Selina solutions

Selina solutions

Grade 7

Rational Numbers | Exercise 2(E)

Question 1

Q1) Evaluate:

(i) \frac{-2}{3}+\frac{3}{4}

(ii) \frac{7}{-27}+\frac{11}{18}

(iii) \frac{-3}{8}+\frac{-5}{12}

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

(v) \frac{-5}{9}+\frac{-7}{12}+\frac{11}{18}

(vi) \frac{7}{-26}+\frac{16}{39}

(vii) -\frac{2}{3}-\left(\frac{-5}{7}\right)

(viii) -\frac{5}{7}-\left(-\frac{3}{8}\right)

(ix) \frac{7}{26}+2+\frac{-11}{13}

(x) -1+\frac{2}{-3}+\frac{5}{6}

Solution 1:

(i) \frac{-2}{3}+\frac{3}{4}

LCM of 3 and 4 is 12.

\frac{-2\times4}{3\times4}=\frac{-8}{12},\frac{3\times3}{4\times3}=\frac{9}{12}

= \frac{-8+9}{12}=\frac{1}{12}

(ii) \frac{7}{-27}+\frac{11}{18}

LCM of 27 and 18 is 54.

\frac{-7\times2}{27\times2}=\frac{-14}{54},\ \frac{11\times3}{18\times3}=\frac{33}{54}

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

(iii) \frac{-3}{8}+\frac{-5}{12}

LCM of 8 and 12 is 24.

= \frac{-3\times3}{8\times3}=\frac{-9}{24},\ \frac{-5\times2}{12\times2}=\frac{-10}{24}

= \frac{-9+\left(-10\right)}{24}=\frac{-19}{24}

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

= \frac{-9}{16}+\frac{5}{12}

LCM of 16 and 12 is 48.

\frac{-9\times3}{16\times3}=\frac{-27}{48},\ \frac{5\times4}{12\times4}=\frac{20}{48}

= \frac{-27+20}{48}=\frac{-7}{48}

(v) \frac{-5}{9}+\frac{-7}{12}+\frac{11}{18}

LCM of 9, 12 and 18 is 36.

\frac{-5\times4}{9\times4}=\frac{-20}{36},\ \frac{-7\times3}{12\times3}=\frac{-21}{36},\ \frac{11\times2}{18\times2}=\frac{22}{36}

= \frac{-20-21+22}{36}=\frac{-19}{36}

(vi) \frac{7}{-26}+\frac{16}{39}

= \frac{-7}{26}+\frac{16}{39}

LCM of 26 and 39 is 78

\frac{-7\times3}{26\times3}=\frac{-21}{78},\ \frac{16\times2}{39\times2}=\frac{32}{78}

= \frac{-21+32}{26}=\frac{11}{78}

(vii) -\frac{2}{3}-\left(\frac{-5}{7}\right)

\frac{-2}{3}-\frac{-5}{7}

LCM of 3 and 7 is 21

= \frac{-2\times7-\left(-5\times3\right)}{21}=\frac{-14-\left(-15\right)}{21}=\frac{-14+15}{21}=\frac{1}{21}

(viii) -\frac{5}{7}-\left(-\frac{3}{8}\right)

LCM of 7 and 8 is 56.

= \frac{-5\times8-\left(-3\times7\right)}{56}=\frac{-40-\left(-21\right)}{56}=\frac{-40+21}{56}=\frac{-19}{56}

(ix) \frac{7}{26}+2+\frac{-11}{13}

LCM of 13, 26 is 26.

= \frac{7+\left(2\times26\right)+\left(-11\times2\right)}{26}=\frac{7+52-22}{26}=\frac{37}{26}

(x) -1+\frac{2}{-3}+\frac{5}{6}

= \frac{-1}{1}+\frac{-2}{3}+\frac{5}{6}

LCM of 1, 3 and 6 is 6.

= \frac{\left(-1\times6\right)+\left(-2\times2\right)+5}{6}=\frac{-6-4+5}{6}=\frac{-5}{6}

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

Learn from an expert tutor.

