1. Which among the following is a unicellular organism that reproduces by budding:

(a) Hydra

(b) Planaria

(c) Yeast

(d) Spirogyra

Answer. (c) Spirogyra

Answer : Yeast is a unicellular organism which reproduces by budding. Spirogyra reproduces by fragmentation; planaria reproduces by regeneration and; hydra (a multicellular organism) reproduces by budding.

2. Which among the following does not reproduce by spore formation:

(a) Penicillium fungus

(b) Yeast fungus

(c) Mucor fungus

(d) Rhizopus fungus

Answer. (b) Yeast fungus

In spore formation method structures called sporangia produces tiny cells called spores. when the spores come in contact with a moist surface , it develops into new individual. example – rhizopus , mucor , penicilium etc. AMOEBA , PARAMAECIUM , PLASMODIUM , YEAST , HYDRA SPYROGYRA ARE NOT PRODUCED BY SPORE FORMATION.

3. The rapid spreading of bread mould on slices of bread are due to:

(i) Presence of large number of spores in air

(ii) Presence of large number of thread-like branched hyphae

(iii) Presence of moisture and nutrients

(iv) Formation of round shaped sporangia

(a) (i) and (iii)

(b) (ii) and (iv)

(c) (i) and (ii)

(d) (iii) and (iv)

Answer. (a) (i) and (iii)

4. The asexual reproduction in the Spirogyra involves:

(a) Breaking up of filaments into smaller bits

(b) Division of a cell into many cells

(c) Division of a cell into two cells

(d) Formation of a large number of buds

Answer. (b) Division of a cell into many cells

Explanation: In spirogyra, asexual reproduction takes place through the process of fragmentation. Spirogyra, a filamentous alga, breaks down its filament into smaller parts, which will grow completely to form a new organism.

5. Reason for the greater similarities among the offsprings produced by asexual reproduction, is:

(i) Asexual reproduction involves only one parent

(ii) Asexual reproduction involves two parents

(iii) Asexual reproduction involves gametes

(iv) Asexual reproduction does not involve gametes

(a) (i) and (ii)

(b) (i) and (iii)

(c) (ii) and (iv)

(d) (i) and (iv)

Answer. (d) (i) and (iv)

6. The process of the division of cell into several cells during reproduction in Plasmodium is termed as:

(a) Fragmentation

(b) Budding

(c) Multiple fission

(d) Binary fission

Answer. (c) Multiple fission

Binary fission is a form of asexual reproduction and cell division where the cell divides into two separate cells.
Budding is a form of asexual reproduction in which a new organism develops from an outgrowth or bud due to cell division at one particular site.
Fragmentation is a form of asexual reproduction in which an organism is split into fragments.
Multiple fission is the process where the nucleus divides into a number of daughter nuclei followed by the division of the cell body into an equal number of parts. This kind of division takes place in Plasmodium.
Thus, the correct answer is option (C), ‘Multiple fission’

7. The number of chromosomes in parents and offsprings of a particular species remains constant due to:

(a) Doubling of chromosomes after zygote formation

(b) Halving of chromosomes during gamete formation

(c) Doubling of chromosomes after gamete formation

(d) Halving of chromosomes after gamete formation

Answer. (b) Halving of chromosomes during gamete formation

Answer: (b) The number of chromosomes in parents and offsprings of a particular species remains constant due to halving of chromosome during gamete formation. The gametes are special type of cells which contain only half the amount of DNA as compared to normal cells of an organism.

8. A Planaria worm is cut horizontally in the middle into two halves P and Q such that the part P contains the whole head of the worm. Another Planaria worm is cut vertically into two halves R and S in such a way that both the cut pieces R and S contain half head each. Which of the cut pieces of the two Planaria worms could regenerate to form the complete respective worms?

(a) Only P

(b) Only R and S

(c) P, Rand S

(d) P, Q, R and S

Answer. (d) P, Q, R and S

In Planaria, each body piece can regenerate into a full, new organism.

9. The number of chromosomes in both parents and offsprings of a particular species remains constant because:
(a) Chromosomes get doubled after zygote formation

(b) Chromosomes get doubled after gamete formation

(c) Chromosomes get halved during gamete formation

(d) Chromosomes get halved after gamete formation

Answer. (c) Chromosomes get halved during gamete formation

Explanation: The number of chromosomes in parents and offsprings of a particular species remains constant due to the halving of chromosomes under the meiosis process during gamete formation.

10. The figure given alongside shows the human male reproductive organs. Which structures make sperms and seminal fluid?

(a) V makes sperms and X makes seminal fluid

(b) W makes sperms and Y makes seminal fluid

(c) X makes sperms and W makes seminal fluid

(d) Y makes sperms and V makes seminal fluid

Answer. (d) Y makes sperms and V makes seminal fluid

Y represents the testes that produce sperms and V represents the prostate gland that produces seminal fluid.

11. An organism capable of reproducing by two asexual reproduction methods one similar to the reproduction in yeast and the other similar to the reproduction in Planaria is:

(a) Spirogyra

(b) Hydra

(c) Bryophyllum

(d) Paramecium

Answer. (b) Hydra

Hydra can reproduce by the method of budding similar to yeast, and also by regeneration as in the case of Planaria.

12. Among the following select the statements that are true regarding the sexual reproduction in flowering plants?

(i) Fertilisation is a compulsory event

(ii) It always results in the formation of zygote

(iii) Offsprings formed are clones

(iv) It requires two types of gametes

(a) (i) nad (iv)

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

(c) (i), (ii) and (iv)

(d) (ii), (iii) and (iv)

Answer. (c) (i), (ii) and (iv)

13. Which among the following are not the functions of testes at puberty?

(i) Formation of germ cells

(ii) Secretion of testosterone

(iii) Development of placenta

(iv) Secretion of estrogen

(a) (i) and (ii)

(b) (i) and (iii)

(c) (ii) and (iv)

(d) (iii) and (iv)

Answer. (d) (iii) and (iv)

14. Which out of the following processes does not lead to the formation of clones:

(a) Fertilisation

(b) Fission

(c) Tissue culture

(d) Fragmentation

Answer. (a) Fertilisation

Fertilisation is the fusion of two gametes in a sexual reproduction. It produces genetically different offsprings and does not lead to the formation of clones.

