Author Archives: Rohit

Ex 5.1 Class 7 Maths Question 1.
Find the complement of each of the following angles :

Solution:

1. Complement of the angle 20° = 90° – 20° = 70°
2. Complement of the angle 63° = 90° – 63° = 27°
3. Complement of the angle 57° = 90° – 57° = 33°

Ex 5.1 Class 7 Maths Question 2.
Find the supplement of each of the following angles :

Solution:
Since, the sum of the measures of an angle and its supplement is 180°, therefore,

1. The supplement of an angle of measure 105° is the angle of (180°-105°), i.e., 75°.
2. The supplement of an angle of measure 87° is the angle of (180°-87°), i.e., 93°.
3. The supplement of an angle of measure 154° is the angle of (180°-154°), i.e., 26°.

Ex 5.1 Class 7 Maths Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary :
(i) 65°, 115°
(ii) 63°, 27°
(iii) 112°, 68°
(iv) 130°, 50°
(v) 45°, 45°
(vi) 80°, 10°
Solution:
(i) 65°, 115°
∵ 65° + 115° = 180°
∴ The given pair of angles are supplementary.

(ii) 63°, 27°
∵ 63° + 27° = 90°
The given pair of angles are complementary.

(iii) 112°, 68°
∵ 112° + 68° = 180°
∴ The given pair of angles are supplementary.

(iv) 130°, 50°
∵ 130° + 50° = 180°
∴ The given pair of angles are supplementary.

(v) 45°, 45°
∵45° + 45° = 90°
∴ The given pair of angles are complementary.

(vi) 80°, 10°
∵ 80° + 10° = 90°
∴ The given pair of angles are complementary.

Ex 5.1 Class 7 Maths Question 4.
Find the angle which is equal to its complement.
Solution:
Let the angle be x°.
Its complement = 90° – x°
According to the question,
x° = 90° – x0
⇒ x° + x° = 90°
⇒ 2x° = 90°
⇒ x° = 90∘3 = 45°

Ex 5.1 Class 7 Maths Question 5.
Find the angle which is equal to its supplement.
Solution:
Let the angle be x°.
Its supplement = 180° – x°
According to the question,
x° = 180° – x°
⇒ x° + x° = 180°
⇒ 2x° = 180°
⇒ x° = 180∘3 = 90°
Hence, the required angle is 90°.

Ex 5.1 Class 7 Maths Question 6.
In the given figure, ∠1 and ∠2 are supplementary angles. If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary?

Solution:
∠2 will increase with the same measure as the decrease in ∠1.

Ex 5.1 Class 7 Maths Question 7.
Can two angles be supplementary if both of them are :

1. acute?
2. obtuse?
3. right?

Solution:

1. No
2. No
3. Yes

Ex 5.1 Class 7 Maths Question 8.
An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?
Solution:
Since the sum of the measure of ah angle and its complement is 90°.
∴ The complement of an angle of measures 45° + x°,
where x > 0 is the angle of [90° – (45° + x°)] = 90° – 45° – x°= 45° – x°.
Clearly, 45° + x° > 45° – x°
Hence, the complement of an angle > 45° is less than 45°.

Ex 5.1 Class 7 Maths Question 9.
In the adjoining figure

1. Is ∠1 adjacent to ∠2?
2. Is ∠AOC adjacent to ∠AOE?
3. Do ∠COE and ∠EOD form a linear pair?
4. Are ∠BOD and ∠DOA supplementary?
5. Is ∠1 vertically opposite to Z4?
6. What is the vertically opposite angle of ∠5

Solution:

1. Yes! ∠ 1 is adjacent to ∠ 2.
2. No! ∠ AOC is not adjacent to ∠ AOE.
3. Yes! ∠ COE and ∠ EOD form a linear pair.
4. Yes! ∠ BOD and ∠ DOA are supplementary.
5. Yes! ∠ 1 is vertically opposite to ∠ 4.
6. The vertically opposite angle of ∠ 5 is ∠ 2 + ∠ 3, i.e., ∠ COB.

Ex 5.1 Class 7 Maths Question 10.
Indicate which pairs of angles are :

1. Vertically opposite angles.
2. Linear pairs.

Solution:

1. ∠1 and ∠4; ∠5 and ∠2 + ∠3
2. ∠4 and ∠5; ∠5 and ∠1

Ex 5.1 Class 7 Maths Question 11.
In the adjoining figure, is ∠1 adjacent to ∠2? Give reasons.

Solution:
No! ∠ 1 is not adjacent to ∠ 2 because their vertex is not common.

Ex 5.1 Class 7 Maths Question 12.
Find the values of the angles x, y, and z in each of the following?

Solution:

Ex 5.1 Class 7 Maths Question 13.
Fill in the blanks:

1. If two angles are complementary, then the sum of their measures is __________
2. If two angles are supplementary, then the sum of their measures is __________
3. Two angles forming a linear pair are __________
4. If two adjacent angles are supplementary, they form a __________
5. If two lines intersect at a point, then the vertically opposite angles are always __________
6. If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are __________

Solution:

1. 90°
2. 180°
3. supplementary
4. linear pair
5. equal
6. obtuse angles

Ex 5.1 Class 7 Maths Question 14.
In the adjoining figure, name the following pairs of angles :

1. Obtuse vertically opposite angles.
2. Adjacent complementary angles.
3. Equal supplementary angles.
4. Unequal supplementary angles.
5. Adjacent angles do not form a linear pair.

Solution:

1. ∠ AOD and ∠ BOC
2. ∠ EOA and ∠ AOB
3. ∠ BOE and ∠ DOE
4. ∠ EOC and ∠ EOA
5. ∠ AOB and ∠ AOE ; ∠ AOE and ∠ EOD ; ∠ EOD and ∠ DOC.

Ex 5.2 Class 7 Maths Question 1.
State the property that is used in each of the following statements?

Solution:
(i) Corresponding angle property.
(ii) Alternate interior angle property.
(iii) Interior angles on the same side of the transversal are supplementary.

Ex 5.2 Class 7 Maths Question 2.
In the adjoining figure, identify :

1. the pairs of corresponding angles.
2. the pairs of alternate interior angles.
3. the pairs of interior angles on the same side of the transversal.
4. the vertically opposite angles

Solution:

1. ∠1, ∠5; ∠2, ∠6; ∠3, ∠7 and ∠4, ∠8 are four pairs of corresponding angles.
2. ∠2, ∠8 and ∠3, ∠5 are two pairs of alternate interior angles.
3. ∠2, ∠5 and ∠3, ∠8 are two pairs of interior angles on the same side of the transversal.
4. ∠1, ∠3; ∠2, ∠4; ∠5, ∠7 and ∠6, ∠8 are four pairs of vertically opposite angles.

Ex 5.2 Class 7 Maths Question 3.
In the adjoining figure p || q. Find the unknown angles

Solution:

Ex 5.2 Class 7 Maths Question 4.
Find the value of x in each of the following figures if l || m :

Solution:
(i) Since, l || m and t is a transversal.
∴ ∠x = (180° – 110°) = 70° [Corresponding angles, Linear pair]

(ii) if l || m and a is a transversal.
Then, ∠x = 1000 [Corresponding angles]

Ex 5.2 Class 7 Maths Question 5.
In the given figure, the arms of two angles are parallel. If ∠ABC = 70°, then find

(i) DGC
(ii) DEF
Solution:

Ex 5.2 Class 7 Maths Question 6.
In the given figures below, decide whether l is parallel to m.

Solution:

Ex 4.1 Class 7 Maths Question 1.
Complete the last column of the table.

