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:
74 comments Page 3 of 8.
Skk said:
9 years ago
Water = X.
Milk = Y.
Can1 contains Water = X/4 and milk is = 3Y/4.
Can2 contains Water = X/2 and Milk is = Y/2.
Total water and milk ratio = 3/5.
But water and milk = X + Y = 12.
Water/milk Ratio = X/4 + X/2/3Y/4 + Y/2= 3/5.
Substitute X = 12 - y.
We will get X = 6 & Y = 6.
Milk = Y.
Can1 contains Water = X/4 and milk is = 3Y/4.
Can2 contains Water = X/2 and Milk is = Y/2.
Total water and milk ratio = 3/5.
But water and milk = X + Y = 12.
Water/milk Ratio = X/4 + X/2/3Y/4 + Y/2= 3/5.
Substitute X = 12 - y.
We will get X = 6 & Y = 6.
MANISH said:
1 decade ago
Initially water and milk percent was 25% water and 75%milk
in 2nd container 50-50%
finally milk was 12 litre
ratio was 3:5 means 3x water and 5x milk s0 5x=12 ;x= 12/5
water 36/5
milk percentage 12/(12+36/5)=62.5
soincrease in milk percentage 12.5%each so omly one option is matching
in 2nd container 50-50%
finally milk was 12 litre
ratio was 3:5 means 3x water and 5x milk s0 5x=12 ;x= 12/5
water 36/5
milk percentage 12/(12+36/5)=62.5
soincrease in milk percentage 12.5%each so omly one option is matching
Meeran said:
1 decade ago
Water : Milk
Can 1 1 : 3
Can 2 1 : 1
Required ratio = 3 : 4.
Therefore ratio of can1+ can2+ added quantity = required quantity.
1+1+1 : 3+1+1 is the only possible way, ratio added must b,
1 : 1.
Hence, option B, 6 : 6 is only possible which can give 1 : 1 ratio.
Can 1 1 : 3
Can 2 1 : 1
Required ratio = 3 : 4.
Therefore ratio of can1+ can2+ added quantity = required quantity.
1+1+1 : 3+1+1 is the only possible way, ratio added must b,
1 : 1.
Hence, option B, 6 : 6 is only possible which can give 1 : 1 ratio.
Ramakrishna said:
10 years ago
In 1st 25% water, 2nd 50% water.
Then resultant ratio given that 3:5, if 3+5 is 100% then 3 is 37.5% water.
That is subtracted from 37.5-25 = 50-37.5 = 12.5.
Then those two values are same so ratio is 1:1 answer 6 lit, 6 lit.
If you do for milk also get same answer.
Then resultant ratio given that 3:5, if 3+5 is 100% then 3 is 37.5% water.
That is subtracted from 37.5-25 = 50-37.5 = 12.5.
Then those two values are same so ratio is 1:1 answer 6 lit, 6 lit.
If you do for milk also get same answer.
Shambhu said:
9 years ago
Let x l & y l taken from can A(1:3) & can B(1:1) resp.
In new 12 of mixture (ratio 3:5).
x + y =12.
Water= 3*12/8 & milk= 5 * 12/8.
Then now, x/4 + y/2 = 3*12/8. (W)
And, 3x/4 + y/2 = 5*12/8. (M)
Solving both equations, we get;
x =6l & y = 6l.
In new 12 of mixture (ratio 3:5).
x + y =12.
Water= 3*12/8 & milk= 5 * 12/8.
Then now, x/4 + y/2 = 3*12/8. (W)
And, 3x/4 + y/2 = 5*12/8. (M)
Solving both equations, we get;
x =6l & y = 6l.
Adithi said:
9 years ago
In the first can water and milk ratio is 1 : 3 and in the second it is 1 : 1.
Mixture is 3 : 5 now sub 1/4 and 3/8 (water) it is 1/8. Next 1/2-3/8 it is 1/8 (milk). So the water and milk ratio is 1 : 1. Hence in total of 12 litres 6 litres milk and 6 litres water.
Mixture is 3 : 5 now sub 1/4 and 3/8 (water) it is 1/8. Next 1/2-3/8 it is 1/8 (milk). So the water and milk ratio is 1 : 1. Hence in total of 12 litres 6 litres milk and 6 litres water.
Mahendra said:
1 decade ago
@moncy The solution to this problem is right
Ans is [B] 6 litres, 6 litre
6 liters from first can gives 1.5 litre of water and 4.5 of milk
6 liters from second can gives 3 litres of water and 3 litres of milk
so, (1.5+3)/(4.5+3)= 4.5/7.5 = 0.6 = 3/5
Ans is [B] 6 litres, 6 litre
6 liters from first can gives 1.5 litre of water and 4.5 of milk
6 liters from second can gives 3 litres of water and 3 litres of milk
so, (1.5+3)/(4.5+3)= 4.5/7.5 = 0.6 = 3/5
Andy said:
5 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.
Vikas said:
1 decade ago
Let x and y liters from the can to be mixed to make 12 liters of milk.
Can1 has x/4 (water) and 3x/4 (milk).
Can2 has y/2 (water) and y/2 (milk).
So new ratio of water to milk is.
(x/3+y/2) / (3x/4+y/2) =3/5.
And x+y=12.
X=6 and y=6.
Can1 has x/4 (water) and 3x/4 (milk).
Can2 has y/2 (water) and y/2 (milk).
So new ratio of water to milk is.
(x/3+y/2) / (3x/4+y/2) =3/5.
And x+y=12.
X=6 and y=6.
Vivek kumar said:
1 decade ago
If we take mixture of both the can be x.
Than (x/4+x/2)/(3x/4 + x/2) = 3x/5x = 3/5.
The ratio comes 3/5 only if we have equal amount of milk(x here). So, according to question,
x+x=12;
2x=12.
x = 6.
12-6 = 6.
So, Answer is 6, 6.
Than (x/4+x/2)/(3x/4 + x/2) = 3x/5x = 3/5.
The ratio comes 3/5 only if we have equal amount of milk(x here). So, according to question,
x+x=12;
2x=12.
x = 6.
12-6 = 6.
So, Answer is 6, 6.
(1)
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