15. The ratio of number of chromosomes in a human zygote and a human sperm is:

(a) 2 : 1

(b) 3 : 1

(c) 1 : 2

(d) 1 : 3

Answer. (a) 2 : 1

Therefore, 23 chromosomes (sperm cell, n) + 23 chromosomes (egg cell, n) = 46 chromosomes (zygote); or 23 pair of chromosomes (zygote, 2n). The answer is 2:1

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1. The movement of a plant part in response to the force of attraction exerted by the earth is called:

(a) Hydrotropism

(b) Geotropism

(c) Chemotropism

(d) Phototropism

Answer. (b) Geotropism

2. A big tree falls in a forest, but its roots are still in contact with the soil. The branches of this fallen tree grow straight up (vertically). This happens in response to:

(a) Water and light

(b) Water and minerals

(c) Gravity and water

(d) Light and gravity

Answer. (d) Light and gravity

3. The main function of the plant hormone called abscisic acid is to:

(a) Increase the length of cells

(b) Promote cell division

(c) Inhibit growth

(d) Promote growth of stem and roots

Answer. (c) Inhibit growth

4. The growth of tendrils in pea plants is due to the:

(a) Effect of sunlight on the tendril cells facing the sun

(b) Effect of gravity on the part of tendril hanging down towards the earth

(c) Rapid cell division and elongation in tendril cells that are away from the support

(d) Rapid cell division and elongation in tendril cells in contact with the support

Answer. (c) Rapid cell division and elongation in tendril cells that are away from the support

5. The plant hormone which triggers the fall of mature leaves and fruits from the plant body is:

(a) Auxin

(b) Gibberellin

(c) Abscisic acid

(d) Cytokinin

Answer. (c) Abscisic acid

6. The stimulus in the process of thigmotropism is:

(a) Touch

(b) Gravity

(c) Light

(d) Chemical

Answer. (a) Touch

7. A growing seedling is kept in a dark room. A burning lamp is placed near to it for a few days. The top part of seedling bends towards the burning candle. This is an example of:

(a) Chemotropism

(b) Hydrotropism

(c) Phototropism

(d) Geotropism

Answer. (c) Phototropism

8. Dandelion flowers open the petals in bright light during the daytime but close the petals in dark at night. This response of dandelion flowers to light is called:

(a) Phototropism

(b) Thigmonasty

(c) Chemotropism

(d) Photonasty

Answer. (d) Photonasty

9. The number of pairs of nerves which arises from the spinal cord is:

(a) 21

(b) 31

(c) 41

(d) 51

Answer. (b) 31

10. Iodine is necessary for the synthesis of which of the following hormone?

(a) Adrenaline

(b) Auxin

(c) Thyroxine

(d) Insulin

Answer. (c) Thyroxine

11. Which of the following controls the involuntary actions in the body?

(a) Medulla in forebrain

(b) Medulla in hindbrain

(c) Medulla in spinal cord

(d) Medulla in midbrain

Answer. (b) Medulla in hindbrain

12. Which of the following control and regulate the life process?

(a) Reproductive and endocrine systems

(b) Respiratory and nervous systems

(c) Endocrine and digestive systems

(d) Nervous and endocrine systems

Answer. (d) Nervous and endocrine systems

13. A doctor advised a person to take injection of insulin because his:

(a) Blood pressure was high

(b) Heart beat was high

(c) Blood sugar was high

(d) Thyroxine level in blood was high

Answer. (c) Blood sugar was high

14. The dramatic changes in body features associated with puberty are mainly because of the secretions of:

(a) Estrogen from testes and testosterone from ovary

(b) Estrogen from adrenal gland and testosterone from pituitary gland

(c) Testosterone from testes and estrogen from ovary

(d) Testosterone from thyroid gland and estrogen from pituitary gland

Answer. (c) Testosterone from testes and estrogen from ovary

15. Electrical impulse travels in a neuron from:

(a) Dendrite → axon → axon end → cell body

(b) Cell body → dendrite → axon → axon end

(c) Dendrite → cell body → axon → axon end

(d) Axon end →axon → cell body → dendrite

Answer. (c) Dendrite → cell body → axon → axon end

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Q1. A dice is thrown. Find the probability of getting an even number.

(A) 2/3

(B) 1

(C) 5/6

(D) 1/2

Answer:  (D)

Explanation: Total number of cases = 6 (1,2,3,4,5,6)

There are three even numbers 2,4,6

Therefore probability of getting an even number is:

P (even) = 3/6

⇒ P (even) = 1/2

Q2. Two coins are thrown at the same time. Find the probability of getting both heads.

(A) 3/4

(B) 1/4

(C) 1/2

(D) 0

Answer:  (B)

Explanation: Since two coins are tossed, therefore total number of cases = 2² = 4

Therefore, probability of getting heads in both coins is:

∴ P (head) = 1/4

Q3. Two dice are thrown simultaneously. The probability of getting a sum of 9 is:

(A) 1/10

(B) 3/10

(C) 1/9

(D) 4/9

Answer:  (C)

Explanation: Total cases = 36

Total cases in which sum of 9 can be obtained are:

(5, 4), (4, 5), (6, 3), (3, 6)

∴ P (9) = 4/36 = 1/9

Q4. 100 cards are numbered from 1 to 100. Find the probability of getting a prime number.

(A) 3/4

(B) 27/50

(C) 1/4

(D) 29/100

Answer:  (C)

Explanation: Total prime numbers from 1 to 100 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

That means 25 out of 100

So probability is:

P (prime) = 25/100

⇒ P (prime) = 1/4

Q5. A bag contains 5 red balls and some blue balls .If the probability of drawing a blue ball is double that of a red ball, then the number of blue balls in a bag is:

(A) 5

(B) 10

(C) 15

(D) 20

Answer:  (B)

Explanation: Let the number of blue balls be x

Then total number of balls will be 5 + x.

According to question,

x/(5 + x) = 2 X (5/5+x)

⇒ x = 10

Q6.  A box of 600 bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. Then the probability that it is non-defective bulb is:

(A) 143/150

(B) 147/150

(C) 1/25

(D) 1/50

Answer:  (B)

Explanation:

P (non-defective bulb) = 1 – P (Defective bulb)

= 1 – (12/600)

= (600 – 12)/600

= 588/600

= 147/150

Q7. Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box randomly, then the probability that the number on card is a perfect square.

