragmentation
(b) Yeast – budding
(c) Money plant – Vegetative propagation

Question 6.
Pick the odd one out from the following on the basis of mode of reproduction and give reason for it, Sugarcane, Potato, Rice, Rose. [NCERT Exemplar]
Answer:
The odd one out is rice.
In the above given pairs as rice reproduces by sexual reproduction and sugarcane, potato and rose reproduces vegetatively.

Question 7.
Give one difference between unisexual and bisexual flower.
Answer:
Unisexual flowers are those which contain either male or female reproductive part and bisexual flower has both reproductive parts (i.e. male and female) on the same flower.

Question 8.
Flowers are colourful and fragrant. Give reason supporting the statement. [HOTS]
Answer:
Flowers are so colourful because they absorb and reflect light energy. Fragrance results from production of volatile chemicals which evaporate.

Question 9.
Boojho had the following parts of a rose plant-a leaf, roots, a branch, a flower, a bud and pollen grains. Which of them can be used to grow a new rose plant? [NCERT Exemplar]
Answer:
Branch can be used to grow a new rose plant. As, rose reproduces by vegetative propagation, i.e. stem cutting method.

Question 10.
Formation of new but similar individuals from parents is characteristic feature of which process?
Answer:
Reproduction is the process of production of new similar organisms from their parents.

Question 11.
Name two plants where vegetative reproduction takes place by roots.
Answer:
Sweet potato and dahlia develops new plants through their roots by the process of vegetative reproduction.

Question 12.
A yellow powdery substance is present in the anther which participates in reproduction process. Name this substance.
Answer:
Pollen grains.

Question 13.
In which part of flower does fertilisation occur?
Answer:
Ovary is the part where male and female gametes fuse together.

Question 14.
Bryophyllum leaves reproduce using which mode of reproduction?
Answer:
Asexually by vegetative propagation.

Question 15.
Which type of pollination does the given figure indicate?

Answer:
The given figure shows self-pollination, as the pollen grains from anther of flower are transferred to the stigma of same flower.

Question 16.
What are the bulb-like projections forming in yeasts?
Answer:
The bulb-like projections formed during reproduction are called buds.

Question 17.
‘Spores’ as a means of asexual reproduction are used by which plants?
Answer:
Spores are produced by fungi, ferns and mosses during unfavourable conditions.

Question 18.
Write the male and female reproductive parts present on the flowers.
Answer:
The male reproductive part is stamen and the female reproductive part is pistil.

Question 19.
Pollen grains are present inside anthers for long time yet they do not die. Why?
Answer:
Pollen grains have a tough protective coat which presents them from drying up.

Question 20.
Name two plants in which pollination occurs by water.
Answer:
Vallisneria and Hydrilla.

Question 21.
Fruit is the ripened ovary of a flower. Explain.
Answer:
After fertilisation, the ovary of a flower develops and becomes a fruit with seeds present inside.

Question 22.
Banana is a fruit without seeds. Give reason.
Answer:
Banana forms from one parent only. There is no seed production.

Reproduction in Plants Class 7 Science Extra Questions Short Answer Type Questions

Question 1.
When you keep food items like bread and fruits outside for a long time especially during the rainy season, you will observe a cottony growth on them. [NCERT Exemplar; HOTS]
(a) What is this growth called?
(b) How does the growth take place?
Answer:
(a) When food items like bread and fruits are kept outside for a long time especially during rainy season, a cottony growth of bread mould, a fungus is observed.
(b) This growth of fungus takes place by spores present in air, which when comes in the contact with moisture in bread germinates and grow to produce new cells.

Question 2.
Collect some flower of different plants like papaya, rose, mustard, lady’s finger, Petunia, cucumber, corn, pea, etc. Group them under following heads.
(a) Which of these plants have unisexual flowers?
(b) Which of these plants have bisexual flower?
(c) What is the other name of unisexual and bisexual flower? [HOTS]
Answer:
(a) Unisexual flowers are papaya, cucumber, com.
(b) Bisexual flowers are rose, mustard, lady’s finger, Petunia, pea.
(c) Unisexual flowers are also called as incomplete flower while bisexual flowers are called hermaphrodite or complete flowers.

Question 3.
In the figure given below, label the part marked (i), (ii) and (iii). [NCERT Exemplar]

Answer:
The parts in the given figure are labelled as follows

Question 4.
Coconut is a large and heavy fruit. How is it adapted for dispersal by water? [NCERT Exemplar]
Answer:
The seeds of some plants that have an outer fibrous or spongy covering are dispersed through water. They have the ability to float in the water and drift along with its flow, e.g. seeds of water lily, lotus, chestnut (singhara) and coconut are dispersed through water. The coconut fruits have a fibrous outer coat which enables them to float in water and carried away by flowing water to far off places.

Question 5.
What is a bud? Which organism reproduce by budding? Given the diagrammatic representation of budding in a plant.
Answer:
Buds are small bulb-like projections of yeast cell.
These are asexual reproducing bodies of yeast.
Diagrammatic representation of budding in yeast Refer to figure on page 178.

Question 6.
Group the seeds given in figure (i) to (iii) according to their means of dispersion.
(a) Seed dispersed by wind.
(b) Seed dispersed by water.
(c) Seed dispersed by animal. [NCERT Exemplar]

Answer:
The seeds and their means of dispersal can be given as follows:
(a) The seed of maple is dispersed by wind. It has winged seed which are light in weight.

(b) Seed of aak or madar has hairy outgrowth which makes it lighter and can be dispersed by wind.

(c) Seed of Xanthium have numerous spines on them which gets attached to the fur of animals. Hence, these are dispersed by animals.

None of the seed given in the figure is dispersed by water

Question 7.
How does male gamete present in pollen grain reaches female gamete present in ovule?
Answer:
After pollination, the pollen grains fall on the surface of stigma and germinate to form a long tube, reaching the ovules inside the ovary, the egg or female gamete is present in the ovule.
The outer surface of pollen grains rupture and male gametes are released to fuse with egg.

Question 8.
How do the plants like sugarcane, potato and rose reproduce when they cannot produce seeds?
Answer:
Sugarcane and rose are propagated by stem cutting that is a method of vegetative propagation, in which stem is capable of growing into a mature independent plants that are identical to their parents.

Potato is an underground modified stem having bud called eyes, which sprout and develop into a new identical plant.
Thus, the plants which cannot produce seeds, can be propagated vegetatively with the help of vegetative parts such as stem, roots, buds and leaves.

Question 9.
Mention the benefits of seed dispersal.
Answer:
Benefits of Seed Dispersals

Seed dispersal avoids overcrowding of young plants around their parent plants.
It helps in preventing competition between the plants and its own seedlings for sunlight, water and minerals.
One of the benefits of seed dispersal is that it enables the plant to grow into new habitats for wider distribution and provides them with better chance of survival.

Question 10.
What is meant by the term fertilisation? List the stepwise manner leading to formation of an embryo.
Answer:

The process in which the male gamete fuses with female gamete to form a new cell (called zygote) is called fertilisation.
When the pollens are deposited on the stigma of the pistil, it begins to germinate. After sometime, a long pollen tube is developed from the pollen grain which passes through the style towards the female gametes in the ovary. The male gametes move down the pollen tube and the tube enters the ovule present inside the ovary.

The tip of pollen tube bursts and the male gamete comes out of the pollen tube. Inside the ovary, the male gametes fuse with the female gametes present in the ovule to form a fertilised egg cell which is called zygote.
The zygote develops into an embryo which is a part of a seed that develops into a new plant.

Question 11.
A student was given a flower. He was asked to pick the different whorls of flower by the forcep. He pulled each part of the flower and laid them on the chart paper in a sequence and named them W, X, Y, Z (from outer to inner whorl). He was unable to name them.
Help the student to name the different parts of a flower. Also help him to tell which part produces male gamete and female gametes. [HOTS]
Answer:
The four whorls of the flower are outermost whorl ‘W’ is green part which is called sepal. Inside sepal the next whorl is X which is coloured and attractive part of the flower called petals. The Y is the inner whorl of flower called stamen. It is the male reproductive part of flower. It consist of two parts, i.e. anther and filament. The anther contains male gametes called pollen grain. The whorl ‘Z’ is the innermost part of the flower called pistil. It is the female reproductive part of flower. It consist of three parts, i.e. stigma, style and ovary. The ovary produces ovule which contains the female gametes or egg cell.

Question 12.
One morning as Paheli strolled in her garden she noticed many small plants, which were not there a week ago. She wondered, where they had come from as nobody had planted them there. Explain the reason for the growth of these plants. [NCERT Exemplar; HOTS]
Answer:
The small plants which were not there in the garden a weak ago have grown up due to seed dispersal. The seeds from the tree may have fallen below or have been dispersed by wind or animals on the ground, which on germination developed into new small plants.

Question 13.
Place a piece of bread in a moist and warm place for few days. Observe it after few days. What will you see?
(a) Name the organism that grows on the bread piece?
(b) What are the thread-like projections called?
(c) What is the knob like structure present on the top of thin stem called?
(d) Which type of reproduction does this organism shows?
(e) From where does the spors comes to the bread piece? [HOTS]
Answer:
(a) When the bread piece is kept in a moist and warm place for few days, bread mould grows on the bread piece.
(b) The thread-like projections are called hyphae or mycelium.
(c) The thin stem having knob-like structure on the top is called sporangia or sporangium which contains hundred of minute spores.
(d) Bread mould shows asexual mode of reproduction.
(e) These spores are present in the air and when favourable conditions arrive, the grow into new plants.

Question 14.
Write how the following seeds are dispersed.
(a) Seeds with wings
(b) Small and light seeds.
(c) Seeds with spines/hooks.
Answer:
The mode of dispersal of the seeds having following properties are as follows:
(a) Seeds with wings-like seeds of drumstick and maple become light weighted and can be blown away by air. Thus, these are dispersed by wind.
(b) Small and light seeds like seeds of grasses and cotton (having hairy growth) are also dispersed through wind.
(c) Seeds of Xanthium, Urena have spines and hooks on them, these hooks or spines are attached to the fur of the animal body. When animals move to other places and rub their body with tree, etc., these seeds fall from their body and get dispersed.. Therefore, these are dispersed through animals.

Question 15.
Why is vegetative propagation a preferred method of asexual reproduction?
Answer:
The following advantages of vegetative reproduction makes it a preferred method

It takes less time to grow and bear flower and fruits than those produced from seeds.
The new plants are the exact copies of parent plant because they are produced from a single parent.

Question 16.
Insects are called agents of pollination. How do they aid in process of pollination?
Answer:
Flowers have nectars that attract insects. Insects suck these nectars as their food. When insects like bee, butterfly, etc., sit on the flower for sucking nectar, the sticky pollen grains get attached to their legs and wings. When these insects again sit on another flower, these pollen grains are transferred to the stigma of that flower from the body of the insects. In this way, insects help in pollination.

Question 17.
What is seed dispersal? What will happen if all the seeds of a plant were to fall at a same place and grow?
Answer:
Plant produces large number of seeds. When these seeds fall down they starts growing. The process by which the seeds are scattered to different place (far and wide from their parents) is called seed dispersal.
The seeds and fruits are dispersed away through various agencies like air, water and animals. Sometimes dispersal takes place by the explosion or bursting of fruits. If all the seeds of a plant were to fall at the same place and grow, there will be a severe competition for sunlight, water, mineral and space. As a result, the survival for the plants will be difficult and the plants who survive will not grow into a healthy plants.

Question 18.
Describe the structure of a flower.
Answer:
The main parts of a flower are

(i) Sepals These are the green leaf-like outermost circle of the flower. All the sepals are together referred to as calyx. The function of calyx is to protect the flower when it is in bud form.

(ii) Petals These are the colourful and most attractive part of flower. These lie inside the sepals. All the petals are together referred to as corolla. These are scented and attract insects for pollination.

(iii) Stamen It is a male reproductive organ of plant. These are the little stalks with swollen top and lies inside the ring of petals. The stamen is made up of two parts, i.e. filament and anther. The stalk of stamen is called filament and the swollen top of stamen is called anther.
Anther contains the pollen grain which have male gamete in it. Pollen grains are exposed when the anther ripens and splits. These appear as the yellow powder like substance which is sticky in nature. Flowers usually have a number of stamens in it.

(iv) Pistil It is the female reproductive part of a flower that lies in the centre of a flower. These are flask-shaped structure which is made up of three parts, i.e. stigma, style and ovary.
The top part of pistil is called stigma. It receives the pollen grains from the anther during pollination. The middle part of the pistil is tube-like structure called style which connects stigma to the ovary.

The swollen bottom part at the base of pistil is called ovary.
The ovary makes ovules and stores them. These ovules contain the female sex cells also called as egg cell. It is the female gamete of flower. Pistil is also called as carpel. The pistil is surrounded by several stamens.
The base of the flower on which all the parts of flower are attached is called receptacle.

Question 19.
Observe the given figure?

(a) Which plant is this? Give the name of the plant.
(b) What does it shows?
(c) From where the new plants are developing?
Answer:
(a) The given figure is of Bryophyllum leaf. It is also called as sprout leaf plant.
(b) It shows vegetative reproduction by leaves.
(c) The leaves of Bryophyllum develops some buds in its margin or edges which grow into new plants, when buried in the soil.

Question 20.
The process of layering is commonly used in jasmine for reproduction. Explain how this process of layering is performed in jasmine. [HOTS]
Answer:
Layering is a method of vegetative reproduction in branches. In this method, a mature branch of parent plant is bent down and covered with soil.
The tip of the branch is kept above the ground. After few days the roots are developed from the branch buried under the soil and develops into a new plant. This method is done in the plants that have long and slender branches, e.g. jasmine.

Reproduction in Plants Class 7 Science Extra Questions Long Answer Type Questions

Question 1.
In the figure of a flower given below, label the parts whose functions are given below and give their names.
(a) The part which contains pollen grains.
(b) The part where the female gamete is formed.
(c) The female reproductive part, where pollen grains germinate.
(d) The colourful part of flower which attracts insects[NCERT Exemplar]

Answer:
The various parts of a flower whose functions are mentioned above are labelled as follows :

Question 2.
Observe the following figure and answer the following questions.
(a) Which type of vegetative propagation has been shown in this figure?
(b) Name two plants where this method of vegetative reproduction takes place.
(c) Is this a sexual or asexual mode of reproduction?
(d) Label the part (A) and (B) in the given figure.

Answer:
(a) The given figure shows grafting method (vegetative reproduction).
(b) Mango and rose are the two plants where this method of vegetative takes place.
(c) It is asexual method of reproduction.
(d) (A) Scion (B) Stock

Question 3.
Explain the process of reproduction in plants, involving the fusion of cells from male and female parts of a flower.
Answer:
When the reproduction in an organism includes two types of gametes, i.e male and female from two different parents, it is called sexual reproduction. The sexual reproduction takes place by the fusion of male and female gametes by the process called fertilisation to form zygote.

Sexual reproduction [fertilisation) in plants The different steps that take place during sexual reproduction in plants are

The pollens are deposited on stigma and begins to germinate.
Pollen tube containing male gametes reaches to the ovary of flower.
The tip of the pollen tubes gets dissolved and male gametes comes out of the pollen tube.
Inside the ovary male gametes fuse with the female gamete or egg present in the ovule.
The fusion of both the gametes will result into a fertilised egg cell which is also called as zygote.

