RD Sharma Solutions Class 8 Rational Numbers Exercise 1.4

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

RD Sharma Solutions Class 7 Chapter 4 Exercise 4.1

Exercise 1.4

Q-1. Simplify each of the following and write as a rational number of the form 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) Given:  \(\frac{3}{4} + \frac{5}{6} + \frac{-7}{8}\)

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

Take the LCM of denominators for the given rational numbers,

= \(\frac{3 \times 6}{4 \times 6} + \frac{5 \times 4}{6 \times 4} – \frac{7 \times 3}{8 \times 3}\)  [Since, the LCM of 4,6, and 8 is 24]

= \(\frac{18}{24} + \frac{20}{24} – \frac{21}{24}\)

= \(\frac{ 18 + 20 – 21 }{ 24 }\)

Therefore,  \(\frac{3}{4} + \frac{5}{6} + \frac{-7}{8}\)= \(\frac{ 17}{ 24 }\)

 

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

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

Take the LCM of denominators for the given rational numbers,

= \(\frac{2 \times 6}{3 \times 6} – \frac{5 \times 3}{6 \times 3} – \frac{7 \times 2}{9 \times 2}\) [Since, the LCM of 3,6, and 9 is 18]

= \(\frac{12}{18} – \frac{15}{18} – \frac{14}{18}\)

= \(\frac{ 12 – 15 – 14  }{ 18 }\)

Therefore, \(\frac{ 2 }{ 3 } + \frac{ -5 }{ 6 } + \frac{ -7 }{ 9 }\)= \(\frac{ -17}{  18  }\)

 

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

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

Take the LCM of denominators for the given rational numbers,

= \(\frac{-11 \times  12}{2 \times 12}  +  \frac{7 \times 4}{6 \times 4} – \frac{5 \times 3}{8 \times 3}\) [Since, the LCM of 2,6, and 8 is 24]

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

= \(\frac{ -132 + 28 – 15  }{ 24 }\)

Therefore, \(\frac{ -11 }{ 2 } + \frac{ 7 }{ 6 } + \frac{ -5 }{ 8 }\)= \(\frac{ -119 }{  24  }\)

 

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

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

Take the LCM of denominators for the given rational numbers,

= \(\frac{-4 \times  6}{5 \times 6}  –  \frac{7 \times 3}{10 \times 3} – \frac{8 \times 2}{15 \times 2}\) [Since, the LCM of 5,10, and 15 is 30]

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

= \(\frac{ -24 – 21 – 16 }{ 30 }\)

Therefore, \(\frac{ -4 }{ 5 } + \frac{ -7 }{ 10 } + \frac{ -8 }{ 15 }\)= \(\frac{ -61 }{  30  }\)

 

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

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

Take the LCM of denominators for the given rational numbers,

= \(\frac{-9 \times  6}{10 \times 6}  +  \frac{22 \times 4}{15 \times 4} – \frac{13 \times 3}{20 \times 3}\) [Since, the LCM of 10, 15, and 20 is 60]

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

= \(\frac{ -54 + 88 – 39}{ 60 }\) = \(\frac{ -5}{60}\)

Therefore, \(\frac{ -9 }{ 10 } + \frac{ 22 }{ 15 } + \frac{ 13 }{ -20 }\)= \( \frac{-1}{60}\)

 

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

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

Take the LCM of denominators for the given rational numbers,

= \(\frac{5  \times  2}{3 \times 2}  –  \frac{3 \times 3}{2 \times 3} – \frac{7 \times 2}{3 \times 2} + \frac{3 \times 6}{1 \times 6} \) [Since, the LCM of 2 and 3 is 6]

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

= \(\frac{ 10 – 9 – 14 + 18  }{ 6 }\)

Therefore,  \(\frac{ 5 }{ 3 }\) + \(\frac{ 3 }{ -2 }\) + \(\frac{ -7 }{ 3 } + 3\) = \(\frac{ 5 }{ 6}\)

 

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) Given: \( \frac{-8}{3} + \frac{-1}{4} + \frac{-11}{6} + \frac{3}{8} – 3 \)

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

Reaarange the given rational numbers in such a way that the same or multiples of a numbers in the denominator is nearby.

