# RS Aggarwal Solutions for Class 8 Maths Chapter 1 - Rational Numbers

RS Aggarwal Solutions for Class 8 Maths Chapter 1- Rational Numbers are provided here. Our expert faculty team has prepared solutions to help you with your exam preparation to acquire good marks in Maths. If you wish to secure an excellent score, solving RS Aggarwal Class 8 Solutions is an utmost necessity. Scoring high marks require a good amount of practice on every topic. These solutions will help you in gaining knowledge and strong command over the subject. Practising the textbook questions will help you in analyzing your level of preparation and knowledge of the concept.

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## Exercise 1A

Q1.Express -3/5 as a rational number whose denominators are:

Q2.express -42/98 as a rational number whose denominator is 7

Q3. Express -48/60 as a rational number whose denominator is 5

Q4.express each of the following rational numbers in the standard form

Q5. Which of the two rational numbers is greater in the given pair?

Q6.which of the two rational numbers is greater in the given pair?

Q7. Fill in the blanks with the correct symbol out of >, =, < :

Q8.Arrange the following rational numbers in ascending order:

Q9.Arrange the following rational numbers in descending order:

Q10. Which of the following statements are true or false?

1. Every whole number is a rational number.Answer: TrueExplanation: By the definition of rational number, p/q where, q &#8800; 0.We know that every whole number can be represented as a/1.From the above two statements we can conclude that every whole number is a rational number.
2. Every integer is a rational number.Answer: TrueExplanation: By the definition of rational number, p/q where, q &#8800; 0.We know that every integer can be represented as a/1From the above two statements we can conclude that every integer is a rational number.
3. 0 is a whole number but it is not a rational number.Answer: FalseExplanation: By the definition of rational number, p/q where, q &#8800; 0.We know that 0 can be represented as 0/1.From the above two statements we can conclude that 0 is a whole number and a rational number.

## Exercise 1B

Q1.Represent each of the following numbers on the number line:

Q2. Represent each of the following numbers on the number line:

Q3. Which of the following statements are true or false?

## Exercise 1C

Q1. Add the following rational numbers:

Q2. Add the following rational numbers:

Q3.Verify the following:

Q4. Verify the following:

Q5. Fill in the blanks:

Q6. Find the additive inverse of each of the following:

Q7. Subtract:

Q8. Using the rearrangement property find the sum:

Q9. The sum of two rational numbers is -2. If one of the numbers is -14/5 , find the other.

Q10. The sum of two rational numbers is -1/2. If one the number is 5/6 . find the other.

Q11. What number should be added to-5/8 so as to get -3/2 ?

Q12. What number should be added to -1 so as to get 5/7 ?

Q13. What number should be subtracted from -2/3 to get -1/6 ?

Q14.

1. Which rational number is its own additive inverse?
2. Is the difference of two rational numbers a rational number?
3. Is addition commutative on rational numbers?
4. Is addition associative on rational numbers?
5. Is subtraction commutative on rational numbers?
6. Is subtraction associative on rational numbers?
7. What is the negative of a negative rational number?Solutions:

## Exercise 1D

Q1. Find each of the following products:

(i) (3/5) × (-7/8)
Solution:

(3/5) × (-7/8)

Here by multiplying numerators and denominators

We get (3× -7) / (5 × 8) => -21/40

(ii) -9/2 × 5/4
Solution:

(-9/2) × (5/4)

Here by multiplying numerators and denominators

We get (-9×5) / (2×4) => -45/8

(iii) (-6/11) × (-5/3)
Solution:

(-6/11) × (-5/3)

Here by multiplying numerators and denominators.

We get (-6×-5) / (11×3) = 30/33

further dividing to lowest terms by common divisor i.e. 3 = 30/33 => 10/11

(iv) (-2/3) × (6/7)
Solution:

(-2/3) × (6/7)

Here by multiplying numerators and denominators.

We get (-2×6) / (3×7) = -12/21.

Further dividing to lowest terms by common divisor i.e. 3 = -12/21 => -4/7

(v) (-12/5) × (10/-3)
Solution:

(-12/5) × (10/-3)

Here by multiplying numerators and denominators.

We get (-12×10) / (5×-3) = -120/-15.

Further dividing to lowest terms by common divisor i.e. 3 = -120/-15 => 40/5

Further dividing to lowest terms by common divisor i.e. 5 = 40/5 => 8

(vi) (25/-9) × (3/-10)
Solution:

(25/-9) × (3/-10)

Here by multiplying numerators and denominators.

We get (25×3) / (-9×-10) = 75/90.

Further dividing to lowest terms by common divisor i.e. 15 = 75/90 => 5/6

(vii) (5/-18) × (-9/20)
Solution:

(5/-18) × (-9/20)

Here by multiplying numerators and denominators.