Question 4.
Name some fruit bearing plants. Now make a table and describe the method of seed dispersal in these fruits as well as the part which helps in the seed dispersal.
Answer:
The method of seed dispersal in the fruits and the parts which help in the seed dispersal:
Name of fruits bearing plants Agents through which seeds are dispersd Parts or seeds which helps in dispersal

Name of fruits bearing plants	Agents through which seeds are dispersd	Parts or seeds which helps in dispersal
Drumstick	Air/wind	Winged seeds
Sunflower	Wind	Hairy seeds
Gokhru	Animal	Spines and Hooks
Coconut	Water	Fibrous coating
Lotus	Water	Thalamus float in water
Poppy	Explosion	Pericarp bursts
Grass	Wind	minute, light weighted seeds

Question 5.
Observe the given figure and answer the following questions.
(a) Name the plant?
(b) Which type of reproduction is seen in this plant?
(c) Is ginger a root or stem?
(d) Label the part of this plant?

Answer:
(a) The given figure is of ginger tuber.
(b) Asexual reproduction is seen in this plant.
(c) Ginger is a stem.
(d) Various parts of this plant can be shown as follows:

Question 6.
Ria went to a plant nursery with her mother. The gardner approached them and asked about their choice of plant. Ria’s mother wanted a flowering plant with fragrance. Gardner showed them a plant and told them that this variety has been prepared by a method of vegetative propagation of stems.
Ria grew curious and asked some questions to gardner.
(a) What is vegetative propagation?
(b) What are methods of producing new plants using stem?
(c) Name any two plants where this method of reproduction is used. Is this a asexual or sexual method?
(d) What values do you observe in Ria? [Value Based Question]
Answer:
(a) Vegetative propagation is the process of reproduction in which new plants are produced from different parts of old plants like stem, roots or leaves.
(b) Cutting and grafting method are two methods where stem is used for vegetative propagation.
(c) Mango and rose are two plants cultivated using vegetative propagation.
This is an asexual method of reproduction.
(d) Ria is curious, inquiring who wants to gain knowledge about process occurring around here.

Question 7.
Ravi was sitting in a garden with his family. His younger sister comes running with different types of flowers and starts to name their parts. She stops as she forgets some names and Ravi noticing this comes to help her.
His sister askes many questions related to flowers and Ravi answers her with all information he has.
(a) What is a flower and mention its function during reproduction?
(b) Flowers help in pollination and fertilisation. Yes/No? Give reason.
(c) Do all flowers have both male and female parts on them specify?
(d) What values are observed in Ravi and his sister? (Value Based Question]
Answer:
(a) A flower is the reproductive part of plant which helps in sexual reproduction.
A flower ensures the occurrence of process of fertilisation.
(b) Yes, flower aids in both pollination and fertilisation. Different colours and fragrance of flowers attract insects to them causing dispersal of pollen grain ensuring pollination.
Fertilisation occurs in the ovary of the flower after pollination of male and female parts.
(c) Number of some flowers can either have male or female parts on them, these flowers are called unisexual.
Some flowers have both, the male and female parts on them so they are called bisexual flowers.
(d) Ravi is aware, sincere, helpful as he helps others and knowledgeable while his sister is inquisitive, ready to acquire new knowledge and observant.

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CHAPTER – 11 Transportation in Animals and Plants | CLASS 7TH | NCERT SCIENCE IMPORTANT QUESTIONS & MCQS | EDUGROWN

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 11 Transportation in Animals and Plants

MCQs

Question 1.
Which of the following is the main circulatory fluid in our body ?
(a) Plasma
(b) Lymph
(c) Blood
(d) None of these

Answer

Answer: (c) Blood

Question 2.
Which one of the following contains haemoglobin?
(a) RBC
(b) WBC
(c) Platelets
(d) None of these

Answer

Answer: (a) RBC

Question 3.
What is the function of WBCs?
(a) Transport of oxygen
(b) Fight against germs
(c) Involved in blood clotting
(d) All of these

Answer

Answer: (b) Fight against germs

Question 4.
Blood platelets help in
(a) formation of urine
(b) excretion of urine
(c) sweating
(d) blood clotting

Answer

Answer: (d) blood clotting

Question 5.
The muscular tube through which stored urine is passed out of the body is called:
(a) kidney
(b) ureter
(c) urethra
(d) urinary bladder

Answer

Answer: (c) urethra

Question 6.
They are pipe-like, consisting of a group of specialised cells. They transport substances and form a two-way traffic in plants. Which of the following terms qualify for the features mentioned above?
(a) Xylem tissue
(b) Vascular tissue
(c) Root hairs
(d) Phloem tissue

Answer

Answer: (d) Phloem tissue

Question 7.
The absorption of nutrients and exchange of respiratory gases between blood and tissues takes place in:
(a) veins
(b) arteries
(c) heart
(d) capillaries

Answer

Answer: (d) capillaries

Question 8.
In which of the following parts of human body are sweat glands absent?
(a) Scalp
(b) Armpits
(c) Lips
(d) Palms

Answer

Answer: (c) Lips

Question 9.
In a tall tree, which force is responsible for pulling water and minerals from the soil?
(a) Gravitational force
(b) Transportation force
(c) Suction force
(d) Conduction force

Answer

Answer: (c) Suction force

Question 10.
Aquatic animals like fish excrete their wastes in gaseous form as
(a) Oxygen
(b) Hydrogen
(c) Ammonia
(d) Nitrogen

Answer

Answer: (c) Ammonia

Match the following:

Column A	Column B
(i) Ammonia	(a) Food
(ii) Urea	(b) Carbon dioxide rich blood
(iii) Uric acid	(c) Water and salts
(iv) Vein	(d) Carry oxygen
(v) Artery	(e) Aquatic animals like fish
(vi) Xylem	(f) Fight against germs
(vii) Phloem	(g) Blood clotting
(viii) Sweat	(h) Absorb water
(ix) Root hair	(i) Humans
(x) Platelets	(j) Oxygen rich blood
(xi) RBCs	(k) Water and minerals
(xii) WBCs	(l) Birds, snakes, lizards

Answer

Answer:

Column A	Column B
(i) Ammonia	(e) Aquatic animals like fish
(ii) Urea	(i) Humans
(iii) Uric acid	(l) Birds, snakes, lizards
(iv) Vein	(b) Carbon dioxide rich blood
(v) Artery	(j) Oxygen rich blood
(vi) Xylem	(k) Water and minerals
(vii) Phloem	(a) Food
(viii) Sweat	(c) Water and salts
(ix) Root hair	(h) Absorb water
(x) Platelets	(g) Blood clotting
(xi) RBCs	(d) Carry oxygen
(xii) WBCs	(f) Fight against germs

Fill in the blanks:

1. Blood is the fluid which flows in the ………………..

Answer

Answer: blood vessels

2. ……………….. binds with oxygen and transports it.

Answer

Answer: Haemoglobin

3. ……………….. fight against germs.

Answer

Answer: WBCs

4. The heart has ……………….. chambers.

Answer

Answer: four

5. A doctor uses the ……………….. to feel the heart beat.

Answer

Answer: stethoscope

6. ……………….. discovered the circulation of blood.

Answer

Answer: William Harvey

7. Filtering the blood periodically through an artificial kidney is called ………………..

Answer

Answer: dialysis

8. Plants absorb water and minerals by the ………………..

Answer

Choose the true and false statements from the following:

1. Blood carries carbon dioxide from the body parts to the lungs.

Answer

Answer: True

2. White blood cells are involved in the clotting of blood.

Answer

Answer: False

3. There are valves present in veins.

Answer

Answer: True

4. Heart acts as a pump for the transport of blood.

Answer

Answer: True

5. Arteries have thin walls while veins have thick walls.

Answer

Answer: False

6. Human heart has two chambers, an atrium and a ventricle.

Answer

Answer: False

7. Partition between the chambers of heart avoid mixing up of blood rich in oxygen with the blood rich in carbon dioxide.

Answer

Answer: True

8. Stethoscope is a device to amplify the sound of the heart.

Answer

Answer: True

9. Rhythmic beating of the various chambers of the heart maintain circulation of blood to the different parts of the body.

Answer

Answer: True

10. When blood reaches the two kidneys, it contains only harmful substances.

Answer

Answer: False

Question 1 .
Name some useful products or materials that are carried by blood.
Answer:
Food, water and oxygen are the products that are carried by blood to every part of the body.

Question 2.
Circulatory system consists of three major organs. Name those organs.
Answer:
The circulatory system consists of three major organs, i.e., blood, blood vessels and heart.

Question 3.
Give the name of blood component which is liquid and contains 90% water.
Answer:
The sticky liquid part of the blood, containing 90% water is called plasma.

Question 4.
Name the organ which is located in the chest cavity with its lower tip slightly tilted towards the left. [NCERT Exemplar; HOTS]
Answer:
The heart is located in the chest cavity with its lower tip slightly tilted towards the left.

Question 5.
RBC contains a red coloured pigment which carries oxygen with it. What is the pigment called?
Answer:
The red coloured pigment of RBC that carries oxygen with it is called haemoglobin.

Question 6.
Veins have valves which allow blood to flow only in one direction. Arteries do not have valves. Yet the blood flows in one direction only. Can you explain why? [NCEAT Examplar; HOTS]
Answer:
Veins have valves to prevent blood from flowing backwards and pooling, whereas arteries pump blood at very high pressures, which naturally prevents back flow

Question 7.
In which form, the oxygen is transported to various body parts by haemoglobin?
Answer:
The red pigment, haemoglobin binds with oxygen to form oxyhaemoglobin which is transported to various body parts.

Question 8.
Certain greenish-blue lines appear just below the skin of our hands and leg. What are these?
Answer:
The greenish-blue lines that appear just below the skin of our hands and legs are veins.

Question 9.
Human blood group is divided into how many groups? Name them.
Answer:
Human blood group is divided into four groups. These are A, B, AB and O.

Question 10.
Waste carbon dioxide and urea are removed from our body by which organs?
Answer:
The carbon dioxide is removed by lungs while urea is removed from the body by kidney.

Question 11.
Blood is a fluid connective tissue. Justify.
Answer:
Blood is a fluid tissue which connects all the parts of body with each other.

Question 12.
Blood performs various functions including protection against infections. How?
Answer:
Blood contains WBC which forms the defense of our body. They eat antigens and fights aganist infections.

Question 13.
Pulse rate can indicate the health states of an individual. How?
Answer:
Pulse rate will increase or decrease from normal rate if a person is not well.

Question 14.
Usually veins carry deoxygenated blood except in one case. Specify.
Answer:
Pulmonary veins carry oxygenated blood from lungs to heart.

Question 15.
State the function of valves ?
Answer:
Valves prevent the back flow of blood between chambers of heart.

Question 16.
Name the functional units of the major excretory organ of humans.
Answer:
Kidney is the major excretory organ and nephrons are its functional units.

Question 17.
Exchange of gases, food and other substances occurs between arteries and veins. How does this exchange happen?
Answer:
Exchange of substances between arteries and veins occurs via capillaries.

Question 18.
Measuring of heartbeats is a significant step during health checkups. Name the instrument used for the same.
Answer:
Stethoscope

Question 19.
Kidneys are the major excretory organs in humans. How will the waste products released will be excreted out if the kidneys are damaged or unfunctional? [HOTS]
Answer:
Artificial ways of waste removal are used like dialysis which are referred to as artificial kidneys.

Question 20.
Arteries have a very thick and elastic walls. Why?
Answer:
Arteries carry blood at a very high pressure due to pumping action of heart hence, the need of thicker walls.

Question 21.
Skin is also considered as an excretory organ. Give reason if you agree. [HOTS]
Answer:
Yes, skin is an excretory organ as it secretes waste products by releasing sweat from the surface.

Question 22.
Heart has three chambers, two ventricles and one atrium. Is it right or wrong?
Answer:
Wrong, the heart has four chambers. Two auricles and two ventricles.

Question 23.
Arteries and veins carry blood to and from the heart. Which of these carry the blood?
(a) Back to the heart from all organs.
(b) Away from heart for distribution in all organs.
Answer:
(a) Veins
(b) Arteries

Question 24.
Urine is called an excretory product. Why?
Answer:
Urine is the mixture of urea and other unwanted salts with water which is needed to be excreted out as its presence in blood can make a person ill.

Question 25.
Sponges and Hydra do not possess any circulatory system then how do they carry out distribution of food and other substances?
Answer:
The water in which these organisms live brings them food and oxygen as it enters their bodies.

Question 26.
If the heartbeats of a person are more than 72-80 beats per minute. What does it signify?
Answer:
The faster heartbeats signify that heart is pumping more blood to the organs as they need increased oxygen and energy supply.

Question 27.
What is the purpose of using stethoscope by doctors?
Answer:
A stethoscope reads heartbeats as diaphragm amplies the rounds of heartbeat when placed on specific areas.

Question 28.
Urinary bladder is the part of human excretory system. What is its role in excretion?
Answer:
Bladder stores the excretory product released after filtration from kidney and excrete it out at specific times.

Transportation in Animals and Plants Class 7 Science Extra Questions Short Answer Type Questions

Question 1.
Arrange the following statements in the correct order in which they occur during the formation and removal of urine in human beings.
(a) Ureters carry urine to the urinary bladder.
(b) Wastes dissolved in water is filtered out as urine in the kidneys.
(c) Urine stored in urinary bladder is passed out through the urinary opening at the end of the urethra.
(d) Blood containing useful and harmful substances reaches the kidneys for filtration.
(e) Useful substances are absorbed back into the blood.
Answer:
The correct order of the formation and removal of urine in human beings is
(d) Blood containing useful and harmful substances reaches the kidneys for filtration.
(e) Useful substances are absorbed back into the blood.
(b) Wastes dissolved in water is filtered out as urine in the kidneys.
(a) Ureters carry urine to the urinary bladder.
(c) Urine stored in urinary bladder is passed out through the urinary opening at the end of the urethra.

Question 2.
Name the tissues of a plant which carries
(a) water and minerals from roots to the leaves.
(b) food from the leaves to the other parts of the plant.
Answer:
The tissue which carries
(a) water and minerals from roots to leaves is xylem.
(b) food from the leaves to the other parts of the plant is phloem.

Question 3.
Look at figure and draw another figure of the same set up as would be observed after a few hours.

Answer:
After the few hours, the figure will be shown as follows

This figure shows that there will be an increase in the level of sugar solution in the potato piece. This increase in the level of sugar solution rises due to water that passes throGgh the wall of potato and goes inside it.

Question 4.
(a) Name the only artery that carries carbon
dioxide rich blood.
(b) Why is it called an artery if it does not carry oxygen-rich blood? [NCERT Exemplar]
Answer:
(a) The only artery that carries carbon dioxide rich blood is pulmonary artery.
(b) The main function of artery is to carry blood away from heart. Also arteries have thick wall and do not contain valves in them. Blood flow in arteries, takes place at high pressure. All these characteristics are found in pulmonary artery. It carries deoxygenated blood from heart to lungs for oxygenation, therefore it is called artery.

