# RS Aggarwal Solutions for Class 7 Maths Exercise 4F Chapter 4 Rational Numbers

RS Aggarwal Solutions for Class 7 Maths Exercise 4F Chapter 4 Rational Numbers in simple PDF are given here. The topics covered in this exercise are the reciprocal or multiplicative inverse of a rational number and division of rational numbers. Students will go through this RS Aggarwal Solutions for Class 7 Maths Chapter 4 Rational Numbers thoroughly with more practice. Students can hence score good marks in Maths by practising RS Aggarwal Class 7 Solutions.

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1. Find the multiplicative inverse or reciprocal of each of the following:
(i) 18

Solution:-

The reciprocal of 18 is (1/18)

(ii) -16

Solution:-

The reciprocal of -16 is (1/-16)

(iii) 13/25

Solution:-

The reciprocal of (13/25) is (25/13)

(iv) -17/12

Solution:-

The reciprocal of (-17/12) is (12/-17)

(v) -6/19

Solution:-

The reciprocal of (-6/19) is (19/-6)

(vi) -3/-5

Solution:-

The reciprocal of (-3/-5) is (-5/-3)

(vii) -1

Solution:-

The reciprocal of -1 is -1

(viii) 0

Solution:-

The reciprocal of 0 is not exist.

2. Simplify:
(i) (4/9) Ã· (-5/12)

Solution:-

We have,

= (4/9) Ã— (12/-5) â€¦ [âˆµ reciprocal of (-5/12) is (12/-5)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (4Ã—12) / (9Ã—-5)

On simplifying,

= (4Ã—4) / (3Ã—-5)

= (-16/15)

(ii) (-8) Ã· (-5/16)

Solution:-

We have,

= (-8/1) Ã— (16/-5) â€¦ [âˆµ reciprocal of (-5/16) is (16/-5)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-8Ã—16) / (1Ã—-5)

= -128/-5

= 128/5

(iii) (-12/7) Ã· (-18)

Solution:-

We have,

= (-12/7) Ã— (1/-18) â€¦ [âˆµ reciprocal of (-18) is (1/-18)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-12Ã—1) / (7Ã—-18)

On simplifying,

= (-4Ã—1) / (7Ã—-6)

= (-2Ã—1) / (7Ã—-3)

= (-2/-21)

= (2/21)

(iv) (-1/10) Ã· (-8/5)

Solution:-

We have,

= (-1/10) Ã— (5/-8) â€¦ [âˆµ reciprocal of (-8/5) is (5/-8)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-1Ã—5) / (10Ã—-8)

On simplifying,

= (-1Ã—1) / (2Ã—-8)

= (-1/-16)

= (1/16)

(v) (-16/35) Ã· (-15/14)

Solution:-

We have,

= (-16/35) Ã— (14/-15) â€¦ [âˆµ reciprocal of (-15/14) is (14/-15)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-16Ã—14) / (35Ã—-15)

On simplifying,

= (-16Ã—2) / (5Ã—-15)

= (-32/-75)

= (32/75)

(vi) (-65/14) Ã· (13/-7)

Solution:-

First we write (13/-7) in standard form

= (13Ã—-1)/ (-7Ã—-1)

= (-13/7)

We have,

= (-65/14) Ã— (7/-13) â€¦ [âˆµ reciprocal of (-13/7) is (7/-13)]

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

= (-65Ã—7) / (14Ã—-13)

On simplifying,

= (-5Ã—1) / (2Ã—-1)

= (-5) / (-2)

= (5/2)

3. Fill in the blanks:
(i) (â€¦â€¦ ) Ã· (-7/5) = (10/19)

Solution:-

Let the required number be (a/b). Then,

(a/b) Ã· (-7/5) = (10/19)

â‡’ (a/b) = (10/19) Ã— (-7/5)

â‡’ (a/b) = (10 Ã— -7) / (19 Ã— 5)

â‡’ (a/b) = (2Ã— -7) / (19Ã—1)

â‡’ (a/b) = (-14/19)

(ii) (â€¦â€¦ ) Ã· (-3) = (-4/15)

Solution:-

Let the required number be (a/b). Then,

(a/b) Ã· (-3/1) = (-4/15)

â‡’ (a/b) = (-4/15) Ã— (-3/1)

â‡’ (a/b) = (-4 Ã— -3) / (15 Ã— 1)

â‡’ (a/b) = (-4Ã— -1) / (5Ã—1)

â‡’ (a/b) = (4/5)

(iii) ( 9/8) Ã· (â€¦. ) = (-3/2)

Solution:-

Let the required number be (a/b). Then,

(9/8) Ã· (a/b) = (-3/2)

â‡’ (a/b) = (9/8) Ã— (2/-3)

â‡’ (a/b) = (9 Ã— 2) / (8 Ã— -3)

â‡’ (a/b) = (3Ã— 1) / (4Ã—-1)

â‡’ (a/b) = (-3/4)

(iv) ( -12) Ã· (â€¦. ) = (-6/5)

Solution:-

Let the required number be (a/b). Then,

(-12/1) Ã· (a/b) = (-6/5)

â‡’ (a/b) = (-12/1) Ã— (5/-6)

â‡’ (a/b) = (-12 Ã— 5) / (1 Ã— -6)

â‡’ (a/b) = (-2Ã— 5) / (1Ã—-1)

â‡’ (a/b) = (-10/-1)

â‡’ (a/b) = 10

4. Divide the sum of (65/12) and (8/3) by their difference

Solution:-

First we have to find the sum of (65/12) + (8/3)

LCM of 12 and 3 is 12

Then,

= (65 Ã— 1) / (12Ã—1) = (65/12)

= (8Ã— 4) / (3Ã— 4) = (32/12)

Now,

= (65 + 32)/12

= (97/12)

Now find the difference of (65/12) â€“ (8/3)

LCM of 12 and 3 is 12

Then,

= (65 Ã— 1) / (12Ã—1) = (65/12)

= (8Ã— 4) / (3Ã— 4) = (32/12)

Now,

= (65 – 32)/12

= (33/12)

Now divide (97/12) Ã· (33/12)

= (97/12) Ã— (12/33)

= (97 Ã— 12) / (12Ã— 33)

= (97 Ã— 1) / (1Ã— 33)

= (97/33)

5. By what number should (-44/9) be divided to get (-11/3)?

Solution:-

Let the required number be x. Then,

= (-44/9) Ã· x = (-11/3)

= (-44/9) Ã— (1/x) = (-11/3)

= (1/ x) = (-11/3) Ã— (9/-44)

= (1/x) = (-11 Ã—9)/ (3 Ã—-44)

= (1/x) = (-1 Ã— 3) / (1 Ã— -4)

= (1/x) = (-3/-4)

Then,

X = (4/3)

6. By what number should (-8/15) be multiplied to get 24?

Solution:-

Let the required number be x. Then,

= X Ã— (-8/15) = 24

= (X) = 24 Ã· (-8/15)

= X = (24/1) Ã— (15/-8)

= X = (24 Ã—15)/ (1 Ã—-8)

= X = (3 Ã— 15) / (1 Ã— -1)

= X = -45

Then,

X = -45

### Access other exercises of RS Aggarwal Solutions For Class 7 Chapter 4 – Rational Numbers

Exercise 4G Solutions