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A man whose bowling average is 12.4 takes 5 wickets for 26

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by deloitte247 » Sat Aug 18, 2018 2:15 pm
Let the wickets claimed by him before the last match = x
Total runs he had given = x * 12.4 = 12.4x
Number of wickets now taken by him = (x + 5)
12 (x + 5) = 12.4x + 26
12x + 60 = 12.4x + 26
12x - 12.4x = 26 - 60
- 0.4x = -34
divide both sides by co - efficient of x
$$=\ -\frac{34}{-0.4}$$
x = 85

Option A is CORRECT.
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by Jeff@TargetTestPrep » Sat Aug 18, 2018 6:50 pm
BrijNath wrote:A man whose bowling average is 12.4, takes 5 wickets for 26 runs and thereby decreasing his average by 0.4. The number of wickets taken by him before the previous match is?

A.85
B.78
C.72
D.64
E.92
Let n = the number of wickets taken by him before the previous match and create the equation:

(12.4n + 26)/(n + 5) = 12

12.4n + 26 = 12n + 60

0.4n = 34

n = 85

Answer: A

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