RD Sharma Solutions Class 8 Rational Numbers Exercise 1.1

RD Sharma Solutions Class 8 Chapter 1 Exercise 1.1

RD Sharma Class 8 Solutions Chapter 1 Ex 1.1 PDF Free Download

Exercise 1.1

Q1. Add the following rational numbers:

(i) \(\frac{-5}{7}\) and \(\frac{3}{7}\)

(ii) \(\frac{-15}{4}\) and \(\frac{7}{4}\)

(iii) \(\frac{-8}{11}\) and \(\frac{-4}{11}\)

(iv) \(\frac{6}{13}\) and \(\frac{-9}{13}\)

Solution:

(i) \(\frac{-5}{7}\) + \(\frac{3}{7}\) = \(\frac{-5 + 3}{7}\) = \(\frac{-2}{7}\)

(ii)  \(\frac{-15}{4}\) + \(\frac{7}{4}\) = \(\frac{-15 + 7}{4}\) = \(\frac{-8}{4}\)= -2

(iii) \(\frac{-8}{11}\) + \(\frac{-4}{11}\) = \(\frac{-8 -4}{11}\)= \(\frac{-12}{11}\)

(iv) \(\frac{6}{13}\) + \(\frac{-9}{13}\)= \(\frac{6 – 9}{13}\)= \(\frac{-3}{13}\)

Q2: Add the following rational numbers:

(i)  \(\frac{3}{4}\) and \(\frac{-5}{8}\)

(ii)  \(\frac{5}{-9}\) and \(\frac{7}{3}\)

(iii) -3  and \(\frac{3}{5}\)

(iv)  \(\frac{-7}{27}\) and \(\frac{11}{18}\)

(v)  \(\frac{31}{-4}\) and \(\frac{-5}{8}\)

(vi)  \(\frac{5}{36}\) and \(\frac{-7}{12}\)

(vii) \(\frac{-5}{16}\) and \(\frac{-7}{24}\)

(viii)  \(\frac{7}{-18}\) and \(\frac{8}{27}\)

Solution:

(i) Clearly, denominators of the given numbers are positive.

The LCM of the denominators 4 and 8 is 8.

Now, we will express \(\frac{3}{4}\) in the form in which it takes the denominator as 8.

\(\frac{3 \times 2}{4 \times 2}\) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Now,

\(\frac{-5}{8}\) + \(\frac{ 6 }{ 8 }\)

= \(\frac{-5 + 6}{8}\) = \(\frac{1}{8}\)

(ii)  \(\frac{5}{-9}\) + \(\frac{7}{3}\)

= \(\frac{-5}{9}\) + \(\frac{7}{3}\)

The LCM of the denominators 9 and 3 is 9.

Now,

We will express \(\frac{7}{3}\) in the form in which it takes denominator as 9.

\(\frac{7 \times 3}{3 \times 3}\) = \(\frac{21}{9}\)

So,

\(\frac{-5}{9}\) + \(\frac{21}{9}\)

= \(\frac{-5 + 21}{9}\) = \(\frac{16}{9}\)

(iii) -3 +  \(\frac{3}{5}\)

\(\frac{- 3}{1}\)\(\frac{3}{5}\)

The LCM of the denominators 1 and 5 is 5.

Now,

We will express  \(\frac{- 3}{1}\) in the form in which it takes denominator as 5.

\(\frac{-3}{1}\)\(\frac{-3 \times  5}{1 \times 5}\) = \(\frac{-15}{5}\)

So,

\(\frac{15}{5}\) + \(\frac{3}{5}\)

= \(\frac{ -15 + 3}{5}\) = \(\frac{-12}{5}\)

(iv)  \(\frac{-7}{27}\)\(\frac{11}{18}\)

The LCM of the denominators 27 and 18 is 54.

Now,

We will express \(\frac{ -7 }{27}\) and \(\frac{11}{18}\) in the form in which it takes denominator as 54.