Question 5.
Name the process and the organ which help in removing the following wastes from the body.
(a) Carbon dioxide
(b) Undigested food
(c) Urine
(d) Sweat [NCERT Exemplar]
Answer:

	Waste	Process	Organ
(a)	Carbon dioxide	Exhalation	Lungs
(b)	Undigested food	Egestion	Large intestine and anus
(c)	Urine	Excretion	Kidneys
(d)	Sweat	Perspiration (sweating)	Sweat glands

Question 6.
Observe given figure and answer the given question.

(a) Name the instrument.
(b) Label the parts A, B and C. [NCERT Exemplar]
Answer:
(a) The name of the given instrument is stethoscope.
(b) Labelled diagram of stethoscope.

Question 7.
What is the relation between the rate of heartbeat and pulse rate? If a pulse rate of an athlete Is 96/min, what will be the number of his heartbeat at the same time? [HOTS]
Answer:
The rhythmic contraction and relaxation of the muscles of the heart is called heartbeat. Whereas, the rhythmical throbbing of the arteries as the blood is pushed forward through them is called pulse. It can be felt in the wrist, temples, etc.
Pulse rate is the number of heartbeats per minutes. The number of heartbeat is equal to the number of pulse per minute.
Therefore, if a pulse rate of an athlete is 96/min then the number of his heartbeat at the same time will also be 96/min

Question 8.
Give one function of each of the following organs,
(a) Blood vessels
(b) Kidney
(c) Blood platelets
(d) Heart
Answer:
The main function of the following organs are as follows:
(a) Blood vessels These run between the heart and the rest of the body. It helps in the transport of blood between heart and various organs of the body.
(b) Kidney It is called as the ‘magic filters’. It helps in the removal of unwanted substances like urea from the blood.
(c) Blood platelets This component of blood helps in blood clotting and prevents the blood loss from the body.
(d) Heart It is a pumping organ which receives blood from the body through veins and pumps it with enough force into the arteries from where it is carried to the various body parts.

Question 9.
Paheli noticed water being pulled up by a motor pump to an overhead tank of a five storeyed building. She wondered how water moves up to great heights in the tall trees standing next to the building. Can you tell why? [NCERT Exemplar; HOTS]
Answer:
When the water is pulled up by a motor-pump to an overhead tank of a five storeyed building, it moves to a great height due to the suction pull. This pull forms the continuous column of water and water rises up to a great height. Similarly, when transpiration occurs in the plants, water is evaporated and this creates a suction pull in the plants.

Due to this suction pressure, water from the soil rises up through the roots of the plants and reaches to a great height in tall plants.

Question 10.
How is transpiration and translocation different from each other.
Answer:
The differences between transpiration and translocation are

Transpiration	Translocation
The evaporation of water from the leaves of plant is called transpiration.	The transport of soluble products of photosynthesis from leaves (from where they are formed) to the other parts, of plants is called translocation.
It takes place through stomata present in the lower surface of leaf.	It occurs in the part of the vascular tissue known as phloem.

Question 11.
Make a table depicting the function of all chambers of the human heart.
Answer:
The human heart is divided into four chambers, i. e. upper two atrium and lower two ventricles. The functions of these chambers can be tabulated as follows

Chamber	Function
Left atrium	Receives oxygenated blood from lungs through pulmonary veins and pours it into left ventricle.
Right atrium	Receives deoxygenated blood from various body parts through superior and inferior vena cava and pours it into right ventricle.
Left ventricle	Pumps oxygenated blood to various parts of body through aorta.
Right ventricle	Pumps deoxygenated blood into lungs through pulmonary artery.

Question 12.
How does the water move from root to leaves?
Answer:
The water moves from root to leaves with the help of specialised cells called vascular tissue. Transport of water and nutrients is done by xylem tissue present in plants.

Question 13.
Observe the given diagram of human heart and label all the parts from A to H.

Answer:

The heart is an organ which beats continuously as a pump for the transport of blood carrying other substances with it, through a network of tubes or blood vessels. The heart pumps blood throughout our life without stopping or relaxing.

Question 14.
The given diagram is of human excretory system. Label the marked parts of it.

Answer:
The various parts of human excretory system are as follows

Question 15.
Paheli says her mother puts ladyfinger and other vegetables in water if they are somewhat dry. She wants to know how water enters into them. [HOTS]
Answer:
By soaking the vegetables in water, the skin of the vegetables becomes moist and water starts moving from one cell to another until the vegetables are fresh again.

Question 16.
Why plants absorb a large quantity of water from the soil, then give it off by transpiration?
Answer:
Plants absorb a large quantity of water from the soil because they need nutrients which are dissolved in the water. The excess water evaporates through the stomata present on the leaf surface by the process of transpiration.

Question 17.
List some animals surrounding your locality group them into following groups.
(a) Animals that excrete ammonia in gaseous forms.
(b) Animals that excrete uric acid in the form of pellets.
(c) Animals that excrete urea in the form urine. [HOTS]
Answer:
Some animals that surround us are fish, frog, birds, tadpole larva, snake, cow, man, rat, monkey, lizard, toad and snail.
These can be grouped as follows
(a) Animals that excrete ammonia in gaseous form (i.e. ammonotelics)-Fish, tadpole larva.
(b) Animals that excrete uric acid in the form of pellets (i.e. uricotelics)—Bird, snake, rat, lizard, snail.
(c) Animals that excrete urea in the form of urine (i.e. ureotelics)-Frog, cow, man, monkey, toad.

Question 18.
Human have two major organs that perform transport of materials. Organ ‘A’ is bean-shaped and dark red in colour lie just above the waist. It helps in’removal of ‘Q’, a waste material from blood. The organ ‘S’ is the opening at the end of the urinary bladder through which the waste material is eliminated.
Organ ‘B’ lies in the chest cavity slightly tilted towards the left side. It pumps continuously and pours liquid ‘C’ into arteries and through very fine tube-like structure ‘D’ distributes the liquid to various parts of the body. What are the name of these organs. [HOTS]
Answer:
Organ ‘A’ is kidney which is bean-shaped and helps in the removal of urea (Q) which is a waste material from the blood. ‘S’ is urethra which is the small opening at the end of urinary bladder. Organ ‘B’ is heart which acts as pump. It pumps liquid blood continuously and pours into arteries, and through capillaries (D) which are fine tube-like structure, the blood is distributed to various parts of the body.

Question 19.
The major function of the arteries is to carry to oxygenated blood throughout the body and that of veins is to carry deoxygenated blood from body parts to heart for purification. There is one artery that carries deoxygenated blood and one vein that carries oxygenated blood. Name the artery and vein. [HOTS]
Answer:
The artery which carries deoxygenated blood or blood rich in CO₂ is pulmonary artery while the pulmonary vein is one which carries oxygenated blood. The pulmonary artery carries deoxygenated blood from heart to lungs while pulmonary vein carries oxygenated blood from lungs to heart.

Question 20.
Boojho’s uncle was hospitalised and put on dialysis after a severe infection in both of his kidneys.
(a) What is dialysis?
(b) When does it become necessary to take such a treatment?
Answer:
The normal functioning of kidney is necessary for good health of a person. But sometime s the kidney may stop working due to infection or injury. This condition of kidney is called kidney failure which may lead to the accumulation of urea in the blood of a person. Since, urea is a toxic substance which must be removed from the blood. Such person having kidney failure cannot survive unless his blood is filtered periodically through the artificial kidney machine to remove urea. The process used for cleaning the blood of a person by separating the waste product urea from it is called dialysis.This machine removes urea and other waste the product periodically.

The long term solution for the patient suffering from kidney failure is kidney transplantation. In this method, the diseased or damaged kidney is removed and matching kidney is donated by a healthy person. The donated kidney is transplanted in its place by performing surgery.

Question 21.
The internal structure of heart has four chambers.
(a) Name the upper chambers of heart.
(b) Name the lower chambers of heart.
Answer:
The vertical section of heart shows that heart is divided into four compartments called as chambers.
(a) The upper two chambers of heart are called atria or atrium.
(b) The lower two chambers of heart are called ventricles.

Question 22.
Explain in brief the main functions of the structural and functional unit of kidney in excretory system.
Answer:
Kidney is the major excretory organ which consists of thousands of tiny filters called nephrons. The major functions of nephron are

To filter blood at high pressure which helps in the separation of nitrogenous waste such as urea from the blood.
It helps in selective re-absorption of some substances (from the initial filtrate which is filtered at a very high pressure). These substances include glucose, amino acid, salts ancf a major amount of water.

Question 23.
What is the special feature present in a human heart which does not allow mixing of blood when oxygen-rich and carbon dioxide-rich blood reach the heart? [NCERT Exemplar]
Answer:
In human, the heart has four chambers. The two upper chambers are called the atria and the two lower chambers are called the ventricles. The partition between the chambers helps to avoid mixing up of blood rich in oxygen with the blood rich in carbon dioxide.

Question 24.
Paheli uprooted a rose plant from the soil. Most of the root tips with root hairs got left behind in the soil. She planted it in a pot with new soil and watered it regularly. Will the plant grow or die? Give reason for your answer. [NCERT Exemplar, HOTS]
Answer:
Possible answers are

Without the root hairs, the roots will not be able to absorb water and nutrients and the plant will die.
The stem of the rose plant may grow new roots and the plant will live.
The rose plant may not be able to survive in a different type of soils.

Transportation in Animals and Plants Class 7 Science Extra Questions Long Answer Type Questions

Question 1.
Priya’s grandfather was taken to the hospital as he was unable to perform excretory processes. Priya heard a nurse talking to her father that her grandfather’s has kidney failure and needs to undergo dialysis. Priya later asked her father as to what is dialysis process and why does grandpa needs it. Her father smiles and tells her all the facts associated with this process.
(a) What do you mean by dialysis?
(b) Why is there a need for dialysis in some people?
(c) Excretion is an important life process. How?
(d) Which is the major excretory organ in humans?
(e) What values do you observe in Priya? [Value Based Question]
Answer:
(a) Dialysis is the process used for cleaning of the blood by separating the waste products in an artificial medium.
(b) Dialysis is needed when the excretory organ of humans, i.e. kidney becomes damaged on unfunctional due to some injury or infection.
(c) Excretion process removes the waste products released in body after the utilisation of food and other components. These products are toxic and may harm us if not removed from our body.
(d) Kidney
(e) Priya is curious, sincere and aware eager to acquire new knowledge.

Question 2.
While learning to ride a bicycle, Boojho lost his balance and fell. He got bruises on his knees and it started bleeding. However, the bleeding stopped after sometime.
(a) Why did the bleeding stop?
(b) What would be the colour of the wounded area and why?
(c) Which type of blood cells are responsible for clotting of blood? [NCERT Exemplar]
Answer:
(a) When a cut or wound starts bleeding after sometime, a clot is formed which plugs the cut and bleeding stops.
(b) Wounded area becomes dark red in colour due to clotting of blood.
(c) The blood clot is formed due to the presence of the cells called platelets in the blood.

Question 3.
Like humans and animals, transportation of water, mineral and nutrients also take place in plants. How?
Answer:
Transport of Substances in Plants
Plants take up water and dissolved minerals from the soil through their roots and transport it to their leaves. The leaves use this water and mineral for synthesising their food by the process called photosynthesis.The food produced by green plants in transported back to all the parts of plant body.
Therefore, it is clear that plants also need a transport system for carrying water, minerals and food through various parts of their body.

Transport of Water and Minerals
Plant root absorbs the water and mineral from the soil. The roots possess root hair which increase the surface area of the root for absorption of water and minerals nutrient that is dissolved in the water. It is moved from roots up to the stem and leaves through the tube-like tissue called as xylem.

Absorption and flow of water is a continuous process through the xylem tissue. Xylem tissues are the continuous network of channels which connect roots to the leaves through the stem and branches. It thus transports water and minerals to the leaves of the entire plant.

Transport of Food Material
The food manufactured in the leaf is transported to different parts of plants. This transportation of food material from leaves to the other parts of plants is carried out by the tissue called phloem and the process of transport of food material is called translocation. The phloem consists of vessels that are known as sieve tubes.

Question 4.
Blood from heart is carried by certain tube-like structure. What are they? Give the structure and functions of different types of blood carrying tubes.
Answer:
These are tubes or pipes that carry blood throughout the body. It runs between the heart and the rest of the body. There are three major types of blood vessels in the body, i.e. arteries, veins and capillaries.

1. Arteries: These carry blood from the heart to all the parts of body. These lie quite deep under our skin and cannot be seen easily. Arteries have thick elastic walls as the blood flows at high pressure due to pumping action from heart through arteries. No valves are present in the arteries. The main artery, i.e. aorta is connected to the left ventricle of the heart. It carries oxygenated blood from the left ventricle to all the parts of body except the lungs. Another artery called pulmonary artery is connected to the right ventricle of the heart and carries deoxygenated blood from the right ventricle to the lungs.
Note: The arteries normally carry oxygenated blood from the heart but one artery called pulmonary artery carries deoxygenated blood from heart to lungs.

2. Veins: These are the blood vessels that carry blood from all the parts of the body back to the heart. These tube-like blood vessels are situated just under the skin and can easily be seen as greenish-blue tubes or lines below the skin. These carry deoxygenated blood from the body parts to heart. Veins have thin walls and blood flows at low pressure through the veins. Therefore, veins have valves in them which allow the blood to flow in one direction and prevent the back flow of blood in veins.
Usually veins carry deoxygenated blood but pulmonary vein that is connected to the left atrium of the heart, carries oxygenated blood from the lungs to the heart.

Functions of Blood
Various functions of blood are

It transports substances like digested food from the small intestine to the other parts of the body.
It carries water to all the parts of the body.
It carries oxygen and C02 during circulation.
It carries waste products like urea from liver to kidney for excretion in urine.
It protects the body from disease.

Question 5.
While riding a bike, Mason fell from it due to loss of balance. He got up and realised that he was bleeding from several wounds badly. He panicked and started to run but Mansi who was looking at him, stopped him and told him to clean his wound with a clean cloth and that blood will stop coming in a while. Mason noticed that he has stopped bleeding after sometime and a hard covering was appearing on his wounds.
(a) Why did the bleeding stop after a while?
(b) What is blood and what type of cells are responsible for clotting?
(c) What of values do you think Mansi have? [Value Based Question]
Answer:
(a) Bleeding stops after sometime because some specialised cells start forming a hard covering called clot at the site of wound.
(b) Bleeding is a fluid connective tissue present in all parts of the body. Platelets are responsible for formation of clot.
(c) Mansi is helpful, knowledgeable and interested in science subject.

Question 6.
Read the following terms given below,
root hairs xylem urethra
arteries kidneys veins
atria capillaries heart
ureter phloem urinary bladder
Group the terms on the basis of the categories given below.
(a) Circulatory system of animals.
(b) Excretory system in human.
(c) Transport of substances in plants. {NCERT Exemplar]
Answer:
The terms on the basis of the categories can be grouped as follows
(a) Circulatory system of human Arteries, atria, capillaries, veins, heart.
(b) Excretory system in human
Urethra, kidneys, ureter, urinary bladder.
(c) Transport of substances in plants Root

Question 7.
(a) What are the different blood groups in human?
(b) Define blood group compatibility.
(c) Make a table to show the donor blood group and recipient blood group.
Answer:
(a) The blood group of an individual human being always remains unchanged throughout their life. Karl Landsteiner described that human blood can be divided into four groups, i.e. A, B, AB and O. These are named on the basis of substance present in the blood (RBC). Every man has one of these four groups of blood which is inherited from parents to offspring and is never changed.

