# RD Sharma Solutions Class 8 Rational Numbers Exercise 1.4

## RD Sharma Solutions Class 8 Chapter 1 Exercise 1.4

### RD Sharma Class 8 Solutions Chapter 1 Ex 1.4 PDF Free Download

Exercise 1.4

Q-1. Simplify each of the following and write as a rational number of the form $\frac{ p }{ q }$:

(i) $\frac{3}{4} + \frac{5}{6} + \frac{-7}{8}$

(ii) $\frac{2}{3} + \frac{-5}{6} + \frac{-7}{9}$

(iii) $\frac{-11}{2} + \frac{7}{6} + \frac{-5}{8}$

(iv) $\frac{-4} {5} + \frac{-7 }{10} + \frac{-8}{15}$

(v) $\frac{-9}{10} + \frac{22}{15} + \frac{13}{-20}$

(vi) $\frac{5}{3} + \frac{3}{-2} + \frac{-7}{3} + 3$

Solution:

(i)   $\frac{3}{4} + \frac{5}{6} + \frac{-7}{8}$

= $\frac{3}{4} + \frac{5}{6} – \frac{7}{8}$

= $\frac{3 \times 6}{4 \times 6} + \frac{5 \times 4}{6 \times 4} – \frac{7 \times 3}{8 \times 3}$

= $\frac{18}{24} + \frac{20}{24} – \frac{21}{20}$

= $\frac{ 18 + 20 – 21 }{ 24 }$

= $\frac{ 17}{ 24 }$

(ii)  $\frac{ 2 }{ 3 } + \frac{ -5 }{ 6 } + \frac{ -7 }{ 9 }$

= $\frac{ 2 }{ 3 } + \frac{ -5 }{ 6 } – \frac{ 7 }{ 9 }$

= $\frac{2 \times 6}{3 \times 6} – \frac{5 \times 3}{6 \times 3} – \frac{7 \times 2}{9 \times 2}$

= $\frac{12}{18} – \frac{15}{18} – \frac{14}{18}$

= $\frac{ 12 – 15 – 14 }{ 18 }$ = $\frac{ -17}{ 18 }$

(iii) $\frac{ -11 }{ 2 } + \frac{ 7 }{ 6 } + \frac{ -5 }{ 8 }$

= $\frac{ -11 }{ 2 } + \frac{ 7 }{ 6 } – \frac{ 5 }{ 8 }$

= $\frac{-11 \times 12}{2 \times 12} + \frac{7 \times 4}{6 \times 4} – \frac{5 \times 3}{8 \times 3}$

= $\frac{-132}{24} + \frac{28}{24} – \frac{15}{24}$

= $\frac{ -132 + 28 – 15 }{ 24 }$ = $\frac{ -119 }{ 24 }$

(iv) $\frac{ -4 }{ 5 } + \frac{ -7 }{ 10 } + \frac{ -8 }{ 15 }$

= $\frac{ -4 }{ 5 } – \frac{ 7 }{ 10 } – \frac{ 8 }{ 15 }$

= $\frac{-4 \times 6}{5 \times 6} – \frac{7 \times 3}{10 \times 3} – \frac{8 \times 2}{15 \times 2}$

= $\frac{-24}{30} – \frac{21}{20} – \frac{16}{30}$

= $\frac{ -24 – 21 – 16 }{ 30 }$ = $\frac{ -61 }{ 30 }$

(v)  $\frac{ -9 }{ 10 } + \frac{ 22 }{ 15 } + \frac{ 13 }{ -20 }$

= $\frac{ -9 }{ 10 } + \frac{ 22 }{ 15 } – \frac{ 13 }{ 20 }$

= $\frac{-9 \times 6}{10 \times 6} + \frac{22 \times 4}{15 \times 4} – \frac{13 \times 3}{20 \times 3}$

= $\frac{-54}{60} + \frac{88}{60} – \frac{39}{60}$

= $\frac{ -54 + 88 – 39}{ 60 }$ = $\frac{ -5}{60}$ = $\frac{-1}{60}$

(vi) $\frac{ 5 }{ 3 }$ + $\frac{ 3 }{ -2 }$ + $\frac{ -7 }{ 3 } + 3$

= $\frac{ 5 }{ 3 } – \frac{ 3 }{ 2 } – \frac{ 7 }{ 3 } + \frac{3}{1}$

= $\frac{5 \times 2}{3 \times 2} – \frac{3 \times 3}{2 \times 3} – \frac{7 \times 2}{3 \times 2} + \frac{3 \times 6}{1 \times 6}$

