Question

A man whose bowling average is$12.4$, takes $5$ wickets for $26$ runs and thereby decreases his average by$0.4$. The number of wickets taken by him before his last match is?
A. $82$
B. $83$
C. $84$
D. $85$

Answer 1

Hint:
Let initially the man has taken $x$ wickets with the bowling speed of $12.4$ that means he takes one wicket after giving the $12.4$ runs. So the total runs scored would be $12.4x$ and in this match he gave $26$ runs. So total runs $ = 12.4x + 26$ and total wickets become $ = x + 5$
Now we know the new bowling average so we can find the value of $x$ easily.

Complete step by step solution:
Here in this question we are given that a man whose bowling average is $12.4$ this means he takes one wicket after giving the $12.4$ runs and now takes $5$ wickets for $26$ runs and thereby decreases his average by $0.4$.
This means that his new bowling average becomes $0.4$ less than the initial bowling average. Therefore new bowling average$ = 12.4 - 0.4 = 12$
Let initially the men has taken $x$ wickets with the bowling speed of $12.4$
So for taking $x$ wicket he has given $12.4x$ runs.
Now his new average is $12$ and has taken $5$ wickets by giving $26$ runs.
Already he has given $12.4x$ so the total runs given by him$ = 12.4x + 26$
As we know he had taken $x$ wickets in the initial match and now $5$
So total number of wickets now$ = x + 5$
We know that the bowling average is equal to the ratio of the number of total runs and the number of wickets taken.
So ${\text{new bowling average}} = \dfrac {{{\text{total runs scored}}}}{{{\text{total wickets taken}}}}$
$
  12 = \dfrac {{12.4x + 26}}{{x + 5}} \\ \Rightarrow
\Rightarrow 12(x + 5) = 12.4x + 26 \\
\Rightarrow 12x + 60 = 12.4x + 26 \\
\Rightarrow 34 = 0.4x \\
\Rightarrow x = \dfrac {{34}}{{0.4}} = 85 \\
 $
So we got the value as $x = 85$

So he has taken $85$ wickets before his last match.

Note:
If the batting average is given that means the number of runs or scores he makes in each match. Batting average is the average of every score in every match.