Ncert solutions

Ncert solutions

Grade 6

Fractions | Exercise 7.6

Question 1

Q1) Solve:

(a) \frac{2}{3}+\frac{1}{7} (b) \frac{3}{10}+\frac{7}{15} (c) \frac{4}{9}+\frac{2}{7} (d) \frac{5}{7}+\frac{1}{3}

(e) \frac{2}{5}+\frac{1}{6} (f) \frac{4}{5}+\frac{2}{3} (g) \frac{3}{4}-\frac{1}{3} (h) \frac{5}{6}-\frac{1}{3}

(i) \frac{2}{3}+\frac{3}{4}+\frac{1}{2} (j) \frac{1}{2}+\frac{1}{3}+\frac{1}{6} (k) 1\frac{1}{3}+3\frac{2}{3}

(l) 4\frac{2}{3}+3\frac{1}{4} (m) \frac{16}{5}-\frac{7}{5} (n) \frac{4}{3}-\frac{1}{2}

Solution 1:

(a) \frac{2}{3}+\frac{1}{7}

LCM of 3 and 7 is 21.

= \frac{\left(2\times7\right)+\left(1\times3\right)}{21}

= \frac{14+3}{21}

= \frac{17}{21}

(b) \frac{3}{10}+\frac{7}{15}

LCM of 10 and 15 is 30.

= \frac{\left(3\times3\right)+\left(7\times2\right)}{30}

= \frac{9+14}{30}

= \frac{23}{30}

(c) \frac{4}{9}+\frac{2}{7}

LCM of 9 and 7 is 63.

= \frac{\left(4\times7\right)+\left(2\times9\right)}{63}

= \frac{28+18}{63}

= \frac{46}{63}

(d) \frac{5}{7}+\frac{1}{3}

LCM of 7 and 3 is 21.

= \frac{\left(5\times3\right)+\left(1\times7\right)}{63}

=\frac{15+7}{21}

=\frac{22}{21}

(e) \frac{2}{5}+\frac{1}{6}

LCM of 5 and 6 is 30.

= \frac{\left(2\times6\right)+\left(1\times5\right)}{30}

= \frac{12+5}{30}

= \frac{17}{30}

(f) \frac{4}{5}+\frac{2}{3}

LCM of 5 and 3 is 15.

= \frac{\left(4\times3\right)+\left(2\times5\right)}{15}

= \frac{12+10}{15}

= \frac{22}{15}

(g) \frac{3}{4}-\frac{1}{3}

LCM of 4 and 3 is 12.

= \frac{\left(3\times3\right)-\left(1\times4\right)}{12}

=\frac{9-4}{12}

= \frac{5}{12}

(h) \frac{5}{6}-\frac{1}{3}

LCM of 6 and 3 is 6.

= \frac{5-\left(1\times2\right)}{6}

= \frac{5-2}{6}

= \frac{3}{6}=\frac{1}{2}

(i) \frac{2}{3}+\frac{3}{4}+\frac{1}{2}

LCM od 3, 4 and 2 is 12.

= \frac{\left(2\times4\right)+\left(3\times3\right)+\left(1\times6\right)}{12}

= \frac{8+9+6}{12}

= \frac{23}{12}

(j) \frac{1}{2}+\frac{1}{3}+\frac{1}{6}

LCM of 2, 3 and 6 is 6.

= \frac{\left(1\times3\right)+\left(1\times2\right)+1}{6}

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

= \frac{6}{6}=1

(k) \frac{4}{3}+\frac{11}{3}

= \frac{4+11}{3}

= \frac{15}{3}=5

(l) \frac{14}{3}+\frac{13}{4}

LCM of 3 and 4 is 12.

= \frac{\left(14\times4\right)+\left(13\times3\right)}{12}

= \frac{56+39}{12}

= \frac{95}{12}=7\frac{11}{12}

(m) \frac{16}{5}-\frac{7}{5}

= \frac{16-7}{5}

= \frac{9}{5}

(n) \frac{4}{3}-\frac{1}{2}

LCM of 3 and 2 is 6.