Solution:

Ex 4.1 Class 7 Maths Question 2.
Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
(b) 7n + 5 = 19 (w = -2)
(c) 7n + 5 = 19 (n = 2)
(d) 4p – 3 = 13 (p = 1)
(e) 4p – 3 = 13 (p = -4)
(f) 4p – 3 = 13 (p = 0)
Solution:

Ex 4.1 Class 7 Maths Question 3.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17 .
(ii) 3m – 14 = 4
Solution:
(i) Let us evaluate the L.H.S. and R.H.S. of the given equation for some values of p and continue to given new values till the L.H.S. becomes equal to the R.H.S.
The given equation is 5p + 2 = 17. We have,
L.H.S. = 5p+ 2 and R.H.S. = 17

Clearly, L.H.S. = R.H.S. for p = 3. Hence, p = 3 is the solution of the given equation.

(ii) Let us evaluate the L.H.S. and R.H.S. of the given equation for some values of m and continue to given new values till the L.H.S. becomes equal to the R.H.S.
The given equation is 3m – 14 = 4, that is, 14 subtracted from 3 times m gives 4.
So, we substitute values which gives 3m > 14.
We have, L.H.S. = 3m – 14 and R.H.S. = 4

Clearly, L.H.S. = R.H.S. for m = 6. Hence, m = 6 is the solution of the given equation.

Ex 4.1 Class 7 Maths Question 4.
Write equations for the following statements:

1. The sum of numbers x and 4 is 9.
2. 2 subtracted from y is 8.
3. Ten times a is 70.
4. The number b divided by 5 gives 6.
5. Three-fourth of t is 15.
6. Seven times m plus 7 gets you 77.
7. One-fourth of a number x minus 4 gives 4.
8. If you take away 6 from 6 times y, you get 60.
9. If you add 3 to one-third of z, you get 30.

Solution:
The equations for the given statements are :

1. x + 4 = 9
2. y – 2 = 8
3. 10 α = 70
4. b5 = 6
5. 34 t = 15
6. 7m + 7 = 77
7. 14 x – 4 = 4
8. 6y – 6 = 60
9. 13 z + 3 = 30

Ex 4.1 Class 7 Maths Question 5.
Write the following equations in statement forms :

1. p + 4 = 15
2. m – 7 = 3
3. 2m = 7
4. m5 = 3
5. 3m5= 6
6. 3p + 4 = 25
7. 4p – 2 = 18
8. p2 + 2 = 8

Solution:
The statements for the given equations are :

1. The sum of numbers p and 4 is 15.
2. The difference of m and 7 is 3.
3. Two times m is 7.
4. The number m divided by 5 gives 3.
5. Three time’s m divided by 5 gives 6.
6. Three times p plus 4 gives 25.
7. Four times p minus 2 gives 18.
8. p divided by 2 plus 2 gives 8.

Ex 4.1 Class 7 Maths Question 6.
Set up an equation in the following cases :
(i) Irfan says that he has 7 marbles more than five times the marbles Parxnit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l.)
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Solution:
(i) Let Parmit has m marbles. Then, five times the marbles Parmit has = 5m
Irfan has 7 marbles more than five times the marbles Parmit has = 5m + 7,
i.e., Irfan has (5m + 7) marbles
But it is given that Irfan has 37 marbles. Therefore, 5m + 7 = 37

(ii) Let the age of Laxmi be y years.
Then, three times Laxmi’s age = 3y years Laxmi’s father is 4 years older than three times Laxmi’s age,
i.e., Age of Laxmi’s father = (3y + 4) years But it is given that Laxmi’s father is 49 years old Therefore, 3y + 4 = 49

(iii) Let the lowest marks be l.
Then, twice the lowest marks = 21
Highest score obtained by a student in her class is twice the lowest marks plus 7, i.e.,
Highest score = 2l + 7
But this is given to be 87
Therefore, 2l + 7 = 87

(iv) Let the base angle be b in degrees. Then the vertex angle is 2b in degrees.
∵ Sum of the angles of a triangle is 180 degrees.
∴ 2 b + b + b = 180° or 4 b = 180°

Ex 4.2 Class 7 Maths Question 1.
Give first the step you will use to separate the variable and then solve the equation :
(a) x – 1 = 0
(b) x + 1= 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = -7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y – 4 = -4
Solution:
(a) We have, x – 1 = 0
In order to solve this equation, we have to get x by itself on the L.H.S. To get x by itself on the L.H.S., we need to shift -1. This can be done by adding 1 to both sides of the given equation.
x – 1 + 1 = 0 + 1 [ Adding 1 to both sides ]
x = 1 [ ∵ -1 + 1 = 0 and 0 + 1 = 1 ]

(b) We have, x + 1 = 0
In order to get x by itself on the L.H.S., we need to shift 1. This can be done by subtracting 1 from both sides of the given equation.
x + 1 – 1 = 0 – 1 [Subtracting 1 from both sides]
⇒ x = -1 [∵ 1 – 1 = 0, 0 – 1 = -1 ]
So, x = -1 is the solution of the given equation.

(c) We have, x – 1 = 5
In order to solve this equation, we have to get x by itself on the L.H.S. To get x by itself on the L.H.S., we need to shift -1. This can be done by adding 1 to both sides of the given equation.

(d) We have, x + 6 = 2
In order to solve this equation, we have to get x by itself on the L.H.S. To get x by itself on the L.H.S., we need to shift 6. This can be done by subtracting 6 from both sides of the given equation.

(e) We have, y – 4 = -7
In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S.,we need to shift -4. This can be done by adding 4 to both sides of the given equation.

(f) We have, y – 4 = 4
In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift -4. This can be done by adding 4 to both sides of the given equation.

(g) We have, y + 4 = 4
In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift 4. This can be done by subtracting 4 from both sides of the given equation.

(h) We have, y + 4 = -4
In order to solve this equation, we have to get y by itself on the L.H.S. To get y by itself on the L.H.S., we need to shift 4. This can be done by subtracting 4 from both sides of the given equation.

Ex 4.2 Class 7 Maths Question 2.
Give first the step you will use to separate the variable and then solve the equation :
(a) 3l = 42
(b) b2 = 6
(c) p7 = 4
(d) 4x = 25
(e) 8y = 36
(f) z3 = 54
(g) a5 = 715
(h) 20t = -10
Solution:
(a) We have, 3l = 42
In order to solve this equation, we have to get l by itself on the L.H.S. For this, 3 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 3.

(b) We have, b2 = 6
In order to solve this equation, we have to get b by itself on the L.H.S. To get b by itself on L.H.S., we have to remove 2 from L.H.S. This can be done by multiplying both sides of the equation by 2. Thus, we have

(c) We have, p7 = 4
In order to solve this equation, we have to get p by itself on the L.H.S. To get p by itself on L.H.S., we have to remove 7 from L.H.S. This can be done by multiplying both sides of the equation by 7. Thus, we have

(d) We have, 4x = 25
In order to solve this equation, we have to get x by itself on the L.H.S. For this, 4 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 4.

(e) We have, 8y = 36
In order to solve this equation, we have to get y by itself on the L.H.S. For this, 8 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 8.

(f) We have, z3 = 54
In order to solve this equation,we have to get z by itself on the L.H.S. To get z by itself on L.H.S., we have to remove 3 from L.H.S. This can be done by multiplying both sides of the equation by 3.
Thus, we have

(g) We have, a5 = 715
In order to solve this equation, we have to get a by itself on the L.H.S. To get a by itself on L.H.S., we have to remove 5 from L.H.S. This can be done by multiplying both sides of the equation by 5.

(h) We have, 20t = -10
In order to solve this equation, we have to get t by itself on the L.H.S. For this, 20 has to be removed from the L.H.S. This can be done by dividing both sides of the equation by 20.