If a person gets injured and heavy blood loss occurs, there is a need to give blood of other person to the patient. The person who gives the blood is called donor while the person who receives the blood is called
recipient.

(b) The process of donation of blood from one person to another is called blood transfusion. Before donation, the blood group must be matched because transfusion of different groups can be dangerous. The RBCs of the patient receiving blood will stick together and may cause death of the patient. This matching of blood group is called blood group compatibility. It can be shown as follows :

(c)

Blood group	Can donate blood to	Can receive blood from
A	A and AB	A and O
B	B and AB	B and O
AB	AB	All the group, i.e. A, B, AB and O
O	All the group, i.e. A, B, AB and 0	O

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RS AGARWAL SOLUTION | CLASS 7TH | CHAPTER-10| Percentage | EDUGROWN

Exercise 10A

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 1

Question 2.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 3

Question 3.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 4

Question 4.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 5

Question 5.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 6

Question 6.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 8

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 9

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 11

Question 9.
Solution:
Let x is the required number
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 12

Question 10.
Solution:
Let x be the required number
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 13

Question 11.
Solution:
10 % of Rs. 90 = 90 x 10100 = Rs. 9
Required amount = Rs. 90 + Rs. 9 = Rs. 99

Question 12.
Solution:
20 % of Rs. 60 = 60×20100 = 12
Required amount = Rs. 60 – 12 = Rs. 48

Question 13.
Solution:
3 % of x = 9
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 14

Question 14.
Solution:
12.5 % of x = 6
⇒ x x 12.5100 = 6
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 15

Question 15.
Solution:
Let x % of 84 = 14
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 16

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10A 17

Exercise 10B

Question 1.
Solution:
Rupesh seemed 495 marks out of 750
Percentage of marks = 495750 x 100 = 66%

Question 2.
Solution:
Monthly salary = Rs. 15625
Increase = 12%
Amount of increase = 15625×12100 = Rs. 1875
New salary = Rs. 15625 + Rs. 1875 = Rs. 17500

Question 3.
Solution:
Excise duty in the beginning = Rs. 950
Reduced duty = Rs. 760
Reduction = Rs. 950 – Rs. 760 = Rs. 190
Reduction percent = 190×100950 = 20%

Question 4.
Solution:
Let x be the total cost of the T.V
96% of x = 10464
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 1
Total cost of T.V = Rs. 10900

Question 5.
Solution:
Let number of students = x
In a school boys = 70%
girls = 100 – 70 = 30%
Now 30% of x = 504
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 2
Number of boys = 1680 – 504 = 1176
and number of total students = 1680

Question 6.
Solution:
Copper required = 69 kg
copper in ore = 12%
Let quantity of ore = x kg
12% of x = 69
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 3
Quantity of ore = 575kg

Question 7.
Solution:
Pass marks = 36%
A students gets marks = 123
But failed by 39 marks
Pass marks = 123 + 39 = 162
Now, 36% of maximum marks = 162
Maximum marks = 162×10036 = 450 marks

Question 8.
Solution:
Let number of apples = x
Number of apples sold = 40% of x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 4
Hence number of apples = 700

Question 9.
Solution:
Let total number of examinees = x
the numbers of examinees who passed = 72% of x
= xx72100
= 18×25
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 5

Question 10.
Solution:
Let the gross value of moped = x
Amount of commission = 5% of x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 6

Question 11.
Solution:
Total gunpowder = 8 kg
Amount of nitre = 75%
amount of sulphur = 10%
Rest of powder which is charcoal = 100 – (75 + 10) = 100 – 85 = 15 = 15%
Amount of charcoal = 8 x 15100 = 120100
= 65 kg = 1 kg 200 grams = 1.2 kg

Question 12.
Solution:
Quantity of chalk = 1 kg or 1000 g
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 7

Question 13.
Solution:
Let total number of days on which the
school open = x
and Sonal’s attendance = 75%
x x 75% = 219
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 8
No. of days on which was school open = 292 days

Question 14.
Solution:
Rate of commission = 3%
Amount of commission = Rs. 42660
Let value of property = x
then 3% of x = 42660
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 9

Question 15.
Solution:
Total votes of the constituency = 60000
Votes polled = 80% of total votes
= 80100 x 60000 = 48000
Votes polled in favour of A = 60% of polled votes
= 60100 x 48000 = 28800
Votes polled in favour of B = 48000 – 28800 = 19200

Question 16.
Solution:
Let original price of shirt = Rs. x
Discount = 12%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 10
Original price of shirt = Rs. 1350

Question 17.
Solution:
Let original price of sweater = x
Rate of increase = 8%
Increased price = x + 8% of x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 11
Hence, original price of sweater = Rs. 1450

Question 18.
Solution:
Let total income = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 12

Question 19.
Solution:
Let the given number = 100
Then increase % = 20%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 14
Decrease = 100 – 96 = 4
Decrease per cent = 4%

Question 20.
Solution:
Let original salary of the officer = Rs. 100
Increase = 20%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 15

Question 21.
Solution:
Rate of commission = 2% on first Rs. 200000
1 % on next Rs. 200000 and 0.5% on remaining price
Sale price of property = Rs.200000 + 200000 +140000 = Rs. 540000
Now commission earned by the
= Rs. 200000 x 2% + Rs. 200000 x 1% + 140000 x 0.5%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 17

Question 22.
Solution:
Let Akhil’s income = Rs. 100
Then income of Nikhil’s will be = Rs. 100 – 20 = Rs. 80
Amount which is more than that of Akhil’s = 100 – 80 = Rs. 20
% age = 20×10080 = 25%

Question 23.
Solution:
Let income of Mr Thomas = Rs. 100
then income of John = Rs. 100 + 20 = Rs. 120
Income of Mr Thomas is less than John = Rs. 120 – 100 = Rs. 20
% age = 20×100120
= 503 = 1623 %

Question 24.
Solution:
Present value of machine = Rs. 387000
Rate of depreciation = 10%
Let 1 year ago the value of machine was = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 18
1 year ago, value of machine = Rs. 430000

Question 25.
Solution:
Present value of car = Rs. 450000
Rate of decreasing of value = 20%
Value after 2 years
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 19

Question 26.
Solution:
Present population = 60000
Rate of increase = 10%
Increased population after 2 years
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 20

Question 27.
Solution:
Let the price of sugar = Rs. 100
and consumption = 100 kg.
Increase price of 100 kg = Rs. 100 + 25 = Rs. 125
Now increased amount on 100 kg = Rs. 125
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10B 21

Exercise 10C

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(b) 34 = 34 x 100 = 75 %

Question 2.
Solution:
(c)
2 : 5 = 25 = 25 x 100 = 40%

Question 3.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 1

Question 4.
Solution:
(c) x% of 75 = 9
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 2

Question 5.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 3

Question 6.
Solution:
(b)
Let x% of 1 day = 36 minutes
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 4

Question 7.
Solution:
(a)
Let x be the required number
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 5

Question 8.
Solution:
(b)
Let x be the required number, then
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 6

Question 9.
Solution:
(d)
Let ore = x, then
5% of x = 400g
xx5100 400
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 7

Question 10.
Solution:
(b)
Let gross value of T.V = x
Commission = 10%
After deducting commission, the value
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 8

Question 11.
Solution:
(b)
Increase in salary = 25%
Let original salary = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 9

Question 12.
Solution:
(c)
Let x be the number of total examinees
No. of examinees passed
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 10

Question 13.
Solution:
(c)
Let total number of apples = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 11

Question 14.
Solution:
(b)
Present value of machine = Rs. 25000
Rate of depreciation = 10% p.a.
Value of machine after one year
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 12

Question 15.
Solution:
(c) Let x be numbers
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 14

Question 16.
Solution:
(c) 60% of 450
= 60100 x 450 = 270

Question 17.
Solution:
(d) Rate of reduction = 6%
Price after reduction = Rs. 658
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 15

Question 18.
Solution:
(b) Boys = 70% of students
No. of girls = 240
Girls percentage = 100 – 70 = 30%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 16

Question 19.
Solution:
(c) Let number = x
11% of x – 7% of x = 18
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 17

Question 20.
Solution:
(a) Let number = x
35% of x + 39 = x
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 18

Question 21.
Solution:
(c) Pass marks = 36%
A students get =145 marks
But failed by 3 5 marks
Then pass marks = 145 + 35 = 180
Maximum marks = 180×10036 = 500

Question 22.
Solution:
(d) Let number be = x
Then decreasing by 40%
RS Aggarwal Class 7 Solutions Chapter 10 Percentage Ex 10C 19

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RS AGARWAL SOLUTION | CLASS 7TH | CHAPTER-9| Unitary Method | EDUGROWN

Exercise 9A

Question 1.
Solution:
Cost of 15 oranges = Rs. 110
Cost of 1 orange = Rs. 11015
and cost of 39 oranges = Rs. 11015 x 39
= Rs. 22 x 13 = Rs. 286

Question 2.
Solution:
In Rs. 260, the sugar is bought = 8 kg
and in Re. 1, the sugar is bought = 8260 kg
Then in Rs. 877.50, the sugar will be bought = 8260 x 877.50 kg
= 8260 x 87750100
= 27 kg

Question 3.
Solution:
In Rs. 6290, silk is purchased = 37 m
and in Re. 1, silk is purchased = 376290 m
and in Rs. 4420, silk will be purchased 37
= 376290 x 4420 m = 26 m

Question 4.
Solution:
Rs. 1110 is wages for = 6 days.
Re. 1 will be wages for = 61110 days
and Rs. 4625 will be wages for
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 1

Question 5.
Solution:
In 42 litres of petrol, a car covers = 357 km
and in 1 litre, car will cover = 35742 km
and in 12 litres, car will cover = 35742 x 12 = 102 km

Question 6.
Solution:
Cost of travelling 900 km is = Rs. 2520
and cost of 1 km will be = Rs. 2520900
andcostof360kmwillbe = Rs. 2520900 x 360 = Rs. 1008

Question 7.
Solution:
To cover a distance of 51 km, time is taken = 45 minutes
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 2

Question 8.
Solution:
If weight is 85.5 kg, then length of iron rod = 22.5 m
If weight is 1 kg, then length of rod will be
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 4

Question 9.
Solution:
In 162 grams, sheets are = 6
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 5

Question 10.
Solution:
1152 bars of soap can be packed in 8 cartons
1 bar of soap coil be packed in
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 6

Question 11.
Solution:
In 44 mm of thickness, cardboards are = 16
In 1 mm of thickness, cardboards will be
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 7

Question 12.
Solution:
If length of shadow is 8.2 m, then
height of flag staff is = 7 m
If length of shadow is 1 m, then height will
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 8

Question 13.
Solution:
16.25 m long wall is build by = 15 men
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 9

Question 14.
Solution:
1350 litres of milk cm be consumed by = 60 patients
1 litres of milk can be consumed by = 601350 patients
and 1710 litres of milk can be consumed
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 10

Question 15.
Solution:
2.8 cm extension is produced by = 150 g.
1 cm extension will be produced by = 1502.8 g
and 19.6 cm extension will be produced by
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9A 11

Exercise 9B

Question 1.
Solution:
48 men can dig a trench in = 14 days
1 man will dig the trench in = 14 x 48 days (less men more days)
28 men will dig the trench m = 14×4828 (more men less days)
= 24 days

Question 2.
Solution:
In 30 days, a field is reaped by = 16 men
In 1 day, it will be reaped by = 16 x 30 (less days, more men)
and in 24 days, it will be reaped by = 16×3024 men (more days, less men)
= 48024
= 20 men

Question 3.
Solution:
In 13 days, a field is grazed by = 45 cows.
In 1 day, the field will be grazed by = 45 x 13 cows (less days, more cows)
and in 9 days the field will be grazed by = 45×139 cows (more days, less cows)
= 5 x 13 = 65 cows

Question 4.
Solution:
16 horses can consume corn in = 25 days
1 horse will consume it in = 25 x 16 days (Less horse, more days)
and 40 horses will consume it in = 25×1640 days (more horses, less days)
= 10 days

Question 5.
Solution:
By reading 18 pages a day, a book is finished in = 25 days
By reading 1 page a day, it will be finished = 25 x 18 days (Less page, more days)
and by reading 15 pages a day, it will be finished in = 25×1815 days (more pages, less days)
= 5 x 6 = 30 days

Question 6.
Solution:
Reeta types a document by typing 40 words a minute in = 24 minutes
She will type it by typing 1 word a minute in = 24 x 40 minutes (Less speed, more time)
Her friend will type it by typing 48 words 24 x 40 a minute in = 24×4048 minutes
(more speed, less time)
= 20 minutes

Question 7.
Solution:
With a speed of 45 km/h, a bus covers a distance in = 3 hours 20 minutes
= 313 = 103 hours
With a speed of 1 km/h it will cover the distance m = 10×453 h
(Less speed, more time)
and with a speed of 36 km/h, it will cover the distance in
= 10x453x36 hr (more speed, less time)
= 256 h
= 416 h
= 4 hr 10 minutes

Question 8.
Solution:
To make 240 tonnes of steel, material is sufficient in = 1 month or 30 days
To make 1 tonne of steel, it will be sufficient in = 30 x 240 days (Less steel, more days)
To make 240 + 60 = 300 tonnes of steel it will be sufficient in = 30×240300 days
= 24 days (more steel, less days)

Question 9.
Solution:
In the beginning, number of men = 210
After 12 days, more men employed = 70
Total men = 210 + 70 = 280
Total period = 60 days.
After 12 days, remaining period = 60 – 12 = 48 days
Now 210 men can build the house in = 48 days
and 1 man can build the house in = 48 x 210 days (less men, more days) .
280 men can build the house in = 48×210280 days
(more men, less days)
= 36 days

Question 10.
Solution:
In 25 days, the food is sufficient for = 630 men
In 1 day, the food will be sufficient for = 630 x 25 men (less days, more men)
and in 30 days, the food will be sufficient for = 630×2530 hr
(more days less men)
= 525 men
Number of men to be transfered = 630 – 525 = 105 men

Question 11.
Solution:
Number of men in the beginning = 120
Number of men died = 30
Remaining = 120 – 30 = 90 men
Total period = 200 days
No. of days passed = 5
Remaining period = 200 – 5 = 195
Now, The food lasts for 120 men for = 195 days
The food will last for 1 man for = 195 x 120 days (Less men, more days)
The food will last for 90 men for = 195×12090
(more men less days)
= 65 x 4 = 260 days

Question 12.
Solution:
Period in the beginning = 28 days
No. of days passed = 4 days.
Remaining period = 28 – 4 = 24 days
The food is sufficient for 24 days for = 1200 soldiers
The food will be sufficient for 1 day for = 1200 x 24 soldiers (Less days, more men)
and the food will be sufficient for 32 days = 1200×2432
= 900 soldiers (more days, less men)
No. of soldiers who left the fort = 1200 – 900 = 300 soldiers

