A batsman scores 98 runs in the 17^{text{th}} | vedclass

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(D) Let the average score of the batsman up to the $16^{	ext{th}}$ match be $x$.Total runs scored in $16$ matches = $16x$.After scoring $98$ runs in

Vedclass · Accepted Answer

$55.5$

Vedclass · Answer

(D) Let the average score of the batsman up to the $16^{	ext{th}}$ match be $x$.Total runs scored in $16$ matches = $16x$.After scoring $98$ runs in the $17^{	ext{th}}$ match,the new total runs = $16x + 98$.The new average after $17$ matches = $\frac{16x + 98}{17}$.According to the problem,the new average is $x + 2.5$.So,$\frac{16x + 98}{17} = x + 2.5$.Multiplying both sides by $17$,we get: $16x + 98 = 17(x + 2.5)$.$16x + 98 = 17x + 42.5$.Rearranging the terms: $x = 98 - 42.5$.$x = 55.5$.Therefore,the average before the $17^{	ext{th}}$ match was $55.5$.

$A$ batsman scores $98$ runs in the $17^{\text{th}}$ match of his career. His average runs per match increases by $2.5$. What is his average before the $17^{\text{th}}$ match?

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