Aptitude - Simplification - Discussion
Discussion Forum : Simplification - General Questions (Q.No. 8)
8.
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:
Let the total number of shots be x. Then,
| Shots fired by A = | 5 | x |
| 8 |
| Shots fired by B = | 3 | x |
| 8 |
| Killing shots by A = | 1 | of | 5 | x | = | 5 | x |
| 3 | 8 | 24 |
| Shots missed by B = | 1 | of | 3 | x | = | 3 | x |
| 2 | 8 | 16 |
 |
3x | = 27 or x = |  |
27 x 16 |  |
= 144. |
| 16 | 3 |
| Birds killed by A = | 5x | = | |
5 | x 144 | |
= 30. |
| 24 | 24 |
Discussion:
40 comments Page 3 of 4.
Danish said:
8 years ago
Superb @Smrithi.
Jas said:
9 years ago
Total shots = 8.
A = 5/8.
B = 3/8.
A succeed 3times in 5/8 shots = 5/24.
B failed 2 times in 3/8 shots = 3/16.
Let total shots = x
B's failed = 3x/16.
Given 3x/16 = 27, x= 144.
Sub in A 5x/24 --> 5x144/24 --> A=30.
A = 5/8.
B = 3/8.
A succeed 3times in 5/8 shots = 5/24.
B failed 2 times in 3/8 shots = 3/16.
Let total shots = x
B's failed = 3x/16.
Given 3x/16 = 27, x= 144.
Sub in A 5x/24 --> 5x144/24 --> A=30.
Taniya said:
8 years ago
Thanks for your explanation @Kajora.
Srinivas said:
8 years ago
Detail explanation of above problem please give me.
SAI RAM said:
8 years ago
Thank you all.
Richa said:
8 years ago
Good job @Kajora.
Dory said:
8 years ago
So if B kills once in 3 shots, miss shots by B = ( 1-(1/3) ) * (3x/8)?
Pollard said:
2 decades ago
Please anybody explain very simply..?
Saif said:
9 years ago
Thank you @Naddy.
Ryan said:
9 years ago
Any other simple method to solve this problem?
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