We get (5×-9) / (-18×20) = -45/-360 = 45/360.

Further dividing to lowest terms by common divisor i.e. 45 = 45/360 => 1/8

(viii) (-13/15) × (-25/26)
Solution:

(-13/15) × (-25/26)

Here by multiplying numerators and denominators.

We get (-13×-25) / (15×26) = 325/390.

Further dividing to lowest terms by common divisor i.e. 5 = 325/390 = 65/78

Further dividing to lowest terms by common divisor i.e. 13 = 65/78 => 5/6

(ix) (16/-21) × (14/5)
Solution:

(16/-21) × (14/5)

Here by multiplying numerators and denominators.

We get (16×14) / (-21×5) = (224 ×-1) / (-105 ×-1) = -224/105.

Further dividing to lowest terms by common divisor i.e. 7 = -224 /105 = -32/15

(x) (-7/6) × (24/1)
Solution:

Here by multiplying numerators and denominators.

We get (-7×24) / (6 ×1) = -168/6.

Further dividing to lowest terms by common divisor i.e. 2 = -168/6 = -84/3.

Further dividing to lowest terms by common divisor i.e. 3 = -84/3 => -28

(xi) (7/24) × (-48)
Solution:

Here by multiplying numerators and denominators.

We get (7×-48) / (24 ×1) = -336/24
Further dividing to lowest terms by common divisor i.e. 6 = -336/24 = -56/4.

Further dividing to lowest terms by common divisor i.e. 6 = -84/6 => -14

(xii) (-13/5) / (-10)
Solution:

Here by multiplying numerators and denominators.

We get (-13×-10) / 5 = 130/5.

Further dividing to lowest terms by common divisor i.e. 5 = 130/5 = 26

1. Q2.verify each of the following:

(i) (3/7) × (-5/9) = (-5/9) × (3/7)
Solution:

Firstly consider LHS

(3/7) × (-5/9) = (3×-5) / (7×9) = -15/63

Further by dividing -15/63 by using common divisor 3 we get, -5/21

RHS: (-5/9) × (3/7)(-5/9) × (3/7) = (-5×3) / (9×7) = -15/63

Further by dividing -15/63 by using common divisor 3 we get, -5/21

∴ LHS = RHS is verified.

(ii) (-8/7) × (13/9) = (13/9) × (-8/7)
Solution:

Firstly consider LHS

(-8/7) × (13/9) = (-8×13) / (7×9) = -104/63

RHS: (13/9) × (-8/7)(13/9) × (-8/7) = (13×-8) / (9×7) = -104/63

∴ LHS = RHS is verified.

(iii) (-12/5) × (7/-36) = (7/-36) × (-12/5)
Solution:

Firstly consider LHS

(-12/5) × (7/-36) = (-12×7) / (5×-36) = -84/-180 = 84/180

RHS: (7/-36) × (-12/5)(7/-36) × (-12/5) = (7×-12) / (-36×5) = -84/-180 = 84/180

∴ LHS = RHS is verified.

(iv) -8 × (-13/12) = (-13/12) × -8
Solution:

Firstly consider LHS

-8 × (-13/12) = (-8×-13) / 12 = 104/12.

Further by dividing 104/12 by using common divisor 4 we get, 26/3

RHS: (-13/12) × -8(-13/12) × -8 = (-13×-8) / 12 = 104/12

Further by dividing 104/12 by using common divisor 4 we get, 26/3

∴ LHS = RHS is verified.

1. Q3.Verify each of the following:

(i) ((5/7) × (12/13)) × (7/18) = (5/7) × ((12/13) × (7/18))
Solution:

Firstly consider LHS

((5/7) × (12/13)) × (7/18)

((5×12) / (7×13)) × (7/18)

(60/91) × (7/18)

((60×7) / (91×18)) = 420/1638

Further by dividing 420/1638 by using common divisor 42 we get, 10/39

RHS: (5/7) × ((12/13) × (7/18))

(5/7) × ((12×7) / (13×18))

(5/7) × (84/234)

(5×84) / (7×234) = 420/1638

Further by dividing 420/1638 by using common divisor 42 we get, 10/39

∴ LHS = RHS is verified.

(ii) (-13/24) × ((-12/5) × (35/36)) = ((-13/24) ×(-12/5)) × (35/36)
Solution:

Firstly consider LHS

(-13/24) × ((-12/5) × (35/36))

(-13/24) × ((-12×35) / (5×36))

(-13/24) × (-420/180)

(-13×-420) / (24×180) = 5460/4320

RHS: ((-13/24) × (-12/5)) × (35/36)

((-13×-12) / (24×5)) × (35/36)

(156/120) × (35/36)

(156×35) / (120×36) = 5460/4320

Further by dividing 5460/4320 by using common divisor 10 we get, 546/432

Further by dividing 546/432 by using common divisor 6 we get, 91/72

∴ LHS = RHS is verified.