Ex 4.2 Class 7 Maths Question 3.
Give the steps you will use to separate the variable and then solve the equation :
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) 20p3 = 40
(d) 3p10 = 6
Solution:

Ex 4.2 Class 7 Maths Question 4.
Solve the following equations :
(a) 10p = 100
(b) 10p + 10 = 100
(c) p4 = 5
(d) −p3 = 5
(e) 3p4 = 6
(f) 3s = -9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Solution:

Ex 4.3 Class 7 Maths Question 1.
Solve the following equations :

Solution:

Ex 4.3 Class 7 Maths Question 2.
Solve the following equations :
(a) 2 (x + 4) = 12
(b) 3 (n – 5) = 21
(c) 3 (n – 5) = -21
(d) -4 (2 + x) = 8
(e) 4 (2 – x) = 8
Solution:

Ex 4.3 Class 7 Maths Question 3.
Solve the following equations :
(a) 4 = 5 (p – 2)
(b) -4 = 5 (p – 2)
(c) 16 = 4 + 3 (t + 2)
(d) 4 + 5 (p – 1) = 34
(e) 0 = 16 + 4 (m – 6)
Solution:

Ex 4.3 Class 7 Maths Question 4.
(a) Construct 3 equations starting with x = 2.
(b) Construct 3 equations starting with x = -2.
Solution:
(a) First equation :
Start with x = 2
Multiply both sides by 3, 3x = 6
Add 2 to both sides, 3x + 2 = 8

Second equation :
Start with x = 2
Multiply both sides by -3, -3x, = -6
Add 8 to both sides, 8 – 3x = 2.

Third equation :
Start with x = 2
Divide both sides by 5, x5 = 25
Subtract 2 from both sides,

(b) First equation :
Start with x = -2
Multiply both sides by 2, 2x = -4
Subtract 3 from both sides, 2x -3 = -7

Second equation :
Start with x = -2
Multiply both sides by – 5, – 5x = 10
Add 10 to both sides, 10 – 5x = 20

Third equation :
Start with x = -2
Divide both sides by 2, x2 = -1
Add 3 to both sides, x2 + 3 = 2

Ex 4.4 Class 7 Maths Question 1.
Set up equations and solve them to find the unknown numbers in the following cases :
(a) Add 4 to eight times a number; you get 60.
(b) One fifth of a number minus 4 gives 3.
(c) If I take three fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from 52 of the number, the result is 23.
Solution:
(a) Let x be the required number. Then, the required equation is

(b) Let x be the required number. Then, the required equation is x5 – 4 = 3.

(c) Let y be the required number. Then, the required equation be 3y4 + 3 = 21

(d) Let the required number be m. Then, the required equation is 2m -11 = 15

(e) Let Munna have x notebooks. Then, the required equation is 50 – 3x = 8

(f) Let the number be x. Then, the required equation is

(g) Let the number be n. Then, the required equation is

Ex 4.4 Class 7 Maths Question 2.
Solve the following :
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°). (c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Solution:

Ex 4.4 Class 7 Maths Question 3.
Solve the following :
(i) Irfan says that he has 7 marbles more than five times the marbles Pannit has. Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit tree planted if the number of non-fruit trees planted was 77?
Solution:

Ex 4.4 Class 7 Maths Question 4.
Solve the following riddle :
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Solution:

Ex 3.1 Class 7 Maths Question 1.
Find the range of heights of any ten students of your class.
Solution:
Let the heights (in cm) of 10 students in the class be 150, 152, 151, 148, 149, 149, 150, 152, 153, 146.
Arranging the heights in ascending order, we have 146, 148, 149, 149, 150, 150, 151, 152, 152, 153.
Range of height of students = 153 – 146 = 7

Ex 3.1 Class 7 Maths Question 2.
Organise the following marks in a class assessment, in a tabular form.
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5,1, 9, 6, 5, 8, 4, 6, 7
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
Solution:
Arranging the marks in a class assessment in a tabular form :

Ex 3.1 Class 7 Maths Question 3.
Find the mean of the first five whole numbers.
Solution:
The first 5 whole numbers are 0, 1, 2, 3, 4
Their mean = 0+1+2+3+45 = 105 = 2

Ex 3.1 Class 7 Maths Question 4.
A cricketer scores the following runs in eight innings: 58, 76,40, 35, 46, 45, 0,100. Find the mean score.
Solution:
Total runs = 58 + 76 + 40 + 35 + 46 + 45 + 0 +100 = 400
Number of observations = 8
∴ Mean = 4008 = 50

Ex 3.1 Class 7 Maths Question 5.
Following table shows the points of each player scored in four games :

Now answer the following questions :
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all four games. How would you find the mean?
(iv) Who is the best performer?
Solution:
(i) A’s average number of points scored per game 14+16+10+104 = 504 = 12.5

(ii) C’s average points per game = 8+11+0+134 = 324 = 8
Since we are comparing the performance, so we divide by 4 to find the mean for C.

(iii) B’s average point per game = 0+8+6+44 = 184 = 4.5
(To find B’s average, we find the sum of all observations and divide this by the number of observations.)

(iv) Since, 12.5 > 8 > 4.5
∴ The best performer is A.

Ex 3.1 Class 7 Maths Question 6.
The marks (out of 100) obtained by a group of students in a science test are 85,76,90,85,39,48,56,95,81 and 75. Find the :
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Solution:
Arranging the marks obtained by a group of students in ascending order, we have 39, 48, 56, 75, 76, 81, 85, 85, 90, and 95.
(i) Highest and the lowest marks obtained are 95 and 39, respectively.
(ii) Range of the marks obtained = 95 – 39 =56
(iii) Mean marks

Ex 3.1 Class 7 Maths Question 7.
The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2540, 2820.
Find the mean enrolment of the school for this period.
Solution:
Sum of the enrolment during six consecutive years
= 1555 + 1670 + 1750 + 2013 + 2540 + 2820
= 12348
Mean enrolment = 123486 = 2058

Ex 3.1 Class 7 Maths Question 8.
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows :

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?
Solution:
(i) Arranging the rainfall during the week in ascending order,
we have 0.0, 0.0, 1.0, 2.1, 5.5, 12.2, 20.5
Range = 20.5 – 0.0 = 20.5

(ii) Sum of the rainfall during the week = 0.0 + 0.0 + 1.0 + 2.1 + 5.5 + 12.2 + 20.5 = 41.3
Mean = 41.37 = 5.9

(iii) For five days, the rainfall was less than the mean rainfall.

Ex 3.1 Class 7 Maths Question 9.
The heights of 10 girls were measured in cm and the results are as follows :
135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height?
Solution:
Arranging the heights (in cm) in the ascending order, we have 128, 132, 135, 139, 141, 143, 146, 149, 150, 151.
(i) Height of the tallest girl is 151 cm.
(ii) Height of the shortest girl is 128 cm.
(iii) Range = (151 – 128) cm = 23 cm.
(iv) Mean height of the girls = 141.4 cm
(v) 5 girls have heights more than the mean height.

Ex 3.2 Class 7 Maths Question 1.
The scores in mathematics test (out of 25) of 15 students is as follows :
19, 25, 23, 20, 9, 20,15, 10, 5,16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
Solution:
Arranging the scores in Mathematics in ascending order,
we have 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25.
Clearly, 20 occurs the maximum number of times.

Ex 3.2 Class 7 Maths Question 2.
The runs scored in a cricket match by 11 players is as follows :
6, 15, 120, 50, 100, 80, 10, 15, 8, 10,15
Find the mean, mode and median of this data. Are the three same?
Solution:

Ex 3.2 Class 7 Maths Question 3.
The weights (in kg) of 15 students of a class are :
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47
(i) Find the mode and median of this data.
(ii) Is there more than one mode?
Solution:
(i) Arranging the weights (in kg) in ascending order,
we have 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50.
Clearly, 38 and 43 occur the maximum number of times.
∴ Modes are 38 and 43.
Here, N = 15, which is odd.