(A) 9/100

(B) 1/10

(C) 3/10

(D) 19/100

Answer:  (B)

Explanation: The perfect square numbers between 2 to 101 are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Total numbers from 2 to 101 =100

So probability of getting a card with perfect square number is:

P (perfect square) = 10/100

⇒ P (perfect square) = 1/10

Q8. What is the probability of getting 53 Mondays in a leap year?

(A) 1/7

(B) 53/366

(C) 2/7

(D) 7/366

Answer:  (C)

Explanation: With 366 days, the number of weeks in a year is

366/7 = 52 (2/7)

i.e., 52 complete weeks which contains 52 Mondays,

Now 2 days of the year are remaining.

These two days can be

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

i.e., there are 7 pairs, in which Monday occurs in 2 pairs,

So probability is:

P (53 Monday) = 2/7

Q9. A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a king of red suit.

(A) 1/26

(B) 3/26

(C) 7/52

(D) 1/13

Answer:  (A)

Explanation: There are total 4 kings in 52 cards, 2 of red colour and 2 of black colour

Therefore, Probability of getting a king of red suit is:

P (King of red suit) = 2/52

⇒ P (King of red suit) = 1/26

Q10. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number 1,2,3……12 ,then the probability that it will point to an odd number is:

(A) 1/6

(B) 1/12

(C) 7/12

(D) 5/12

Answer:  (A)

Explanation: The odd numbers in 1,2,3……..12 are:

1,3,5,7,9,11

Therefore probability that an odd number will come is:

P (odd number) = 6/12

⇒ P (odd number) = 1/2

Q11. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Aryan wins if all the tosses give the same result i.e. three heads or three tails and loses otherwise. Then the probability that Aryan will lose the game.

(A) 3/4

(B) 1/2

(C) 1

(D) 1/4

Answer:  (A)

Explanation: Total outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Favourable outcomes for losing game are

HHT, HTH, THH, HTT, THT, TTH

Therefore probability of losing the game is:

P (Losing the game) = 6/8

⇒ P (Losing the game) = 3/4

Q12. Riya and Kajal are friends. Probability that both will have the same birthday isthe same birthday is:

(A) 364/365

(B) 31/365

(C) 1/365

(D) 1/133225

Answer:  (C)

Explanation:

Riya may have any one of 365 days of the year as her birthday. Similarly Kajal may have any one of 365days as her birthday.

Total number of ways in which Riya and Kajal may have their birthday are:

365 × 365

Then Riya and Kajal may have same birthday on any one of 365 days.

Therefore number of ways in which Riya and Kajal may have same birthday are:

= 365/365 X 365

= 1/365

Q13. A number x is chosen at random from the numbers -2, -1, 0 , 1, 2. Then the probability that x² < 2 is?

(A) 1/5

(B) 2/5

(C) 3/5

(D) 4/5

Answer:  (C)

Explanation: We have 5 numbers −2,−1,0,1,2

Whose squares are 4,1,0,1,4

So square of 3 numbers is less than 2

Therefore Probability is:

P (x² < 2) = 3/5

Q14. A jar contains 24 marbles. Some are red and others are white. If a marble is drawn at random from the jar, the probability that it is red is 2/3, then the number of white marbles in the jar is:

(A) 10

(B) 6

(C) 8

(D) 7

Answer:  (C)

Explanation: Let the number of white marbles be x.

Since only two colour marbles are present, and total probability we know of all the events is equal to 1.

P (white) = 1 – P (red)

x/24 = 1 – (2/3)

⇒ x/24 = 1/3

⇒ x = 8

So there are 8 white marbles.

Q15. A number is selected at random from first 50 natural numbers. Then the probability that it is a multiple of 3 and 4 is:

(A) 7/50

(B) 4/25

(C) 1/25

(D) 2/25

Answer:  (D)

Explanation: The numbers that are multiple of 3(from first 50 natural numbers) are:

3, 6, 9, 12, 15, 18………………..48

The numbers that are multiple of 4 (from first 50 natural numbers) are:

4, 8, 12, 16…………………….48

The numbers that are multiples of 3 and 4 both are the multiples of 3×4=12 as both 3 and 4 are co-prime.

So common multiples are:

12, 24, 36, 48

Therefore probability is:

P (multiple of 3 and 4) = 4/50

⇒ P (multiple of 3 and 4) = 2/25

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(A) lower limits of the classes

(B) upper limits of the classes

(C) midpoints of the classes

(D) frequencies of the class marks.

Answer:  (C)

Explanation: We know that d_i = x_i – a_i. i.ed_i’s are the deviations from the midpoints of the classes.

Q2. While computing mean of the grouped data, we assume that the frequencies are:

(A) evenly distributed over all the classes                                         

(B) centered at the class marks of the classes

(C) centered at the upper limits of the classes

(D) centered at the lower limits of the classes

Answer:  (B)

Explanation: In computing the mean of grouped data, the frequencies are centred at the class marks of the classes

(A) 0

(B) – 1

(C) 1

(D) 2

Answer:  (A)

Explanation:

Q4. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency of a grouped data gives its:

(A) Mean

(B) Median

(C) Mode

(D) All of these

Answer:  (B)

Explanation: Since the intersection point of less than type ogive and more than ogive gives the median on the abscissa.

Q5. For the following distribution,

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

The sum of lower limits of median class and modal class is:

(A) 15

(B) 25

(C) 30

(D) 35

Answer:  (B)

Explanation:

Class	Frequency	Cumulative Frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Now N/2 = 66/2 = 33 which lies in the interval 10 – 15.Therefore lower limit of the median class is 10.

The highest frequency is 20 which lies in the interval 15 – 20. Therefore, lower limit of modal class is 15.

Hence required sum is 10 + 15 = 25

Q6.  If the arithmetic mean of x, x + 3, x + 6, x + 9 and x + 12 is 10, then x = ?

(A) 1

(B) 2

(C) 6

(D) 4

Answer:  (D)

Explanation:

According to question

Q7. If the mean of first n natural numbers is 5n/9, then n =?

(A) 6

(B) 7

(C) 9

(D) 10

Answer:  (C)

Explanation:

But according to question,

Q8. If 35 is removed from the data, 30, 34, 35, 36, 37, 38, 39, 40 then the median increases by:

(A) 2

(B) 1.5

(C) 1

(D) 0.5

Answer:  (D)

Explanation: We have

30, 34, 35, 36, 37, 38, 39, 40

The data has 8 observations, so there are two middle terms, 4^th and 5^th term i.e. 36 and 37.