Exercise 9C

Objective Questions.
Marks (✓) against the correct answer in each of the following :
Question 1.
Solution:
(c)
Weight of 4.5 m rod = 17.1 kg
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9C 1

Question 2.
Solution:
(d) None of these 0.8 cm represent the map = 8.8 km
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9C 2

Question 3.
Solution:
(c) In 20 minutes, Raghu covers = 5 km
in 1 minutes, he will cover = 520 km
and in 50 minutes, he will cover
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9C 3

Question 4.
Solution:
(d)
No. of men in the beginning = 500
More men arrived = 300
No. of total men = 500 + 300 = 800
For 500 men, provision are for = 24 days
For 1 man, provision will be = 24 x 500 days (less men, more days)
and for 800 men, provision will be = 24800 x 500 days
(more men less days)
= 15 days

Question 5.
Solution:
(b) Total cistern = 1
Filled in 1 minute = 45
Unfilled = 1 – 45 = 15
45 of cistern is filled in = 1 minutes = 60 seconds
1 full cistern can be filled in = 60×54 = 75 seconds
More time = 75 – 60 = 15 seconds

Question 6.
Solution:
(a)
15 buffaloes can eat as much as = 21 cows
1 buffalo will eat as much as = 2115 cows
35 buffaloes will eat as much as
= 21×3515 cm = 49 cows

Question 7.
Solution:
(b) 4 m long shadow is of a tree of height = 6 m
1 m long shadow of flagpole will of height = 64 m
50 m long shadow, the height of pole 6 will be = 64 x 50 = 75 m

Question 8.
Solution:
(b) 8 men can finish the work in = 40 days
1 man will finish it in=40 x 8 days (less men, more days)
8 + 2 = 10 men will finish it in = 40×810 days
(more men, less days)
= 32 days

Question 9.
Solution:
(b)
16 men can reap a field in = 30 days
1 man will reap the field in = 30 x 16 days
and 20 men will reap the field in = 30×1620 = 24 days

Question 10.
Solution:
(c) 10 pipe can fill tank in = 24 minutes
1 pipe will fill it in = 24 x 10 minutes (less pipe, more time)
and 10 – 2 = 8 pipes will fill the tank in
= 24×108 = 30 minutes

Question 11.
Solution:
(d) 6 dozen or 6 x 12 = 72 eggs
Cost of 72 eggs is = Rs. 108
Cost of 1 egg will be = Rs. 10872
and cost of 132 eggs will be 108
= Rs. 10872 x 132 = Rs. 198

Question 12.
Solution:
(b) 12 workers take to complete the work = 4 hrs.
1 worker will take = 4 x 12 hrs. (less worker, more time)
15 workers will take = 4×1215 hrs. (more workers, less time)
= 165 hr. = 3 hrs. 12 min

Question 13.
Solution:
(a) 27 days – 3 days = 24 days
Men = 500 + 300 = 800
For 500 men, provision is sufficient = 24 days
For 1 man, provision will be = 24 x 500 (less man, more days)
and for 500 + 300 = 800 men provision
will be sufficient = 24×500800 = 15 days
(more men, less days)

Question 14.
Solution:
(c) No. of rounds of rope = 140
Radius of base of cylinder = 14 cm
Radius of second cylinder of cylinder = 20 cm
If radius is 14 cm, then rounds of rope are = 140
If radius is 1 cm, then round = 140 x 14 (less radius more rounds)
and if radius is 20 cm, then rounds will
be = 140×1420 = 98 (more radius less rounds)

Question 15.
Solution:
(d) A worker makes toy in 23 hr= 1
He will make toys in 1 hr = 1 x 32
and will make toys in 223 hrs. = 1 x 32 x 223
= 11 (more time more toys)

Question 16.
Solution:
(d) A wall is constructed in 8 days by = 10 men
It will be constructed in 1 day by = 10 x 8 men (less time, more men)
10 x 8
RS Aggarwal Class 7 Solutions Chapter 9 Unitary Method Ex 9C 4
More men required = 160 – 10 = 150

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RS AGARWAL SOLUTION | CLASS 7TH | CHAPTER-8| Ratio and Proportion | EDUGROWN

Exercise 8A

Question 1.
Solution:
(i) 24 : 40
HCF of 24, 40 = 8
24 : 40 = 24 ÷ 8 : 40 ÷ 8 = 3 : 5 (Dividing by 8)
(ii) 13.5 : 15 or 135 : 150
HCF of 135 and 150 = 15
135 ÷ 15 : 150 ÷ 15 (Dividing by 15)
= 9 : 10
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 1
HCF of 25, 65, 80 = 5
Dividing by 5,
5 : 13 : 16

Question 2.
Solution:
(i) 75 paise : 3 rupees = 75 paise : 300 paise
(converting into same kind)
HCF of 75, 300 = 75
75 : 300 = 75 ÷ 75 : 300 ÷ 75 (Dividing by 75) = 1 : 4
(ii) 1 m 5 cm : 63 cm = 105 cm : 63 cm
(converting into same kind)
HCF of 105 and 63 = 21
105 ÷ 21 : 63 ÷ 21 (Dividing by 21)
= 5 : 3
(iii) 1 hour 5 minutes : 45 minutes = 65 minutes : 45 minutes
(converting into minutes)
13 : 9 (dividing by 5)
= 13 : 9
(iv) 8 months : 1 year = 8 months : 12 months
(converting into the same kind)
HCF of 8 and 12 = 4
Dividing by 4
8 ÷ 4 : 12 ÷ 4
= 2 : 3
(v) 2 kg 250 g : 3 kg = 2250 g : 3000 g (converting into the same kind)
HCF of 2250 and 3000 = 750
Dividing by 750,
2250 ÷ 750 : 3000 ÷ 750 = 3 : 4
(vi) 1 km : 750 m = 1000 m : 750 m
(converting into metre)
= 4 : 3 (dividing by 250)
= 4 : 3

Question 3.
Solution:
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 2

Question 4.
Solution:
A : B = 5 : 8, B : C = 16 : 25
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 3

Question 5.
Solution:
A : B = 3 : 5, B : C = 10 : 13
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 4
A : B : C = 6 : 10 : 13

Question 6.
Solution:
A : B = 5 : 6 and B : C = 4 : 7
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 5

Question 7.
Solution:
Total amount = Rs. 360
Sum of ratios = 7 + 8 = 15
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 6

Question 8.
Solution:
Total amount = Rs. 880
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 7

Question 9.
Solution:
Total amount = Rs. 5600
Ratio in A : B : C = 1 : 3 : 4
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 8

Question 10.
Solution:
Let x be added to each term Then
9 + x : 16 + x = 2 : 3
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 9

Question 11.
Solution:
Let x be subtracted from each term Then
(17 – x) : (33 – x) = 7 : 15
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 10

Question 12.
Solution:
Ratio in two numbers = 7 : 11
Let first number = 7x
Then second number = 11x
Then adding 7 to each number, the ratio is 2 : 3
7x+711x+7 = 23
By cross multiplying:
3 (7x + 7) = 2 (11x + 7)
⇒ 21x + 21 = 22x + 14
⇒ 21 – 14 = 22x – 21x
⇒ x = 7
First number = 7x = 7 x 7 = 49
and second number = 11x = 11 x 7 = 77
Hence numbers are 49, 77

Question 13.
Solution:
The ratio in two numbers = 5 : 9
Let the first number = 5x
Then second number = 9x
By subtracting 3 from each number the ratio is 1 : 2
5x–39x–3 = 12
By cross multiplication,
2 (5x – 3) = 1 (9x – 3)
⇒ 10x – 6 = 9x – 3
⇒ 10x – 9x = -3 + 6
⇒ x = 3
First number = 5x = 5 x 3 = 15
and second .number = 9x = 9 x 3 = 27
Hence numbers are 15, 27

Question 14.
Solution:
Ratio in two numbers = 3 : 4
LCM = 180
Let first number = 3x
Then second number = 4x
Now LCM = 3 x 4 x x = 12x
12x = 180
⇒ x = 15
Numbers will be 3 x 15 = 45 and 4 x 15 = 60

Question 15.
Solution:
Ratio in present ages of A and B = 8 : 3
Let A’s age = 8x
Then B’s age = 3x
6 years hence,
A’s age will be = 8x + 6
and B’s will be = 3x + 6
8x+63x+6 = 94
(By cross multiplication)
4 (8x + 6) = 9 (3x + 6)
⇒ 32x + 24 = 27x + 54
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s present age = 8x = 8 x 6 = 48 years
and B’s age = 3x = 3 x 6 = 18 years

Question 16.
Solution:
Ratio in copper and zinc = 9 : 5
Let alloy = x gm
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 12

Question 17.
Solution:
Ratio in boys and girls = 8 : 3
and total number of girls = 375
Let number of boys = 8x
Then number of girls = 3x
3x = 375
⇒ x = 125
Number of boys = 8x = 8 x 125 = 1000

Question 18.
Solution:
Ratio in income and savings = 11 : 2
Let income = 11x
Then savings = 2x
But savings = Rs. 2500
2x = 2500
⇒ x = 1250
Then income = 1250 x 11 = Rs. 13750
and expenditure = Total income – savings = 13750 – 2500 = Rs. 11250

Question 19.
Solution:
Total amount = Rs. 750
Ratio in rupee, 50 P and 25 P coins =5 : 8 : 4
Let number of rupees = 5x
Number of 50 P coins = 8x
and number of 25 coins = 4x
According to the condition,
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 13
Number of 1 Re coins = 5x = 5 x 75 = 375
Number of 50 P coins = 8x = 8 x 75 = 600
and number of 25 P coins = 4x = 4 x 75 = 300

Question 20.
Solution:
(4x + 5) : (3x + 11) = 13 : 17
4x+53x+11 = 1317
By cross multiplication,
68x + 85 = 39x + 143
⇒ 68x – 39x = 143 – 85
⇒ 29x = 58
x = 2
Hence x = 2

Question 21.
Solution:
x : y = 3 : 4
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 14

Question 22.
Solution:
x : y = 6 : 11
xy = 611
Now (8x – 3y) : (3x + 2y)
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 15

Question 23.
Solution:
Sum of two numbers = 720
Ratio of two numbers = 5 : 7
Let first number = 5x
Then second number = 7x
5x + 7x = 720
⇒ 12x = 720
⇒ x = 60
First number = 5x = 5 x 60 = 300
and second number = 7x = 7 x 60 = 420

Question 24.
Solution:
(i) (5 : 6) or (7 : 9)
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 16

Question 25.
Solution:
(i) (5 : 6), (8 : 9), (11 : 18)
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8A 19

Exercise 8B

Question 1.
Solution:
We know that a, b, c, d are in proportion if ad = bc
Now 30, 40, 45, 60 are in proportion
if 30 x 60 = 40 x 45
if 1800= 1800
which is true
30, 40, 45, 60 are in proportion.

Question 2.
Solution:
We know that if a, b, c, d are in proportion if ad = bc
Now 36, 49, 6, 7 are in proportion
if 36 x 7 = 49 x 6
if 252 = 294
But 252 ≠ 294
36, 49, 6, 7 are not in proportion

Question 3.
Solution:
2 : 9 : : x : 27
9 x x = 2 x 27
x = 2×279 = 2 x 3 = 6

Question 4.
Solution:
8 : x : : 16 : 35
x x 16 = 8 x 35
x = 8×356 = 352 = 17.5

Question 5.
Solution:
x : 35 : : 48 : 60
x x 60 = 35 x 48
x = 7 x 4 = 28

Question 6.
Solution:
Let x be the fourth proportional, then
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8B 1
x = 352 = 17.5
Fourth proportional = 17.5

Question 7.
Solution:
36, 54, x are in continued proportion
36 : 54 : : 54 : x
⇒ 36 x x = 54 x 54
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8B 2

Question 8.
Solution:
27, 36, x are in continued proportion
27 : 36 :: 36 : x
27 x x = 36 x 36
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8B 3

Question 9.
Solution:
Let x be the third proportional, then
(i) 8 : 12 : : 12 : x
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8B 4

Question 10.
Solution:
Third proportional = 28 then
7 : x :: x : 28
⇒ 7 x 28 = x x x
⇒ x² = 28 x 7 = 196
⇒ x = √196 = 14
x = 14

Question 11.
Solution:
Let x be the mean proportional, then
(i) 6 : x :: x : 24
⇒ x² = 6 x 24 = 144
x = √144 = 12
Mean proportional = 12
(ii) 3 : x : : x : 27
⇒ x² = 3 x 27 = 81
x = √81 = 9
Mean proportional = 9
(iii) 0.4 : x :: x : 0.9
⇒ x² = 0.4 x 0.9
x = √o.36 = 0.6
Mean proportional = 0.6

Question 12.
Solution:
Let x be added to each of the given numbers then
5 + x, 9 + x, 7 + x, 12 + x are in proportion
5+x9+x = 7+x12+x
By cross multiplication :
(5 + x) (12 + x) = (7 + x) (9 + x)
⇒ 60 + 5x + 12x + x² = 63 + 7x + 9x + x²
⇒ 60 + 17x + x² = 63 + 16x + x²
⇒ 17x + x² – 16x – x² = 63 – 60
⇒ x = 3
Required number = 3

Question 13.
Solution:
Let x be subtracted from each of the given number, then
10 – x, 12 – x, 19 – x and 24 – x are in proportion
10–x12–x = 19–x24–x
By cross multiplication :
(10 – x) (24 – x) = (19 – x) (12 – x)
⇒ 240 – 10x – 24x + x² = 228 – 19x – 12x + x²
⇒ 240 – 34x + x² = 228 – 31x + x²
⇒ -34x + x² + 31x – x² = 228 – 240
⇒ -3x = -2
⇒ 3x = 12
⇒ x = 4
Required number = 4

Question 14.
Solution:
Scale of map = 1 : 5000000
Distance between two town on the map = 4 cm
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8B 6

Question 15.
Solution:
Height of a tree = 6 cm
and its shadow at same time = 8 m
Shadow of a pole = 20 m
Let height of pole = x m
6 : 8 = x : 20
⇒ x= 6×208 = 15 m
Height of pole = 15 m

Exercise 8C

Objective questions :
Mark (✓) against the correct answers in each of the following :
Question 1.
Solution:
(d) a : b = 3 : 4, b : c = 8 : 9
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 1

Question 2
Solution:
(a) A : B = 2 : 3, B : C = 4 : 5
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 2

Question 3.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 3

Question 4.
Solution:
(b) 15% of A = 20% of B
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 4

Question 5.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 5

Question 6.
Solution:
(b) A : B = 5 : 7, B : C = 6 : 11
LCM of 7, 6 = 42
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 6

Question 7.
Solution:
(c) 2A = 3B = 4C = x
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 7

Question 8.
Solution:
(a)
A3 = B4 = C5 = 1(suppose)
A = 3, B = 4, C = 5
A : B : C = 3 : 4 : 5

Question 9.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 8

Question 10.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 9

Question 11.
Solution:
(c) (3a + 5b) : (3a – 5b) = 5 : 1
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 11

