Solve:
(i) $3-\frac{2}{5}$
(ii) $4+\frac{7}{8}$
(iii) $\frac{3}{5}+\frac{2}{7}$
(iv) $\frac{9}{11}-\frac{4}{15}$
(v) $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$
(vi) $2\frac{2}{3}+3\frac{1}{2}$
(vii) $8\frac{1}{2}-3\frac{5}{8}$

AcademicMathematicsNCERTClass 7

To do:

We have to solve the given expressions.

Solution:

(i) $3-\frac{2}{5}$

$=\frac{3}{1}-\frac{2}{5}$ 

$=\frac{3\times5-2}{5}$                      (L.C.M. of $1$ and $5$ is $5$)

$=\frac{15-2}{5}$

$=\frac{13}{5}$

(ii) $4+\frac{7}{8}$

$=\frac{4}{1}+\frac{7}{8}$

$=\frac{4\times8+7}{8}$             (L.C.M. of $1$ and $8$ is $8$)

$=\frac{32+7}{8}$

$=\frac{39}{8}$

(iii) $\frac{3}{5}+\frac{2}{7}$

$=\frac{3\times7+2\times5}{35}$     (L.C.M. of $5$ and $7$ is $35$)

$=\frac{21+10}{35}$

$=\frac{31}{35}$

(iv) $\frac{9}{11}-\frac{4}{15}$

$=\frac{9\times15-4\times11}{165}$        (L.C.M. of $11$ and $15$ is $165$)   

$=\frac{135-44}{165}$

$=\frac{91}{165}$

(v) $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$

$=\frac{7\times1+2\times2+3\times5}{10}$       (L.C.M. of $10, 5$ and $2$ is $10$)

$=\frac{7+4+15}{10}$

$=\frac{26}{10}$

(vi) $2\frac{2}{3}+3\frac{1}{2}$

$=\frac{2\times3+2}{3}+\frac{3\times2+1}{2}$

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

$=\frac{8}{3}+\frac{7}{2}$

$=\frac{8\times2+7\times3}{6}$            (L.C.M. of $3$and $2$ is $6$)

$=\frac{16+21}{6}$

$=\frac{37}{6}$

(vii) $8\frac{1}{2}-3\frac{5}{8}$

$=\frac{8\times2+1}{2}-\frac{3\times8+5}{8}$

$=\frac{16+1}{2}-\frac{24+5}{8}$

$=\frac{17}{2}-\frac{29}{8}$

$=\frac{17\times4-29\times1}{8}$       (L.C.M. of $2$ and $8$ is $8$)

$=\frac{68-29}{8}$

$=\frac{39}{8}$

raja
Updated on 10-Oct-2022 13:32:37

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