The median is the mean of both these terms.

Median = (36 + 37)/2

Median = 36.5

When 35 is removed from given data as 30, 35, 36, 37, 38, 39, 40 then the number of observations becomes 7.

Now the median is the middle most i.e 4^th term which is equal to 37.

Therefore median is increased by 37 – 36.5 = 0.5

Q9. The Median when it is given that mode and mean are 8 and 9 respectively, is:

 (A) 8.57

(B) 8.67

 (C) 8.97

(D) 9.24

Answer:  (B)

Explanation: By Empirical formula:

Mode = 3median – 2 mean

8 = 3medain – 2 X 9

8 = 3median – 18

3median = 8 + 18

Median = 26/3

Median = 8.67

(A) 3

(B) 4

(C) 5

(D) 6

Answer:  (D)

Explanation: According to question,

Q11. In a hospital, weights of new born babies were recorded, for one month. Data is as shown:

Weight of new born baby (in kg)	1.4 – 1.8	1.8 – 2.2	2.2 – 2.6	2.6 – 3.0
No of babies	3	15	6	1

Then the median weight is:

(A) 2kg

(B) 2.03kg

(C) 2.05 kg

(D) 2.08 kg

Answer:  (C)

Explanation: Construct a table as follows:

Class-interval	Frequency (fi)	Midpoint (xi)	Cumulative Frequency (cf)
1.4-1.8	3	1.6	3
1.8-2.2	15	2	18
2.2-2.6	6	2.4	24
2.6-3.0	1	2.8	25

Since N/2 = 25/2 = 12.5

12.5 is near to cumulative frequency value 18

So median class interval is 1.8 – 2.2

∴Median = l + [(N/2 – cf)/f]/h

Here

Hence median weight is 2.05 kg.

Q12. In a small scale industry, salaries of employees are given in the following distribution table:

Salary (in Rs.)	4000 – 5000	5000-6000	6000-7000	7000-8000	8000-9000	9000-10000
Number of employees	20	60	100	50	80	90

Then the mean salary of the employee is:

(A) Rs. 7350

(B) Rs.  7400

(C) Rs. 7450

(D) Rs. 7500

Answer: (C)

Explanation:

Therefore mean is:

Q13.  For one term, absentee record of students is given below. If mean is 15.5, then the missing frequencies x and y are:

Number of days	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40	TOTAL
Total Number of students	15	16	x	8	y	8	6	4	70

(A) x = 4 and y = 3

(B) x = 7 and y = 7

(C) x = 3 and y = 4

(D) x = 7 and y = 6

Answer:  (D)

Explanation: Construct a table as follows:

Class-interval	Frequency (fi)	Midpoint (xi)	fixi
0-5	15	2.5	37.5
5 – 10	16	7.5	120
10 – 15	x	12.5	12.5x
15 – 20	8	17.5	140
20 – 25	y	22.5	22.5y
25 -30	8	27.5	220
30 – 35	6	32.5	195
35 – 40	4	37.5	150
TOTAL	70		12.5x+22.5y+862.5

mean = (12.5x + 22.5y + 862.5)/70

⇒ 15.5 = (12.5x +22.5y + 862.5)/70

⇒ 15.5 X 70 = 12.5x +22.5y + 862.5

⇒ 12.5x + 22.5y = 222.5

⇒ 125x + 225y = 2225

⇒ 5x + 9y = 89                        …..(i)

Also,

x + y + 57 = 70

x + y = 13         ……(ii)

Multiplying equation (ii) by 5 and then subtracting from (i) as,

Substituting the value of y in equation (ii), we get

x + y = 13

⇒ x + 6 = 13

⇒ x = 7

Hence x = 7 and y = 6

Q14. Pocket expenses of a class in a college are shown in the following frequency distribution:

Pocket expenses	0-200	200-400	400-600	600-800	800-1000	1000-1200	1200-1400
Number of students	33	74	170	88	76	44	25

Then the median for the above data is:

(A) 485.07

(B) 486.01

(C) 487.06

(D) 489.03

Answer:  (C)

Explanation:

Class-interval	Frequency (fi)	Midpoint (xi)	fixi	cf
0-200	33	100	3300	33
200-400	74	300	22200	107
400-600	170	500	85000	277
600-800	88	700	61600	365
800-1000	76	900	68400	441
1000-1200	44	1100	48400	485
1200-1400	25	1300	32500	510
	510		321400

Since N/2 = 510/2 = 255

255 is near to cumulative frequency value 277.

So median class interval is 400-600

Here,

l = 400

N/2 = 255

cf = 107

f = 170

h = 100

Therefore,

Answer:  (B)

Explanation: We have

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1. A cylindrical pencil sharpened at one edge is the combination of

(A) a cone and a cylinder

(B) frustum of a cone and a cylinder

(C) a hemisphere and a cylinder

(D) two cylinders.

Answer: (A)

Explanation: The shape of a sharpened pencil is:

2. A cone is cut through a plane parallel to its base and then the cone that is for medon one side of that plane is removed. The new part that is left over on the other side of the plane is called

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

Answer:  (A)

Explanation: Observe figure

3. During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase

(B) decrease

(C) remain unaltered

(D) be doubled

Answer:  (C)

Explanation: During conversion of one solid shape to another, the volume of the new shape will remain unaltered.

4. A right circular cylinder of radius r cm and height h cm (h>2r) just encloses a sphere of diameter

(A) r cm

(B) 2r cm

(C) h cm

(D) 2h cm

Answer: (B)

Explanation: Because the sphere is enclosed inside the cylinder, therefore the diameter of sphere is equal to the diameter of cylinder which is 2r cm.

5. A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that 1/8th space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

(A) 142244

(B) 142396

(C) 142496

(D) 142596

Answer:  (A)

Explanation:

6. A metallic spherical shell of internal and external diameters 4 cm and 8 cm respectively, is melted and recast into the form of a cone with base diameter 8cm. The height of the cone is

(A) 12cm

(B) 14cm

(C) 15cm

(D) 18cm

Answer:  (B)

Explanation:Since volume will remain same, therefore,

7. A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is

(A) 21cm

(B) 23cm

(C) 25cm

(D) 19cm

Answer:  (A)

Explanation: Since volume will remain the same, therefore,

8. If two solid hemispheres of same base radii r, are joined together along their bases, then curved surface area of this new solid is

(A) 4πr²

(B) 6πr²

(C) 3πr²

(D) 8πr²

Answer:  (A)

Explanation: Because curved surface area of a hemisphere is and here we join two solid hemispheres along their bases of radii r, from which we get a solid sphere.