(iii) ((-9/5) × (-10/3)) × (21/-4) = (-9/5) × ((-10/3) × (21/-4))

Solution:

Firstly consider LHS

((-9/5) × (-10/3)) × (21/-4)

((-9×-10) / (5×3)) × (21/-4)

(90/15) × (21/-4)

((90×21) / (15×-4)) = (1890/-60) × -1 = -1890/60

Further by dividing -1890/60 by using common divisor 10 we get, -189/6

Further by dividing -189/6 by using common divisor 3 we get, -63/2

RHS: (-9/5) × ((-10/3) × (21/-4))

(-9/5) × ((-10×21) / (3×-4))

(-9/5) × (-210/-12)

(-9×-210) / (5×-12) = (1890/-60) × -1 = -1890/60

Further by dividing -1890/60 by using common divisor 10 we get, 189/6

Further by dividing -189/6 by using common divisor 3 we get, -63/2

∴ LHS = RHS is verified.

1. Q4. Fill in the blanks:

(i) (-23/17) × (18/35) = (18/35) × (….)
Solution:

we know the commutative law states that, a × b = b × a

By using the above law

(-23/17) × (18/35) = (18/35) × (-23/17)

(ii) -38 × (-7/19) = (-7/19) × (…)
Solution:

we know the commutative law states that, a × b = b × a

By using the above law

-38 × (-7/19) = (-7/19) × (-38)

(iii) ((15/7) × (-21/10)) × (-5/6) = (…) × ((-21/10) × (-5/6))
Solution:

we know the associative law states that, (a × b) × c = a × (b × c)

By using the above law

((15/7) × (-21/10)) × (-5/6) = (15/7) × ((-21/10) × (-5/6))

(iv) (-12/5) × ((4/15) × (25/-16)) = ((-12/5) × (4/5)) × (…)
Solution:

we know the associative law states that, (a × b) × c = a × (b × c)

By using the above law

(-12/5) × ((4/15) × (25/-16)) = ((-12/5) × (4/5)) × (25/-16)

1. Q5. Find the multiplicative inverse (i.e. reciprocal) of:We know that the multiplicative inverse of a number (a/b) is (b/a)

(i) 13/25
Solution:

The multiplicative inverse of (13/25) is (25/13)

(ii) -17/12
Solution:

The multiplicative inverse of (-17/12) is (12/-17)

We put it in the standard form i.e. (12/-17) × -1 we get, -12/17

(iii) -7/24
Solution:

The multiplicative inverse of (-7/24) is (24/-7)

We put it in the standard form i.e. (24/-7) × -1 we get, -24/7

(iv) 18
Solution:

The multiplicative inverse of 18 is (1/18)

(v) -16
Solution:

The multiplicative inverse of -16 is (1/-16)

We put it in the standard form i.e. (1/-16) × -1 we get, -1/16

(vi) (-3)/(-5)
Solution:

The multiplicative inverse of (-3/-5) is (-5/-3)

We put it in the standard form i.e. (-5/-3) × -1 we get, 5/3

(vii) -1
Solution:

The multiplicative inverse of -1 is -1

(viii) 0/2
Solution:

The multiplicative inverse of 0/2 is 2/0

Since 2/0 is undefined.

The multiplicative inverse of 0/2 is undefined

(ix) 2/-5
Solution:

The multiplicative inverse of (2/-5) is (-5/2)

(x) -1/8
Solution:

The multiplicative inverse of (-1/8) is (8/-1)