(ii) Yes, there are more than one mode.

Ex 3.2 Class 7 Maths Question 4.
Find the mode and median of the following data :
13, 16, 12, 14, 19, 12, 14, 13, 14.
Solution:
Arranging the data in ascending order, we have
12, 12, 13, 13, 14, 14, 14, 16, 19
Clearly, 14 occurs the maximum number of times.

Ex 3.2 Class 7 Maths Question 5.
Tell whether the statement is true or false :

1. The mode is always one of the numbers in a data.
2. The mean is one of the numbers in a data.
3. The median is always One of the numbers in a data.
4. The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9.

Solution:

1. True
2. False
3. True
4. False

Ex 3.3 Class 7 Maths Question 1.
Use the bar graph to answer the following questions.

(a) Which is the most popular pet?
(b) How many students have dog as a pet?

Solution:
Clearly, from the given bar graph :

(a) The most popular pet is cat.
(b) Eight children have dog as a pet.

Ex 3.3 Class 7 Maths Question 2.
Read the bar graph which shows the number of books sold by a bookstore during five consecutive years and answer the following questions :
(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?

Solution:
Clearly, from the given graph, we have
(i) Number of books sold in the year
1989 : 170 (approx.)
1990 : 475 (approx.)
1992 : 225 (approx.)
(ii) In the year 1990, about 475 books were sold. In the year 1992, about 225 books were sold.
(iii) Fewer than 250 books were sold in the years 1989 and 1992.
(iv) It can be estimated using the height of the bar such that height of 1 cm = 100 books.

Ex 3.3 Class 7 Maths Question 3.
Number of children in six different classes are given below. Represent the data on a bar graph.

(a) How would you choose a scale?
(b) Answer the following questions :
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of class sixth to the student of class eighth.
Solution:
(a) Start the scale at 0. The greatest value in the data is 135, so end the scale at a value greater than 135, such as 140. Use equal divisions along the axes, such as increments of 10. Here, we take 1 unit for 10 children. The bar graph is as under :

(b) (i) The fifth class has the maximum number of children. The minimum number of children are in class tenth.
(ii) Ratio of students of class sixth to eighth is 120 : 100 = 6 : 5.

Ex 3.3 Class 7 Maths Question 4.
The performance of a student in 1st Term and 2nd Term is given. Draw a double bar graph choosing appropriate scale and answer the following :

1. In which subject has the child improved his performance the most?
2. In which subject is the improvement the least?
3. Has the performance gone down in any subject?

Solution:
The graph is as under :

1. In Mathematics the child has improved his performance the most.
2. In Social Science the child has improved his performance the least.
3. Yes, the performance has gone down in Hindi.

Ex 3.3 Class 7 Maths Question 5.
Consider this data collected from a survey of a colony.

(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Solution:
(i) The graph is as under :

It is inferred that more people prefer cricket and less athletics.
(ii) Most popular sport is cricket.
(iii) Watching is more preferred than participating.

Ex 3.3 Class 7 Maths Question 6.
Take the data giving the minimum and the maximum temperature of various cities given in the beginning of this chapter. Plot a double bar graph using the data and answer the following :

1. Which city has the largest difference in the minimum and maximum temperature on the given date?
2. Which is the hottest city and which is the coldest city?
3. Name two cities where maximum temperature of one was less than the minimum temperature of the other.
4. Name the city which has the least difference between its minimum and the maximum temperature.

Solution:
The graph is as under :

1. The city Jammu has the largest difference in the minimum and maximum temperature on the given date.
2. Jammu is the hottest city and Bangalore is the coldest city.
3. The name of the two cities where maximum temperature of one was less than the minimum temperature of other are Bangalore and Jaipur or Bangalore and Ahmedabad.
4. Mumbai has the least difference between its minimum and the maximum temperature.

Ex 3.4 Class 7 Maths Question 1.
Tell whether the following is certain to happen, impossible, can happen but not certain.

1. You are older today than yesterday.
2. A tossed coin will land heads up.
3. A die when tossed shall land up with 8 on top.
4. The next traffic light seen will be green.
5. Tomorrow will be a cloudy day.

Solution:

1. Certain to happen
2. Can happen but not certain
3. Impossible
4. Can happen but not certain
5. Can happen but not certain

Ex 3.4 Class 7 Maths Question 2.
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with number 2 ?
(ii) What is the probability of drawing a marble with number 5 ?
Solution:
Out of 6 marbles, one can be drawn in 6 ways. So, total number of events = 6

(i) The marble with number 2 can be obtained only in one way.
∴ Required probability = 16

(ii) The marble with number 5 can be obtained only in one way.
∴ Required probability = 16

Ex 3.4 Class 7 Maths Question 3.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Solution:
On tossing of a coin, the possible outcomes are head (H) or tail (T).
Required probability = 12

Ex 2.1 Class 7 Maths Question 1.
Solve :

Solution:

Ex 2.1 Class 7 Maths Question 2.
Arrange the following in descending order :

Solution:
We observe that the given fractions neither have a common denominator nor a common numerator. So, first we convert them into like fractions, i.e., fractions having common denominator. For this, we first find the L.C.M. of the denominators of the given fractions.

Ex 2.1 Class 7 Maths Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

Solution:

⇒ The sum of the numbers in each row, in each column and along the diagonals is the same.
So, the given square is a magic square.

Ex 2.1 Class 7 Maths Question 4.
A rectangular sheet of paper is 12 12 cm long and 10 23 cm wide.Find its perimeter.
Solution:

Ex 2.1 Class 7 Maths Question 5.
Find the perimeters of (i) ∆ ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Solution:

Ex 2.1 Class 7 Maths Question 6.
Salil wants to put a picture in a frame. The picture is 7 35 cm wide. To fit in the frame, the picture cannot be more than 7 310 cm wide. How much should the picture be trimmed?
Solution:

Ex 2.1 Class 7 Maths Question 7.
Ritu ate 35 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?
Solution:

Ex 2.1 Class 7 Maths Question 8.
Michael finished colouring a picture in 712 hour. Vaibhav finished colouring the same picture in 34 hour. Who worked longer? By what fraction was it longer?
Solution:

Ex 2.2 Class 7 Maths Question 1.
Which of the drawings (a) to (d) show :

Solution:
(i) → (d)
(ii) → (b)
(iii) → (a)
(iv) → (c)

Ex 2.2 Class 7 Maths Question 2.
Some pictures (a) to (c) are given below. Tell which of them show :

Solution:
(i)  → (c)
(ii) → (a)
(iii) → (b)

Ex 2.2 Class 7 Maths Question 3.
Multiply and reduce to lowest form and convert into a mixed fraction.

Solution:

Ex 2.2 Class 7 Maths Question 4.
Shade :
(i) 12 of the circles in box (a)
(ii) 23 of the triangles in box (b)
(iii) 35 of the squares in box (c).

Ex 2.2 Class 7 Maths Question 5.
Find :

Solution:

Ex 2.2 Class 7 Maths Question 6.
Multiply and express as a mixed fraction :

Solution:

Ex 2.2 Class 7 Maths Question 7.
Find :

Solution:

Ex 2.2 Class 7 Maths Question 8.
Vidya and Pratap went for a picnic. Their mother gave them water bottle that contained 5 litres of water. Vidya consumed 25 of the water. Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Solution:

Ex 2.2 Class 7 Maths Question 1.
Which of the drawings (a) to (d) show :

Solution:
(i) → (d)
(ii) → (b)
(iii) → (a)
(iv) → (c)

Ex 2.2 Class 7 Maths Question 2.
Some pictures (a) to (c) are given below. Tell which of them show :

Solution:
(i)  → (c)
(ii) → (a)
(iii) → (b)

Ex 2.2 Class 7 Maths Question 3.
Multiply and reduce to lowest form and convert into a mixed fraction.

