Aptitude - Alligation or Mixture - Discussion
Discussion Forum : Alligation or Mixture - General Questions (Q.No. 4)
4.
A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
Answer: Option
Explanation:
Let the cost of 1 litre milk be Re. 1
| Milk in 1 litre mix. in 1st can = | 3 | litre, C.P. of 1 litre mix. in 1st can Re. | 3 |
| 4 | 4 |
| Milk in 1 litre mix. in 2nd can = | 1 | litre, C.P. of 1 litre mix. in 2nd can Re. | 1 |
| 2 | 2 |
| Milk in 1 litre of final mix. = | 5 | litre, Mean price = Re. | 5 |
| 8 | 8 |
By the rule of alligation, we have:
| C.P. of 1 litre mixture in 1st can C.P. of 1 litre mixture in 2nd can | ||||||||
|
Mean Price
|
|
||||||
|
|
|||||||
 Ratio of two mixtures = |
1 | : | 1 | = 1 : 1. |
| 8 | 8 |
| So, quantity of mixture taken from each can = |  |
1 | x 12 |  |
= 6 litres. |
| 2 |
Discussion:
76 comments Page 3 of 8.
Andy said:
6 years ago
Three parts milk and 1 part water in one container and 1 part milk and 1 part water in the second container.
Representing the milk 3x/4+1x/2 = 7.5(5/8*12) Solving x you get 6.
Representing the water 1x/4+1x/2 = 4.5(3/8*12) Solving x you get 6.
Representing the milk 3x/4+1x/2 = 7.5(5/8*12) Solving x you get 6.
Representing the water 1x/4+1x/2 = 4.5(3/8*12) Solving x you get 6.
Sabbir said:
6 years ago
You can solve the math in this way as well,
In can1 there is 25% (1/4) water and 75% (3/4) milk.
Again, in can2 there is 50% (1/2) water and 50% (1/2) milk.
So, total volume of water = 1/4+1/2 = 3/4.
Total volume of milk = 3/4+1/2 = 5/4.
Now, as per the question, the ratio of water and milk must be 3x and 5x.
So, we can write, 3/4:5/4 = 3x:5x.
Or, 3/4* 5x = 5/4 * 3x.
Or, 15x/4 = 15x/4.
That's to say the volume of water and milk must be the same. Hence the answer will be 6 liters.
In can1 there is 25% (1/4) water and 75% (3/4) milk.
Again, in can2 there is 50% (1/2) water and 50% (1/2) milk.
So, total volume of water = 1/4+1/2 = 3/4.
Total volume of milk = 3/4+1/2 = 5/4.
Now, as per the question, the ratio of water and milk must be 3x and 5x.
So, we can write, 3/4:5/4 = 3x:5x.
Or, 3/4* 5x = 5/4 * 3x.
Or, 15x/4 = 15x/4.
That's to say the volume of water and milk must be the same. Hence the answer will be 6 liters.
(3)
Dinesh said:
6 years ago
How will you find the quantity of mixture?
Prakhar said:
6 years ago
Best explanation, thanks @Ajay.
Amin said:
7 years ago
Why we take 12 litres of milk as x+y=12?
Amit Ambadkar said:
7 years ago
Let's see 2 methods to solve the problem.
Let 1st can contain x lit of mixture(means 0.25x lit water and 0.75x lit milk).
2nd can contain y lit of mixture(means 0.5y lit milk and 0.5y lit water).
After mixing both the ratio of water to milk is 3:5.
So,
(0.25x+0.5y):(0.75x+0.5y) = 3:5.
After solving x:y=1:1,
And we have x+y=12,
So x=y=6.
Let 1st can contain x lit of mixture(means 0.25x lit water and 0.75x lit milk).
2nd can contain y lit of mixture(means 0.5y lit milk and 0.5y lit water).
After mixing both the ratio of water to milk is 3:5.
So,
(0.25x+0.5y):(0.75x+0.5y) = 3:5.
After solving x:y=1:1,
And we have x+y=12,
So x=y=6.
Guru Prasad said:
7 years ago
Alternate method:
-
CAN:1---> 25% water and the remaining 75% milk.
CAN:2---> 50% water and 50% milk.
Given ratio is 3:5. since only the quantity of milk is to be estimated therefore there is 5/8 of milk present in the mixture. converting this into percentage it is 62.50%.
By using allegation:-
{75% = M & 50% = N and D = 62.50%}--------> General Assumption.
M - D = 75% - 62.50%
= 12.50%.
D - N = 62.50% - 50%,
= 12.50%.
(M - D) : (D - N) = 1 : 1.
Therefore the quantity of milk to be mixed from each of the containers in order to 12 litres of milk is;
(1/2) * 12
= 6 litres. since the obtained ratio is 1:1 therefore from container-1 it is 6 litres and from container-2 it is also 6 litres.
-
CAN:1---> 25% water and the remaining 75% milk.
CAN:2---> 50% water and 50% milk.
Given ratio is 3:5. since only the quantity of milk is to be estimated therefore there is 5/8 of milk present in the mixture. converting this into percentage it is 62.50%.
By using allegation:-
{75% = M & 50% = N and D = 62.50%}--------> General Assumption.
M - D = 75% - 62.50%
= 12.50%.
D - N = 62.50% - 50%,
= 12.50%.
(M - D) : (D - N) = 1 : 1.
Therefore the quantity of milk to be mixed from each of the containers in order to 12 litres of milk is;
(1/2) * 12
= 6 litres. since the obtained ratio is 1:1 therefore from container-1 it is 6 litres and from container-2 it is also 6 litres.
(2)
Nani said:
7 years ago
Nice explanation, thanks a lot @Vivek Kumar.
Sravani said:
8 years ago
So in this problem,
5/8 is taken as the total mixture is in the ratio 3:5(water:milk).
so as we are calculating for milk quantity in the aligation process, the milk quantity is according to ratio 5/3+5=5/8.
Hope you understand this.
5/8 is taken as the total mixture is in the ratio 3:5(water:milk).
so as we are calculating for milk quantity in the aligation process, the milk quantity is according to ratio 5/3+5=5/8.
Hope you understand this.
Sridharsan said:
8 years ago
Actually, we need 12 ltrs of milk, here where is 12 ltr of milk, 6+6=12 (it is a mix, not purely milk).
Please explain in detail.
Please explain in detail.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers