# RS Aggarwal Solutions Class 7 Ex 1A

Q1) Divide:

(i) 65 by -13

(ii) -84 by 12

(iii) -76 by 19

(iv) -132 by 12

(v) -150 by 25

(vi) -72 by -18

(vii) -105 by -21

(viii) -36 by -1

(ix) 0 by -31

(x) -63 by 63

(xi) -23 by -23

(xii) -8 by 1

(i) 65 ÷ (-1 3) = 65÷-13 = -5

(ii) (-84)÷ 12 = -84÷12 = -7

(iii) (-76) ÷ 19 = -76÷19 = -4

(iv) (-132) ÷ 12 = -132÷12 = -11

(v) (-150) ÷ 25 = -150÷25 = -6

(vi) (-72) ÷(-18) = -72÷-18 = 4

(vii) (-105)÷ (-21) = -105÷-21 = 5

(viii) (-36) ÷ (-1) = -36÷-1 = 36

(ix) 0 ÷ (-31) = 0÷-31 = 0

(x) (-63) ÷63 = -63÷63 = -1

(xi) (-23) ÷ (-23) = -23÷-23 = 1

(xii) (-8) ÷ 1 = -8÷1 = -8

Q2) Fill in the blanks:

(i) 72 ÷ (…….)= -4

(ii) -36 ÷ (…….)= -4

(iii) (……..)÷ (-4)=24

(iv) (…….)÷ 25 – 0

(v) (……) ÷(-1)=36

(vi) (…….) ÷1 = -37

(vii) 39 ÷ (…….)= -1

(viii) 1 + (…….) =-1

(ix) -1+(…….)= -1

(i) 72= (x) = -4

72x = -4

x = -18

(ii)

-36 ÷ (x) = -4

=> -36x = -4

x = -36-4 = 9

(iii)

(x)÷ (-4) = 24

=>x – 4 = 24

=>x = 24x(-4)

= -96

(iv)

(x)÷ 25 = 0

x / 25 = 0

x = 25×0 = 0

(v)

(x)÷(-1)= 36

=>36x(-1) = -36

(vi)

(x)÷ 1 = -37

x÷1= -37

=>x = -37×1

= -37

(vii)

39÷ (x) = -1

=>39x = -1×39 = -39

(viii)

1 ÷ (x)= -1

=>1x = -1×1 = -1

(ix) -1 ÷ (x) = -1

=> -1x =-1

=> x = -1-1= 1

Q3) Write (T) for true and (F) for false for each of the following statements:

(i) 0 ÷ (-4) = 0

(iv) (-8)÷1= -8

(ii) (-6)÷ 0 = 0

(iii) (-5) ÷ (-1) = -5

(iv) (-8) ÷ 1 = -8

(v) (-1) ÷ (-1) = -1

(vi) (-9) ÷ (-1) = 9

(i) True (T). Dividing zero by any integer gives zero.

(ii) False (F). Division by zero gives an indefinite number.

(iii) False (F). -5-1 = 5

(iv) True (T). -81= -8

(v) False (F). -1-1 = 1

(vi) True (T). -9-1 = 9

EXERCISE – 1D

Mark (✓) against the correct answer in each of the following:

1)1)6-(-8)=?

(a) -2 (b) 2 (c) 14 (d) none of these

(c) 14

Given:

6 – (-8)

=6+8 = 14

2)2)-9- (-6)= ?

(a) -15 (b) -3 (c) 3 (d) none of these

(b) -3

Given:

-9 – (-6)

= -9 + 6

= -3.

3)By how much does 2 exceed -3?

(a) -1 (b) 1(c) -5 (d) 5

(d) 5

We can see that

-3+5=2

Hence, 2 exceeds -3 by 5.

4)What must be subtracted from -1 to get -6?

(a) 5 (b) -5 (c) 7 (d) -7

(a) 5

Let the number to be subtracted be x.

To find the number, we have:

-1 – x = -6

x = -1 + 6 = 5

5)How much less than -2 is -6?

(a) 4 (b) -4 (c) 8 (d) -8

(c) 4

We can see that

(-2)- (-6)= (-2)+ 6 = 4

Hence, -6 is four (4) less than -2.

6) On subtracting 4 from -4, we get

(a) 8 (b) -8 (c) 0 (d) None of these

(b) -8

Subtracting 4 from -4, we get:

(-4) – 4 = -8

7) By how much does -3 exceed -5 ?

(a) -2 (b) 2 (c) 8 (d) -8

(b) 2

Required number = (-3) – (-5)

= 5 – 3 = 2

8) What must be subtracted from -3 to get -9 ?

(a) -6 (b) 12 (c) 8 (d) -8

(c) 6

(-3) – x = -9

x=(-3)+9=6

Hence, 6 must be subtracted from -3 to get -9.

9)On subtracting 6 from -5, we get:

(a) 1 (b) 11 (c) -11 (d) None of these.

(c) -11

Subtracting 6 from -5, we get:

(-5) – 6=-11

10)On subtracting -13 from -8, we get

(a) -21 (b) 21 (c) 5 (d) -5

(c) 5

Subtracting —13 from —8, we get:

(-8) — (-13)

= —8 + 13

=5

11) (-36) ÷ (-9) = ?

(a) 4 (b) -4 (c) None of these

(a) 4

(-36) ÷ (-9) = 4

Here, the negative signs in both the numerator and denominator got cancelled with each other.

12) 0 ÷ (-5) = ?

(a) -5 (b) 0 (c) Not defined

(b) 0

Dividing zero by any integer gives zero as the result.

13) (-8) ÷ 0 = ?

(a) -8 (b) 0 (c) Not defined

(c) not defined

Dividing any integer by zero is not defined.

14) Which of the following is a true statement ?

(a) -11 > -8 (b) -11 < -8 (c) -11 and -8 cannot be compared

(b) -11 < -8

Negative integers decrease with increasing magnitudes.

15) The sum of two integers is 6. If one of them is -3, then the other is

(a) -9 (b) 9 (c) 3 (d) -3

(b) 9

Let the other integer be a. Then we have:

-3 + a = 6

a = 6 – (-3) = 9

16) The sum of two integers is 14, If one of them is -8,then the other is

(a) 22 (b) -22 (c) 6 (d) -6

(a) -10

Let the other integer be a. Then, we have:

6 + a = -4

a = -4 – 6 = -10

Hence, the other integer is -10.

17) The sum of two integers is 14. If one of them is -8, then the other is,

(a) 22 (b) -22 (c) 6 (d) -6

(a) 22

Let the other integer be a Then. we have:

-8 + a = 14

a = 14 + 8 = 22

Hence, the other integer is 22.

18) The additive inverse of -6 is

(a) $frac{1}{6}$ (b) $frac{-1}{6}$ (c) 6 (d) 5

(c) 6

The additive inverse of any integer a is -a.

Thus, the additive inverse of -6 is 6.

19)( -15 ) x 8 + ( -15 ) x 2 = ?

(a) 150 (b) -150 (c) 90 (d) -90

(b) -150

We have (-15) x 8 + (-15) x 2

= (-15) x (8 + 2) [Associative property]

= -150

20) (-12) x 6 – (-12) x 4 = ?

(a) 24 (b) -24 (c) 120 (d) -120

(b) -24

We have (-12) x 6 – (-12) x 4

= (-12) x (6 – 4) [Associative property]

= -24

21) (-27) x (-16) + (-27) x (-14) = ?

(a) -810 (b) 810 (c) -54 (d) 54

(b) 810

(-27) x (-16) + (-27) x (-14)

= (-27) x (-16 + (-14)) [Associative property]

=(-27) x (-30)

= 810

22) 30 x (-23) + 30 x 14 = ?

(a) -270 (b) 270 (c) 1110 (d) -1110

(a) -270

30x(-23)+ 30×14

= 30 x (-23 + 14) [Associative property]

= 30 x (-9)

= -270

23) The sum of two integers is 93. If one of them is -59, then the other one is

(a) 34 (b) -34 (c) 152 (d) -152

(c) 152

Let the other integer be a. Then, we have:

-59 +a= 93

a= 93 + 59=152

24) (?) + (-18) = -5

(a) -90 (b) 90 (c) None of these

(b) 90

x÷ (-18) = -5

x = -18 * -5= 90′

#### Practise This Question

While travelling in a bus, what is the direction in which the traffic light appears to move?