NCERT Solutions Class 7 Maths Chapter 8 Rational Numbers Exercise 8.2

 

NCERT Solutions Class 7 Maths Chapter 8 Rational Numbers Exercise 8.2

 

Introduction:

NCERT Class 7 Maths Chapter 8 Rational Number Exercise 8.1

NCERT Class 7 Maths Chapter 8 Rational Number Exercise 8.2

Class 7 Maths Exercise 8.1 (Page-141)

 

Q1. Find the sum:

(i) \(\displaystyle \frac{{5}}{4}\) + (\(\displaystyle \frac{{-11}}{4}\))

(ii) \(\displaystyle \frac{{5}}{3}\) + \(\displaystyle \frac{{3}}{5}\)

(iii) (\(\displaystyle \frac{{-9}}{10}\)) + \(\displaystyle \frac{{22}}{15}\)

(iv) \(\displaystyle \frac{{-3}}{-11}\) + \(\displaystyle \frac{{5}}{9}\)

(v) \(\displaystyle \frac{{-8}}{19}\) + \(\displaystyle \frac{{-2}}{57}\)

(vi) \(\displaystyle \frac{{-2}}{3}\) + 0

(vii)  -2 \(\displaystyle \frac{{1}}{3}\) +  4 \(\displaystyle \frac{{3}}{5}\)

Solution: 

(i) \(\displaystyle \frac{{5}}{4}\) + (\(\displaystyle \frac{{-11}}{4}\))

= \(\displaystyle \frac{{5}}{4}\) \(\displaystyle \frac{{-11}}{4}\)

= \(\displaystyle \frac{{-6}}{4}\)

= \(\displaystyle \frac{{-3}}{2}\)

 

(ii)  \(\displaystyle \frac{{5}}{3}\) + \(\displaystyle \frac{{3}}{5}\)

Taking L.C.M to make denominator same

= \(\displaystyle \frac{{5×5}}{3×5}\) + \(\displaystyle \frac{{3×3}}{5×3}\)

= \(\displaystyle \frac{{25}}{15}\) + \(\displaystyle \frac{{9}}{15}\)

= \(\displaystyle \frac{{34}}{15}\)

 

(iii)  (\(\displaystyle \frac{{-9}}{10}\)) + \(\displaystyle \frac{{22}}{15}\)

Taking L.C.M to make denominator same

  = (\(\displaystyle \frac{{-9×3}}{10×3}\)) + \(\displaystyle \frac{{22×2}}{15×2}\)

  = (\(\displaystyle \frac{{-27}}{30}\)) + \(\displaystyle \frac{{44}}{30}\)

= \(\displaystyle \frac{{17}}{30}\)

 

(iv) \(\displaystyle \frac{{-3}}{-11}\) + \(\displaystyle \frac{{5}}{9}\)

= \(\displaystyle \frac{{3}}{11}\) + \(\displaystyle \frac{{5}}{9}\)

Taking L.C.M to make denominator same

= \(\displaystyle \frac{{3×9}}{11×9}\) + \(\displaystyle \frac{{5×11}}{9×11}\)

= \(\displaystyle \frac{{27}}{99}\) + \(\displaystyle \frac{{55}}{99}\)

= \(\displaystyle \frac{{82}}{99}\)

 

(v)  \(\displaystyle \frac{{-8}}{19}\) + \(\displaystyle \frac{{-2}}{57}\)

= \(\displaystyle \frac{{-8}}{19}\)  \(\displaystyle \frac{{-2}}{57}\)

Taking L.C.M to make denominator same

= \(\displaystyle \frac{{-8×3}}{19×3}\)  \(\displaystyle \frac{{-2×1}}{57×1}\)

= \(\displaystyle \frac{{-24}}{57}\)  \(\displaystyle \frac{{-2}}{57}\)

= \(\displaystyle \frac{{-26}}{57}\)

 

(vi) \(\displaystyle \frac{{-2}}{3}\) + 0

=  \(\displaystyle \frac{{-2}}{3}\)

 

(vii)   -2 \(\displaystyle \frac{{1}}{3}\) +  4 \(\displaystyle \frac{{3}}{5}\)

converting the given mixed fraction into improper fraction

= \(\displaystyle \frac{{-7}}{3}\)+ \(\displaystyle \frac{{23}}{5}\)

Taking L.C.M to make denominator same

= \(\displaystyle \frac{{-7×5}}{3×5}\) + \(\displaystyle \frac{{23×3}}{5×3}\)

= \(\displaystyle \frac{{-35}}{15}\) + \(\displaystyle \frac{{69}}{15}\)