\(\frac{ -7 }{ 27 }\) = \(\frac{ -7 \times  2 }{ 27 \times  2 }\) = \(\frac{-14}{54}\)

\(\frac{11}{18}\) = \(\frac{11 \times  3}{ 18 \times 3}\) = \(\frac{33}{54}\)

So,

\(\frac{ -14 }{ 54 }\) + \(\frac{ 33 }{ 54 }\)

= \(\frac{-14 + 33}{54}\) = \(\frac{19}{54}\)

(v) \(\frac{ 31 }{ -4 }\) + \(\frac{ -5 }{ 8 }\)

= \(\frac{ 31 }{ -4 }\) = \(\frac{ – 31 }{ 4 }\)

The LCM of the denominators 4 and 8 is 8.

Now,

We will express \(\frac{ -31 }{ 4 }\) in the form in which it takes denominator as 8.

\(\frac{ -31 }{ 4 }\) = \(\frac{ -31 \times  2 }{ 4 \times  2 }\) = \(\frac{ -62 }{ 8 }\)

So,

\(\frac{ -62 }{ 8 }\) + \(\frac{ -5 }{ 8 }\)

= \(\frac{ -62 – 5 }{ 8 }\) = \(\frac{ -67 }{ 8 }\)

(vi) \(\frac{ 5 }{ 36 }\) + \(\frac{ -7 }{ 12 }\)

The LCM of the denominator 12 and 36 is 36.

Now,

We will express \(\frac{ -7 }{ 12 }\) in the form in which it takes denominator as 36.

\(\frac{ -7 }{ 12 }\) = \(\frac{ -7 \times  3 }{ 12 \times 3 }\) = \(\frac{ -21 }{ 36 }\)

So,

\(\frac{ -21 }{ 36 }\) + \(\frac{ 5 }{ 36 }\)

= \(\frac{ -21 + 5 }{ 36 }\) = \(\frac{ -16 }{ 36 }\) = \(\frac{ -4 }{ 9 }\)

(vii) \(\frac{ -5 }{ 16 }\) and \(\frac{ 7 }{ 24 }\)

The LCM of the denominators 16 and 26 is 48.

Now,

We will express \(\frac{ -5 }{ 16 }\) and \(\frac{ 7 }{ 24 }\) in the form in which it takes denominator as 48.

\(\frac{ -5 }{ 16 }\) = \(\frac{ -5 \times 3 }{ 16 \times 3 }\) = \(\frac{ -15 }{ 48 }\)

\(\frac{ 7 }{ 24 }\) = \(\frac{ 7 \times 2 }{ 24 \times 2 }\) = \(\frac{ 14 }{ 48 }\)

So,

\(\frac{ -15 }{ 48 }\) + \(\frac{ 14 }{ 48 }\)

= \(\frac{ -15 + 14 }{ 48 }\) = \(\frac{ -1 }{ 48 }\)

(viii) \(\frac{ 7 }{ -18 }\) + \(\frac{ 8 }{ 27 }\)

\(\frac{ 7 }{ -18 }\) = \(\frac{ -7 }{ 18 }\)

The LCM of the denominator 18 and 27 is 54.

Now,

We will express \(\frac{ -7 }{ 18 }\) and \(\frac{ 8 }{ 27 }\) in the form in which it takes denominator as 54.

\(\frac{ -7 }{ 18 }\) = \(\frac{ -7 \times 3 }{ 18 \times 3 }\)= \(\frac{ -21 }{ 54 }\)

\(\frac{ 8 }{ 27 }\) = \(\frac{ 8 \times 2  }{ 27 \times  2 }\) = \(\frac{ 16 }{ 54 }\)

So,

\(\frac{ -21 }{ 54 }\) + \(\frac{ 16 }{ 54 }\)

= \(\frac{ -21 + 16 }{ 54 }\) = \(\frac{ -5 }{ 54 }\)

Q-3. Simplify:

(i)  \(\frac{ 8 }{ 9 }\) + \(\frac{ -11 }{ 6 }\)

(ii)  3 + \(\frac{ 5 }{ -7 }\)

(iii)  \(\frac{ 1 }{ -12 }\) + \(\frac{ 2 }{ -15 }\)

(iv)  \(\frac{ -8 }{ 19 }\) + \(\frac{ -4 }{ 57 }\)

(v)  \(\frac{ 7 }{ 9 }\) + \(\frac{ 3 }{ -4 }\)