Solution:

Ex 2.2 Class 7 Maths Question 4.
Shade :
(i) 12 of the circles in box (a)
(ii) 23 of the triangles in box (b)
(iii) 35 of the squares in box (c).

Ex 2.2 Class 7 Maths Question 5.
Find :

Solution:

Ex 2.2 Class 7 Maths Question 6.
Multiply and express as a mixed fraction :

Solution:

Ex 2.2 Class 7 Maths Question 7.
Find :

Solution:

Ex 2.2 Class 7 Maths Question 8.
Vidya and Pratap went for a picnic. Their mother gave them water bottle that contained 5 litres of water. Vidya consumed 25 of the water. Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
Solution:

Chapter 2 Fractions and Decimals Exercise 2.3

Ex 2.3 Class 7 Maths Question 1.
Find :

Solution:

Ex 2.3 Class 7 Maths Question 2.
Multiply and reduce to lowest form (if possible) :

Solution:

Ex 2.3 Class 7 Maths Question 3.
Multiply the following fractions :

Solution:

Ex 2.3 Class 7 Maths Question 4.
Which is greater?

Solution:

Ex 2.3 Class 7 Maths Question 5.
Saili plants 4 saplings in a row, in her garden. The distance between two adjacent saplings is 34 m. Find the distance between the first and the last sapling.
Solution:
Let four saplings be planted in a row at the points A, B, C and D such that AB = BC = CD = 34 m

Ex 2.3 Class 7 Maths Question 6.
Lipika reads a book for 1 34 hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution:

Ex 2.3 Class 7 Maths Question 7.
A car runs 16 km using 1 litre of petrol. How much distance will it cover using 2 34 litres of petrol?
Solution:

Ex 2.3 Class 7 Maths Question 8.
(a) (i) Provide the number in the box ( ), such that 23 × ( ) = 1030
(ii) The simplest form of the number obtained in ( ) is __________
(b) (i) Provide the number in the box ( ) such that 35 × ( ) = 2475
(ii) The simplets form of the number obtained in ( ) is __________
Solution:

Ex 2.4 Class 7 Maths Question 1.
Find :

Solution:

Ex 2.4 Class 7 Maths Question 2.
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

Solution:

Ex 2.4 Class 7 Maths Question 3.
Find :

Solution:

Ex 2.4 Class 7 Maths Question 4.
Find:

Solution:

Ex 2.5 Class 7 Maths Question 1.
Which is greater?
(i) 0.5 or 0.05
(ii) 0.7 or 0.5
(iii) 7 or 0.7
(iv) 1.37 or 1.49
(v) 2.03 or 2.30
(vi) 0.8 or 0.88
Solution:
(i) To compare 0.5 and 0.05
Here, the whole number part in each decimal is equal to zero. So, whole number parts are equal. Therefore, we will compare the decimal parts. We observe that the extreme left digits in the decimal parts are 5 and 0, respectively such that 5 > 0.
∴ 0.5 > 0.05

(ii) To compare 0.7 and 0.5
Here, the whole number parts are equal, so we compare the decimal parts. In the decimal parts, the extreme left digits are 7 and 5, respectively such that 7 > 5.
∴ 0.7 > 0.5

(iii) To compare 7 and 0.7
The given decimals have distinct whole numbers, so we compare whole number parts only. These are 7 and 0, respectively such that 7 > 0.
∴ 7 > 0.7

(iv) To compare 1.37 and 1.49
Here, the whole number parts are equal, so we compare the decimal parts. In the decimal parts, the extreme left digits are 3 and 4, respectively such that 3 < 4; i.e., 4 > 3.
∴ 1.49 > 1.37

(v) To compare 2.03 and 2.30
Here, the whole number parts are equal, so we compare the decimal parts. In the decimal parts, the extreme left digits are 0 and 3, respectively such that 0 < 3, i.e., 3 > 0.
∴ 2.30 > 2.03

(vi) To compare 0.8 and 0.88
Here, the whole number parts are equal, so we compare the decimal parts. In decimal part also, digits up to tenths place are equal. So, compare the digits at hundredths place. The digits at hundredths place are 0 and 8 such that 8 > 0.
∴ 0.88 > 0.8

Ex 2.5 Class 7 Maths Question 2.
Express as rupees using decimals :
(i) 7 paise
(ii) 7 rupees 7 paise
(iii) 77 rupees 77 paise
(iv) 50 paise
(v) 235 paise
Solution:

Ex 2.5 Class 7 Maths Question 3.
(i) Express 5 cm in metre and kilometre
(ii) Express 35 mm in cm, in and km
Solution:

Ex 2.5 Class 7 Maths Question 4.
Express in kg :
(i) 200g
(ii) 3470 g
(iii) 4 kg 8 g
Solution:

Ex 2.5 Class 7 Maths Question 5.
Write the following decimal numbers in the expanded form :
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Solution:

Ex 2.5 Class 7 Maths Question 6.
Write the place value of 2 in the following decimal numbers :

1. 2.56
2. 21.37
3. 10.25
4. 9.42
5. 63.352

Solution:

1. Place value of 2 in 2.56 is ones.
2. Place value of 2 in 21.37 is tens.
3. Place value of 2 in 10.25 is tenths.
4. Place value of 2 in 9.42 is hundredths.
5. Place value of 2 in 63.352 is thousandths.

Ex 2.5 Class 7 Maths Question 7.
Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?

Solution:

Ex 2.5 Class 7 Maths Question 8.
Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Solution:
Shyama bought apples = 5 kg 300 g
Shyama bought mangoes = 3 kg 250 g
Toal fruits bought by Shyama
= 5 kg 300 g + 3 kg 250 g
= 5.300 kg + 3.250 kg
= 8.550 kg

Sarala bought oranges = 4 kg 800 g
Sarala bought banana = 4 kg 150 g
Total fruits bought by Sarala
= 4 kg 800 g + 4 kg 150 g
= 4.800 kg +4.150 kg
= 8.950 kg

Since, 8.950 > 8.550
∴ Sarala bought more fruits.

Ex 2.5 Class 7 Maths Question 9.
How much less is 28 km than 42.6 km?
Solution:
Required difference = 42.6 km – 28 km = 14.6 km

Ex 2.6 Class 7 Maths Question 1.
Find :
(i) 0.2 × 6
(ii) 8 × 46
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.86
Solution:

Ex 2.6 Class 7 Maths Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:

Ex 2.6 Class 7 Maths Question 3.
Find :

1. 1.3 × 10
2. 36.8 × 10
3. 15a 7 × 10
4. 168.07 × 10
5. 31.1 × 100
6. 156.1 × 100
7. 3.62 × 100
8. 43.07 × 100
9. 0.5 × 10
10. 0.08 × 10
11. 0.9 × 100
12. 0.03 × 1000

Solution:

1. 1.3 × 10 = 13
2. 36.8 × 10 = 368
3. 153.7 × 10 = 1537
4. 168.07 × 10 = 1680.7
5. 31.1 × 100 = 3110
6. 156.1 × 100 = 15610
7. 3.62 × 100 = 362
8. 43.07 × 100 = 4307
9. 0.5 × 10 = 5
10. 0.08 × 10 = 0.8
11. 0.9 × 100 = 90
12. 0.03 × 1000 = 30

Ex 2.6 Class 7 Maths Question 4.
A two wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:
Distance covered in one litre of petrol = 55.3 km
Distance covered in 10 litres of petrol = (55.3 × 10) km = 553 km