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

Take the LCM of denominators for the given rational numbers inside the brackets,

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

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

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

Again, take the LCM of denominators for the given rational numbers,

= \(\frac{ -27 \times 4}{6 \times 4 } – \frac{ -1 \times 3 }{ 8 \times 3 } – \frac{ 3\times 24 }{ 1 \times 24 }\) [Since, the LCM of 6, 8 and 1 is 24]

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

= \(\frac{ -108 + 3 – 72 }{ 24 }\)

= \(\frac{ -177 }{ 24 }\)

Therefore, \( \frac{-8}{3} + \frac{-1}{4} + \frac{-11}{6} + \frac{3}{8} – 3 \)= \(\frac{ -59 }{ 8 }\)

 

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

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

Reaarange the given rational numbers in such a way that the same or multiples of a numbers in the denominator is nearby.

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

Take the LCM of denominators for the given rational numbers inside the brackets,

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

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

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

= \(  \frac{ 1 }{ 21 } + \frac{ 1 }{ 1 } – \frac{ 7 }{ 9 }\)

Again, take the LCM of denominators for the given rational numbers,

= \(\frac{ 1 \times 3 }{ 21 \times 3 } + \frac{ 1 \times 63 }{ 1 \times 63 } – \frac{ 7 \times 7 }{ 9 \times 7 }\) [Since, the LCM of 21, 1 and 1 is 63]

= \(\frac{ 3 }{ 63 } + \frac{ 63 }{ 63 } – \frac{ 49 }{ 63 }\)

= \(\frac{ 3 + 63 – 49 }{ 63 }\)

Therefore, \( \frac{ 6 }{ 7 } + 1 + \frac{ -7 }{ 9 } + \frac{ 19 }{ 21 } + \frac{ -12 }{ 7 }  \)= \(\frac{ 17 }{ 63 }\)

 

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

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

Reaarange the given rational numbers in such a way that the same or multiples of a numbers in the denominator is nearby.

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

Take the LCM of denominators for the given rational numbers inside the brackets,

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

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

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

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

Again, take the LCM of denominators for the given rational numbers,

= \(\frac{69 \times 3 }{ 8 \times 3 } – \frac{ 29 \times 4 }{ 6 \times 4 } + \frac{ 6 \times 24 }{ 1 \times 24 }\) [Since, the LCM of 6, and 8  is 24]

= \(\frac{ 207 }{ 24 } – \frac{ 116 }{ 24 } + \frac{ 144 }{ 24 }\)

= \(\frac{ 207 -116 + 144 }{ 24 }\)

Therefore, \( \frac{ 15 }{ 2 }  + \frac{ 9 }{ 8 } + \frac{ -11 }{ 3 } + 6 + \frac{ -7 }{ 6 }  \)= \(\frac{ 235 }{ 24 }\)

 

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

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

Reaarange the given rational numbers in such a way that the same or multiples of a numbers in the denominator is nearby.

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

Take the LCM of denominators for the given rational numbers inside the brackets,

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

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

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

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

Again, take the LCM of denominators for the given rational numbers,

= \(\frac{-7 \times 35 }{ 4 \times 35 } + \frac{ 11 \times 10 }{ 14 \times 10 } – \frac{ 37 \times 14 }{ 10 \times 14 } \) [Since, the LCM of 4, 14 and 10 is 140]

= \(\frac{ 245 }{ 140 }  + \frac{ 110 }{ 140 } – \frac{ 518 }{ 140 } \)

= \(\frac{ 245 + 110 – 518 }{ 140 }\)

Therefore, \( \frac{ -7 }{ 4 }  + 0 +  \frac{ -9 }{ 5 } + \frac{ 19 }{ 10 } + \frac{ 11 }{ 14 }  \)= \(\frac{ – 121 }{ 24 }\)

 

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

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

Reaarange the given rational numbers in such a way that the same or multiples of a numbers in the denominator is nearby.