(vi)  \(\frac{ 5 }{ 26 }\) + \(\frac{ 11 }{ -39 }\)

(vii)  \(\frac{ -16 }{ 9 }\) + \(\frac{ -5 }{ 12 }\)

(viii)  \(\frac{ -13 }{ 8 }\) + \(\frac{ 5 }{ 36 }\)

(ix)  0 + \(\frac{ -3 }{ 5 }\)

(x)  1 + \(\frac{ -4 }{ 5 }\)

Solution:

(i) \(\frac{ 8 }{ 9 }\) + \(\frac{ -11 }{ 6 }\)

The LCM of the denominator 9 and 6 is 18.

Now,

We will express \(\frac{ 8 }{ 9 }\) and \(\frac{ -11 }{ 6 }\) in the form in which it takes denominator as 18.

\(\frac{ 8 }{ 9 }\) = \(\frac{ 8 \times 2 }{ 9 \times 2 }\) = \(\frac{ 16 }{ 18 }\)

\(\frac{ -11 }{ 6 }\) = \(\frac{ -11 \times 3 }{ 6 \times 3 }\) = \(\frac{ -33 }{ 18 }\)

So,

\(\frac{ 16 }{ 18 }\) + \(\frac{ -33 }{ 18 }\)

= \(\frac{ 16 – 33 }{ 18 }\) = \(\frac{ -17 }{ 18 }\)

(ii)  3 + \(\frac{ 5 }{ -7 }\)

\(\frac{ 5 }{ -7 }\) = \(\frac{ -5 }{ 7 }\)

The LCM of the denominator 1 and 7 is 7.

Now,

We will express \(\frac{ 3 }{ 1 }\) in the form in which it takes denominator as 7.

\(\frac{ 3 }{ 1 }\) = \(\frac{ 3 \times 7 }{ 1 \times 7 }\) = \(\frac{ 21 }{ 7 }\)

So,

\(\frac{ 21 }{ 7 }\) + \(\frac{ -5 }{ 7 }\)

= \(\frac{ 21 – 5 }{ 7 }\) = \(\frac{ 16 }{ 7 }\)

(iii)  \(\frac{ 1 }{ -12 }\) + \(\frac{ 2 }{ -15 }\)

\(\frac{ 1 }{ -12 }\) = \(\frac{ – 1 }{ 12 }\)

\(\frac{ 2 }{ -15 }\) = \(\frac{ -2 }{ 15 }\)

The LCM of the denominators 12 and 15 is 60.

Now,

We will express \(\frac{ -1 }{ 12 }\) and \(\frac{ -2 }{ 15 }\) in the form in which it takes denominator as 60.

\(\frac{ -1 }{ 12 }\) = \(\frac{ -1 \times  5 }{ 12 \times 5 }\) = \(\frac{ -5 }{ 60 }\)

\(\frac{ -2 }{ 15 }\) = \(\frac{ -2 \times 4 }{ 15 \times 4 }\) = \(\frac{ -8 }{ 60 }\)

So,

\(\frac{ -5 }{ 60 }\) + \(\frac{ -8 }{ 60 }\)

= \(\frac{ -5 – 8 }{ 60 }\) = \(\frac{ -13 }{ 60 }\)

(iv)  \(\frac{ -8 }{ 19 }\) + \(\frac{ -4 }{ 57 }\)

The LCM of the denominator of 19 and 57 is 57.

Now,

We will express \(\frac{ -8 }{ 19 }\) in the form in which it takes denominator as 57.

\(\frac{ -8 }{ 19 }\) = \(\frac{ -8 \times 3 }{ 19 \times 3 }\) = \(\frac{ -24 }{ 57 }\)

So,

\(\frac{ -24 }{ 57 }\) + \(\frac{ -4 }{ 57 }\)

= \(\frac{ -24 – 4 }{ 57  }\) = \(\frac{ -28 }{ 57 }\)

(v)  \(\frac{ 7 }{ 9 }\) + \(\frac{ 3 }{ -4 }\)

\(\frac{ 3 }{ -4 }\) = \(\frac{ -3 }{ 4 }\)

The LCM of the denominator 9 and 4 is 36.