We put it in the standard form i.e. (8/-1) × -1 we get, -8/1 = -8

1. ## Exercise 1E

1. Simplify:

1. 4/9 ÷ -5/12
Solution: Firstly convert ÷ to ×4/9 ÷ -5/12 = 4/9 × 12/-54×12 / 9×-5 = 48/-45By multiplying -1 on both numerator and denominator we get,48/-45 = 48×-1/-45×-1 = -48/45By dividing further by 3 into lowest terms we get,-48/45 = -16/15
2. -8 ÷ -7/16
Solution: Firstly convert ÷ to ×-8/1 ÷ -7/16 = -8/1 × 16/-7-8×16 / 1×-7 = -128/-7By multiplying -1 on both numerator and denominator we get,-128/-7 = -128×-1/-7×-1 = 128/7
3. -12/7 ÷ (-18)
Solution: Firstly convert ÷ to ×-12/7 ÷ (-18/1) = -12/7 × 1/-18-12×1 / 7×-18 = -12/-126By multiplying -1 on both numerator and denominatorwe get,-12/-126 = -12×-1 / -126×-1 = 12/126By dividing further by 6 into lowest terms we get,12/126 = 2/21
4. -1/10 ÷ -8/5
Solution: Firstly convert ÷ to ×-1/10 ÷ -8/5 = -1/10 × 5/-8-1×5 / 10×-8 = -5/-80By multiplying -1 on both numerator and denominatorwe get,-5/-80 = -5×-1/-80×-1 = 5/80By dividing further by 5 into lowest terms we get,5/80 = 1/16
5. -16/35 ÷ -15/14
Solution: Firstly convert ÷ to ×-16/35 ÷ -15/14 = -16/35 × 14/-15-16×14 / 35×-15 = -224/-525By multiplying -1 on both numerator and denominatorwe get,-224/-525 = -224×-1/-525×-1 = 224/525By dividing further by 7 into lowest terms we get,224/525 = 32/75
6. -65/14 ÷ 13/7
Solution: Firstly convert ÷ to ×-65/14 ÷ -13/7 = -65/14 × 7/13-65×7 / 14×13 = -455/182By dividing further by 7 into lowest termswe get,-455/182 = -65/26Dividing further by 13 into lowest terms we get,-65/26 = -5/2

2. Verify whether the given statement is true or false:

1. 13/5 ÷ 26/10 = 26/10 ÷ 13/5
Solution: Firstly consider LHS(13/5) ÷ (26/10)Convert ÷ to ×13/5 ÷ 26/10 = 13/5 × 10/2613×10 / 5×26 = 130/130 = 1RHS: (26/10) ÷ (13/5)Convert ÷ to ×26/10 ÷ 13/5 = 26/10 × 5/1326×5 / 10×13 = 130/130 = 1∴ LHS = RHS, The given statement is True.
2. -9 ÷ 3/4 = 3/4 ÷ (-9)
Solution: Firstly consider LHS-9/1 ÷ 3/4Convert ÷ to ×-9/1 ÷ 3/4 = -9/1 × 4/3-9×4 / 1×3 = -36/3 = -12RHS: 3/4 ÷ (-9)Convert ÷ to ×3/4 ÷ (-9/1) = 3/4 × 1/-93×1 / 4×(-9) = 3/-36 = 1/-12∴ LHS ≠ RHS, The given statement is false.
3. -8/9 ÷ -4/3 = -4/3 ÷ -8/9
Solution: Firstly consider LHS-8/9 ÷ -4/3Convert ÷ to ×-8/9 ÷ -4/3 = -8/9 × 3/-4-8×3 / 9×(-4) = -24/-36 = 2/3RHS: -4/3 ÷ -8/9Convert ÷ to ×-4/3 ÷ -8/9 = -4/3 × 9/-8-4×9 / 3×(-8) = -36/-24 = 3/2∴ LHS ≠ RHS, The given statement is false.
4. -7/24 ÷ 3/-16 = 3/-16 ÷ -7/24
Solution: Firstly consider LHS-7/24 ÷ 3/-16Convert ÷ to ×-7/24 ÷ 3/-16 = -7/24 × -16/3-7×(-16) / 24×3 = 112/72 = 14/9RHS: 3/-16 ÷ -7/24Convert ÷ to ×3/-16 ÷ -7/24 = 3/-16 × 24/-73×24 / -16×-7 = 72/112 = 9/14∴ LHS ≠ RHS, The given statement is false.

3. Verify whether the given statement is true or false:

1. (5/9 ÷ 1/3) ÷ 5/2 = 5/9 ÷ (1/3 ÷ 5/2)
Solution: Firstly consider LHS(5/9 ÷ 1/3) ÷ 5/2Convert ÷ to ×(5/9 × 3/1) ÷ 5/2(5×3 / 9×1) ÷ 5/215/9 ÷ 5/2Convert ÷ to ×15/9 × 2/515×2 / 9×5 = 30/45By dividing further by 15 into lowest terms we get,30/45 = 2/3RHS: 5/9 ÷ (1/3 ÷ 5/2)Convert ÷ to ×5/9 ÷ (1/3 × 2/5)5/9 ÷ (1×2 / 3×5)5/9 ÷ 2/15Convert ÷ to ×5/9 × 15/25×15 / 9×2 = 75/18

By dividing further by 3 into lowest terms we get,

75/18 = 25/6

∴ LHS ≠ RHS, The given statement is false.