Hence the curved surface area of new solid = 2πr² + 2πr² = 4πr²

9. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is

(A) 4πrh + 4πr²

(B) 4πrh − 4πr²

(C) 4πrh + 2πr²           

(D) 4πrh − 2πr²

Answer:  (C)

Explanation: Since the total surface area of cylinder of radius r and height h = 2πrh + 2πr².

When one cylinder is placed over the other cylinder of same height and radius,

Then height of new cylinder = 2h

And radius of the new cylinder = r

Therefore total surface area of new cylinder

= 2πr (2h) + 2πr²

= 4πrh + 2πr²

10. The radii of the top and bottom of a bucket of slant height 45cm are 28cm and 7 cm respectively. The curved surface area of the bucket is:

(A) 4950 cm²

(B) 4951 cm²

(C) 4952 cm²

(D) 4953 cm²

Answer: (A)

Explanation:

Curved Surface area of the bucket = π (R + r) l

⇒ Curved surface area of the bucket = π (28 + 7) X 45

⇒ Curved surface area of the bucket = 4950 cm²

11. A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2 cm. The capacity of the capsule is

(A) 0.36 cm³

(B) 0.35 cm³

(C) 0.34 cm³

(D) 0.33 cm³

Answer:  (A)

Explanation:

Since diameter of the cylinder = diameter of the hemisphere = 0.5cm

Radius of cylinder r = radius of hemisphere r = 0.5/2 = 0.25 cm

Observe the figure,

Total length of capsule = 2cm

Capacity of capsule is:

12. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

Answer:  (C)

Explanation:

Therefore diameter of each solid sphere = 2cm

13. The diameters of the two circular ends of the bucket are 44 cm and 24 cm. The height of the bucket is 35 cm. The capacity of the bucket is

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

Answer:  (A)

Explanation: Since shape of bucket is like Frustum,

Therefore, volume of bucket

14. Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is:

(A) 3 : 4

(B) 4 : 3

(C) 9 : 16

(D) 16 : 9

Answer: (D)

Explanation: According to question,

Therefore ratio of surface area is:

15. A mason constructs a wall of dimensions 270cm× 300cm × 350cm with the bricks each of size 22.5cm × 11.25cm × 8.75cm and it is assumed that 1/8 space is covered by the mortar. Then the number of bricks used to construct the wall is:

(A) 11100

(B) 11200

(C) 11000

(D) 11300

Answer:  (B)

Explanation: According to question,

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1. If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

(A) R₁ + R₂ = R

(B) R₁² + R₂² = R²

(C) R₁ + R₂ < R

(D) R₁² + R₂² < R²

Answer: (B)

Explanation: According to given condition,

Area of circle = Area of first circle + Area of second circle

2. If the circumference of a circle and the perimeter of a square are equal, then

(A) Area of the circle = Area of the square

(B) Area of the circle > Area of the square

(C) Area of the circle < Area of the square

(D) Nothing definite can be said about the relation between the areas of the circle and square.

Answer:  (B)

Explanation: According to given condition

Circumference of a circle = Perimeter of square

Hence Area of the circle > Area of the square

3. Area of the largest triangle that can be inscribed in a semi-circle of radius r units, in square units is:

(A) r²

(B) 1/2r²

(C) 2 r²

(D) √2r²

Answer:  (A)

Explanation: The triangle inscribed in a semi-circle will be the largest when the perpendicular height of the triangle is the same size as the radius of the semi-circle.

Consider the following figure:

4. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is:

(A) 22:7

(B) 14:11

(C) 7:22

(D) 11:14

Answer: (B)

Explanation: Perimeter of circle = Perimeter of square

5. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be

(A) 10 m

(B) 15 m

(C) 20 m

(D) 24 m

Answer: (A)

Explanation: Area of first circular park, whose diameter is 16m

= πr² = π (16/2)² = 64π m²

Area of second circular park, whose diameter is 12m

= πr² = π (12/2)² = 36π m²

According to question,

Area of new circular park =

6. The area of the circle that can be inscribed in a square of side 6 cm is

(A) 36 π cm²

(B) 18 π cm²

(C) 12 π cm²

(D) 9 π cm²

Answer:  (D)

Explanation: Given,

Side of square = 6 cm

Diameter of a circle = side of square = 6cm

Therefore, Radius of circle = 3cm

Area of circle

= πr²

= π (3)²

= 9π cm²

7. The area of the square that can be inscribed in a circle of radius 8 cm is

(A) 256 cm²

(B) 128 cm²

(C) 642 cm²

(D) 64 cm²

Answer:  (B)

Explanation: Radius of circle = 8 cm

Diameter of circle = 16 cm = diagonal of the square

Therefore side of square = diagonal/√2

= 16/√2

Therefore, are of square is = (side)² = (16/√2)²

= 256/2

= 128 cm²

8. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is

(A) 56 cm

(B) 42 cm

(C) 28 cm

(D) 16 cm

Answer:  (C)

Explanation: According to question,

Circumference of circle = Circumference of first circle + Circumference of second circle

πD = πd₁ + πd₂

D = 36 + 20

D = 56cm

9. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm respectively, is

(A) 31 cm

(B) 25 cm

(C) 62 cm

(D) 50 cm

Answer:  (D)

Explanation: According to question

Therefore diameter = 2 × 25 = 50cm

10. If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then

(A) the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

(B) the angle of the corresponding sector of the first circle is equal the angle of the corresponding sector of the other circle.

(C) the angle of the corresponding sector of the first circle is half the angle of the corresponding sector of the other circle.

(D) the angle of the corresponding sector of the first circle is 4 times the angle of the corresponding sector of the other circle.

Answer:  (A)

Explanation: According to Question,

11. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?