= \(\displaystyle \frac{{34}}{15}\)

 

Q2. Find:

(i) \(\displaystyle \frac{{7}}{24}\) - \(\displaystyle \frac{{17}}{36}\)

(ii) \(\displaystyle \frac{{5}}{63}\) - (\(\displaystyle \frac{{-6}}{21}\))

(iii) \(\displaystyle \frac{{-6}}{13}\) - (\(\displaystyle \frac{{-7}}{15}\))

(iv) \(\displaystyle \frac{{-3}}{8}\) - \(\displaystyle \frac{{7}}{11}\)

(v)  -2 \(\displaystyle \frac{{1}}{9}\) - 6

Solution: 

(i) \(\displaystyle \frac{{7}}{24}\) - \(\displaystyle \frac{{17}}{36}\)

Taking L.C.M to make denominator same

= \(\displaystyle \frac{{7×3}}{24×3}\) - \(\displaystyle \frac{{17×2}}{36×2}\)

= \(\displaystyle \frac{{21}}{72}\) - \(\displaystyle \frac{{34}}{72}\)

= - \(\displaystyle \frac{{13}}{72}\)

 

(ii) \(\displaystyle \frac{{5}}{63}\) - (\(\displaystyle \frac{{-6}}{21}\))

Taking L.C.M to make denominator same

= \(\displaystyle \frac{{5×1}}{63×1}\) - \(\displaystyle \frac{{-6×3}}{21×3}\)

= \(\displaystyle \frac{{5}}{63}\) - (\(\displaystyle \frac{{-18}}{63}\))

= \(\displaystyle \frac{{5}}{63}\) + \(\displaystyle \frac{{18}}{63}\)

= \(\displaystyle \frac{{23}}{63}\)

 

(iii) \(\displaystyle \frac{{-6}}{13}\) - (\(\displaystyle \frac{{-7}}{15}\))

Taking L.C.M to make denominator same

= \(\displaystyle \frac{{-6×15}}{13×15}\) - \(\displaystyle \frac{{-7×13}}{15×13}\)

= \(\displaystyle \frac{{-90}}{195}\) - (\(\displaystyle \frac{{-91}}{195}\))

= \(\displaystyle \frac{{-90}}{195}\) + \(\displaystyle \frac{{91}}{195}\)

= \(\displaystyle \frac{{1}}{195}\)

 

(iv) \(\displaystyle \frac{{-3}}{8}\) - \(\displaystyle \frac{{7}}{11}\)

Taking L.C.M to make denominator same

= \(\displaystyle \frac{{-3×11}}{8×11}\) - \(\displaystyle \frac{{7×8}}{11×8}\)

= \(\displaystyle \frac{{-33}}{88}\) - \(\displaystyle \frac{{56}}{88}\)

= \(\displaystyle \frac{{-89}}{88}\)

 

(v) -2 \(\displaystyle \frac{{1}}{9}\) - 6

Converting mixed fraction into improper fraction

= \(\displaystyle \frac{{-19}}{9}\) - 6

= \(\displaystyle \frac{{-19}}{9}\) - \(\displaystyle \frac{{6}}{1}\)

= \(\displaystyle \frac{{-19×1}}{9×1}\) - \(\displaystyle \frac{{6×9}}{1×9}\)

= \(\displaystyle \frac{{-19}}{9}\) - \(\displaystyle \frac{{54}}{9}\)

= \(\displaystyle \frac{{-73}}{9}\)

 

Q3. Find the product:

(i) \(\displaystyle \frac{{9}}{2}\) × (\(\displaystyle \frac{{-7}}{4}\))

(ii) \(\displaystyle \frac{{3}}{10}\) × (-9)

(iii) \(\displaystyle \frac{{-6}}{5}\) × \(\displaystyle \frac{{9}}{11}\)

(iv) \(\displaystyle \frac{{3}}{7}\) × ( \(\displaystyle \frac{{-2}}{5}\))

(v) \(\displaystyle \frac{{3}}{11}\) × \(\displaystyle \frac{{2}}{5}\)

(vi) \(\displaystyle \frac{{3}}{-5}\) × \(\displaystyle \frac{{-5}}{3}\)

Solution: 

(i) \(\displaystyle \frac{{9}}{2}\) × (\(\displaystyle \frac{{-7}}{4}\))

= \(\displaystyle \frac{{-63}}{8}\)

 

(ii)  \(\displaystyle \frac{{3}}{10}\) × (-9)

= \(\displaystyle \frac{{3×-9}}{10}\)

= \(\displaystyle \frac{{-27}}{10}\)

 