Now,

We will express \(\frac{ 7 }{ 9 }\) and \(\frac{ -3 }{ 4 }\) in the form in which it takes denominator as 36.

\(\frac{ 7 }{ 9 }\) = \(\frac{ 7 \times 4 }{ 9 \times 4 }\) = \(\frac{ 28}{ 36 }\)

\(\frac{ -3 }{ 4 }\) = \(\frac{ -3 \times 9 }{ 4 \times 9 }\) = \(\frac{ -27 }{ 36 }\)

So,

\(\frac{ 28 }{ 36 }\) + \(\frac{ -27 }{ 36 }\)

= \(\frac{ 28 – 27 }{ 36 }\) = \(\frac{ 1 }{ 36 }\)

(vi)  \(\frac{ 5 }{ 26 }\) + \(\frac{ 11 }{ -39 }\)

\(\frac{ 11 }{ -39 }\) = \(\frac{ -11 }{ 39 }\)

The LCM of the denominator 26 and 39 is 78.

Now,

We will express \(\frac{ -3 }{ 4 }\) and \(\frac{ -11 }{ 39 }\) in the form in which it takes denominator as 78.

\(\frac{ 5 }{ 26 }\) = \(\frac{ 5 \times 3 }{ 26 \times 3 }\) = \(\frac{ 15 }{ 78 }\)

\(\frac{ -11 }{ 39 }\) = \(\frac{ -11 \times 2 }{ 39 \times  2 }\) = \(\frac{ -22 }{ 78 }\)

So,

\(\frac{ 15 }{ 78 }\) + \(\frac{ -22 }{ 78 }\)

= \(\frac{ 15 – 22 }{ 78 }\) = \(\frac{ -7 }{ -78 }\)

(vii)  \(\frac{ -16 }{ 9 }\) + \(\frac{ -5 }{ 12 }\)

The LCM of the denominator 9 and 12 is 36.

Now,

We will express \(\frac{ -16 }{ 9 }\)  and \(\frac{ -5 }{ 12 }\) in the form in which it takes denominator as 36.

\(\frac{ -16 }{ 9 }\) = \(\frac{ -16 \times 4 }{ 9 \times 4 }\) = \(\frac{ -64 }{ 36 }\)

\(\frac{ -5 }{ 12 }\) = \(\frac{ -5 \times  3}{ 12 \times 3 }\) = \(\frac{ -15 }{ 36 }\)

So,

\(\frac{ -64 }{ 36 }\) + \(\frac{ -15 }{ 36 }\)

=\(\frac{ -64 – 15 }{ 36 }\) = \(\frac{ -79 }{ 36 }\)

(viii)  \(\frac{ -13 }{ 8 }\) + \(\frac{ 5 }{ 36 }\)

The LCM of the denominator 8 and 36 is 72.

Now,

We will express \(\frac{ -13 }{ 8 }\) and \(\frac{ 5 }{ 36 }\) in the form in which it takes denominator as 72.

\(\frac{ -13 }{ 8 }\) = \(\frac{ -13 \times  9 }{ 8 \times  9 }\) = \(\frac{ -117 }{ 72 }\)

\(\frac{ 5 }{ 36 }\) = \(\frac{ 5 \times 2 }{ 36 \times 2 }\) = \(\frac{ 10 }{ 72 }\)

So,

\(\frac{ -117 }{ 72 }\) + \(\frac{ 10 }{ 72 }\)

= \(\frac{ -117 + 10 }{ 72 }\) = \(\frac{ -107 }{ 72 }\)

(ix)  0 + \(\frac{ -3 }{ 5 }\)

= \(\frac{ -3 }{ 5 }\)

(x)  1 + \(\frac{ -4 }{ 5 }\)

The LCM of the denominator 1 and 5 is 5.

Now,

We need to express  \(\frac{ 1 }{ 1 }\) in the form in which it takes denominator as 5.