2. ((-16) ÷ 6/5) ÷ -9/10 = (-16) ÷ (6/5 ÷ -9/10)Solution: Firstly consider LHS
((-16) ÷ 6/5) ÷ -9/10Convert ÷ to ×(-16/1 × 5/6) ÷ -9/10(-16×5 / 1×6) ÷ -9/10-80/6 ÷ -9/10Convert ÷ to ×-80/6 × 10/-9-80×10 / 6×-9 = -800/-54 = 800/54By dividing further by 2 into lowest terms we get,800/54 = 400/27RHS: (-16) ÷ (6/5 ÷ -9/10)Convert ÷ to ×-16 ÷ (6/5 × 10/-9)-16 ÷ (6×10 / 5×-9)-16 ÷ 60/-45Convert ÷ to ×-16/1 × -45/60-16×-45 / 1×60 = 720/60

By dividing further by 60 into lowest terms we get,

720/60 = 12

∴ LHS ≠ RHS, The given statement is false.

3. (-3/5 ÷ -12/35) ÷ 1/14 = -3/5 ÷ (-12/35 ÷ 1/14)Solution: Firstly consider LHS(-3/5 ÷ -12/35) ÷ 1/14Convert ÷ to ×(-3/5 × 35/-12) ÷ 1/14(-3×35 / 5×-12) ÷ 1/14-105/-60 ÷ 1/14 = 105/60 ÷ 1/14Convert ÷ to ×105/60 × 14/1105×14 / 60×1 = 1470/60By dividing further by 30 into lowest terms we get,1470/60 = 49/2RHS: -3/5 ÷ (-12/35 ÷ 1/14)Convert ÷ to ×-3/5 ÷ (-12/35 × 14/1)-3/5 ÷ (-12×14 / 35×1)-3/5 ÷ -168/35Convert ÷ to ×-3/5 × 35/-168

-3×35 / 5×-168 = -105/-840 = 105/840

By dividing further by 5 into lowest terms we get,

105/840 = 21/168

dividing further by 3 into lowest terms we get,

21/168 = 7/56 = 1/8

∴ LHS ≠ RHS, The given statement is false.

4. The product of two rational numbers is -9. If one of the number is -12, find the other.

Solution: We know that the given details,

The product of two rational numbers = -9

One rational number = -12

Let us consider the other number as x

Now, solving

-12 × x = -9

-12x = -9

x = -9/-12 = ¾

∴ the other rational number is ¾.

5. The product of two rational numbers is -16/9. If one of the number is -4/3, find the other.

Solution: We know that the given details,

The product of two rational numbers = -16/9

One rational number = -4/3

Let us consider the other number as x

Now, solving

-4/3 × x = -16/9

x = -16/9 ÷ -4/3

Convert ÷ to ×

x = -16/9 × 3/-4

x = -16×3 / 9×-4

x = -48/-36

By dividing further by 12 into lowest terms we get,

48/36 = 4/3

∴ the other rational number is 4/3.

6. By what rational number should we multiply -15/56 to get -5/7?

Solution: let us consider x to be multiplied to the number given i.e.

-15/56 × x = -5/7

x = -5/7 ÷ -15/56

Convert ÷ to ×

x = -5/7 × 56/-15

x = -5×56 / 7×-15

x = -280/-105

By dividing further by 35 into lowest terms we get,

280/105 = 8/3

∴ 8/3 has to be multiplied to -15/56 to get -5/7.

## Exercise 1F

1. Find the rational number between ¼ and 1/3

Solution: Let us consider the rational number as x

So to find the rational number between ¼ and 1/3

By using the formula x= ½ (a/b + c/d)

x = ½(1/4 + 1/3)

by taking the LCM for 4 and 3 we get,

x = ½((1×3 + 1×4)/12)

= ½((3+4)/12)

= ½(7/12)

= ½ × 7/12

= 7/24

∴ the rational number between ¼ and 1/3 is 7/24.

2. Find the rational number between 2 and 3

Solution: Let us consider the rational number as x

So to find the rational number between 2 and 3

By using the formula x= ½ (a/b + c/d)

x = ½(2 + 3)

x = ½(5)

= 5/2

∴ the rational number between 2 and 3 is 5/2.

3. Find the rational number between -1/3 and 1/2

Solution: Let us consider the rational number as x

So to find the rational number between -1/3 and 1/2

By using the formula x= ½ (a/b + c/d)

x = ½(-1/3 + 1/2)

by taking the LCM for 3 and 2 we get,

x = ½((-1×2 + 1×3)/6)

= ½((-2+3)/6)

= ½(1/6)

= ½ × 1/6

= 1/12

∴ the rational number between -1/3 and 1/2 is 1/12.

