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?
18 litres
24 litres
32 litres
42 litres
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 1 of 10.

Shiv said:   4 years ago
Only wine 8 liters
and 3 times operation.
So, 8 * 3=24.
(17)

CHETHAN K P said:   3 years ago
Wine:water = 16 : 65
Total =16 + 65 = 81.
Now wine: total = 16:81.
(16)

Ajai said:   4 years ago
@All

Here is the explanation for Why it was divided by x?

Here the answer,
16/81 is a wine part of total volume initial x.
So that's why :(16/81)(x) --->step 2
And x goes LHS to denominator x.
(5)

Surya said:   7 years ago
Let the quantity of the wine in the cask originally be x litres. 8 litres of wine is removed from it and replaced with water. This operation is done a total 4 times. After 4 operations,

Wine : Water = 16 : 65 or Wine : Total = 16 : 81.

[(x-8)/x]4 = 16/81
(x-8)/x = 2/3
3x - 24 = 2x
x = 24
The correct option is D.
(3)

Salaj Mondal said:   4 years ago
x^4 = 81
=>x=3
y^4=16
=>y=2
x - y
= 3 - 2
= 1.

1 ratio means 8 litre.
so, 3 ratio means 24 litre.
(3)

Ayush Nigam said:   9 years ago
I think most of you understood the solution but for those who didn't let me explain:

First of all, There is a formula which says:

Suppose container contains x of liquid from which if y units are taken out (of the whole liquid) and then replaced with water, and this process is simply repeated n no of times, then the quantity of original liquid present in the mixture now is:

x[(1- y/x )^n].

Ok , now as the question says here we need to find the original quantity of the wine, let it be x.

Now the quantity of liquid (not wine) removed is = 8 which is 'y' for the formula.

This thing is done total of 4 times.

So now the quantity of the wine present in the final mixture will be:

x[(1 - 8/x) ^4]. No problem till here.

Now as the problem says, the ratio of wine to water in the final mixture, we have calculated wine but we can't calculate the amount of water in the final mixture.
So, what to do next.

We are given the ratio of wine to water in the mixture and remember the water was being replaced each time, so total quantity of the mixture is still 'x'.

Thus, if 16/65 was the given ratio of wine to water then, 16/(16+65)

will be the ratio of wine to total mixture ie:

x[(1-8/x)^n] / x = 16/81

I hope everyone get it now.
(3)

Barath s said:   4 years ago
For those who don't understand why it is divided by x?

Here in 16/81, 81 can be taken as total mixture are 81y is the total capacity of the cask. So x is the total wine that was kept at first which is nothing but the total capacity of the cask.
(2)

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.
(1)

Aniket said:   8 years ago
What if instead of wine it would have been 6 litres of solution of wine and water for second time?
(1)

Rajesh said:   5 years ago
Final wine = x{1 - 8/x}^4 ----> 1.

where x is the original wine quantity.
(As from formula)
Now given final wine to water ratio i.e.
Wine: water = 16:65.

Therefore final wine (in litres) = (16÷81)*(x) where ----> 2
81 = 16 +65 represnting total volume.
X = also representing initial total volume.
Equating 1 and 2 we will get the answer.
(1)


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