(iii) \(\displaystyle \frac{{-6}}{5}\) × \(\displaystyle \frac{{9}}{11}\)

= \(\displaystyle \frac{{-6×9}}{5×11}\)

= \(\displaystyle \frac{{-54}}{55}\)

 

(iv) \(\displaystyle \frac{{3}}{7}\) × ( \(\displaystyle \frac{{-2}}{5}\))

= \(\displaystyle \frac{{3×-2}}{7×5}\)

= \(\displaystyle \frac{{-6}}{35}\)

 

(v) \(\displaystyle \frac{{3}}{11}\) × \(\displaystyle \frac{{2}}{5}\)

= \(\displaystyle \frac{{3×2}}{11×5}\)

= \(\displaystyle \frac{{6}}{55}\)

 

(vi)  \(\displaystyle \frac{{3}}{-5}\) × \(\displaystyle \frac{{-5}}{3}\)

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

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

= 1

Q4. Find the value of :

(i) (-4) ÷ \(\displaystyle \frac{{2}}{3}\)

(ii) \(\displaystyle \frac{{-3}}{5}\) ÷ 2

(iii) \(\displaystyle \frac{{-4}}{5}\) ÷ (-3)

(iv) \(\displaystyle \frac{{-1}}{8}\) ÷ \(\displaystyle \frac{{3}}{4}\)

(v) \(\displaystyle \frac{{-2}}{13}\) ÷ \(\displaystyle \frac{{1}}{7}\)

(vi) \(\displaystyle \frac{{-7}}{12}\) ÷ (\(\displaystyle \frac{{-2}}{13}\))

(vii) \(\displaystyle \frac{{3}}{13}\) ÷ (\(\displaystyle \frac{{-4}}{65}\))

Solution: 

(i)  (-4) ÷ \(\displaystyle \frac{{2}}{3}\)

= (-4) × \(\displaystyle \frac{{3}}{2}\)

= \(\displaystyle \frac{{-4×3}}{2}\)

= \(\displaystyle \frac{{-12}}{2}\)

= -6

(ii) \(\displaystyle \frac{{-3}}{5}\) ÷ 2

= \(\displaystyle \frac{{-3}}{5}\) × \(\displaystyle \frac{{1}}{2}\)

= \(\displaystyle \frac{{-3×1}}{5×2}\)

= \(\displaystyle \frac{{-3}}{10}\)

 

(iii)  \(\displaystyle \frac{{-4}}{5}\) ÷ (-3)

= \(\displaystyle \frac{{-4}}{5}\) × \(\displaystyle \frac{{1}}{-3}\)

= \(\displaystyle \frac{{-4×1}}{5×-3}\)

= \(\displaystyle \frac{{-4}}{-15}\)

= \(\displaystyle \frac{{4}}{15}\)

 

(iv) \(\displaystyle \frac{{-1}}{8}\) ÷ \(\displaystyle \frac{{3}}{4}\)

= \(\displaystyle \frac{{-1}}{8}\) × \(\displaystyle \frac{{4}}{3}\)

= \(\displaystyle \frac{{-1×4}}{8×3}\)

= \(\displaystyle \frac{{-4}}{24}\)

= \(\displaystyle \frac{{-1}}{6}\)

 

(v) \(\displaystyle \frac{{-2}}{13}\) ÷ \(\displaystyle \frac{{1}}{7}\)

= \(\displaystyle \frac{{-2}}{13}\) × \(\displaystyle \frac{{7}}{1}\)

= \(\displaystyle \frac{{-14}}{13}\)

 

(vi) \(\displaystyle \frac{{-7}}{12}\) ÷ (\(\displaystyle \frac{{-2}}{13}\))

= \(\displaystyle \frac{{-7}}{12}\) × (\(\displaystyle \frac{{13}}{-2}\))

= \(\displaystyle \frac{{-7×13}}{12×-2}\)

= \(\displaystyle \frac{{-91}}{-24}\)

= \(\displaystyle \frac{{91}}{24}\)

 

(vii) \(\displaystyle \frac{{3}}{13}\) ÷ (\(\displaystyle \frac{{-4}}{65}\))

= \(\displaystyle \frac{{3}}{13}\) × (\(\displaystyle \frac{{65}}{-4}\))

= \(\displaystyle \frac{{3×65}}{13×-4}\)

= \(\displaystyle \frac{3}{{-4}}\) × \(\displaystyle \frac{65}{{13}}\)

= \(\displaystyle \frac{3}{{-4}}\) × 5

= \(\displaystyle \frac{15}{{-4}}\)

= \(\displaystyle \frac{-15}{{4}}\)