= $\frac{10}{6} – \frac{9}{6} – \frac{21}{6} + \frac{9}{3}$

= $\frac{ 10 – 9 – 14 + 18 }{ 6 }$ = $\frac{ 5 }{ 6}$

Q-2. Express each of the following as a rational number of the form $\frac{ p }{ q }$:

(i) $\frac{-8}{3} + \frac{-1}{4} + \frac{-11}{6} + \frac{3}{8} – 3$

(ii) $\frac{6}{7} + 1 + \frac{-7}{9} + \frac{19}{21} + \frac{-12}{7}$

(iii)  $\frac{15}{2} + \frac{9}{8} + \frac{-11}{3} + 6 + \frac{-7}{6}$

(iv) $\frac{-7}{4} + 0 + \frac{-9}{5} + \frac{19}{10} + \frac{11}{14}$

(v) $\frac{-7}{4} + \frac{5}{3} + \frac{-1}{2} + \frac{-5}{6} + 2$

Solution:

(i) $\frac{-8}{3} + \frac{-1}{4} + \frac{-11}{6} + \frac{3}{8} – 3$

= $\frac{-8}{3} – \frac{1}{4} – \frac{11}{6} + \frac{3}{8} – \frac{3}{1}$

= $\left (\frac{-8}{3} – \frac{11}{6} \right ) – \left (\frac{1}{4} -\frac{3}{8} \right ) – \frac{3}{1}$

= $\left (\frac{-8 \times 2}{3 \times 2} – \frac{11}{6} \right ) – \left (\frac{1 \times 2}{4 \times 2} -\frac{3}{8} \right ) – \frac{3}{1}$

= $\left (\frac{ -16 }{ 6 } – \frac{ 11 }{ 6 } \right ) – \left (\frac{ 2 }{ 8 } -\frac{ 3 }{ 8 } \right ) – \frac{ 3 }{ 1 }$

= $\frac{ -27 }{ 6 } – \frac{ -1 }{ 8 } – \frac{ 3 }{ 1 }$

= $\frac{ -27 \times 4}{6 \times 4 } – \frac{ -1 \times 3 }{ 8 \times 3 } – \frac{ 3\times 24 }{ 1 \times 24 }$

= $\frac{ -108 }{ 24 } + \frac{ 3 }{ 24 } – \frac{ 72 }{ 24 }$

= $\frac{ -108 + 3 – 72 }{ 24 }$

= $\frac{ -177 }{ 24 }$ = $\frac{ -59 }{ 8 }$

(ii) $\frac{ 6 }{ 7 } + 1 + \frac{ -7 }{ 9 } + \frac{ 19 }{ 21 } + \frac{ -12 }{ 7 }$

= $\frac{ 6 }{ 7 } + \frac{1}{1} – \frac{ 7 }{ 9 } + \frac{ 19 }{ 21 } – \frac{ 12 }{ 7 }$

= $\left ( \frac{ 6 }{ 7 } + \frac{ 19 }{ 21 } – \frac{ 12 }{ 7 } \right ) + \frac{ 1 }{ 1 } – \frac{ 7 }{ 9 }$

= $\left ( \frac{ 6 \times 3 }{ 7 \times 3 } + \frac{ 19 }{ 21 } – \frac{ 12 \times 3 }{ 7 \times 3 } \right ) + \frac{ 1 }{ 1 } – \frac{ 7 }{ 9 }$

= $\left ( \frac{ 18 }{ 21 } + \frac{ 19 }{ 21 } – \frac{ 36 }{ 21 } \right ) + \frac{ 1 }{ 1 } – \frac{ 7 }{ 9 }$

= $\left ( \frac{ 18 + 19 – 36 }{ 21 } \right ) + \frac{ 1 }{ 1 } – \frac{ 7 }{ 9 }$

= $\frac{ 1 }{ 21 } + \frac{ 1 }{ 1 } – \frac{ 7 }{ 9 }$

= $\frac{ 1 \times 3 }{ 21 \times 3 } + \frac{ 1 \times 63 }{ 1 \times 63 } – \frac{ 7 \times 7 }{ 9 \times 7 }$

= $\frac{ 3 }{ 63 } + \frac{ 63 }{ 63 } – \frac{ 49 }{ 63 }$

= $\frac{ 3 + 63 – 49 }{ 63 }$ = $\frac{ 17 }{ 63 }$

(iii) $\frac{ 15 }{ 2 } + \frac{ 9 }{ 8 } + \frac{ -11 }{ 3 } + 6 + \frac{ -7 }{ 6 }$

= $\frac{ 15 }{ 2 } + \frac{ 9 }{ 8 } – \frac{ 11 }{ 3 } + \frac{ 6 }{ 1 } – \frac{ 7 }{ 6 }$