4. Find two rational number between -3 and -2

Solution: Let us consider the rational number as x

So to find the rational number between -3 and -2

By using the formula x= ½ (a/b + c/d)

x = ½(-3 + (-2))

= ½(-3-2)

= ½(-5) = -5/2

Now we find the rational number between -5/2 and -2. Since -5/2 also lies between -3 and -2.

Let us consider another rational number as y

So let us use the formula again

y= ½(-5/2 + (-2))

= ½ ((-5/2)-2)

= ½ ((-5-4)/2)

= ½ (-9/2)

= ½ × -9/2

= -9/4

∴ Two rational numbers between -3 and 2 is -5/2 and -9/4.

## Exercise 1G

1. From a rope 11m long, two pieces of lengths $2\frac{3}{5}$m and $3\frac{3}{10}$m are cut off. What is the length of the remaining rope?

Solution:

2. we know that the given details areThe length of the rope =11mLength of the 1st piece = $2\frac{3}{5}$mLength of the 2nd piece = $3\frac{3}{10}$mTotal length of the pieces = length of 1st piece + length of 2nd piece

= $2\frac{3}{5}$ + $3\frac{3}{10}$

=(5×2+3)/5 + (10×3+3)/10

=13/5 + 33/10

By taking the LCM for 5 and 10 is 10

= (13×2 + 33×1)/10

= (26+33)/10

=59/10m

Length of the remaining rope = length of the rope – total length of the pieces

= 11m – 59/10m

By taking the LCM

= (11×10 -59×1)/10

= (110-59)/10

=51/10

= $5\frac{1}{10}$m

∴ The length of the remaining rope is $5\frac{1}{10}$m.

2. A drum full of rice weighs $40\frac{1}{6}$kg. If the empty drum weighs $13\frac{3}{4}$kg, find the weight of rice in the drum.

Solution:

3. we know that the given details areA drum full of rice weighs = $40\frac{1}{6}$kgEmpty drum weighs = $13\frac{3}{4}$kgWeight of rice in the drum = drum full of rice – weight of empty drum= $40\frac{1}{6}$ – 13 ¾

= 241/6 – 55/4

By taking the LCM for 6 and 4 is 12

= (241×2 – 55×3)/12

= (482 – 165)/12

= 317/12

= $26\frac{5}{12}$kg

∴ The weight of rice in the drum =$26\frac{5}{12}$kg.

3. A basket contains three types of fruits weighing $19\frac{1}{3}$kg in all. If $8\frac{1}{9}$kg of these be apples, $3\frac{1}{6}$kg be oranges and the rest pears, what is the weight of the pears in the basket?

Solution:

4. we know that the given details areWeight of three types of fruits = $19\frac{1}{3}$kgWeight of apples = $8\frac{1}{9}$kgWeight of oranges = $3\frac{1}{6}$kgWeight of pears = Weight of three types of fruits – (Weight of apples + Weight of oranges)

= $19\frac{1}{3}$ – ($8\frac{1}{9}$ + $3\frac{1}{6}$)

= 58/3 – (73/9 + 19/6)

= 58/3 – (73×2 + 19×3)/18 (by taking LCM for 9 and 6 is 18)

= 58/3 – (146 + 57)/18

= 58/3 – 203/18

By taking LCM for 3 and 18 is 18

= (58×6 – 203×1)/18

= (348 – 203)/18

= 145/18

= $8\frac{1}{18}$kg

∴ The weight of pears is $8\frac{1}{18}$kg.

4. On one day a rickshaw puller earned ₹160. Out of his earnings he spent ₹$26\frac{3}{5}$ on tea and snacks. ₹50 ½ on food and ₹$16\frac{2}{5}$ on repairs of the rickshaw. How much did he save on that day?

Solution:

5. we know that the given details areTotal earnings = ₹160Earnings spent on tea and snacks = ₹$26\frac{3}{5}$Earnings spent on food = ₹$50\frac{1}{2}$Earnings spent on repairs = ₹$16\frac{2}{5}$

∴ Total expenditure = Earnings spent on tea and snacks + Earnings spent on food + Earnings spent on repairs

= ₹$26\frac{3}{5}$ + ₹50 ½ + ₹$16\frac{2}{5}$

= 133/5 + 101/2 + 82/5

By taking LCM for 5 and 2 is 10

= (133×2 + 101×5 + 82×2)/10

= (266 + 505 +164)/10

= 935/10

Total Savings = Total Earnings – Total Expenditure

= ₹160 – ₹935/10

By taking 10 as the LCM

= (160×10 – 935×1)/10

= (1600 – 935)/10

= 665/10 = 66 ½

∴ Total Savings on that day is ₹66 ½.

