A cricketer has a certain average for 10 innings. In eleventh inning, he scored 108 runs, thereby in

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A cricketer has a certain average for $10$ innings. In the eleventh inning, he scored $108$ runs, thereby increasing his average by $6$ runs. His new average is

A

$48$

B

$52$

C

$55$

D

$60$

Solution

Let his average in the tenth innings be $x$.

His score in $11^{th}$ innings $=108$ runs

$\therefore$ Average in eleventh innings $=\frac{x \times 10+108}{11}=x+6$

$10 x+108=(x+6) 11$

$11 x-10 x=108-66$

$x=42$

New average $=x+6=42+6=48$ runs

Standard 13

Quantitative Aptitude

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