= $\left (\frac{ 15 }{ 2 } + \frac{ 9 }{ 8 } \right ) – \left (\frac{ 11 }{ 3 } + \frac{ 7 }{ 6 } \right ) + \frac{ 6 }{ 1 }$

= $\left (\frac{15 \times 4}{2 \times 4} + \frac{ 9 }{ 8 } \right ) – \left (\frac{11 \times 2}{3 \times 2} + \frac{ 7 }{ 6 } \right ) + \frac{6}{1}$

= $\left (\frac{ 60 }{ 8 } + \frac{ 9 }{ 8 } \right ) – \left (\frac{ 22 }{ 6 } + \frac{ 7 }{ 6 } \right ) + \frac{ 6 }{ 1 }$

= $\frac{ 60 + 9 }{ 8 } – \frac{ 22 + 7 }{ 6 } + \frac{ 6 }{ 1 }$

= $\frac{ 69 }{ 8 } – \frac{ 29 }{ 6 } + \frac{ 6 }{ 1 }$

= $\frac{69 \times 3 }{ 8 \times 3 } – \frac{ 29 \times 4 }{ 6 \times 4 } + \frac{ 6 \times 24 }{ 1 \times 24 }$

= $\frac{ 207 }{ 24 } – \frac{ 116 }{ 24 } + \frac{ 144 }{ 24 }$

= $\frac{ 207 -116 + 144 }{ 24 }$ = $\frac{ 235 }{ 24 }$

(iv) $\frac{ -7 }{ 4 } + 0 + \frac{ -9 }{ 5 } + \frac{ 19 }{ 10 } + \frac{ 11 }{ 14 }$

= $\frac{ -7 }{ 4 } – \frac{ 9 }{ 5 } + \frac{ 19 }{ 10 } + \frac{ 11 }{ 14 }$

= $\frac{ -7 }{ 4 } + \frac{ 11 }{ 14 }- (\frac{ 9 }{ 5 } + \frac{ 19 }{ 10 } )$

= $\frac{-7}{4} + \frac{ 11 }{ 14 } ) – (\frac{9 \times 2}{5 \times 2} + \frac{ 19 }{ 10 } )$

= $\frac{ -7 }{ 4 } + \frac{ 11 }{ 14 } – \left (\frac{ 18 }{ 10 } + \frac{ 19 }{ 10 } \right )$

= $\frac{ -7 }{ 4 } + \frac{ 11 }{ 14 } – \frac{ 18 + 19 }{ 10 }$

= $\frac{ -7 }{ 4 } + \frac{ 11 }{ 14 } – \frac{ 37 }{ 10 }$

= $\frac{-7 \times 35 }{ 4 \times 35 } + \frac{ 11 \times 10 }{ 14 \times 10 } – \frac{ 37 \times 14 }{ 10 \times 14 }$

= $\frac{ 245 }{ 140 } + \frac{ 110 }{ 140 } – \frac{ 518 }{ 140 }$

= $\frac{ 245 + 110 – 518 }{ 140 }$ = $\frac{ – 121 }{ 24 }$

(v) $\frac{ -7 }{ 4 } + \frac{ 5 }{ 3 } + \frac{ -1 }{ 2 } + \frac{ -5 }{ 6 } + 2$

= $\frac{ -7 }{ 4 } + \frac{ 5 }{ 3 } – \frac{ 1 }{ 2 } – \frac{ 5 }{ 6 } + \frac{ 2 }{ 1 }$

= $\left (\frac{ -7 }{ 4 } – \frac{ 1 }{ 2 } \right ) + \left (\frac{ 5 }{ 3 } – \frac{ 5 }{ 6 } \right ) + \frac{2}{1}$

= $\left (\frac{-7 }{ 4 } – \frac{1 \times 2 }{2 \times 2} \right ) + \left (\frac{5 \times 2}{3 \times 2} -\frac{ 5 }{ 6 } \right ) + \frac{ 2 }{ 1 }$

= $\left (\frac{ -7 }{ 4 } – \frac{ 2 }{ 4 } \right ) + \left (\frac{ 10 }{ 6 } -\frac{ 5 }{ 6 } \right ) + \frac{ 2 }{ 1 }$

= $\frac{ -7 – 2 }{ 4 } + \frac{ 10 – 5 }{ 6 } + \frac{ 2 }{ 1 }$

= $\frac{ -9 }{ 4 } + \frac{ 5 }{ 6 } + \frac{ 2 }{ 1 }$

= $\frac{ -9 \times 3}{4 \times 3 } + \frac{ 5 \times 2 }{ 6 \times 2 } + \frac{ 2 \times 12 }{ 1 \times 12 }$