5. Find the cost of $3\frac{2}{6}$meters of cloth at ₹66 ¾ per meter.

Solution:

6. we know that the given details areCost of the cloth per meter = ₹66 ¾Total meters = $3\frac{2}{5}$metersTotal cost of cloth = Cost of the cloth per meter × Total meters= ₹66 ¾ × $3\frac{2}{5}$

= 267/4 × 17/5

= 4335/20

Further we can divide by 5 we get,

4335/20 = 867/4 = 216 ¾

∴ Cost of the cloth is ₹216 ¾.

6. A car moving at an average speed of $60\frac{2}{5}$km/hr. How much distance will it cover in 6 ¼ hours?

Solution:

7. we know that the given details areSpeed of the car = $60\frac{2}{5}$ km/hr.Total hours = 6 ¼ hoursBy using the formula, speed = distance/timeDistance of the car = speed of the car × time taken

= $60\frac{2}{5}$× 6 ¼

=302/5 × 25/4

= 7550/20

= 755/2 = 377 ½

∴ Distance covered = 377 ½ km.

7. Find the area of a rectangular park which is $36\frac{3}{5}$m long and $16\frac{2}{3}$m broad.

Solution:

8. we know that the given details areLength of the park = $36\frac{3}{5}$mBreadth of the park = $16\frac{2}{3}$mBy using the formula, area of rectangle = length × breadthArea of rectangular park = $36\frac{3}{5}$ × $16\frac{2}{3}$

= 183/5m × 50/3m

= 9150/15m2 = 610m2

∴ Area of the park = 610m2

8. Find the area of a square plot of land whose each side measures 8 ½ meters.

Solution:

9. we know that the given details areOne side measures = $8\frac{1}{2}$mBy using the formula,Area of square = side × sideArea of square plot = 8 ½m × 8 ½m

= 17/2m × 17/2m

= 289/4m2 = 72 ¼ m2

∴ Area of square plot is 72 ¼ m2

9. One liter of petrol costs ₹63 ¾. What is the cost 34 liters of petrol?

Solution:

10. we know that the given details areCost of 1 liter petrol = ₹63 ¾Cost of 34 liters petrol = 34 × cost of 1 liter petrol= 34 × 63 ¾= 34 × 255/4

= 8670/4

We can further divide by 2 we get,

= 4335/2

= 2167 ½

∴ Cost of 34 liters petrol is ₹2167 ½

10. An aeroplane covers 1020km in an hour. How much distance will it cover in $4\frac{1}{6}$hours?

Solution:

11. we know that the given details areDistance covered in 1 hour = 1020kmDistance covered in $4\frac{1}{6}$ hour = $4\frac{1}{6}$ × distance covered in 1 hour= 4 1/6 × 1020= 25/6 × 1020

=25500/6

= 4250km

∴ Distance covered in $4\frac{1}{6}$hours = 4250km.

## Exercise 1H

1. (-5/16 + 7/12) = ?

1. -7/48
2. 1/24
3. 13/48
4. 1/3

Solution:

By solving (-5/16 + 7/12) we get,

(-5/16 + 7/12)

By taking LCM for 16 and 12 is 48

(-5/16 + 7/12) = (-5×3 + 7×4)/48= (-15 + 28)/48= 13/48

1. 2. (8/-15 + 4/-3) = ?
1. 28/15
2. -28/15
3. -4/5
4. -4/15

Solution:

By solving (-5/16 + 7/12) we get,(8/-15 + 4/-3)

Firstly we have multiply both numerator and denominator by -1

(8×-1/-15×-1 + 4×-1/-3×-1) = (-8/15 + -4/3)

By taking LCM for 15 and 3 is 15

(-8/15 + -4/3) = (-8×1 + -4×5)/15= (-8 + (-20))/15= (-8-20)/15= -28/15

1. 3. (7/-26 + 16/39) = ?
1. 11/78
2. -11/78
3. 11/39
4. -11/39

Solution:

By solving (7/-26 + 16/39) we get,(7/-26 + 16/39)

Firstly we have multiply both numerator and denominator by -1

(7×-1/-26×-1 + 16/39) = (-7/26 + 16/39)

By taking LCM for 26 and 39 is 78

(-7/26 + 16/39) = (-7×3 + 16×2)/78= (-21 + 32)/78= 11/78

1. 4. (3 + 5/-7) = ?
1. -16/7
2. 16/7
3. -26/7
4. -8/7

Solution:

Firstly we have multiply both numerator and denominator by -1

(3/1 + 5/-7) = (3/1 + 5×-1/-7×-1)= (3/1+ (-5/7))

By taking LCM for 1 and 7 is 7

(3/1+ (-5/7)) = (3×7 + (-5×1))/7= (21 + (-5))/7= (21 – 5)/7= 16/7

1. 5. (31/-4 + -5/8) = ?
1. 67/8
2. 57/8
3. -57/8
4. -67/8

Solution:

