Verbal Reasoning - Arithmetic Reasoning - Discussion
Discussion Forum : Arithmetic Reasoning - Section 2 (Q.No. 7)
7.
A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed
Answer: Option
Explanation:
Discussion:
5 comments Page 1 of 1.
Preethi said:
4 years ago
Thanks @Sano Vai.
(1)
Krutika said:
5 years ago
Well explained, Thanks @Donald.
Sano Vai said:
6 years ago
B: 1 miss in 2 shots so, 27 misses in 54 shots,
while B shots 3, A shots 5 so, in 54 shots of B, A hits 5/3*54 = 90.
A: 3 shots 1 kills so, 90 shots(90*1/3) = 30 kills.
while B shots 3, A shots 5 so, in 54 shots of B, A hits 5/3*54 = 90.
A: 3 shots 1 kills so, 90 shots(90*1/3) = 30 kills.
(2)
Raju said:
8 years ago
@Donald.
It's nice explanation, thanks.
It's nice explanation, thanks.
Donald said:
1 decade ago
B has an accuracy of 50%
so in order to miss 27 times he would have fired 54 shots (27*2)
For every 3 shots fired by three A will fire 5 shots
So (54/3*5)= 90 shots fired by A
A has an accuracy of 1/3
So A has killed (90*1/3)
30 Birds: Answer = 30 Birds
so in order to miss 27 times he would have fired 54 shots (27*2)
For every 3 shots fired by three A will fire 5 shots
So (54/3*5)= 90 shots fired by A
A has an accuracy of 1/3
So A has killed (90*1/3)
30 Birds: Answer = 30 Birds
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