= \frac{\left(4\times2\right)-\left(1\times3\right)}{6}

= \frac{8-3}{6}

= \frac{5}{6}

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Question 1

Q1) Solve:

(a) \frac{2}{3}+\frac{1}{7} (b) \frac{3}{10}+\frac{7}{15} (c) \frac{4}{9}+\frac{2}{7} (d) \frac{5}{7}+\frac{1}{3}

(e) \frac{2}{5}+\frac{1}{6} (f) \frac{4}{5}+\frac{2}{3} (g) \frac{3}{4}-\frac{1}{3} (h) \frac{5}{6}-\frac{1}{3}

(i) \frac{2}{3}+\frac{3}{4}+\frac{1}{2} (j) \frac{1}{2}+\frac{1}{3}+\frac{1}{6} (k) 1\frac{1}{3}+3\frac{2}{3}

(l) 4\frac{2}{3}+3\frac{1}{4} (m) \frac{16}{5}-\frac{7}{5} (n) \frac{4}{3}-\frac{1}{2}

Solution 1:

(a) \frac{2}{3}+\frac{1}{7}

LCM of 3 and 7 is 21.

= \frac{\left(2\times7\right)+\left(1\times3\right)}{21}

= \frac{14+3}{21}

= \frac{17}{21}

(b) \frac{3}{10}+\frac{7}{15}

LCM of 10 and 15 is 30.

= \frac{\left(3\times3\right)+\left(7\times2\right)}{30}

= \frac{9+14}{30}

= \frac{23}{30}

(c) \frac{4}{9}+\frac{2}{7}

LCM of 9 and 7 is 63.

= \frac{\left(4\times7\right)+\left(2\times9\right)}{63}

= \frac{28+18}{63}

= \frac{46}{63}

(d) \frac{5}{7}+\frac{1}{3}

LCM of 7 and 3 is 21.

= \frac{\left(5\times3\right)+\left(1\times7\right)}{63}

=\frac{15+7}{21}

=\frac{22}{21}

(e) \frac{2}{5}+\frac{1}{6}

LCM of 5 and 6 is 30.

= \frac{\left(2\times6\right)+\left(1\times5\right)}{30}

= \frac{12+5}{30}

= \frac{17}{30}

(f) \frac{4}{5}+\frac{2}{3}

LCM of 5 and 3 is 15.

= \frac{\left(4\times3\right)+\left(2\times5\right)}{15}

= \frac{12+10}{15}

= \frac{22}{15}

(g) \frac{3}{4}-\frac{1}{3}

LCM of 4 and 3 is 12.

= \frac{\left(3\times3\right)-\left(1\times4\right)}{12}

=\frac{9-4}{12}

= \frac{5}{12}

(h) \frac{5}{6}-\frac{1}{3}

LCM of 6 and 3 is 6.

= \frac{5-\left(1\times2\right)}{6}

= \frac{5-2}{6}

= \frac{3}{6}=\frac{1}{2}

(i) \frac{2}{3}+\frac{3}{4}+\frac{1}{2}

LCM od 3, 4 and 2 is 12.

= \frac{\left(2\times4\right)+\left(3\times3\right)+\left(1\times6\right)}{12}

= \frac{8+9+6}{12}

= \frac{23}{12}

(j) \frac{1}{2}+\frac{1}{3}+\frac{1}{6}

LCM of 2, 3 and 6 is 6.

= \frac{\left(1\times3\right)+\left(1\times2\right)+1}{6}

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

= \frac{6}{6}=1

(k) \frac{4}{3}+\frac{11}{3}

= \frac{4+11}{3}

= \frac{15}{3}=5

(l) \frac{14}{3}+\frac{13}{4}

LCM of 3 and 4 is 12.

= \frac{\left(14\times4\right)+\left(13\times3\right)}{12}

= \frac{56+39}{12}

= \frac{95}{12}=7\frac{11}{12}

(m) \frac{16}{5}-\frac{7}{5}

= \frac{16-7}{5}

= \frac{9}{5}

(n) \frac{4}{3}-\frac{1}{2}

LCM of 3 and 2 is 6.

= \frac{\left(4\times2\right)-\left(1\times3\right)}{6}

= \frac{8-3}{6}

= \frac{5}{6}

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