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 4 of 10.
Sundar said:
9 years ago
How can did wine left ratio is 16:65 to change 16:81 please explain anybody?
Vivek kumar said:
6 years ago
From where 16:81 came, there is 16:65? can anyone explain.
Santosh said:
5 years ago
Let x be initial volume(in lit) of wine.
By formula final amount of wine after n(=4) iteration = x{1 - 8/x}^4 ---->1.
Given the final ratio of wine to water is 16:65.
So the volume of wine in the final iteration is {16/(16+65)}x -------> 2.
Equate 1 and 2,
And we get x=24.
By formula final amount of wine after n(=4) iteration = x{1 - 8/x}^4 ---->1.
Given the final ratio of wine to water is 16:65.
So the volume of wine in the final iteration is {16/(16+65)}x -------> 2.
Equate 1 and 2,
And we get x=24.
Sadiq said:
5 years ago
@Mahesh.
After 4 operations, wine and water ratio is 16:65.
Just assume that original wine was (16+65)=81, which we assumed by x.
After 4 operations, wine and water ratio is 16:65.
Just assume that original wine was (16+65)=81, which we assumed by x.
Vivek Rajput said:
5 years ago
@Sanjay(first comment) x is the total quantity.
x(1 - (8/x))4
--------------- = 16/81.
x
the numerator is the quantity of wine left and the denominator is the total quantity.
x(1 - (8/x))4
--------------- = 16/81.
x
the numerator is the quantity of wine left and the denominator is the total quantity.
Mahesh said:
5 years ago
The resulting ratio is 16:65.
How we get 16/81?
Please explain.
How we get 16/81?
Please explain.
Prashanth reddy said:
6 years ago
Here cask initially contains x lit of wine.
Each time the operation is repeated the same amount of water is replaced by wine.
So, after any number of operations, the final quantity in cask is always x.
Now wine:water = 16:65.
Wine:total mixture = 16:81.
Wine left after final operation:total mixture=16:8.
Each time the operation is repeated the same amount of water is replaced by wine.
So, after any number of operations, the final quantity in cask is always x.
Now wine:water = 16:65.
Wine:total mixture = 16:81.
Wine left after final operation:total mixture=16:8.
Amrita said:
6 years ago
This will be more understandable by using the fomula (Q/T)=(1-R/T)^n.
Here Q=16(present quantity,from the ratio).
T=81(total quantity of solution,16+65),
R=8(replacing quantity).
n=4.
So,
(16/81)=1-8/T)^4,
We have to find the value of T In RHS,
So by taking the fourth root of 16/8 we get 2/3,
Now by solving the eq 2/3=(1-8/T).
T=24.
Here Q=16(present quantity,from the ratio).
T=81(total quantity of solution,16+65),
R=8(replacing quantity).
n=4.
So,
(16/81)=1-8/T)^4,
We have to find the value of T In RHS,
So by taking the fourth root of 16/8 we get 2/3,
Now by solving the eq 2/3=(1-8/T).
T=24.
Prashant Shrestha said:
6 years ago
How come x(1-8/x)^4 = [x(1-8/x)^4] /x?
Rachana said:
6 years ago
@Vivek Kumar.
We have a formula for this to be solved.
(Q/T) = (1- (R/T) )^n.
Where
Q = Quantity of any liquid,
T = Total quantity of soln,
n = no.of operations,
R = Replacing quantity.
We have a formula for this to be solved.
(Q/T) = (1- (R/T) )^n.
Where
Q = Quantity of any liquid,
T = Total quantity of soln,
n = no.of operations,
R = Replacing quantity.
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