By solving (31/-4 + -5/8) we get,

Firstly we have multiply both numerator and denominator by -1

(31/-4 + -5/8) = (31×-1/-4×-1 + -5/8)= (-31/4 + -5/8)

By taking LCM for 4 and 8 is 8

(-31/4 + -5/8) = (-31×2 + -5×1)/8= (-62 + (-5))/8= (-62-5)/8= -67/8

1. 6. What should be added to 7/12 to get -4/15?
1. 17/20
2. -17/20
3. 7/20
4. -7/20

Solution:

Let us consider x as a number to be added

7/12 + x = -4/15x = -4/15 – 7/12

By taking LCM for 15 and 12 is 60

x = (-4×4 – 7× 5)/60= (-16 – 35)/60= -51/60

We can further divide by 3 we get,-51/60 = -17/20

1. 7. (2/3 + (-4/5) + 7/15 + (-11/20)) = ?
1. -1/5
2. -4/15
3. -13/60
4. -7/30

Solution:

we solve for,(2/3 + (-4/5) + 7/15 + (-11/20))

By taking LCM for 3, 5, 15 and 20 is 60

(2×20 + (-4×12) + 7×4 + (-11×3))/60 = (40 + (-48) + 28 + (-33))/60= (40 – 48 +28 – 33)/60= (68-81)/60= -13/60

1. 8. The sum of two numbers is -4/3. If one of the number is -5, what is the other?
1. -11/3
2. 11/3
3. -19/3
4. 19/3

Solution:

let us consider one of the number as x-5 + x = -4/3x = -4/3 + 5/1

By taking LCM for 3 and 1 is 3 we get,

x = (-4×1 + 5×3)/3= (-4+15)/3= 11/3

1. 9. What should be added to -5/7 to get -2/3?
1. -29/21
2. 29/21
3. 1/21
4. -1/21

Solution:

let us consider one of the number as

x-5/7 + x = -2/3x = -2/3 – (-5/7)= -2/3 + 5/7

By taking LCM for 3 and 7 is 21 we get,

x = (-2×7 + 5×3)/21= (-14+15)/21= 1/21

1. 10. What should be subtracted from -5/3 to get 5/6?
1. 5/2
2. 3/2
3. 5/4
4. -5/2

Solution:

let us consider one of the number as x-5/3 –x = 5/6-5/3 – 5/6 = xx = -5/3 – 5/6

By taking LCM for 3 and 6 is 6 we get,x = (-5×2 – 5×1)/6= (-10-5)/6= -15/6

Further we can divide by 3 we get,-15/6 = -5/2

1. 11. (-3/7)-1=?
1. 7/3
2. -7/3
3. 3/7
4. None of these

Solution:

We know that for any real number a≠0 then, a-1= 1/a

∴ By using the formula, (-3/7)-1 = 1/ (-3/7)= 7/-3

By multiplying both numerator and denominator by -1 we get,

7/-3 = (7×-1/-3×-1) = -7/3

## RS Aggarwal Solutions for Class 8 Maths Chapter 1 – Rational Numbers

Chapter 1- Rational Numbers contains 8 exercises and the RS Aggarwal solutions present on this page provides the solutions for the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

• Rational numbers – the operations of addition, subtraction and multiplication
• Representation of Rational numbers on the Real Line
• The operations addition and multiplication are
• Closure property
• Commutative for rational numbers
• Associative for rational numbers
• Rational number 0 is the additive identity for rational numbers
• Existence of Additive Inverse and Multiplicative Inverse
• Rational number 1 is the multiplicative identity for rational numbers
• Distributivity of rational numbers: For all rational numbers a, b and c, a (b + c) = ab + ac and a (b – c) = ab – ac
• Division of Rational Numbers
• An alternative method of finding a large number of rational numbers between two given rational numbers
• Word problems

Exercise 1A

Exercise 1B

Exercise 1C

Exercise 1D

Exercise 1E

Exercise 1F

Exercise 1G

Exercise 1H

### Chapter Brief of RS Aggarwal Solutions for Class 8 Maths Chapter 1 – Rational Numbers

RS Aggarwal Solutions for Class 8 Maths Chapter 1 – Rational Numbers ensures that the students are thorough and familiar with the concepts. The solutions are designed in such a way that they are easy to understand and solve. Since we use rational numbers in our daily life likewise to calculate age, mass, simple arithmetic, date, etc. It makes so much fun in learning rational numbers and their fractions. In the RS Aggarwal Solutions provided at BYJU’S, answers for all the exercise problems are given which enhances familiarity with the concepts.