Ex 2.6 Class 7 Maths Question 5.
Find:
(i) 2.5 × 0.3
(ii) 0.1 × 51.7
(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vi) 11.2 × 0.15
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 100.01 × 1.1
Solution:

Ex 2.7 Class 7 Maths Question 1.
Find :
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v) 651.2 ÷ 4
(vi) 14.49 ÷ 7
(vii) 3.96 ÷ 4
(viii) 0.80 ÷ 5
Solution:

Ex 2.7 Class 7 Maths Question 2.
Find :

1. 48 ÷ 10
2. 52.5 ÷ 10
3. 0.7 ÷ 10
4. 361 ÷ 10
5. 697 ÷ 10
6. 272.23 ÷ 10
7. 0.56 ÷ 10

Solution:

1. 4.8 ÷ 10 = 0.48
2. 52.5 ÷ 10 = 5.25
3. 0.7 ÷ 10 = 0.07
4. 33.1 ÷ 10 = 3.31
5. 272.23 ÷ 10 = 27.223
6. 0.56 ÷ 10 = 0.056
7. 3.97 ÷ 10 = 0.397

Ex 2.7 Class 7 Maths Question 3.
Find :

1. 2.7 ÷ 100
2. 0.3 ÷ 100
3. 0.78 ÷ 100
4. 432.6 ÷ 100
5. 23.6 ÷ 100
6. 98.53 ÷ 100

Solution:

1. 2.7 ÷ 100 = 0.027
2. 0.3 ÷ 100 = 0.003
3. 0.78 ÷ 100 = 0.0078
4. 432.6 ÷ 100 = 4.326
5. 23.6 ÷ 100 = 0.236
6. 98.53 ÷ 100 = 0.9853

Ex 2.7 Class 7 Maths Question 4.
Find :

1. 7.9 ÷ 1000
2. 26.3 ÷ 1000
3. 38.53 ÷ 1000
4. 128.9 ÷ 1000
5. 0.5 ÷ 1000

Solution:

1. 7.9 ÷ 1000 = 0.0079
2. 26.3 ÷ 1000 = 0.0263
3. 38.53 ÷ 1000 0.03853
4. 128.9 ÷ 1000 = 0.1289
5. 0.5 ÷ 1000 = 0.0005

Ex 2.7 Class 7 Maths Question 5.
Find :
(i) 7 ÷ 3.5
(ii) 36 ÷ 0.2
(iii) 3.25 ÷ 0.5
(iv) 30.94 ÷ 0.7
(v) 0.5 ÷ 0.25
(vi) 7.75 ÷ 0.25
(vii) 76.5 ÷ 0.15
(viii) 37.8 ÷ 1.4
(ix) 2.73 ÷ 1.3
Solution:

Ex 2.7 Class 7 Maths Question 6.
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?
Solution:
Distance covered in 2.4 litres of petrol = 43.2 km
Distance covered in 1 litre of petrol = (43.2 ÷ 2.4) km

Ex 1.1 Class 7 Maths Question 1.
Following number line shows the temperature in degree Celsius (°C) at different places on a particular day.

(a) Observe this number line and write the temperature of the places marked on it.
(b) What is the temperature difference between the hottest and the coldest places among the above?
(c) What is the temperature difference between Lahulspiti and Srinagar?
(d) Can we say temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?
Answer.
(a) From the given number line, we find that the temperature of the indicated places as under :

(b) Temperature difference between the hottest and the coldest places
= Temperature of Bangalore – Temperature of Lahul-Spiti
= 22°C-(-8°C)
= 22°C + 8°C
= 30°C

(c) The temperature difference between Lahulspiti and Srinagar
= -2 °C – (-8 °C)
= -2 °C +8 °C = 6 °C

(d) Temperature of Srinagar and Shimla together
= Temperature of Srinagar + Temperature of Shimla
= -2° + 5°C
= 3°C
Temperature at Shimla = 5°C
Temperature at Srinagar = – 2°C.

Therefore, we can say that the temperature of Srinagar and Shimla together is less than the temperature at Shimla but the temperature of Srinagar and Shimla together is not less than the temperature at Srinagar.

Ex 1.1 Class 7 Maths Question 2.
In a quiz, positive marks are given for correct answers, and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, 5, -10, 15 and 10, what was his total at the end?
Solution.
Jack’s scores in five successive rounds were given as 25, -5, -10, 15 and 10.
Jack’s total score
=25 +(-5)+ (-10)+ 15+ 10
= 25-5-10+15 + 10 = 50-15 = 35

Ex 1.1 Class 7 Maths Question 3.
At Srinagar, the temperature was -5 °C on Monday and then it dropped by 2 C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4 °C. What was the temperature on this day?
Solution.
Temperature at Srinagar on Monday = – 5°C
The drop-in temperature at Srinagar on Tuesday = 2°C
∴Temperature at Srinagar on Tuesday = – 5°C – 2°C = – 7°C
Rise in temperature at Srinagar on Wednesday = 4°C
Temperature at Srinagar on Wednesday
= – 7°C + 4°C
= – (7 – 4)°C
= -3°C.

Ex 1.1 Class 7 Maths Question 4.
A plane is flying at the height of 5000 m above sea level. At a particular point, it is exactly above a submarine floating 1200 m below sea level. What is the vertical distance between them?

Solution.
Vertical distance between the plane and the submarine
= 5000 m + 1200 m = 6200 m

Ex 1.1 Class 7 Maths Question 5.
Mohan deposits ₹ 2,000 in his bank account and withdraws ₹ 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.
Solution.
Amount deposited = + ₹2000
Balance in Mohan’s account after the withdrawal
= ₹ 2000 – ₹ 1642
= ₹ (2000 – 1642)
= ₹ 358.

Ex 1.1 Class 7 Maths Question 6.
Rita goes 20 km towards the east from point A to the point B. From B, she moves 30 km towards the west along the same road. If the distance towards east is represented by a positive integer, then how will you represent the distance travelled towards the west? By which integer will you represent her final position from A?

Solution.
The distance towards west will be represented by a negative integer.
Rita’s movement is shown as under:

Since, Rita moves 20 km towards east from a point A, so she reaches B, and then from B she moves 30 km towards the west along the same road and reaches C. Thus, her final position from A will be represented by the integer -10.

Ex 1.1 Class 7 Maths Question 7.
In a magic square each row, column, and diagonal have the same sum. Check, which of the following is a magic square?

Solution.
In square (i) :
Row 1 : 5 + (-1) + (-4) = 5-1-4 =0
Row 2 : (-5) + (-2) + 7 =-5-2 + 7 = 0
Row 3 : 0 + 3 + (-3) =0+3-3 =0
Column 1 : 5 + (-5) + 0= 5- 5 + 0= 0
Column 2 : (-1) + (-2) + 3 = -1 -2 + 3 = 0
Column 3 : (-4) + 7 + (-3) = -4 + 7- 3 = 0
Diagonal 1 : 5 + (-2) + (-3) = 5 – 2 – 3 =0
Diagonal 2 : (-4) + (-2) + 0 = -4 – 2 + 0 = -6
∵ The sum of digits along with the diagonal 2 ≠ 0.
Thus, it is not a magic square.
In square (ii) :
Row 1 : 1 + (-10) + 0 = 1-10+0 = -9
Row 2 : (-4) +(-3) +(-2) = -4-3-2 = -9
Row 3 : (-6) + 4 + (-7) = -6 + 4 – 7 = -9
Column 1 : 1 + (-4) + (-6) = 1- 4- 6 = -9
Column 2 : (-10) + (-3) + 4 = -10-3 + 4 = -9
Column 3 : 0 + (-2) + (-7) = 0-2-7 =-9
Diagonal 1 : 1 + (-3) + (-7) = -9
Diagonal 2: 0 + (-3) + (-6) = 0- 3- 6 = -9
∵ Each row, column, and diagonal have the same sum.
Thus, it is a magic square.