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

Take the LCM of denominators for the given rational numbers inside the brackets,

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

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

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

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

Again, take the LCM of denominators for the given rational numbers,

= \(\frac{ -9 \times 3}{4 \times 3 } + \frac{ 5 \times 2 }{ 6 \times 2 } + \frac{ 2 \times 12 }{ 1 \times 12 }\) [Since, the LCM of 4, 6 and 1 is 12]

= \(\frac{ -27 }{ 12 } + \frac{ 10 }{ 12 } + \frac{ 24 }{ 12 }\)

= \(\frac{ -27 + 10 + 24 }{ 12 }\)

Therefore, \( \frac{ -7 }{ 4 }  +  \frac{ 5 }{ 3 } + \frac{ -1 }{ 2 } + \frac{ -5 }{ 6 } + 2 \)= \(\frac{ 7}{ 12 }\)

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) Given: \(\frac{ -3 }{ 2 } + \frac{ 5 }{ 4 } – \frac{ 7 }{ 4 }\)

Take the LCM of denominators for the given rational numbers,

Since, the LCM of 2 and 4 is 4

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

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

= \(\frac{ -6 + 5 – 7 }{ 4 }\)

Therefore, \(\frac{ -3 }{ 2 } + \frac{ 5 }{ 4 } – \frac{ 7 }{ 4 }\) = \(\frac{ -8 }{ 4 }\) = -2

 

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

Take the LCM of denominators for the given rational numbers,

Since, the LCM of 3 and 6 is 6

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

= \(\frac{ 10 }{ 6 } – \frac{ 7 }{ 6 } – \frac{ 4 }{ 6 }\)

= \(\frac{ 10 – 7 – 4  }{ 6 }\)

Therefore, \(\frac{ 5 }{ 3 } – \frac{ 7 }{ 6 } +  \frac{ -2 }{ 3 }\)= \(\frac{ -1 }{ 6 }\)

 

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

Take the LCM of denominators for the given rational numbers,

Since, the LCM of 3, 4 and 6 is 12

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

= \(\frac{ 15 }{ 12 } – \frac{ 14 }{ 12 } + \frac{ 8 }{ 12 }\)

= \(\frac{ 15 – 14 – 8  }{ 12 }\)

Therefore,  \(\frac{ 5 }{ 4 } – \frac{ 7 }{ 6 } –  \frac{ -2 }{ 3 }\) = \(\frac{ -5 }{ 12 }\)

 

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

Take the LCM of denominators for the given rational numbers,

Since, the LCM of 5, 7 and 10 is 70

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

= \(\frac{ -28 }{ 70 } + \frac{ 21 }{ 70 } + \frac{ 40 }{ 70 }\)

= \(\frac{ -28 + 21 + 40 }{ 70 }\)

Therefore, \(\frac{ -2 }{ 5 } – \frac{ -3 }{ 10 } –  \frac{ -4 }{ 7 }\) = \(\frac{ 33 }{ 70 }\)

 

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

Take the LCM of denominators for the given rational numbers,

Since, the LCM of 5, 6, and 15 is 30

\(\frac{ 5 \times 5 }{ 6 \times 5 } – \frac{ 2 \times 6 }{ 5 \times 6 } +  \frac{ 2 \times 2 }{ 15 \times 2 }\)

= \(\frac{ 25 }{ 30 } – \frac{ 12 }{ 30 } + \frac{ 4 }{ 30 }\)

= \(\frac{ 25 – 12 + 4 }{ 30 }\)

Therefore, \(\frac{ 5 }{ 6 } + \frac{ -2 }{ 5 } –  \frac{ -2 }{ 15 }\) = \(\frac{ 17 }{ 30 }\)

 

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

Take the LCM of denominators for the given rational numbers,

Since, the LCM of 8, 9 and 36 is 72

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

= \(\frac{ 27 }{ 72 } + \frac{ 16 }{ 72 } – \frac{ 10 }{ 72 }\)

= \(\frac{ 27 + 16 – 10 }{ 72 }\)

= \(\frac{ 33 }{ 72 }\)

Therefore,  \(\frac{ 3 }{ 8 } – \frac{ -2 }{ 9 } +  \frac{ -5 }{ 36 }\)= \( \frac{ 11 }{ 24 }\)

Leave a Comment

Your email address will not be published. Required fields are marked *