Question

$16$ children are to be divided into two groups $A$ and $B$ of $10$ and $6$ children. The average percent marks obtained by the children of group $A$ is $75$ and the average percent marks of all the $16$ children is $76$. What is the average percent marks of children of group $B$?

• A. $77\frac{1}{3}$
• B. $77\frac{2}{3}$
• C. $78\frac{1}{3}$
• D. $78\frac{2}{3}$

Answer 1

Answer: (B)

$77\frac{2}{3}$

Given that group A has $10$ children and

group B has $6$ children

Given that average percent marks obtained by group A is $75$

Therefore, $\frac{75}{100}=\frac{\text{sum of the marks obtained by group A}}{10}$

$⟹\text{sum of the marks obtained by group A}=7.5$ ------(1)

Given that average percent marks obtained by group A and B is $76$

Therefore, $\frac{76}{100}=\frac{\text{sum of the marks obtained by group A and B}}{16}$

$⟹\text{sum of the marks obtained by group A and B}=12.16$ ------(2)

sum of the marks obtained by group B is $12.16-7.5=4.66$

Therefore, the average percent marks obtained by group B is $\frac{\text{sum of the marks obtained by group B}}{6}\times 100$

$=\frac{4.66}{6}\times 100=77\frac{2}{3}$