6 frac{1}{7} + 15 frac{2}{3} + 11 frac{1}{6} | vedclass

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Question

(C) Let the missing value be $x$.Given equation: $6 \frac{1}{7} + 15 \frac{2}{3} + 11 \frac{1}{6} = 33 \frac{1}{21} - x$.Convert mixed fractions to im

Vedclass · Accepted Answer

$\frac{1}{14}$

Vedclass · Answer

(C) Let the missing value be $x$.Given equation: $6 \frac{1}{7} + 15 \frac{2}{3} + 11 \frac{1}{6} = 33 \frac{1}{21} - x$.Convert mixed fractions to improper fractions:$6 \frac{1}{7} = \frac{43}{7}$,$15 \frac{2}{3} = \frac{47}{3}$,$11 \frac{1}{6} = \frac{67}{6}$,$33 \frac{1}{21} = \frac{694}{21}$.Sum of the left side: $\frac{43}{7} + \frac{47}{3} + \frac{67}{6}$.The least common multiple of $7, 3, 6$ is $42$.$= \frac{43 	imes 6 + 47 	imes 14 + 67 	imes 7}{42} = \frac{258 + 658 + 469}{42} = \frac{1385}{42}$.Now,$x = \frac{694}{21} - \frac{1385}{42}$.$= \frac{1388 - 1385}{42} = \frac{3}{42} = \frac{1}{14}$.Thus,the missing value is $\frac{1}{14}$.

$6 \frac{1}{7} + 15 \frac{2}{3} + 11 \frac{1}{6} = 33 \frac{1}{21} - ?$

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