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:
42 comments Page 2 of 5.
Kaviya said:
4 years ago
Thanks for explaining @Smirthi.
(3)
Abhinav said:
1 decade ago
B kills once in 2 shots.
No. of failure shots = 1.
No. of succeeded shots = 1.
When no. of failure shots = 27 then no. of succeeded shots = 27,
So, total no. of shots by B = 27+27 = 54.
When B shoots 3 shot then A shoots 5 shots.
So, When B shoots 1 shot then A shoots 5/3 shot.
So, When B shoots 54 shot then A shoots 54*5/3 = 90 shots.
When A shoots 3 it killed 1.
So, when A shoots 90 it killed = 90/3 = 30.
No. of failure shots = 1.
No. of succeeded shots = 1.
When no. of failure shots = 27 then no. of succeeded shots = 27,
So, total no. of shots by B = 27+27 = 54.
When B shoots 3 shot then A shoots 5 shots.
So, When B shoots 1 shot then A shoots 5/3 shot.
So, When B shoots 54 shot then A shoots 54*5/3 = 90 shots.
When A shoots 3 it killed 1.
So, when A shoots 90 it killed = 90/3 = 30.
(2)
Mufassil said:
4 years ago
Thanks for explaining the answer @Hary.
(2)
Sapna said:
8 years ago
Thanks @Sudip.
(1)
Panduri Alekhya said:
4 weeks ago
How 5/8 and 3/8 come? Can anyone explain it?
(1)
Siva said:
8 years ago
Well explained, Thanks @Smrithi.
(1)
Malay giri said:
8 years ago
A-27*5=135,
b-27*3=81,
a-b=135-81,
=54/3=18,
A-18*5=90,
A=90/3=30.
b-27*3=81,
a-b=135-81,
=54/3=18,
A-18*5=90,
A=90/3=30.
(1)
Sonu said:
3 years ago
Thanks @Sudip.
(1)
Rahul said:
1 decade ago
It is given that if A fires 5 shots B fires only 3 times.
It is given that p(A)=1/3 and p(B)=1/2.
It is also given that A missed 27 times so he fired totally 54 times.
From the above we can calculate that totally A fired 90 times.
So total successful firings by A=90*1/3=30.
It is given that p(A)=1/3 and p(B)=1/2.
It is also given that A missed 27 times so he fired totally 54 times.
From the above we can calculate that totally A fired 90 times.
So total successful firings by A=90*1/3=30.
(1)
Deepali said:
9 months ago
Every time B fires 3 shots, A fires 5 shots
That’s a firing ratio: A: B = 5 : 3, every time B fires 3, A fires 5.
But B, from 3 shots, he misses 1 shot every time he fires and kills 2, so the total he misses is 27.
In simple language, B misses 1 shots and kills 2, We have given no of shots he missed, but it doesn't include the shots he used to kill, which is 2.
To get the total no of shots of B, 27x2=54 shots.
To calculate A shots, which is 5:3, 54/3=18 times A fired, and 1 time firing consists of 5 shots.
So, 18 x 5 = 90.
The total no of shots A fired is 90, kills once in 3 shots, so 90/3= 60.
That’s a firing ratio: A: B = 5 : 3, every time B fires 3, A fires 5.
But B, from 3 shots, he misses 1 shot every time he fires and kills 2, so the total he misses is 27.
In simple language, B misses 1 shots and kills 2, We have given no of shots he missed, but it doesn't include the shots he used to kill, which is 2.
To get the total no of shots of B, 27x2=54 shots.
To calculate A shots, which is 5:3, 54/3=18 times A fired, and 1 time firing consists of 5 shots.
So, 18 x 5 = 90.
The total no of shots A fired is 90, kills once in 3 shots, so 90/3= 60.
(1)
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