(A) 300

(B) 400

(C) 450

(D) 500

Answer: (D)

Explanation:

12. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m, then the area of the field in which the cow can graze is:

(A) 154 m²

(B) 156 m²

(C) 158 m²

(D) 160 m²

Answer:  (A)

Explanation:Figure according to question is:

Area of the field in which cow can graze= Area of a sector AFEG

= (θ/360) X πr²

= (90/360) X π (14)²

= (1/4) X (22/7) X 196

= 154 m²

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13. The area of the shaded region in Fig., where arcs drawn with centres P, Q, Rand S intersect in pairs at mid-points A, B, C and D of the sides PQ, QR, RS and SP, respectively of a square PQRS, is:

(A) 25.25 cm²

(B) 27.45 cm²

(C) 29.65 cm²

(D) 30.96 cm²

Answer:  (D)

Explanation:

14. Area of a sector of central angle 120° of a circle is 3π cm². Then the length of the corresponding arc of this sector is:

(A) 5.8cm

(B) 6.1cm

(C) 6.3cm

(D) 6.8cm

Answer:  (C)

Explanation:

Given that

Area of a sector of central angle 120° of a circle is 3π cm²

15. A round table cover has six equal designs as shown in the figure. If the radius of thecover is 28 cm, then the cost of making the design at the rate of Rs. 0.35 per cm² is:

(A) Rs.146.50

(B) Rs.148.75

(C) Rs.152.25

(D) Rs.154.75

Answer:  (B)

Explanation: The area of the hexagon will be equal to six equilateral triangles with each side equal to the radius of circle.

Area of given hexagon = Area of 6 equilateral triangles.

= 6 X (√3/4) X (side)²

= 6 X (√3/4) X (28)²

= 1999.2 cm²                           (Taking √3 = 1.7)        

Area of circle = πr²

= π × 28²

= 2464 cm²

So, area of designed portion = 2464 – 1999.2 = 464.8 cm²

Cost of making design = 464.8 × 0.35

= Rs. 162.68

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1. To divide a line segment AB in the ratio 5:7, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is:

(A) 8

(B) 10

(C) 11

D) 12

Answer: (D)

Explanation: We know that to divide a line segment in the ratio m : n, first draw a ray AX which makes an acute angle BAX , then marked m+n points at equal distances from each other.

Here m = 5, n = 7

So minimum number of these point = m + n = 5 + 7 = 12

2. To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A₁, A₂, A₃, ….are located at equal distances on the ray AX and the point B is joined to

(A) A₁₂                                                 

(B) A₁₁

(C) A₁₀                                               

(D) A₉

Answer: (B)

Explanation: Here minimum 4+7=11 points are located at equal distances on the ray AX and then B is joined to last point, i.e., A_11.

3. To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the points A₁, A₂, A₃, … and B₁, B₂, B₃, … are located at equal distances on ray AX and BY, respectively. Then the points joined are

(A) A₅ and B₆                                      

(B) A₆ and B₅

(C) A₄ and B₅                                      

(D) A₅ and B₄

Answer:  (A)

Explanation:   Observe the following figure:

4. To construct a triangle similar to a given ΔABC with its sides 3/7 of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B₁, B₂, B₃, … on BX at equal distances and next step is to join:

(A) B₁₀ to C                            

(B) B₃ to C

(C) B₇ to C              

(D) B₄ to C

Answer: (C)

Explanation: Here we locate points B₁,B₂,B₃,B₄,B₅,B₆ and B₇ on BX at equal distances and in next step join the last point B₇ to C

5. To construct a triangle similar to a given ΔABC with its sides 8/5 of the corresponding sides of ΔABC draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is:

(A) 5                                                  

(B) 8

(C) 13                             

(D) 3

Answer: (B)

Explanation: To construct a triangle similar to a given triangle with its sides m/n of the corresponding sides of given triangle ,the minimum number of points to be located at equal distance is equal to the greater of m and n in m/n.

Here, m/n = 8/5

So the minimum number of points to be located at equal distance on ray BX is 8.

6. To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be:

(A) 135° 

(B) 90°

(C) 60°

(D) 120⁰

Answer:  (D)

Explanation: The angle between them should be 120⁰ because the figure formed by the intersection point of pair of tangents, the two end points of those two radii (at which tangents are drawn) and the centre of circle, is a quadrilateral. Thus the sum of the opposite angles in this quadrilateral must be 180^o.

7. To divide a line segment AB in the ratio p : q (p, q are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is

(A) greater of p and q            

(B) p + q

(C) p + q – 1                            

(D) pq

Answer:  (B)

Explanation: We know that to divide a line segment in the ratio m : n, first draw a ray AX which makes an acute angle BAX , then mark m + n points at equal distances from each other.

Here m = p, n = q

So minimum number of these points = m + n = p + q

8. To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is:

(A) 105°                                             

(B) 70°

(C) 140°                                     

(D) 145°

Answer:  (D)

Explanation: The angle between them should be 145⁰ because the figure formed by the intersection point of pair of tangents, the two end points of those two radii (at which tangents are drawn) and the centre of circle, is a quadrilateral. Thus the sum of the opposite angles in this quadrilateral must be 180^o.

9. By geometrical construction, it is possible to divide a line segment in the ratio:

Answer:  (A)

Explanation:

10. A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of ___________ from the centre.

(A) 5cm                                              

(B) 2cm

(C) 3cm                                            

(D) 3.5cm

Answer:  (A)

Explanation: The pair of tangents can be drawn from an external point only, so its distance from the centre must be greater than radius. Since only 5cm is greater than radius of 3.5cm. So the tangents can be drawn from the point situated at a distance of 5cm from the centre.

11. To divide a line segment AB in the ratio 5:6, draw a ray AX such that ∠BAX is an acute angle, then drawa ray BY parallel to AX and the points A₁, A₂, A₃,…. and B₁, B₂, B₃,…. are located to equal distances on ray AX and BY, respectively. Then, the points joined are

(A) A₅ and B₆                                             

(B) A₆ and B₅    

(C) A₄ and B₅                                              

(D) A₅ and B₄

Answer: (A)

Explanation:

To divide line segment AB in the ratio 5:6.

Steps of construction

1. Draw a ray AX making an acute ∠BAX.

2. Draw a ray BY parallel to AX by taking ∠ABY equal to ∠BAX.

3. Divide AX into five (m = 5) equal parts AA₁, A₁A₂, A₂A₃, A₃A₄ and A₄A₅

4. Divide BY into six (n = 6) equal parts and BB₁, B₁B₂, B₂B₃, B₃B₄, B₄B₅ and B₅B₆.

5. Join B₆ A₅. Let it intersect AB at a point C.

Then, AC : BC = 5 : 6

12. A rhombus ABCD in which AB = 4cm and ABC = 60^o, divides it into two triangles say, ABC and ADC. Construct the triangle AB’C’ similar to triangle ABC with scale factor 2/3. Select the correct figure.

