Question

The average(arithmetic mean) of six numbers is $3$. If the average of the least and the greatest of these numbers is $5$, which set of four numbers can't represent the other four numbers?

• A. $-4,0,5,7$
• B. $0,0,0,8$
• C. $\frac{1}{2},\frac{3}{2},1,5$
• D. $1,1,1,1$
• E. $1,2,2,3$

Answer 1

Answer: (D)

$1,1,1,1$

Given: The average of six numbers is $3$. If the average of the least and greatest number is $5$.

Then total the least and greatest number =$5\times 2=10$

(A) if number are as per (A)

The average $=$ $\frac{-4+0+5+7+10}{6}=3$

(B) if number are as per (B)

The average $=$ $\frac{0+0+0+8+10}{6}=3$

(C) if number are as per option (C)

$\frac{\frac{1}{2}+\frac{3}{2}+1+5+10}{6}=3$

(D) if number are as per (D)

The average $=$ $\frac{1+1+1+1+10}{6}=\frac{14}{6}=2.33$

(E) if number are as per (E)

The average $=$ $\frac{1+2+2+3+10}{6}=3$

In all options, the average of option C is not $2$, the answer is (C).