Ex 1.1 Class 7 Maths Question 8.
Verify a – (-b) = a + b for the following values of a and b:

1. a = 21, b = 18
2. a = 118, b = 125
3. a = 75, b = 84
4. a = 28, b = 11

Solution.

1. L.H.S. = a – (-b) = 21 – (-18) = 21 +18 = 39
   R.H.S. = a + b = 21 +18 =39
   ∴ L.H.S. = R.H.S.
2.  L.H.S. = a – (-b) = 118 – (-125) = 118 +125 = 243
   R.H.S. = a + b = 118+125 = 243
   ∴ L.H.S. = R.H.S.
3.  L.H.S. = a – (-b) = 75 – (-84) = 75+ 84 = 159
   R.H.S. = a + b =75 + 84 = 159
4.  L.H.S. = a – (-b) = 28 – (-11) = 28+ 11 = 39
   R.H.S. = a + b = 28 + 11 = 39

Ex 1.1 Class 7 Maths Question 9.
Use the sign of >, < or = in the box to make the statements true.
(a) (-8) + (-4) Undefined control sequence \boxed (-8) – (-4)
(b) (-3) + 7 – (19) Undefined control sequence \boxed 15 – 8 +(-9)
(c) 23 – 41 + 11 Undefined control sequence \boxed 23 – 41 – 11
(d) 39+ (-24) – (15) Undefined control sequence \boxed 36 + (-52) – (-36)
(e) -231 + 79 + 51 Undefined control sequence \boxed -399 + 159 + 81
Solution.

Ex 1.1 Class 7 Maths Question 10.
A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.
(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?
(ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?

(iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following :
(a) -3 + 2-… = -8
(b) 4 – 2 +… = 8.
In (a), the sum (-8) represents going down by eight steps. So, what will the sum 8 in (b) represent?
Solution.
(i) To reach the water level his jump will be as follows:
(- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) = -8.
Hence in 11 jumps, he will reach the water level.

(ii) To reach back to the top step his jumps will be as follows:
4 + (-2) + 4 + (-2) + 4 = 8
Therefore, he will be out of the tank in 5 jumps.

(iii) (a) – 3 + 2 + (- 3) + 2 + (- 3) + 2 + (- 3) + 2 + (T 3) + 2 + (- 3) = -8
(6) 4 – 2 + 4 – 2 + 4 = 8
The sum 8 in (b) will represent going up.

Ex 1.2 Class 7 Maths Question 1.
Write down a pair of integers whose:
(a) sum is -7
(b) difference is -10
(c) sum is 0
Solution.
(a) A pair of integers whose sum is -7 can be (-1) and (-6).
∵ (-1) + (-6) = -7
(b) A pair of integers whose difference is -10 can be (-11) and (-1)
∵ -11 – (-1) = -11+1 = -10
(c) A pair of integers whose sum is 0 can be 1 and (-1).
∵ (-1) + (1) = 0.

Ex 1.2 Class 7 Maths Question 2.
(a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and a positive integer whose sum is -5.
(c) Write a negative integer and a positive integer whose difference is -3.
Solution.
(a) A pair of negative integers whose difference gives 8 can be -12 and -20.
∵ (-12) – (-20) = -12+20 =8 .
(b) A negative integer and a positive integer whose sum is -5 can be -13 and 8.
∵ (-13) + 8 = -13 +8 = -5
(c) A negative integer and a positive integer whose difference is -3 can be -1 and 2.
∵ (-1) – 2 = – 1 -2 = -3

Ex 1.2 Class 7 Maths Question 3.
In a quiz, team A scored -40, 10, 0 and team B scored 10, 0, -40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
Solution.
Total scores of team A = (-40) + 10 +0
= -40 + 10 + 0 = -30
and, total scores of team B = 10 + 0 + (-40)
= 10 + 0 – 40 = -30
Since, the total scores of each team are equal.
∴ No team scored more than the other but each have equal score.
Yes, integers can be added in any order and the result remains unaltered. For example, 10 +0 +(-40) = -30 = -40 +0 +10

Ex 1.2 Class 7 Maths Question 4.
Fill in the blanks to make the following statements true:
(i) (-5) + (-8) = (-8) + (………)
(ii) -53 + ……. = -53
(iii) 17+ …… = 0
(iv) [13 + (-12)] + (……) = 13 + [(-12) + (-7)]
(v) (-4) + [15 + (-3)] = [-4 + 15] + ……
Solution.
(i) (-5) + (-8) = (-8) + (-5)
(ii) -53 + 0 = -53
(iii) 17 + (-17) = 0
(iv) [13 + (-12)] + (-7) = (13) + [(-12) + (-7)]
(v) (-4) + [15 + (-3)] = [(-4) + 15] + (-3)

Ex 1.3 Class 7 Maths Question 1.
Find each of the following products :
(a) 3 x (-1)
(b) (-1) x 225
(c) (-21) x (-30)
(d) (-316) x (-1)
(e) (-15) x O x (-18)
(f) (-12) x (-11) x (10)
(g) 9 x (-3) x (-6)
(h) (-18) x (-5) x (-4)
(i) (-1) x (-2) x (-3) x 4 Sol. (a) 3 x (-1) = – (3 x 1) = -3
(j) (-3) x (-6) x (-2) x (-1)
Solution.
(a) 3 x (-1) = – (3 x 1) = -3
(b) (-1) x 225 = – (1 x 225) = -225
(c) (-21) x (-30) = 21 x 30 =630
(d) (-316) x (-1) = 316 x 1 = 316
(e) (-15) x 0 x (-18) = [(-15) x 0] x (-18) = 0 x (-18) = 0
(f) (-12) x (-11) x (10) = [(-12) x (-11)] x (10)
= (132) x (10) =1320
(g) 9 x (-3) x (-6) = [9 x (-3)] x (-6) = (-27) x (-6) = 162
(h) (-18) x (-5) x (-4) = [(-18) x (-5)] x (-4)
= 90 x (-4) – -360
(i) (-1) x (-2) x (-3) x 4 = [(-1) x (-2)] x [(-3) x 4]
= (2)x (-12) = -24
(j) (-3) x (-6) x (-2) x (-1) = [(-3) x (-6)] x [(-2) x (-1)] = (18) x (2) = 36

Ex 1.3 Class 7 Maths Question 2.
Verify the following:
(a) 18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]
(b) (-21) x [(-4) + (-6)] = [(-21) x (-4)] + [(-21) x (-6)]
Solution.
(a) We have,
18 x [7 + (-3)] = 18 x 4 = 72
and, [18 x 7] + [18 x (-3)] = 126 – 54 =72
18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]
(b) We have,
(-21) x [(-4) + (-6)] = (-21) x (-4 -6)
= (-21)(-10) = 210 and, [(-21) x (-4) + [(-21) x (-6)]
= 84+126 =210
∴ (-21) x [(-4) + (-6)] = [(-21) x (-4)] + [(-21) x (-6)]

Ex 1.3 Class 7 Maths Question 3.
(i) For any integer a, what is (-1) x a equal to?
(ii) Determine the integer whose product with (-1) is
(a) -22
(b) 37
(c) 0
Solution.
(i) For any integer a, (-1) x a = -a.
(ii) We know that the product of any integer and (-1) is the additive inverse of integer.
The integer whose product with (-1) is
(a) additive inverse of -22, t. e., 22.
(b) additive inverse of 37, i.e., -37.
(c) additive inverse of 0, i.e., 0.