Answer:  (A)

13. A triangle ABC is such that BC = 6 cm, AB = 4 cm and AC = 5 cm. For the triangle similar to this triangle with its sides equal to (3/4)th of the corresponding sides of ΔABC, correct figure is:

Answer:  (D)

14. For ∆ABC in which BC = 7.5 cm, ∠B =45⁰ and AB – AC = 4, select the correct figure.

(D) None of these

Answer:  (B)

15. Draw the line segment AB = 5 cm. From the point A draw a line segment AD = 6cm making an angle of 60⁰ with AB. Draw a perpendicular bisector of AD. Select the correct figure.

(D) None of these

Answer:  (A)

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1. If angle between two radii of a circle is 130º, then the angle between the tangents at the ends of the radii is: 

(A) 90º

(B) 50º

(C) 70º

(D) 40º

Answer: (B)

Explanation: If the angle between two radii of a circle is 130º, then the angle between tangents is 180º − 130º = 50º. (By the properties of circles and tangents)

2. The length of tangent from an external point P on a circle with centre O is

(A) always greater than OP

(B) equal to OP

(C) always less than OP

(D) information insufficient

Answer:  (C)

Explanation: Observe figure

The angle between tangent and radius is 90º. This implies OP is the hypotenuse for the right triangle OQP right angled at Q and hypotenuse is the longest side in aright triangle, therefore the length of tangent from an external point P on a circle with centre O is always less than OP.     

3. If angle between two tangents drawn from a point P to a circle of radius ‘a’and centre ‘O’ is 90°, then OP =

(A) 2a√2

(B) a√2

(C) a/√2

(D) 5a√2

Answer:  (B)

Explanation: From point P, two tangents are drawn

OT = a (given)

Also line OP bisects the RPT

Therefore,

 TPO =  RPO = 45º

Also

OT is perpendicular to PT

In right triangle OTP

sin 45° = OT/OP

⇒ 1/√2 = a/OP

⇒ OP = a√2

4. The length of tangent from an external point on a circle is

(A) always greater than the radius of the circle.

(B) always less than the radius of the circle.

(C) may or may not be greater than the radius of circle

(D) None of these.

Answer:  (C)

Explanation: Observe the figure,

In figure OQ is the radius and QP is the tangent.

For right triangle OQP, radius and tangents are two smaller sides, smaller than hypotenuse OP.

Also the length of tangent depends upon the distance of external point from circle. Thus,

The length of tangent from an external point on a circle may or may not be greater than the radius of circle.

5. In the following figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to

(A) 4 cm

(B) 2 cm

(C) 2√3 cm

(D) 4√3 cm

Answer: (C)

Explanation: Join OA

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore,

∠OAT = 90°

In triangle OAT

cos30° = AT/OT

√3/2 = AT/4

AT = 2√3 cm

6. At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is

(A) 4 cm

(B) 5 cm

(C) 6 cm

(D) 8 cm

Answer:  (D)

Explanation: Observe the figure,

Since OC = OA= radius =5cm

Therefore

OE = AE – AO

OE = 8 – 5

OE = 3 cm

In right triangle OEC

OC² = OE² + CE²

⇒ 5² = 3² + CE²

⇒ CE² = 25 – 9

⇒ CE² = 16

⇒ CE = 4

Therefore length of chord CD = 2×4 = 8cm

7. In the following figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to

(A) 100°

(B) 80°

(C) 90°

(D) 75°

Answer:  (A)

Explanation: From figure using properties of circle and tangents

∠OPQ = 90° – 50°

∠OPQ = 40°

OP = OQ = radius

So (∠E) = ∠OQP = 40°

Now In ∆POQ

∠POQ = 180° – (∠OPQ + ∠OQP)

∠POQ = 180° – (40° + 40°)

∠POQ = 100°

8. In the following figure, PA and PB are tangents from a point P to a circle with centre O. Then the quadrilateral OAPB must be a

(A) Square

(B) Rhombus

(C) Cyclic quadrilateral

(D) Parallelogram

Answer:  (C)

Explanation: Since tangent and radius intersect at right angle,

So,

∠OAP = ∠OBP = 90°

⇒ ∠OAP + ∠OBP = 90° + 90° = 180°

Which are opposite angles of quadrilateral OAPB

So the sum of remaining two angles is also 180°

Therefore Quad OAPB is cyclic Quadrilateral.

9. If d₁, d₂ (d₂ > d₁) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, then

(A) d₂² = c² + d₁²

(B) d₂² = c² – d₁²

(C) d₁² = c² + d₂²

(D) d₁² = c² – d₂²

Answer:  (A)

Explanation:

Let AB be a chord of a circle which touches the other circle at C. Then ΔOCB is right triangle.

By Pythagoras theorem

10. If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents at A and B is:

(A) 60°

(B) 120°

(C) 80°

(D) 100°

Answer:  (A)

Explanation: Observe the figure

Using properties of circles and tangents, angle between tangents is:

= 180° – 60°

= 120°

11. If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to

(A) 2√3 cm

(B) 6√3 cm

(C) 3√3 cm

(D) 3 cm

Answer:  (C)

Explanation: Let P be an external point from which pair of tangents are drawn as shown in the figure given below:

Join OA and OP

Also OP is a bisector line of ∠APC

∠APO = ∠CPO = 30°

OA ⊥ AP

 Therefore, in triangle OAP

tan30° = OA/AP

1/√3 = 3/AP

AP = 3√3cm

12. In the following figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to

(A) 20°

(B) 40°

(C) 35°

(D) 45°

Answer:  (B)

Explanation: In the given figure

13. In the adjoining figure, Δ ABC is circumscribing a circle. Then, the length of BC is

(A) 7 cm

(B) 8 cm                               

(C) 9 cm

(D) 10 cm

Answer:  (D)

Explanation: Since lengths of tangents from same external point are equal.

