Find:
$(i)$. $\frac{7}{24\ }- \frac{17}{36}$
$(ii)$. $\frac{5}{63}-\ (-\frac{6}{21})$
$(iii)$. $-\frac{6}{13}\ -\ (-\frac{7}{15})$
$(iv)$. $-\frac{3}{8}-\frac{7}{11}$
$(v)$. $-2\frac{1}{9}\ -\ 6$

AcademicMathematicsNCERTClass 7

Given:

$(i)$. $\frac{7}{24\ }- \frac{17}{36}$

$(ii)$. $\frac{5}{63}-\ (-\frac{6}{21})$

$(iii)$. $-\frac{6}{13}\ -\ (-\frac{7}{15})$

$(iv)$. $-\frac{3}{8}-\frac{7}{11}$

$(v)$. $-2\frac{1}{9}\ -\ 6$


To do: To solve the given expression.


Solution: 

 $(i)$. $\frac{7}{24\ }- \frac{17}{36}$

$=\frac{7\times 3}{24\times 3}-\frac{17\times 2}{36\times 2}$       [LCM of $24$ and $36$ is $72$ ]

$=\frac{21-34}{72}$

$=-\frac{13}{72}$

$(ii)$. $\frac{5}{63}-\ (-\frac{6}{21})$

$=\frac{5\times 1}{63\times 1}-(-\frac{6\times 3}{21\times 3})$    [LCM of 63 and 21 is 63]


$=\frac{5-\left(-18\right)}{63}$

$=\frac{23}{63}$

$(iii)$. $-\frac{6}{13}\ -\ (-\frac{7}{15})$

$=-\frac{6\times 15}{13\times 15}-(-\frac{7\times 13}{15\times 13})$    [LCM of 13 and 15 is 195]

$=-\frac{90}{195}+\frac{91}{195}$

$=\frac{-90+91}{195}$

$=\frac{1}{195}$

$(iv)$. $-\frac{3}{8}-\frac{7}{11}$

$=-\frac{3\times 11}{8\times 11}-\frac{7\times 8}{11\times 8}$        [LCM of 8 and 11 is 88]

$=-\frac{33}{88}-\frac{56}{88}$

$=\frac{-33-56}{88}$

$=-\frac{89}{88}$

$=-1\frac{1}{88}$           

$(v)$.  $-2\frac{1}{9}\ -\ 6$

$=\frac{-2\times 9+1}{9}-6$

$=-\frac{18+1}{9}-\frac{6}{1}$

$=-\frac{19}{9}-\frac{6}{1}$

$=-\frac{19\times 1}{9\times 1}-\frac{6\times 9}{1\times 9}$     [LCM of 9 and 1 is 9]

$=-\frac{19}{9}-\frac{54}{9}$

$=\frac{-19-54}{9}$

$=\frac{-73}{9}$

$=-8\frac{1}{9}$

raja
Updated on 10-Oct-2022 13:35:25

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