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