Book a free class now
subject-cta

Question 1

Q1) Evaluate:

(i) \frac{-2}{3}+\frac{3}{4}

(ii) \frac{7}{-27}+\frac{11}{18}

(iii) \frac{-3}{8}+\frac{-5}{12}

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

(v) \frac{-5}{9}+\frac{-7}{12}+\frac{11}{18}

(vi) \frac{7}{-26}+\frac{16}{39}

(vii) -\frac{2}{3}-\left(\frac{-5}{7}\right)

(viii) -\frac{5}{7}-\left(-\frac{3}{8}\right)

(ix) \frac{7}{26}+2+\frac{-11}{13}

(x) -1+\frac{2}{-3}+\frac{5}{6}

Solution 1:

(i) \frac{-2}{3}+\frac{3}{4}

LCM of 3 and 4 is 12.

\frac{-2\times4}{3\times4}=\frac{-8}{12},\frac{3\times3}{4\times3}=\frac{9}{12}

= \frac{-8+9}{12}=\frac{1}{12}

(ii) \frac{7}{-27}+\frac{11}{18}

LCM of 27 and 18 is 54.

\frac{-7\times2}{27\times2}=\frac{-14}{54},\ \frac{11\times3}{18\times3}=\frac{33}{54}

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

(iii) \frac{-3}{8}+\frac{-5}{12}

LCM of 8 and 12 is 24.

= \frac{-3\times3}{8\times3}=\frac{-9}{24},\ \frac{-5\times2}{12\times2}=\frac{-10}{24}

= \frac{-9+\left(-10\right)}{24}=\frac{-19}{24}

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

= \frac{-9}{16}+\frac{5}{12}

LCM of 16 and 12 is 48.

\frac{-9\times3}{16\times3}=\frac{-27}{48},\ \frac{5\times4}{12\times4}=\frac{20}{48}

= \frac{-27+20}{48}=\frac{-7}{48}

(v) \frac{-5}{9}+\frac{-7}{12}+\frac{11}{18}

LCM of 9, 12 and 18 is 36.

\frac{-5\times4}{9\times4}=\frac{-20}{36},\ \frac{-7\times3}{12\times3}=\frac{-21}{36},\ \frac{11\times2}{18\times2}=\frac{22}{36}

= \frac{-20-21+22}{36}=\frac{-19}{36}

(vi) \frac{7}{-26}+\frac{16}{39}

= \frac{-7}{26}+\frac{16}{39}

LCM of 26 and 39 is 78

\frac{-7\times3}{26\times3}=\frac{-21}{78},\ \frac{16\times2}{39\times2}=\frac{32}{78}

= \frac{-21+32}{26}=\frac{11}{78}

(vii) -\frac{2}{3}-\left(\frac{-5}{7}\right)

\frac{-2}{3}-\frac{-5}{7}

LCM of 3 and 7 is 21

= \frac{-2\times7-\left(-5\times3\right)}{21}=\frac{-14-\left(-15\right)}{21}=\frac{-14+15}{21}=\frac{1}{21}

(viii) -\frac{5}{7}-\left(-\frac{3}{8}\right)

LCM of 7 and 8 is 56.

= \frac{-5\times8-\left(-3\times7\right)}{56}=\frac{-40-\left(-21\right)}{56}=\frac{-40+21}{56}=\frac{-19}{56}

(ix) \frac{7}{26}+2+\frac{-11}{13}

LCM of 13, 26 is 26.

= \frac{7+\left(2\times26\right)+\left(-11\times2\right)}{26}=\frac{7+52-22}{26}=\frac{37}{26}

(x) -1+\frac{2}{-3}+\frac{5}{6}

= \frac{-1}{1}+\frac{-2}{3}+\frac{5}{6}

LCM of 1, 3 and 6 is 6.

= \frac{\left(-1\times6\right)+\left(-2\times2\right)+5}{6}=\frac{-6-4+5}{6}=\frac{-5}{6}

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

Learn from an expert tutor.

Book a free class now
subject-cta
linkedin instgram facebook youtube
2020 © Quality Tutorials Pvt Ltd All rights reserved
linkedin instgram facebook youtube