If $p$, $q$, $r$ be three positive numbers such that $p > q > r$ when the smallest numbers is added to the difference of the rest two numbers, then the average of the resultant number and the original numbers except to the smallest number is 21 more than the average of all the three original numbers. The value of than the average of all the three original numbers. The value of $(p-q)$ is:
7
14
63
42
None of these
Solution
$\dfrac{[r+(p-q)]+p+q}{3} = 21 +\dfrac{p + q + r}{3}$
$\Rightarrow \dfrac{2p+r}{3}- 21 = \dfrac{p + q + r}{3}$
$\dfrac{p-q}{3} = 21$
$\Rightarrow p-q = 63$