Ncert solutions

Grade 7

# Question 1

Q1) Solve:

\left(i\right)\ 2-\frac{3}{5}\ \left(ii\right)\ 4+\frac{7}{8}\ \left(iii\right)\ \frac{3}{5}+\frac{2}{7}\ \left(iv\right)\ \frac{9}{11}-\frac{4}{5}\ \left(v\right)\ \frac{7}{10}+\frac{2}{5}+\frac{3}{2}\ \left(vi\right)\ 2\frac{2}{3}+3\frac{1}{2}\ \left(vii\right)8\frac{1}{2}-3\frac{5}{8}

Solution 1

i) 2\ -\ \frac{3}{5}=\ \frac{2}{1}-\ \frac{3}{5}=\ \frac{2\times5-3\times1}{1\times5}=\ \frac{10-3}{5}=\ \frac{7}{5}=1\frac{2}{5}

ii) 4+\frac{7}{8}=\ \frac{4}{1}+\frac{7}{8}=\ \frac{4\times8+1\times7}{1\times8}=\ \frac{32+7}{8}=\ \frac{39}{8}=\ 4\ \frac{7}{8}

iii) \frac{3}{5}+\frac{2}{7}=\ \frac{3\times7+2\times5}{5\times7}=\ \frac{31}{35}

iv) \frac{9}{11}-\frac{4}{5}=\ \frac{9\ \times15-4\times11\ }{11\times15}=\ \frac{91}{165}

v) \frac{7}{10}+\ \frac{2}{5}\ +\ \frac{3}{2}\ =\ \frac{7+4+15}{10} (LCM of 10, 5 and 2)

=\ \frac{26}{10}=\ \frac{26\div2}{10\div2}=\ \frac{13}{5}=\ 2\ \frac{3}{5}

vi) 2\ \frac{2}{3}+\ 3\ \frac{1}{2}=\ \frac{8}{3}+\frac{7}{2}=\ \frac{8\times2+3\times7}{3\times2}\ =\ \frac{16+21}{6}

\frac{37}{6}=\ 6\ \frac{1}{6}

vii) 8\ \frac{1}{2}-3\ \frac{5}{8}=\ \frac{17}{2}-\frac{29}{8}\ =\ \frac{17\times4-29\times1}{8}

\frac{68-29}{8}=\ \frac{39}{8}=\ 4\ \frac{7}{8}

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