= $\frac{ -27 }{ 12 } + \frac{ 10 }{ 12 } + \frac{ 24 }{ 12 }$

= $\frac{ -27 + 10 + 24 }{ 12 }$ = $\frac{ 7}{ 12 }$

Q-3. Simplify:

(i) $\frac{-3}{2} + \frac{5}{4} – \frac{7}{4}$

(ii) $\frac{5}{3} – \frac{7}{6} + \frac{-2}{3}$

(iii) $\frac{5}{4} – \frac{7}{6} – \frac{-2}{3}$

(iv) $\frac{-2}{5} – \frac{-3}{10} – \frac{-4}{7}$

(v) $\frac{5}{6} + \frac{-2}{5} – \frac{-2}{15}$

(vi) $\frac{3}{8} – \frac{-2}{9} + \frac{-5}{36}$

Solution:

(i) $\frac{ -3 }{ 2 } + \frac{ 5 }{ 4 } – \frac{ 7 }{ 4 }$

Taking the LCM of the denominators:

$\frac{ -3 \times 2 }{ 2 \times 2 } + \frac{ 5 }{ 4 } – \frac{ 7 }{ 4 }$

= $\frac{ -6 }{ 4 } + \frac{ 5 }{ 4 } – \frac{ 7 }{ 4 }$

= $\frac{ -6 + 5 – 7 }{ 4 }$

= $\frac{ -8 }{ 4 }$ = -2

(ii) $\frac{ 5 }{ 3 } – \frac{ 7 }{ 6 } + \frac{ -2 }{ 3 }$

Taking the LCM of the denominators:

$\frac{ 5 \times 2 }{ 3 \times 2 } – \frac{ 7 }{ 6 } + \frac{ -2 \times 2 }{ 3 \times 2 }$

= $\frac{ 10 }{ 6 } – \frac{ 7 }{ 6 } – \frac{ 4 }{ 6 }$

= $\frac{ 10 – 7 – 4 }{ 6 }$ = $\frac{ -1 }{ 6 }$

(iii) $\frac{ 5 }{ 4 } – \frac{ 7 }{ 6 } – \frac{ -2 }{ 3 }$

Taking the LCM of the denominators:

$\frac{ 5 \times 3 }{ 4 \times 3 } – \frac{ 7 \times 2 }{ 6 \times 2 } + \frac{ 2 \times 4 }{ 3 \times 4 }$

= $\frac{ 15 }{ 12 } – \frac{ 14 }{ 12 } + \frac{ 8 }{ 12 }$

= $\frac{ 15 – 14 – 8 }{ 12 }$ = $\frac{ -5 }{ 12 }$

(iv) $\frac{ -2 }{ 5 } – \frac{ -3 }{ 10 } – \frac{ -4 }{ 7 }$

Taking the LCM of the denominators:

$\frac{ -2 \times 14 }{ 5 \times 14 } + \frac{ 3 \times 7 }{ 10 \times 7 } + \frac{ 4 \times 10 }{ 7 \times 10 }$

= $\frac{ -28 }{ 70 } + \frac{ 21 }{ 70 } + \frac{ 40 }{ 70 }$

= $\frac{ -28 + 21 + 40 }{ 70 }$ = $\frac{ 33 }{ 70 }$

(v) $\frac{ 5 }{ 6 } + \frac{ -2 }{ 5 } – \frac{ -2 }{ 15 }$

Taking the LCM of the denominators:

$\frac{ 5 \times 5 }{ 6 \times 5 } – \frac{ 2 \times 6 }{ 5 \times 6 } + \frac{ 2 \times 2 }{ 15 \times 2 }$

= $\frac{ 25 }{ 30 } – \frac{ 12 }{ 30 } + \frac{ 4 }{ 30 }$

= $\frac{ 25 – 12 + 4 }{ 30 }$ = $\frac{ 17 }{ 30 }$

(vi) $\frac{ 3 }{ 8 } – \frac{ -2 }{ 9 } + \frac{ -5 }{ 36 }$

Taking the LCM of the denominators:

$\frac{ 3 \times 9 }{ 8 \times 9 } + \frac{ 2 \times 8 }{ 9 \times 6 } – \frac{ 5 \times 2 }{ 36 \times 2 }$

= $\frac{ 27 }{ 72 } + \frac{ 16 }{ 72 } – \frac{ 10 }{ 72 }$

= $\frac{ 27 + 16 – 10 }{ 72 }$

= $\frac{ 33 }{ 72 }$ = $\frac{ 11 }{ 24 }$

#### Practise This Question

In a swimming race 3 swimmers compete . The probability of A and B wining is same and twice that of C.What is the probability that B or C wins. Assuming no two finish the race at the same time.