Therefore,

BZ=BL=4cm

Also

AZ=AM=3cm

This gives

MC= 9 – 3 = 6 cm

Similarly LC = MC = 6cm

So length of BC = BL + LC = 4cm + 6cm = 10cm    

14. In the following figure, AB is a chord of the circle and AOC is its diameter such that ACB = 50°. If AT is the tangent to the circle at the point A, then BAT is equal to

(A) 65°

(B) 60°

(C) 50°

(D) 40°

Answer:  (C)

Explanation: Here

∠ABC = 90 (Angle in Semicircle)

In ∆ACB

∠A + ∠B + ∠C = 180°

∠A = 180° – (90° + 50°)

∠A = 40°

Or ∠OAB = 40°

Therefore, ∠BAT = 90° – 40° = 50°

15. In the following figure, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ, then ∠RQS.

(A) 30°

(B) 60°

(C) 90°

(D) 120°

Answer:  (A)

Explanation: Since PQ = PR

(Lengths of tangents from same external point are equal)

Therefore,

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NCERT Solution : Circles

MCQs: Circles

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1.If the length of the shadow of a tower is increasing, then the angle of elevation of the sun

(A) is also increasing                      

(B) is decreasing

(C) remains unaffected

(D) Don’t have any relation with length of shadow

Answer: (B)

Explanation: Observe the following figure, Let A represents sun, then as the length of shadow increases from DC to DB , the angle of elevation decreases from 60 to 30.

2. The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will

(A) also get doubled                                                      

(B) will get halved

(C) will be less than 60 degree                   

(D) None of these

Answer: (C)

Explanation: According to Question:

3. If the height of a tower and the distance of the point of observation from its foot,both, are increased by 10%, then the angle of elevation of its top

(A) increases                                                                                     

(B) decreases

(C) remains unchanged                                                

(D) have no relation.

Answer:  (C)

Explanation: Since          

tan θ = h/x

Where h is height and x is distance from tower,

If both are increased by 10%, then the angle will remain unchanged.

4. A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall will be

(A) 7.5m                                                                              

(B) 7.7m

(C) 8.5m                                                                              

(D) 8.8m

Answer:  (A)

Explanation: Given that the height of ladder is 15m

Let height of vertical be = h

And the ladder makes an angle of elevation 60° with the wall

In triangle QPR

5. An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high.Determine the angle of elevation of the top of the tower from the eye of the observer.

(A) 30°                                                                                                             

(B) 45°

(C) 60°                                                                                                 

(D) 90°

Answer:  B

Explanation: Let the angle of elevation of the tower from the eye of observer be θ.

Given that:

AB = 22m, PQ = 1.5m = MB

QB = PM = 20.5m

AM = AB − MB = 22 − 1.5 = 20.5m

Now in triangle APM

6. The angles of elevation of the top of a tower from two points distant s and t from its foot are complementary. Then the height of the tower is:

(A) st                                                                                    

(B) s²t²

(C) √st                                                                 

(D) s/t

Answer:  (C)

Explanation: Let the height of tower be h.

Construct figure according to given information as,

7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Then the height of tower is:

(A) 20√3                                             

(B) 25√3

(C) 10√3                                             

(D) 30√3

Answer:  (B)

Explanation: Given condition can be represented as follows where SQ is the pole.

Let the height be h and RQ = x m

Then from figure:

8. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is

(A) equal to the angle of depression of its reflection.

(B) double to the angle of depression of its reflection

(C) not equal to the angle of depression of its reflection

(D) information insufficient

Answer:  (C)

Explanation: Observe the figure,

We know that if P is a point above the lake at a distance d, then the reflection of the point in the lake would be at the same distance d.

Also the angle of elevation and depression from the surface of the lake is same.

Here the man is standing on a platform 3m above the surface , so its angle of elevation to the cloud and angle of depression to the reflection of the cloud cannot be same.

9. If a pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is

(A) 60°                                                                   

(B) 45°

(C) 30°                                                 

(D) 90°

Answer:  (A)

Explanation: According to Question:

Therefore,

tan θ = 6/2√3

⇒ tan θ = √3

⇒ tan θ = tan60°

⇒ θ = 60°

10. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower

(A) 10 (√3 + 1)                                

(B) 5√3

(C) 5 (√3 + 1)                                                   

 (D) 10√3

Answer:(A)

Explanation: Since after moving towards the tower the angle of elevation of the top increases by 15°.

Therefore angle becomes 30° + 15° = 45°

Observe the figure,

11. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Then the height of the tower is:

Answer:  (B)

Explanation: Observe the figure,

12. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60°, then the distance between the two towers is:

(A) 10√3 m                                                                        

(B) 15√3 m

(C) 12√3 m                                                                        

(D) 36 m

Answer:  (A)

Explanation: Observe the figure,

Let the distance between two towers be x m.

From figure,

tan60° = 30/x

⇒ √3 = 30/x

⇒ x = 30/√3

⇒ x = 10√3m

13. The angle of elevation of the top of a vertical tower from a point on the ground is60°. From another point 10 m vertically above the first, its angle of elevation is45°. Find the height of the tower.

(A) 5 (√3 + 3) m                                                              

(B) (√3 +3) m

(C) 15 (√3 +3)                                                  

(D) 5√3

Answer:  (A)

Explanation: According to Question:

14. A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be A and B respectively. Then the height of the other house is:

Answer:  (B)

Explanation: Observe the figure,

Let the height of another house be H m and distance between two houses is x m.

From figure,

15. There are two windows in a house. A window of the house is at a height of 1.5 m above the ground and the other window is 3 m vertically above the lower window. Ram and Shyam are sitting inside the two windows. At an instant, the angle of elevation of a balloon from these windows are observed as 45° and 30° respectively. Find the height of the balloon from the ground.

(A) 7.598m                                                                         

(B) 8.269m

(C) 7.269m                                                                         

(D) 8.598 m

Answer:  (D)

Explanation: Let PQ be the ground level, Ram be sitting at A, Shyam be sitting at B and the balloon be at C from the ground.

Then

AP = 1.5m

And

AB = 3m

AP = DQ = 1.5m and BA = ED = 3m

Let the height of balloon from ground be h,

Then CE = (h − 4.5)m

In right triangle ADC

In right triangle CEB

Important Link

Quick Revision Notes : Some Applications of Trigonometry

NCERT Solution : Some Applications of Trigonometry

MCQs: Some Applications of Trigonometry

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