Question 12.
Solution:
(c) 7 : x :: 35 : 45
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 12
x = 9

Question 13.
Solution:
(b) Let x to be added to each term of 3 : 5
Then 3+x5+x = 56
By cross multiplication
18 + 6x = 25 + 5x
6x – 5x = 25 – 18
x = 7
7 is to be added

Question 14.
Solution:
(d) Ratio in two numbers = 3 : 5
Let first number = 3x
Then second number = 5x
According to the condition,
3x+105x+10 = 57
(By cross multiplication)
25x + 50 = 21x + 70
25x – 21x = 70 – 50
4x = 20
x = 5
First number = 3 x 5 = 15
and second number = 5 x 5 = 25
Sum of numbers = 15 + 25 = 40

Question 15.
Solution:
(a)
Let x be subtracted from each of the term
15–x19–x = 34
⇒ 4 (15 – x) = 3 (19 – x)
⇒ 60 – 4x = 57 – 3x
⇒ -4x + 3x = 57 – 60
⇒ -x = -3
x = 3
Required number = 3

Question 16.
Solution:
(a)
Amount = Rs. 420
and ratio = 3 : 4
Sum of ratios = 3 + 4 = 7
A’s share = 420×37 = Rs. 60 x 3 = Rs. 180

Question 17.
Solution:
(d)
Let number of boys = x, then
x : 160 : : 8 : 5
⇒ x x 5 = 160 x 8
x = 160×85 = 32 x 8 = 256
Number of total students of the school = 256 + 160 = 416

Question 18.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 8 Ratio and Proportion Ex 8C 13

Question 19.
Solution:
(c)
Let x be the third proportional to 9 and 12 then
9 : 12 :: x : 12
⇒ 9 x x = 12 x 12
⇒ x = 12×129 = 1449 = 16
Third proportional = 16

Question 20.
Solution:
Answer = (b)
Mean proportional of 9 and 16 = √(9 x 16) = √144 = 12

Question 21.
Solution:
(a)
Let age of A = 3x
and age of B = 8x
6 years hence, their ages will be 3x + 6 and 8x + 6
3x+68x+6 = 49
⇒ 9 (3x + 6) = 4 (8x + 6)
⇒ 27x + 54 = 32x + 24
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s age = 3x = 3 x 6 = 18 years

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RS AGARWAL SOLUTION | CLASS 7TH | CHAPTER-7| Linear Equations in One Variable | EDUGROWN

Exercise 7A

Solve the following equations. Check your result in each case.
Question 1.
Solution:
3x – 5 = 0
Adding 5 to both sides
3x – 5 + 5 = 0 + 5
⇒ 3x = 5
⇒ x = 53
Check:
L.H.S. = 3x – 5
= 3 x 53 – 5
= 5 – 5
= 0
= R.H.S.
Hence x = 53

Question 2.
Solution:
8x – 3 = 9 – 2x
⇒ 8x + 2x = 9 + 3 (By transposing)
⇒ 10x = 12
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 1

Question 3.
Solution:
7 – 5x = 5 – 7x
⇒ – 5x + 7x = 5 – 7 (By transposing)
⇒ 2x = -2
x = -1
Check:
L.H.S. = 7 – 5x = 7 – 5(-1) = 7 + 5 = 12
R.H.S. = 5 – 7x = 5 – 7(-1) = 5 + 7 = 12
L.H.S. = R.H.S.
Hence x = -1

Question 4.
Solution:
3 + 2x = 1 – x
⇒ 2x + x = 1 – 3 (By transposing)
⇒ 3x = -2
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 2

Question 5.
Solution:
2(x – 2) + 3(4x – 1) = 0
⇒ 2x – 4 + 12x – 3 = 0
⇒ 2x + 12x = 4 + 3 (By transposing)
⇒ 14x = 7
⇒ x = 714 = 12
Check : L.H.S. = 2(x – 2) + 3 (4x -1)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 3

Question 6.
Solution:
5 (2x – 3) – 3(3x – 7) = 5
⇒ 10x – 15 – 9x + 21 = 5
⇒ 10x – 9x – 15 + 21 = 5
⇒ 10x – 9x = 5 + 15 – 21 (By transposing)
⇒ x = 20 – 21 = -1
⇒ x = -1
Check:
L.H.S. = 5 (2x – 3) – 3(3x – 7)
= 5[2 x (-1) -3] -3[3 (-1) -7] = 5[-2 – 3] – 3[-3 – 7]
= 5 x (-5) -3 x (-10)
= -25 + 30
= 5 = R.H.S.
Hence x = -1

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 4

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 6
L.H.S. = R.H.S.
Hence x = 48

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 7

Question 10.
Solution:
3x + 2(x + 2) = 20 – (2x – 5)
⇒ 3x + 2x + 4 = 20 – 2x + 5
⇒ 5x + 4 = 25 – 2x
⇒ 5x + 2x = 25 – 4 (By transposing)
⇒ 7x = 21
⇒ x = 3
Check:
L.H.S.= 3x + [2(x + 2)] = 3 x 3 + 2(3 + 2) = 9 + 2 x 5 = 9 + 10 = 19
R.H.S. = 20 – (2x – 5) = 20 – (2 x 3 – 5) = 20 – (6 – 5) = 20 – 1 = 19
L.H.S. = R.H.S.
Hence x = 3

Question 11.
Solution:
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
⇒ 13y – 52 – 3y + 27 – 5y – 20 = 0
⇒ 13y – 3y – 5y – 52 + 27 – 20 = 0
⇒ 13y – 8y – 72 + 27 = 0
⇒ 5y – 45 = 0
⇒ 5y = 45 (By transposing)
⇒ y = 9
Check:
L.H.S. = 13(y – 4) – 3(y – 9) – 5(y + 4)
= 13(9 – 4) – 3(9 – 9) – 5(9 + 4)
= 13 x 5 – 3 x 0 – 5 x 13
= 65 – 0 – 65 = 0 = R.H.S.
Hence y = 9

Question 12.
Solution:
2m+53 = 3m – 10
⇒ 2m + 5 = 3 (3m – 10) (By cross multiplication)
⇒ 2m + 5 = 9m – 30
⇒ 2m – 9m = -30 – 5
⇒ -7m = -35
⇒ m = 5
m = 5
Check:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 8
R.H.S. = 3m – 10 = 3 x 5 – 10 = 15 – 10 = 5
L.H.S. = R.H.S.
Hence m = 5

Question 13.
Solution:
6(3x + 2) – 5(6x – 1) = 3(x – 8) – 5(7x – 6) + 9x
⇒ 18x + 12 – 30x + 5 = 3x – 24 – 35x + 30 + 9x
⇒ 18x – 30x + 12 + 5 = 3x – 35x + 9x – 24 + 30
⇒ -12x + 17 = -23x + 6
⇒ – 12x + 23x = 6 – 17
⇒ 11x = -11
x = – 1
Check:
L.H.S. = 6(3x + 2) – 5(6x – 1)
= 6[3x (-1) + 2] – 5[6 x (-1) x -1]
= 6[-3 + 2] – 5[-6 – 1]
= 6 x (-1) – 5 x (-7)
= -6 + 35 = 29
R.H.S. = 3(x – 8) – 5 (7x – 6) + 9x
= 3[-1 – 8] -5 [7 x (-1) – 6] + 9 (-1)
= 3 x (-9) – 5 [-7 – 6] – 9
= -27 – 5(-13) – 9
= -27 + 65 – 9
= 65 – 36 = 29 .
L.H.S. = R.H.S.
Hence x = -1

Question 14.
Solution:
t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t – 4)
⇒ t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12t
⇒ t – 2t + 10t – 10 = 8t – 3t + 18
⇒ 9t – 10 = 5t + 18
⇒ 9t – 5t = 18 + 10 (By transposing)
⇒ 4t = 28
⇒ t = 7
Check:
L.H.S. = t – [2t + 5] -5[1 – 2t]
= 7 – [2 x 7 + 5] – 5[1 – 2 x 7]
= 7 – [14 + 5] – 5 [1 – 14]
= 7 – 19 – 5(-13)
= 7 – 19 + 65
= 72 – 19 = 53
R.H.S. = 2[3 + 4t) – 3(t – 4)
= 2 (3 + 4 x 7) – 3(7 – 4)
= 2(3 + 28) – 3(3)
= 2(31) – 9 = 62 – 9 = 53
L.H.S. = R.H.S.
Hence t = 7 Ans.

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 9

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 11

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 12

Question 18.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 14

Question 19.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 16

Question 20.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 18

Question 21.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 19

Question 22.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 21

Question 23.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 22

Question 24.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 25

Question 25.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 27

Question 26.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 29

Question 27.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 31

Question 28.
Solution:
0.18 (5x – 4) = 0.5x + 0.8
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 32

Question 29.
Solution:
2.4 (3 – x) – 0.6 (2x – 3) = 0
⇒ 7.2 – 2.4x – 1.2x + 1.8 = 0
⇒ -2.4x – 1.2x = – (7.2 + 1.8).
L.H.S. = 2.4 (3 – x) – 0.6 (2x – 3)
⇒ 2.4 (3 – 2.5) – 0.6 (2 x 2.5 – 3)
⇒ 2.4 (0.5) – 0.6 (5 – 3)
⇒ 1.2 – 0.6 x 2 = 1.2 – 1.2 = 0 = R.H.S.
Hence x = 2.5

Question 30.
Solution:
0.5x – (0.8 – 0.2x) = 0.2 – 0.3x
⇒ 0.5x – 0.8 + 0.2x = 0.2 – 0.3x
⇒ 0.5x + 0.2x + 0.3x = 0.2 + 0.8
⇒ 1.0x = 1.0
⇒ x = 1
Check :
L.H.S. = 0.5x – (0.8 – 0.2x)
= 0.5 x 1 – (0.8 – 0.2 x 1)
= 0.5 – (0.8 – 0.2) = 0.5 – 0.6 = -0.1
R.H.S. = 0.2 – 0.3x = 0.2 – 0.3 x 1 = 0.2 – 0.3 = -0.1
L.H.S. = R.H.S.
Hence x = 1

Question 31.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 34

Question 32.
Solution:
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A 35

Exercise 7B

Question 1.
Solution:
Let the required number = x
Then 2x – 7 = 45
2x = 45 + 7 = 52
x = 26
Required number = 26

Question 2.
Solution:
Let the required number = x Then
3x + 5 = 44
⇒ 3x = 44 – 5 = 39
x = 13
Required number = 13

Question 3.
Solution:
Let the required fraction = x
then 2x + 4 = 265
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 1

Question 4.
Solution:
Let the required number = x
and half of .the number = x2
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 2

Question 5.
Solution:
Let the required number = x
Two third of the number = 23 x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 4

Question 6.
Solution:
Let the required number = x
Then, 4x = x + 45
⇒ 4x – x = 45
⇒ 3x = 45
⇒ x = 15
Required number = 15

Question 7.
Solution:
Let the required number = x
Then x – 21 = 71 – x
⇒ x + x = 71 + 21
⇒ 2x = 92
⇒ x = 46

Question 8.
Solution:
Let the original number = x
Then 23 of the number = 23 x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 5

Question 9.
Solution:
Let the second number = x
then first number = 25 x
their sum = 70
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 6

Question 10.
Solution:
Let the required number = x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 7

Question 11.
Solution:
Let the required number = x
Fifth part of the number = x5
Fourth part of the number = x4
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 9

Question 12.
Solution:
Let first natural number = x then
next number = x + 1
x + x + 1 = 63
⇒ 2x = 63 – 1 = 62
x = 31
first number = 31
and second number = 31 + 1 = 32
Numbers are 31, 32

Question 13.
Solution:
Let first odd number = 2x + 1
second odd number = 2x + 3
2x + 1 + 2x + 3 = 76
⇒ 4x + 4 = 76
⇒ 4x = 76 – 4 = 72
x = 18
First number = 2x + 1 = 2 x 18 + 1 = 36 + 1 = 37
Second number = 2x + 3 = 2 x 18 + 3 = 36 + 3 = 39
Numbers are 37, 39

Question 14.
Solution:
Let first positive even number = 2x
Second number = 2x + 2
Third number = 2x + 4
2x + 2x +2 + 2x + 4 = 90
⇒ 6x + 6 = 90
⇒ 6x = 90 – 6 = 84
x = 14
First even number = 2x = 2 x 14 = 28
Second number = 2x + 2 = 2 x 14 + 2 = 28 + 2 = 30
Third number = 30 + 2 = 32
Required numbers are 28, 30, 32

Question 15.
Solution:
Sum of two numbers = 184
Let first number = x
Then second number = 184 – x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 10
First part = 72
Second part = 184 – 72 = 112
Hence parts are 72, 112

Question 16.
Solution:
Total number of notes = 90
Let number of notes of Rs. 5 = x
Then number of notes of Rs.10 = 90 – x
Then x x 5 + (90 – x) x 10 = 500
⇒ 5x + 900 – 10x = 500
⇒ -5x = 500 – 900 = -400
x = 8
Number of 5 rupees notes = 80
and ten rupees notes = 90 – 80 = 10

Question 17.
Solution:
Amount of coins = Rs. 34
Let 50 paisa coins = x
then 25 paisa coins = 2x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 11
Number of 50 paisa coins = 34
and number of 25 paisa coins = 2x = 2 x 34 = 68

Question 18.
Solution:
Let present age of Raju’s cousin = x years
then age of Raju = (x – 19) years
After 5 years,
Raju’s age = x – 19 + 5 = (x – 14) years
and his cousin age = x + 5
(x – 14) : (x + 5) = 2 : 3
⇒ x–14x+5 = 23
⇒ 3(x – 14) = 2 (x + 5) (By cross multiplication)
⇒ 3x – 42 = 2x + 10
⇒ 3x – 2x = 10 + 42
⇒ x = 52
Raju’s age = x – 19 = 52 – 19 = 33 years
and his cousin age = 52 years.

