Aptitude - Alligation or Mixture - Discussion
Discussion Forum : Alligation or Mixture - General Questions (Q.No. 14)
14.
8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally?
Answer: Option
Explanation:
Let the quantity of the wine in the cask originally be x litres.
| Then, quantity of wine left in cask after 4 operations = |  |
x |  |
1 - | 8 |  |
4 |  litres. |
| x |
 |
|
x(1 - (8/x))4 | |
= | 16 |
| x | 81 |
 |
|
1 - | 8 | |
4 | = | |
2 | |
4 |
| x | 3 |
| |
|
x - 8 | |
= | 2 |
| x | 3 |
3x - 24 = 2x
x = 24.
Discussion:
96 comments Page 8 of 10.
Dileep said:
1 decade ago
@Reeya.
I didn't get you. what you are saying.
From the given information we can write it as,
x(1-8/x)^4 = 16/81.
i.e x = 24.
I didn't get you. what you are saying.
From the given information we can write it as,
x(1-8/x)^4 = 16/81.
i.e x = 24.
Reeya said:
1 decade ago
I think the question is wrong. I know it can never be said as an excuse, but if the the ratio of wine remaining to that of water is 16:65, then we can take the denominator, wine originally as x.
i.e., (x (1-8/x) ^4) / (x) = 16/ (16+65) = 16/81.
Correct me if I am wrong.
i.e., (x (1-8/x) ^4) / (x) = 16/ (16+65) = 16/81.
Correct me if I am wrong.
Pr!!!!! said:
1 decade ago
The ratio of wine to that of water is given as 16:81.
The initial quantity of wine is not given and it is assumed as 'x'.
And so the ratio becomes 16x:81.
And hence for simplification 'x' came in denominator.
The initial quantity of wine is not given and it is assumed as 'x'.
And so the ratio becomes 16x:81.
And hence for simplification 'x' came in denominator.
Jana said:
1 decade ago
Can anyone please explain me how "x" came in the denominator in the 3rd step?
Richa said:
1 decade ago
let the quantity of the wine in the cask originally be x litres.
Then, quantity of wine left in cask after 4 operations = x(1 -8/x)^4 litres.
x(1 - (8/x))^4 = 16x/81
(1 -8/x)= 2/3
(x - 8)/x =2/3
3x - 24 = 2x
x = 24.
Then, quantity of wine left in cask after 4 operations = x(1 -8/x)^4 litres.
x(1 - (8/x))^4 = 16x/81
(1 -8/x)= 2/3
(x - 8)/x =2/3
3x - 24 = 2x
x = 24.
Atchiya said:
1 decade ago
I think performs operation more than 3 times means it includes the first time which is in process + 3 more times. So 4 times.
Please can anyone explain hoe that X came else explain this in alternative easy method.
Please can anyone explain hoe that X came else explain this in alternative easy method.
Erik said:
1 decade ago
There can be no doubt about the fact that they got the last ratio wrong in the question.
@rattle, sorry but you are right if the replacement stopped after the first round. After that when on the second round 8 litres is taken from the cask, it includes some wine and some water and then 8 litres of water in put in. This continues twice more. Sorry but we cannot assume x to be the amount of water in the cask after mixing is complete.
@rattle, sorry but you are right if the replacement stopped after the first round. After that when on the second round 8 litres is taken from the cask, it includes some wine and some water and then 8 litres of water in put in. This continues twice more. Sorry but we cannot assume x to be the amount of water in the cask after mixing is complete.
Tez said:
1 decade ago
Let the quantity of the wine in the cask originally be x liters.
Then, quantity of wine left in cask after 4 operations.
=[x (1- (8/8) ) ^4].
Hence, [{x (1- (8/x) ^4) ) }/x]=16/97.
=>[ (x-8) /x]^4=16/97.
=> (x-8) /x=2/3.1 (since 97^1/4=3.1).
=>x=22.5 Liters.
Answer should be 22.5 Liters.
Then, quantity of wine left in cask after 4 operations.
=[x (1- (8/8) ) ^4].
Hence, [{x (1- (8/x) ^4) ) }/x]=16/97.
=>[ (x-8) /x]^4=16/97.
=> (x-8) /x=2/3.1 (since 97^1/4=3.1).
=>x=22.5 Liters.
Answer should be 22.5 Liters.
Shaswat said:
1 decade ago
@Prateek.
Here it is given that '8 liters are drawn from a cask full of wine' not 'wine and water'
so you cannot subtract i.e.,
Amt. of water left is given by x- x(1-8/x)^4.
So how can you subtract ?
Here it is given that '8 liters are drawn from a cask full of wine' not 'wine and water'
so you cannot subtract i.e.,
Amt. of water left is given by x- x(1-8/x)^4.
So how can you subtract ?
Prateek said:
1 decade ago
I agree with Sahil.
Another method:
Amt. of wine left is given by x(1-8/x)^4.
Amt. of water left is given by x- x(1-8/x)^4.
So x(1-8/x)^4/[x-x(1-8/x)^4] = 16:81
With this method, i got answers to a couple of questions which were same as this question. But unable to get the answer to this one. I think my method is perfectly alright.
Another method:
Amt. of wine left is given by x(1-8/x)^4.
Amt. of water left is given by x- x(1-8/x)^4.
So x(1-8/x)^4/[x-x(1-8/x)^4] = 16:81
With this method, i got answers to a couple of questions which were same as this question. But unable to get the answer to this one. I think my method is perfectly alright.
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