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 7 of 10.
Pandit said:
1 decade ago
Lets,
The volume of the drum is x.
In 1st operation 8 liter is drawn.
Remaining pure wine is x-8.
Now. If we draw whole volume = the volume of wine is drawn is x-8.
""""""""""''' 1 unit volume="""""""""""""""""""""""""" (x-8) /x.
""""""""""'''8 unit volume=""""""""""""""""""""""""""8*(x-8) /x.
The volume of wine remaining after 1st operation is (x-8) -{8* (x-8) /x} = { (x-8) ^2}/x...
Similarly in 4th draw pure wine is x (1-8/x) ^4.
The volume of the drum is x.
In 1st operation 8 liter is drawn.
Remaining pure wine is x-8.
Now. If we draw whole volume = the volume of wine is drawn is x-8.
""""""""""''' 1 unit volume="""""""""""""""""""""""""" (x-8) /x.
""""""""""'''8 unit volume=""""""""""""""""""""""""""8*(x-8) /x.
The volume of wine remaining after 1st operation is (x-8) -{8* (x-8) /x} = { (x-8) ^2}/x...
Similarly in 4th draw pure wine is x (1-8/x) ^4.
PRAVU said:
1 decade ago
@Sanjay.
Pure wine is divided by total amount in the vessel, i.e. x.
So, pure wine: water = 16:65 OR it can be said as pure wine:total = 16:(16+65) = 16:81
Note that in each operation the total is always x litre.
Pure wine is divided by total amount in the vessel, i.e. x.
So, pure wine: water = 16:65 OR it can be said as pure wine:total = 16:(16+65) = 16:81
Note that in each operation the total is always x litre.
PRAVU said:
1 decade ago
Note that in each drawing, we take out 8/x part of the remaining wine and every time the total (wine +water) is x litre.
In 1st draw, pure wine is (x-8) i.e. x(1-8/x).
In 2nd draw, pure wine is (x-8)-(8/x)(x-8) i.e x(1-8/x)^2.
Similarly in 4th draw pure wine is x(1-8/x)^4.
Wine:water=16:65 or wine:total = 16:81.
(note: 16+65 = 81).
So, x(1-x)^4/x = 16/81.
In 1st draw, pure wine is (x-8) i.e. x(1-8/x).
In 2nd draw, pure wine is (x-8)-(8/x)(x-8) i.e x(1-8/x)^2.
Similarly in 4th draw pure wine is x(1-8/x)^4.
Wine:water=16:65 or wine:total = 16:81.
(note: 16+65 = 81).
So, x(1-x)^4/x = 16/81.
Murali said:
1 decade ago
After n operations the wine left is {(x-8)^n}/x^n-1.
However, at the end of every cycle the total quantity is x. Therefore quantity of water left after n operations is x-quantity of wine left= x-{(x-8)^n}/x^n-1.
Now, dividing quantity of wine left by quantity of water left we get: {(x-8)^n}/{x^n-(x-8)^n} = 16/65. Substitute each one of the options n see which one matches rather than going to solve this complex expression.
However, at the end of every cycle the total quantity is x. Therefore quantity of water left after n operations is x-quantity of wine left= x-{(x-8)^n}/x^n-1.
Now, dividing quantity of wine left by quantity of water left we get: {(x-8)^n}/{x^n-(x-8)^n} = 16/65. Substitute each one of the options n see which one matches rather than going to solve this complex expression.
Yogi damle said:
1 decade ago
If the answer is 24 liters then we cannot take out 8 liters of wine 4 times.
Kola.shiva said:
1 decade ago
Simple method.
8 liters of original wine drawn from the cask.
So question they had given at last the ratio is 16:65.
So we add recent 8 liters to present ratio of wine.
16+8 = 24. Simple.
8 liters of original wine drawn from the cask.
So question they had given at last the ratio is 16:65.
So we add recent 8 liters to present ratio of wine.
16+8 = 24. Simple.
(1)
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?
Shanjev said:
1 decade ago
Why the quantity of wine left in cask after 4 op is DIVIDED BY X, can anyone explain me?
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.
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)}.
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