Question 19.
Solution:
Let present age of son = x years
Age of father = (x + 30) years
12 years after,
Father’s age = x + 30 + 12 = (x + 42) years
and son’s age = (x + 12) years
(x + 42) = 3(x + 12)
⇒ x + 42 = 3x + 36
⇒ 3x + 36 = x + 42
⇒ 3x – x = 42 – 36
⇒ 2x = 6
⇒ x = 3
Son’s age = 3 years
Father’s age = 3 + 30 = 33 years

Question 20.
Solution:
Ratio in present ages of Sonal and Manoj = 7 : 5
Let Sonal’s age = 7x
then Manoj’s age = 5x
10 years hence,
Sonal’s age will be = 7x + 10
and Manoj’s age = 5x + 10
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 12
Sonal’s present age = 7x = 7 x 5 = 35 years
and Manoj’s age = 5x = 5 x 5 = 25 years

Question 21.
Solution:
Five years ago,
Let Son’s age = x years
and father’s age = 7x years
Present age of son = (x + 5) years
and age of father = (7x + 5) years
5 years hence,
father’s age = 7x + 5 + 5 = 7x + 10
and Son’s age = x + 5 + 5 = x + 10
(7x + 10) = 3(x + 10)
⇒ 7x + 10 = 3x + 30
⇒ 7x – 3x = 30 – 10
⇒ 4x = 20
⇒ x = 5
Father present age = 7x + 5 = 7 x 5 + 5 = 35 + 5 = 40 years
and son’s age = x + 5 = 5 + 5 = 10 years

Question 22.
Solution:
Let age of Manoj 4 years ago = x
then his present age = x + 4
After 12 years his age will be = x + 4 + 12 = x + 16
x + 16 = 3(x)
x + 16 = 3x
⇒ 16 = 3x – x
⇒ 2x = 16
x = 8
His present age = 8 + 4 = 12 years

Question 23.
Solution:
Let total marks = x
Pass marks = 40% of x = 40×100 = 25 x
No. of marks got by Rupa = 185
No. of marks by which she failed = 15
Pass marks = 185 + 15 = 200
25 x = 200
⇒ x = 200×52 x
⇒ x = 500
Hence total marks = 500

Question 24.
Solution:
Sum of digits = 8
Let units digit = x
Then tens digit = 8 – x
and number will be x + 10 (8 – x) ….(i)
By adding 18, the digits are reversed then
units digit = 8 – x
and tens digit = x
Number = (8 – x) = 10x
According to the condition,
(8 – x) + 10x = 18 + x + 10 (8 – x)
⇒ 8 – x + 10x = 18 + x + 80 – 10x
⇒ 10x – x – x + 10x = 18 + 80 – 8
⇒ 18x = 90
⇒ x = 5
Number is
x + 10(8 – x) = 5 + 10(8 – 5) = 5 + 10 x 3 = 35

Question 25.
Solution:
Cost of 3 tables and 2 chairs = 1850
Cost of table = Rs. 75 + cost of a chair
Let cost of chair = Rs. x,
then Cost of table = Rs. 75 + x
According to the condition,
3 (75 + x) + 2x = 1850
⇒ 225 + 3x + 2x = 1850
⇒ 225 + 5x = 1850
⇒ 5x = 1850 – 225 = 1625
x = 325
Cost of chair = Rs. 325
and cost of table = Rs. 325 + 75 = Rs. 400

Question 26.
Solution:
S.P of article = Rs. 495
gain = 10%
Let cost price = Rs. x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 13

Question 27.
Solution:
Perimeter of field = 150 m
Length + Breadth = 1502 = 75 m
[Perimeter = 2(l + b)]
Let length = x Then breadth = 75 – x
Then x = 2(75 – x)
⇒ x = 150 – 2x
⇒ x + 2x = 150
⇒ 3x = 150
⇒ x = 1503 = 50
Length = 50 m
and breadth = 75 – 50 = 25 m

Question 28.
Solution:
Perimeter of an isosceles triangle = 55 m
Let the third side of an isosceles triangle = x
Then each equal side = (2x – 5) m
According to the condition,
x + 2 (2x – 5) = 55
⇒ x + 4x – 10 = 55
⇒ 5x = 55 + 10
⇒ 5x = 65
⇒ x = 13
and 2x – 5 = 2 x 13 – 5 = 21 m
Sides will be 13m, 21m, 21m

Question 29.
Solution:
Sum of two complementary angles = 90°
Let first angle = x
then second = 90° – x
x – (90 – x) = 8
⇒ x – 90 + x = 8
⇒ 2x = 8 + 90
⇒ 2x = 98
⇒ x = 49
first angle = 49°
and second angle = 90° – 49° = 41°
Hence angles are 41°, 49°

Question 30.
Solution:
Sum of two supplementary angles = 180°
Let first angle = x
Then second angle = 180° – x
x – (180° – x) = 44°
⇒ x – 180° + x = 44°
⇒ 2x = 44° + 180° = 224°
⇒ 2x = 224°
⇒ x = 112°
First angle = 112°
and second angle = 180° – 112° = 68°
Hence angles are 68°, 112°

Question 31.
Solution:
In an isosceles triangle
Let each equal base angles = x
Then vertex angle = 2x
According to the condition,
x + x + 2x = 180° (sum of angles of a triangle)
⇒ 4x = 180°
⇒ x = 45°
Then vertex angle = 2x = 2 x 45° = 90°
Angles of the triangle are 45°, 45° and 90°

Question 32.
Solution:
Let length of total journey = x km
According to the condition,
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 14
⇒ 39x + 80 = 40x
⇒ 40x – 39x = 80
⇒ x = 80
Total journey = 80km

Question 33.
Solution:
No. of days = 20 Let no. of days he worked = x
Then he will receive amount = x x Rs. 120 = Rs. 120x
No. of days he did not work = 20 – x
Fine paid = (20 – x) x Rs. 10 = Rs. 10(20 -x)
120x – 10 (20 – x) = 1880
⇒ 120x – 200 + 10x = 1880
⇒ 130x = 1880 + 200 = 2080
x = 16
No. of days he remained absent = 20 – x = 20 – 16 = 4 days

Question 34.
Solution:
Let value of property = x
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7B 15

Question 35.
Solution:
Solution = 400 mL
Quantity of alcohol = 15% of 400 mL
= 400×15100 = 60 mL
Let pure alcohol added = x mL
Total solution = 400 + x
and total alcohol = (x + 60)
Now (400 + x) x 32% = x + 60
⇒ (400 + x) x 32100 = x + 60
⇒ 32 (400 + x) = 100 (x + 60)
⇒ 12800 + 32x = 100x + 6000
⇒ 12800 – 6000 = 100x – 32x
⇒ 6800 = 68x
⇒ x = 6800
Pure alcohol added = 100 mL

Exercise 7C

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 1

Question 2.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 2

Question 3.
Solution:
(a)
2n + 5 = 3 (3n – 10)
⇒ 2n + 5 = 9n – 30
⇒ 9n – 2n = 5 + 30
⇒ 7n = 35
⇒ n = 5

Question 4.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 4

Question 5.
Solution:
(c)
8 (2x – 5) – 6 (3x – 7) = 1
⇒ 16x – 40 – 18x + 42 = 1
⇒ -2x + 2 = 1
⇒ -2x = 1 – 2 = -1
x = 12

Question 6.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 5

Question 7.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 6

Question 8.
Solution:
(b)
Let first whole number=x
Then second number = x + 1
and sum = 53
x + x + 1 = 53
⇒ 2x = 53 – 1
⇒ 2x = 52
⇒ x = 26
Smaller number = 26

Question 9.
Solution:
Let first even number = 2x
Then second number = 2x + 2
and sum = 86
2x + 2x + 2 – 86
⇒ 4x = 86 – 2 = 84
⇒ x = 21
Larger even number = 2x + 2 = 2 x 21 + 2 = 42 + 2 = 44

Question 10.
Solution:
(b)
Let first odd number = 2x + 1
Second number = 2x + 3
2x + 1 + 2x + 3 = 36
⇒ 4x + 4 = 36
⇒ 4x = 36 – 4 = 32
⇒ x = 8
Smaller number = 2x + 1 = 2 x 8 + 1 = 16 + 1 = 17

Question 11.
Solution:
(d)
Let number = x
2x + 9 = 31
⇒ 2x = 31 – 9 = 22
⇒ x = 11

Question 12.
Solution:
(a)
Let number = x then
3x + 6 = 24
⇒ 3x = 24 – 6 = 18
⇒ x = 6
Number = 6

Question 13.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7C 7

Question 14.
Solution:
(b)
Let first angle = x
Then second = 90° – x
x – (90° – x) = 10
⇒ x – 90° + x = 10°
⇒ 2x = 10° + 90° = 100°
x = 50°
Second angle = 90° – 50° = 40°
Larger angle = 50°

Question 15.
Solution:
(b)
Let first angle = x
Then second = 180° – x
x – (180° – x) = 20°
⇒ x – 180° + x = 20°
⇒ 2x = 20° + 180° = 200°
x = 100°
Second angle = 180° – 100° = 80°
Smaller angle = 80°

Question 16.
Solution:
(c)
Let age of A = 5x
Then age of B = 3x
After 6 years,
A’s age = 5x + 6
and B’s age = 3x + 6
5x+63x+6 = 75
⇒ 25x + 30 = 21x + 42
⇒ 25x – 21x = 42 – 30
⇒ 4x = 12
⇒ x = 3
A’s age = 5x = 5 x 3 = 15 years

Question 17.
Solution:
(b)
Let the number = x
According to the condition,
5x = 80 + x
⇒ 5x – x = 80
⇒ 4x = 80
⇒ x = 20
Number = 20

Question 18.
Solution:
(c)
Let width of rectangle = x m
Then length = 3x m
Perimeter = 96 m
2 (x + 3x) = 96
⇒ x + 3x = 962 = 48
⇒ 4x = 48
⇒ x = 12
Length = 3x = 12 x 3 = 36 m

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RS AGARWAL SOLUTION | CLASS 7TH | CHAPTER-6| Algebraic Expressions | EDUGROWN

Exercise 6A

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 1

Question 2.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 7

Question 3.
Solution:
Sum of (a + 3b – 4c), (4a – b + 9c) and (-2b + 3c – a)
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 8
Now subtract (2a – 3b + 4c) from 4a + 8c
= 4a + 8c – (2a – 3b + 4c)
= 4a + 8c – 2a + 3b – 4c
= 4a – 2a + 3b + 8c – 4c
= 2a + 3b + 4c

Question 4.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 9

Question 5.
Solution:
Sum of (8a – 6a² + 9) and (-10a – 8 + 8a²)
= 8a – 6a² + 9 + (-10a) – 8 + 8a²
= 8a – 10a – 6a² + 8a² + 9 – 8
= -2a + 2a² + 1
Now -3 – (-2a + 2a² + 1)
= (-3 + 2a – 2a² – 1)
= -4 + 2a – 2a²

Question 6.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6A 10

Exercise 6B

Find the products:
Question 1.
Solution:
3 x 8 a²⁺⁴ = 24a⁶

Question 2.
Solution:
(-6x³) x (5x²) = -6 x 5x²⁺³ = -30x⁵

Question 3.
Solution:
(-4ab) x (-3a²bc)
= (-4) x (-3) a. a².b. b. c
= 12.a²⁺¹. b¹⁺¹.c
= 12a³b²c

Question 4.
Solution:
= (2a²b³) x (-3a³b)
= 2 x (-3) a². a³. b³. b. b
= -6a²⁺³.b³⁺¹
= -6a⁵.b⁴

Question 5.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 1

Question 6.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 2

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 3

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 4

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 5

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 6

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 7

Question 12.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 9

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 10

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 11

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 12

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 14

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 15

Question 18.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 16

Question 19.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 18

Question 20.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 20

Question 21.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 21

Find the products given below and in each case verify the result for a = 1, b = 2 and c = 3 :
Question 22.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 23

Question 23.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 24

Question 24.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 27

Question 25.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6B 29

Exercise 6C

Find each of the following products:
Question 1.
Solution:
4a(3a + 7b) = 4a x 3a + 4a x 7b = 12a² + 28ab

Question 2.
Solution:
5a(6a – 3b) = 5a x 6a – 5a x 3b = 30a² – 15ab

Question 3.
Solution:
8a²(2a + 5b) = 8a² x 2a + 8a² x 5b = 16a³ + 40a²b

Question 4.
Solution:
9x² (5x + 7) = 9x² x 5x + 9x² x 7 = 45x³ + 63x²

Question 5.
Solution:
ab(a² – b²) = ab x a² – ab x b² = a³b – ab³

Question 6.
Solution:
2x² (3x – 4x²) = 2x² x 3x – 2x² x 4x² = 6x³ – 8x⁴

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 1

Question 8.
Solution:
-1 7x² (3x – 4) = -17x² x 3x – 17x² x (-4) = -51x³ + 68x²

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 2

Question 10.
Solution:
-4x²y (3x² – 5y)
= -4x²y x 3x² – 4x²y x (-5y)
= -12x²y + 20x²y²

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 3

Question 12.
Solution:
9t² (t + 7t³) = 9t² x t + 9t² x 7t³ = 9t³ + 63t⁵

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 4

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 5

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 6

Question 16.
Solution:
24x² (1 – 2x)
= 24x² x 1 – 24x² x 2x
= 24x² – 48x³
If x = 2, then
24x² – 48x³
= 24(2)² – 48(2)³
= 24 x 4 – 48 x 8
= 96 – 384
= -288

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6C 7

Question 18.
Solution:
s (s² – st) = s x s² – s x st = s³ – s²t
If s = 2, t = 3, then
s³ – s²t = (2)³ – (2)² x 3 = 8 – 4 x 3 = 8 – 12 = -4

Question 19.
Solution:
-3y (xy + y²) = -3y x xy + (-3y) x y² = -3xy² – 3y³
if x = 4, y = 5, then
-3xy² – 3y³
= -3(4)(5)² – 3(5)³ = -3 x 4 x 25 – 3 x 125 = -300 – 375 = -675

Simplify each of the following:
Question 20.
Solution:
a(b – c) + b(c – a) + c(a – b) = ab – ac + bc – ab + ac – bc = 0

Question 21.
Solution:
a(b – c) – b(c – a) – c(a – b) = ab – ac – bc + ab – ac + bc = 2ab – 2ac

Question 22.
Solution:
3x² + 2(x + 2) – 3x (2x + 1)
= 3x² + 2x + 4 – 6x² – 3x = 3x² – 6x² + 2x – 3x + 4 = -3x² – x + 4

Question 23.
Solution:
x (x + 4) + 3x (2x² – 1) + 4x² + 4
= x² + 4x + 6x³ – 3x + 4x² + 4
= 6x³ + x² + 4x² + 4x – 3x + 4
= 6x³ + 5x² + x + 4

Question 24.
Solution:
2x² + 3x (1 – 2x³) + x (x + 1)
= 2x² + 3x – 6x⁴ + x² + x
= – 6x⁴ + 2x² + x² + 3x + x
= – 6x⁴ + 3x² + 4x

Question 25.
Solution:
a²b (a – b²) + ab(4ab – 2a²) – a³b (1 – 2b)
= a³b – a²b³ – 2a³b² + 4a²b³ – 2a³b² – a³b + 2a³b²
= a³b – a³b + 2a³b² – a²b³ + 4a²6³
= 3a²b³

Question 26.
Solution:
4st (s – t) – 6s² (t – t²) – 3t² (2s² – s) + 2st(s – t)
= 4s²t – 4st² – 6s²t + 6s²t² – 6s²t² + 3st² + 2s²t – 2st²
= 4s²t – 6s²t + 2s²t – 4st² + 3st² – 2st² + 6s²t² – 6s²t²
= 6s²t – 6s²t – 6st² + 3st² + 6s²t² – 6s²t²
= – 3st²

Exercise 6D

Find each of the following products.
Question 1.
Solution:
(5x + 7) (3x + 4)
= 5x (3x + 4) + 7 (3x + 4)
= 5x x 3x + 5x x 4 + 7 x 3x + 7 x 4
= 15x² + 20x + 21x + 28
= 15x² + 41x + 28

Question 2.
Solution:
(4x – 3) (2x + 5)
= 4x (2x + 5) – 3 (2x + 5)
= 4x x 2x + 4x x 5 – 3 x 2x -3 x 5
= 8x² + 20x – 6x – 15
= 8x² + 14x – 15

