Question

The average of twenty results is $18$, that of first $9$ is $14$ and that of last $9$ is $17$. The average of middle two results is

• A. $45$
• B. $50.5$
• C. $40.5$
• D. $51.5$

Answer 1

Answer: (C)

$40.5$

Given that average of $20$ results is $18$

Let the results be $r_1,r_2,...r_{20}$

Therefore, $\frac{r_1+r_2+...+r_{20}}{20}=18$

$⟹r_1+r_2+...+r_{20}=18\times 20=360$ -------(1)

Average of first $9$ is $14$

$⟹\frac{r_1+r_2+...+r_9}{9}=14$

$⟹r_1+r_2+...+r_9=14\times 9=126$ -------(2)

Average of last $9$ is $17$

$⟹\frac{r_{12}+r_{13}+...+r_{20}}{9}=17$

$⟹r_{12}+r_{13}+...+r_{20}=17\times 9=153$ -------(3)

substituting (2) and (3) in (1) we get

$126+r_{10}+r_{11}+153=360$

$⟹r_{10}+r_{11}+279=360$

$⟹r_{10}+r_{11}=81$

Average of middle two $r_{10},r_{11}$ is $\frac{r_{10}+r_{11}}{2}=\frac{81}{2}=40.5$