\(\frac{ 1 }{ 1 }\) =  \(\frac{ 1 \times 5 }{ 1 \times 5 }\) =  \(\frac{ 5}{ 5 }\)

So,

\(\frac{ 5}{ 5 }\) + \(\frac{ -4}{ 5 }\)

= \(\frac{ 5 – 4 }{ 5 }\) = \(\frac{ 1 }{ 5 }\)

Q-4. Add and express the sum as a mixed fraction:

(i)  \(\frac{ -12}{ 5 }\) and \(\frac{ 43}{ 10 }\)

(ii)  \(\frac{ 24}{ 7 }\) and \(\frac{ -11 }{ 4 }\)

(iii)  \(\frac{ -31 }{ 6 }\) and \(\frac{ -27}{ 8 }\)

(iv)  \(\frac{ 101 }{ 6 }\) and \(\frac{ 7 }{ 8 }\)

Solution:

(i)  We have:

\(\frac{ -12 }{ 5 }\) and \(\frac{ 43 }{ 10 }\)

The LCM of the denominator 5 and 10 is 10.

Now,

We will express \(\frac{ -12 }{ 5 }\) in the form in which it takes denominator as 10.

\(\frac{ -12 }{ 5 }\) = \(\frac{ -12 \times  2 }{ 5 \times 2  }\) = \(\frac{ -24 }{ 10 }\)

So,

\(\frac{ -24 }{ 10 }\) + \(\frac{ 43 }{ 10 }\)

= \(\frac{ -24 + 43 }{ 10 }\) = \(\frac{ 19 }{ 10 }\)

(ii) We have:

\(\frac{ 24 }{ 7 }\) and \(\frac{ -11 }{ 4 }\)

The LCM of the denominator 7 and 4 is 28.

Now,

We will express \(\frac{ 24 }{ 7 }\) and \(\frac{ -11 }{ 4 }\) in the form in which it takes denominator as 10.

\(\frac{ 24 }{ 7 }\) = \(\frac{ 24 \times 4 }{ 7 \times 4  }\) = \(\frac{ 96 }{ 28 }\)

\(\frac{ -11 }{ 4 }\) = \(\frac{ -11 \times 7 }{ 4 \times  7  }\) = \(\frac{ -77 }{ 28 }\)

So,

\(\frac{ 96 }{ 28 }\) + \(\frac{ -77 }{ 28 }\)

= \(\frac{ 96 – 77 }{ 28 }\) = \(\frac{ 19 }{ 28 }\)

(iii)  We have:

\(\frac{ -31 }{ 6 }\) and \(\frac{ -27 }{ 8 }\)

The LCM of the denominator 6 and 8 is 24.

Now,

We will express \(\frac{ -31}{ 6 }\) and \(\frac{ -27 }{ 8 }\) in the form in which it takes denominator as 24.

\(\frac{ -31 }{ 6 }\) = \(\frac{ -31 \times 4 }{ 6 \times 4 }\) = \(\frac{ -124 }{ 24 }\)

\(\frac{ -27 }{ 8 }\) = \(\frac{ -27 \times 3 }{ 8 \times 3 }\) =\(\frac{ -81 }{ 24 }\)

So,

\(\frac{ -124 }{ 24 }\) + \(\frac{ -81 }{ 24 }\)

= \(\frac{ -124 – 81 }{ 24 }\)

= \(\frac{ -205 }{ 24 }\) = \(-8 \frac{ 13 }{ 24 }\)

(iv) We have:

\(\frac{ 101 }{ 6 }\) and \(\frac{ 7 }{ 8 }\)

The LCM of the denominator 6 and 8 is 24.

Now,

We will express \(\frac{ 101 }{ 6 }\) and \(\frac{ 7 }{ 8 }\) in the form in which it takes denominator as 24.

\(\frac{ 101 }{ 6 }\) = \(\frac{ 101 \times 4 }{ 6 \times 4 }\) = \(\frac{ 404 }{ 24 }\)

\(\frac{ 7 }{ 8 }\) = \(\frac{ 7 \times 3 }{ 8 \times 3 }\) = \(\frac{ 21 }{ 24 }\)

So,

\(\frac{ 404 }{ 24 }\) + \(\frac{ 21 }{ 24 }\)

= \(\frac{ 404 + 21 }{ 24 }\)

= \(\frac{ 425 }{ 24 }\) = \(17 \frac{ 17 }{ 24 }\)


Practise This Question

Which of the following pair of linear equations has infinite solutions?