Question 3.
Solution:
(x – 6) (4x + 9)
= x (4x + 9) – 6 (4x + 9)
= x x 4x + x x 9 – 6 x 4x – 6 x 9
= 4x² + 9x – 24x – 54
= 4x² – 15x – 54

Question 4.
Solution:
(5y – 1) (3y – 8)
= 5y x 3y – 5y x 8 – 1 x 3y – 8 x (-1)
= 15y² – 40y – 3y + 8
= 15y² – 43y + 8

Question 5.
Solution:
(7x + 2y) (x + 4y)
= 7x (x + 4y) + 2y (x + 4y)
= 7x x x + 7x x 4y + 2y x x + 2y x 4y
= 7x² + 28xy + 2xy + 8y²
= 7x² + 30xy + 8y²

Question 6.
Solution:
(9x + 5y) (4x + 3y)
= 9x (4x + 3y) + 5y (4x + 3y)
= 9x x 4x + 9x x 3y + 5y x 4x + 5y x 3y
= 36x² + 27xy + 20xy + 15y²
= 36x² + 47xy +15y²

Question 7.
Solution:
(3m – 4n) (2m – 3n)
= 3m (2m – 3n) – 4n (2m – 3n)
= 3m x 2m – 3m x 3n – 4n x 2m – 4n x (-3n)
= 6m² – 9mn – 8mn + 12n²
= 6m² – 17mn + 12n²

Question 8.
Solution:
(0.8x – 0.5y) (1.5x – 3y)
= 0.8x (1.5x – 3y) – 0.5y (1.5x – 3y)
= 0.8x x 1.5x – 0.8x x 3y – 0.5y x 1.5x – 0.5y x (-3y)
= 1.20x² – 2.4xy – 0.75xy + 1.5y²
= 1.2x² – 3.15xy + 1.5y²

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6D 1

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6D 3

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6D 4

Question 12.
Solution:
(x² – a²) (x – a)
= x² (x – a) – a² (x – a)
= x² x x – x² x a – a² x x – a² (-a)
= x³ – x²a – xa² + a³

Question 13.
Solution:
(3p² + q²) (2p² – 3q²)
= 3p² (2p² – 3q²) + q² (2p² – 3q²)
= 3p² x 2p² – 3p² x 3q² + q² x 2p² – q² x 3q²
= 6q⁴ – 9p²q² + 2p²q² – 3q⁴
= 6p⁴ – 7p²q² – 3q⁴

Question 14.
Solution:
(2x² – 5y²) (x² + 3y²)
= 2x² (x² + 3y²) – 5y² (x² + 3y²)
= 2x² x x² + 2x² x 3y² – 5y² x x² – 5y² x 3y²
= 2x⁴ + 6x²y² – 5x²y² – 15y⁴
= 2y⁴ + x²y² – 15y⁴

Question 15.
Solution:
(x³ – y³) (x² + y²)
= x³ (x² + y²) – y³ (x² + y²)
= x³ x x² + x³ x y² – y³ x x² – y³ x y²
= x⁵ + x³y² – x²y³ – y⁵

Question 16.
Solution:
(x⁴ + y⁴) (x² – y²)
= x⁴ (x² – y²) + y⁴ (x² – y²)
= x⁴ x x² – x⁴ x y² + y⁴ x x² – y⁴ x y²
= x⁶ – x⁴y² + x²y⁴ – y⁶

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 6 Algebraic Expressions Ex 6D 6

Question 18.
Solution:
(x² – y²) (x + 2y)
= x² (x + 2y) – y² (x + 2y)
= x² x x + x² x 2y – y² x x – y² x 2y
= x³ + 2x²y – xy² – 2y³

Question 19.
Solution:
(2x + 3y – 5) (x + y)
= 2x (x + y) + 3y (x + y) – 5 (x + y)
= 2x x x + 2x x y + 3y x x + 3y x y – 5 x x – 5 x y
= 2x² + 2xy + 3xy + 3y² – 5x – 5y
= 2x² + 5xy + 3y² – 5x – 5y

Question 20.
Solution:
(3x + 2y – 4) (x – y)
= 3x (x – y) + 2y (x – y) – 4 (x – y)
= 3x x x – 3x x y + 2y x x – 2y x y – 4 x x – 4 x (-y)
= 3x² – 3xy + 2xy – 2y² – 4x + 4y
= 3x² – xy – 2y² – 4x + 4y

Question 21.
Solution:
(x² – 3x + 7) (2x + 3)
= x² (2x + 3) – 3x (2x + 3) + 7 (2x + 3)
= x² x 2x + x² x 3 – 3x x 2x – 3x x 3 + 7 x 2x + 7 x 3
= 2x³ + 3x² – 6x² – 9x + 14x + 21
= 2x³ – 3x² + 5x + 21

Question 22.
Solution:
(3x² + 5x – 9) (3x – 9)
= 3x² (3x – 9) + 5x (3x – 9) – 9 (3x – 9)
= 3x² x 3x – 3×2 x 9 + 5x x 3x + 5x x (-9) – 9 x 3x – 9 x (-9)
= 9x³ – 27x² + 15x² – 45x – 27x + 81
= 9x³ – 12x² – 72x + 81

Question 23.
Solution:
(9x² – x + 15) (x² – 3)
= 9x² (x² – 3) – x (x² – 3) + 15 (x² – 3)
= 9x² x x² – 9x² x 3 – x x x² + x x 3 + 15 x x² – 15 x 3
= 9x⁴ – 27x² – x³ + 3x + 15x² – 45
= 9x⁴ – x³ – 12x² + 3x – 45

Question 24.
Solution:
(x² + xy + y²) (x – y)
= x² (x – y) + xy (x – y) + y² (x – y)
= x² x x – x² x y + xy x x – xy x y + y² x x – y² x y
= x³ – x²y + x²y – xy² + xy² – y²
= x³ – y³

Question 25.
Solution:
(x² – xy + y²) (x + y)
= x² (x + y) – xy (x + y) + y² (x + y)
= x³ + x²y – x²y – xy² + xy² + y³
= x³ + y³

Question 26.
Solution:
(x² – 5x + 8) (x² + 2)
= x² (x² + 2) – 5x (x² + 2) + 8 (x² + 2)
= x² x x² + x² x 2 – 5x x x² – 5x x 2 + 8 x x² + 8 x 2
= x⁴ + 2x² – 5x³ – 10x + 8x² + 16
= x⁴ – 5x³ + 2x² + 8x² – 10x + 16
= x⁴ – 5x³ + 10x² – 10x + 16

Simplify:
Question 27.
Solution:
(3x + 4) (2x – 3) + (5x – 4) (x + 2)
= [3x (2x – 3) + 4 (2x – 3)] + [5x (x + 2) – 4 (x + 2)]
= [3x x 2x – 3x x 3 + 4 x 2x – 4 x 3] + [5x x x + 5x x 2 – 4 x x – 4 x 2]
= [6x² – 9x + 8x – 12] + [5x² + 10x – 4x – 8]
= (6x² – x – 12) + (5x² + 6x – 8)
= 6x² – x – 12 + 5x² + 6x – 8
= 6x² + 5x² – x + 6x – 12 – 8
= 11x² + 5x – 20

Question 28.
Solution:
(5x – 3) (x + 4) – (2x + 5) (3x – 4)
= [5x x x + 5x x 4 – 3 x x – 3 x 4] – [2x x 3x + 2x x (-4) + 5 x 3x + 5x (-4)]
= [5x² + 20x – 3x – 12] – [6x² – 8x + 15x – 20]
= (5x² + 17x – 12) – (6x² + 7x – 20)
= 5x² + 17x – 12 – 6x² – 7x + 20
= 5x² – 6x² + 17x – 7x – 12 + 20
= -x² + 10x + 8

Question 29.
Solution:
(9x – 7) (2x – 5) – (3x – 8) (5x – 3)
= [9x (2x – 5) -7 (2x – 5)] – [3x (5x – 3) -8 (5x – 3)]
= [9x x 2x – 9x x 5 – 7 x 2x – 7 x (-5)] – [3x x 5x – 3x x 3 – 8 x 5x – 8 x (-3)]
= [18x² – 45x – 14x + 35] – [15x² – 9x – 40x + 24]
= 18x² – 45x – 14x + 35 – 15x² + 9x + 40x – 24
= 18x² – 15x² – 45x – 14x + 9x + 40x + 35 – 24
= 3x² – 59x + 49x + 11
= 3x² – 10x + 11

Question 30.
Solution:
(2x + 5y) (3x + 4y) – (7x + 3y) (2x + y)
= [2x (3x + 4y) + 5y (3x + 4y)] – [7x (2x + y) + 3y (2x + y)]
= [2x x 3x + 2x x 4y + 5y x 3x + 5y x 4y] – [7x x 2x + 7x x y + 3y x 2x + 3y x y]
= [6x² + 8xy + 15xy + 20y²] – [14x² + 7xy + 6xy + 3y²]
= [6x² + 23xy + 20y²] – [14x² + 13xy + 3y^2]
= 6x² + 23xy + 20y² – 14x² – 13xy – 3y²
= 6x² – 14x² + 23xy – 13xy + 20y² – 3y²
= -8x² + 10xy + 11y²

Question 31.
Solution:
(3x² + 5x – 7) (x – 1) – (x² – 2x + 3) (x + 4)
= [3x² (x – 1) + 5x (x – 1) – 7 (x – 1)] – [x² (x + 4) – 2x (x + 4) + 3 (x + 4)]
= [3x² x x – 3x² x 1 + 5x x x – 5x x (-1) – 7 x x -7 x (-1)] -[x² x x + x² x 4 – 2x x x – 2x x 4 + 3 x x + 3 x 4]
= [3x³ – 3x² + 5x² – 5x – 7x + 7] – [x³ + 4x² – 2x² – 8x + 3x + 12]
= [3x³ + 2x² – 12x + 7] – [x³ + 2x² – 5x + 12]
= 3x³ + 2x² – 12x + 7 – x³ – 2x² + 5x – 12
= 3x³ – x³ + 2x² – 2x² – 12x + 5x – 12 + 7
= 2x³ – 7x – 5

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RS AGARWAL SOLUTION | CLASS 7TH | CHAPTER-5| Exponents | EDUGROWN

Exercise 5A

Question 1.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 1

Question 2.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 2

Question 3.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 4

Question 4.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 6

Question 5.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 7

Question 6.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 8

Question 7.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 9

Question 8.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 12

Question 9.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 14

Question 10.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 18

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 20

Question 12.
Solution:
Product of two numbers = 4
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 21

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 22

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 23

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 24

Question 16.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 26

Question 17.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 27

Question 18.
Solution:
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5A 29

Exercise 5B

Question 1.
Solution:
We can write in standard form
(i) 538 = 5.38 x 10²
(ii) 6428000 = 6.428000 x 10⁶ = 6.428 x 10⁶
(iii) 82934000000 = 8.2934000000 x 10¹⁰ = 8.2934 x 10¹⁰
(iv) 940000000000 = 9.40000000000 x 10¹¹ = 9.4 x 10¹¹
(v) 23000000 = 2.3000000 x 10⁷ = 2.3 x 10⁷

Question 2.
Solution:
(i) Diameter of Earth = 12756000 m = 1.2756000 x 10⁷m = 1.2756 x 10⁷ m
(ii) Distance between Earth and Moon = 384000000 m = 3.84000000 x 10⁸ = 3.84 x 10⁸
(iii) Population of India in March 2001 = 1027000000 = 1.027000000 x 10⁹ = 1.027 x 10⁹
(iv) Number of stars in a galaxy = 100000000000 = 1.00000000000 x 10¹¹ = 1 x 10¹¹
(v) The present age of universe = 12000000000 years = 1.2000000000 x 10¹⁰ years = 1.2 x 10¹⁰ years

Question 3.
Solution:
The expanded form will be
(i) 684502 = 6 x 10⁵ + 8 x 10⁴ + 4 x 10³ + 5 x 10² + 2
(ii) 4007185 = 4 x 10⁶ + 0 x 10⁵ + 0 x 10⁴ + 7 x 10³ + 1 x 10² + 8 x 10¹ + 5 x 10⁰
(iii) 5807294 = 5 x 10⁶ + 8 x 10⁵ + 0 x 10⁴ + 7 x 10³ + 2 x 10² + 9 x 10¹ + 4 x 10⁰
(iv) 50074 = 5 x 10⁴ + 0 x 10³ + 0 x 10² + 7 x 10¹ + 4 x 10⁰

Question 4.
Solution:
(i) 6 x 10⁴ + 3 x 10³ + 0 x 10² + 7 x 10¹ + 8 x 10⁰
= 60000 + 3000 + 0 + 70 + 8
= 63078
(ii) 9 x 10⁶ + 7 x 10⁵ + 0 x 10⁴ + 3 x 10³ + 4 x 10² + 6 x 10¹ + 2 x 10⁰
= 9000000 + 700000 + 0 + 3000 + 400 + 60 + 2
= 9703462
(iii) 8 x 10⁵ + 6 x 10⁴ + 4 x 10³ + 2 x 10² + 9 x 10¹ + 6 x 10⁰
= 800000 + 60000 + 4000 + 200 + 90 + 6
= 864296

Exercise 5C

OBJECTIVE QUESTIONS
Mark (✓) tick against the correct answer in each of the following :
Question 1.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 1

Question 2.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 2

Question 3.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 3

Question 4.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 4

Question 5.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 5

Question 6.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 6

Question 7.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 7

Question 8.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 8

Question 9.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 9

Question 10.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 10

Question 11.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 11

Question 12.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 12

Question 13.
Solution:
(d)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 13

Question 14.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 14

Question 15.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 15

Question 16.
Solution:
(a)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 16

Question 17.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 17

Question 18.
Solution:
(b)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 18

Question 19.
Solution:
(c)
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 19

Question 20.
Solution:
(c)
Required number = 10^-1 ÷ (-8)^-1
RS Aggarwal Class 7 Solutions Chapter 5 Exponents Ex 5C 20

Question 21.
Solution:
(c)
The number which is in standard form is 2.156 x 10⁶

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 14 Electric Current and Its Effects

Electric Current and Its Effects Class 7 Science Extra Questions Short Answer Type Questions

Electric Current and Its Effects Class 7 Science Extra Questions Long Answer Type Questions

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 13 Motion and Time

Motion and Time Class 7 Science Extra Questions Short Answer Type Questions

Motion and Time Class 7 Science Extra Questions Long Answer Type Questions

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 12 Reproduction in Plants

Reproduction in Plants Class 7 Science Extra Questions Short Answer Type Questions

Reproduction in Plants Class 7 Science Extra Questions Long Answer Type Questions

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 11 Transportation in Animals and Plants

Transportation in Animals and Plants Class 7 Science Extra Questions Short Answer Type Questions

Transportation in Animals and Plants Class 7 Science Extra Questions Long Answer Type Questions

Exercise 10A

Exercise 10B

Exercise 10C

Exercise 9A

Exercise 9B

Exercise 9C

Exercise 8A

Exercise 8B

Exercise 8C

Exercise 7A

Exercise 7B

Exercise 7C

Exercise 6A

Exercise 6B

Exercise 6C

Exercise 6D

Exercise 5A

Exercise 5B

Exercise 5C

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