Ex 1.3 Class 7 Maths Question 4.
Starting from (-1) x 5, write various products showing some pattern to show (-1) x (-1) = 1.
Solution.
(-1) x 5 = -5
(-1) x 4 = -4 = [-5 – (-1)] = -5 +1
(-1) x 3 = -3 = [-4 – (-1)] = -4 +1
(-1) x 2 = -2 = [-3 – (-1)] = -3 +1
(-1) x 1 = -1 = [-2 – (-1)] = -2 +1
(-1) x 0 = 0 = [-1 – (-1)] = -1 +1
(-1) x (-1) =[0 – (-1)] = 0 + 1 = 1

Ex 1.3 Class 7 Maths Question 5.
Find the product, using suitable properties :
(a) 26 x (-48) + (-48) x (-36)
(b) 8 x 53 x (-125)
(c) 15 x (-25) x (-4) x (-10)
(d) (-41) x 102
(e) 625 x (-35) +(-625) x 65
(f) 7 x (50-2)
(g) (-17) x (-29)
(h) (-57) x (-19) + 57
Solution.
(a) We have, 26 x (-48) + (-48) x (-36)
= (-48) x 26 + (-48) x (-36)
= (-48) x [26 + (-36)]
= (-48) x (26 – 36)
=(-48) x (-10)= 480
(b) We have,
8 x 53 x (-125) = [8 x (-125)] x 53
= (-1000) x 53 = -53000
(c) We have,
15 x (-25) x (-4) x (-10)
=15 x [(-25) x (-4)] x (-10)
= 15 x (100) x (-10)
= (15 x 100) x (-10)
= 1500 x (-10) = -15000
(d) We have,
(-41) x 102 = (-41) x (100 +2)
= (-41) x 100 + (-41) x 2 = -4100 – 82 = -4182
(e) We have, 625 x (-35) + (-625) x 65
= 625 x (-35) + (625) x (-65)
= 625 x [(-35)+ (-65)]
= 625 x (-100) = -62500
(f) 7 x (50-2) = 7 x 50 – 7 x 2
= 350 -14 =336
(g) (-17) x (-29) = (-17) x [(-30) + 1]
= (-17) x (-30) + (-17) x 1 = 510 – 17 = 493
(h) (-57) x (-19)+ 57 =57 x 19 + 57 x 1
= 57 x (19 +1)
= 57 x 20 = 1140

Ex 1.3 Class 7 Maths Question 6.
A certain freezing process requires that room temperature be lowered from 40 °C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?
Solution.
Initial room temperature = 40 X
Temperature lowered every hour = (-5) °C
Temperature lowered in 10 hours = (-5) x 10 °C = -50 °C
∴ Room temperature after 10 hours = 40 X – 50 X = -10 °C

Ex 1.3 Class 7 Maths Question 7.
In a class test containing 10 questions, 5 marks are awarded for every correct answer and (-2) marks are awarded for every incorrect, answer and 0 for questions not attempted.
(i) Mohan gets four correct and six incorrect answers. What is his score?
(ii) Reshma gets five correct answers and five incorrect answers, what is her score?
(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?
Solution.
(i) Marks awarded for one correct answer = 5
Marks scored for 4 correct answer = 5 x 4 = 20
Marks awarded for one incorrect answer = (-2)
Marks scored for 6 incorrect answer = (-2) x 6 = -12
Hence, Mohan’s score = 20 – 12 = 8 marks.
(ii) Reshma’s score for 5 correct answers = 5 x 5 = 25 marks
Reshma’s score for 5 incorrect answers = (-2) x 5 = -10 marks
Hence, Reshma’s score = 25-10 =15 marks
(iii) Heena’s score for 2 correct and 5 incorrect answers
= (5 x 2) + {(-2) x 5}
= 10+ (-10) = 10 – 10 =0.

Ex 1.3 Class 7 Maths Question 8.
A cement company earns a profit of ? 8 per bag of white cement sold and a loss of ? 5 per bag of grey cement sold.
(a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?
(b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags?
Solution.
Profit on sale of 1 bag of white cement = ₹ 8
Loss on sale of 1 bag of grey cement = – ₹ 5
(a) Profit on sale of 3000 bags of white cement
= ₹ (3000 x 8)
= ₹ 24,000
Loss on sale of 5000 bags of grey cement = ₹ (5000 x -5)
= – ₹ 25,000
Difference between the two = ₹ 24,000 – ₹ 25,000 = – ₹ 1,000
Hence, there is a loss of ₹ 1000.
(b) Loss on the sale of 6400 bags of grey cement = ₹ (6400 x 5) = ₹ 32,000
In order to have neither profit nor loss, the profit on the sale of white cement should be ? 32,000.
Number of white cement bags sold
=TotalprofitProfitperbag
=320008
Hence, 4000 bags of white cement should be sold to have neither profit nor loss.
Replace the blank with an integer to make it a true statement.
(a) (-3) x = 27
(b) 5 x = -35
(c) 7 x (-8) = -56
(d) (-11) x (-12) = 132
Solution.
(a) (-3) x (-9) = 27
(b) 5 x (-7) = (-35)
(c) 7 x (-8) = (-56)
(d) (-11) x (-12) =132

Ex 1.4 Class 7 Maths Question 1.
Evaluate each of the following:

Solution.

Ex 1.4 Class 7 Maths Question 2.
values of a, b and c.
(a) a = 12, b = -4, c = 2
(b) a = (-10), b = 1, c = 1
Solution.

Ex 1.4 Class 7 Maths Question 3.
Fill in the blanks :

Solution.

Ex 1.4 Class 7 Maths Question 4.
Write five pairs of integers (a, b) such that a÷b = -3. One such pair is (6, – 2) because 6÷(−2)=(−3) .
Solution.
Five pairs of integers (a, b) such that a + b = -3 are : (-6,2), (-9, 3), (12,-4), (21,-7), (-24, 8)
Note : We may write many such pairs of integers.

Ex 1.4 Class 7 Maths Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2 °C per hour until mid-night, at what time would the temperature be 8 °C below zero? What would be the temperature at mid-night?
Solution.
Difference in temperatures +10 °C and -8
= [10 – (-8)] °C = (10 + 8)° C = 18 °C
Decrease in temperature in one hour = 2°C
Number of hours taken to have temperature 8 °C below zero =TotaldecreaseDecreaseinonehour
=182
So, at 9 P.M., the temperature will be 8 °C below zero
Temperature at mid-night = 10 °C – (2 x 12) °C
= 10°C – 24 °C = -14 °C

Ex 1.4 Class 7 Maths Question 6.
In a class test (+ 3) marks are given for every correct answer and (- 2) marks are given for every incorrect answer and no marks for not attempting any question, (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores – 5 marks in this test though she has got 7 correct, answers. How many questions has she attempted incorrectly?
Solution.
(i) Marks given for 12 correct answers at the rate of + 3 marks for each answer = 3 x 12 = 36 Radhika’s score = 20 marks
∴ Marks deducted her for incorrect answers = 20 – 36 = -16
Marks given for one incorrect answer = -2
Number of incorrect answers =(−16)÷(−2)=8
(ii) Marks given for 7 correct answers at the rate of + 3 marks for each answer = 3 x 7 = 21 Mohini’s score = -5
∴ Marks deducted for incorrect answers
= – 5 – 21 = -26
Marks given for one incorrect answer = -2
∴ Number of incorrect answers =(−26)÷(−2)=13

Ex 1.4 Class 7 Maths Question 7.
An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Answer.
Difference in heights at two positions = 10 m – (- 350 m) = 360 m
Rate of descent = 6 m/minute
∴ Time taken =(360)÷(6) minutes
= 60 minutes = 1 hour
Hence, the elevator will take 1 hour to reach – 350 m.