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 3 of 10.
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.
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.
Jana said:
1 decade ago
Can anyone please explain me how "x" came in the denominator in the 3rd step?
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.
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.
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.
Karthiga said:
1 decade ago
The numerator 16(assume 16 liters) represents the quantity of remaining wine.
The denominator (16+65=81)represents
{the quantity of remaining wine +quantity of water added=initial quantity of wine(x)}.
The denominator (16+65=81)represents
{the quantity of remaining wine +quantity of water added=initial quantity of wine(x)}.
Shovik said:
1 decade ago
The real problem is that some of the initial steps are missing out here.
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] = 16:65.
=> x(1-8/x)^4/x = 16:81.
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] = 16:65.
=> x(1-8/x)^4/x = 16:81.
Shanjev said:
1 decade ago
Why the quantity of wine left in cask after 4 op is DIVIDED BY X, can anyone explain me?
Amit said:
1 decade ago
After the first step of adding 8 liters of water, the cask contains a mix of wine and water so the next time we take 8 liters out, it could have 75 % wine or 35 % wine or any other amount. So, how can we calculate the final proportion of wine?
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