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 1 of 8.
Moncy said:
1 decade ago
If it is 6litres of milk and 6litres of water making 12litres of the mix,then does not it mean that milk and water is 50% and 50%??how will it be in the ratio 3:5 to satisfy the requirement??Please explain this..
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
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
Nagendramurthy said:
1 decade ago
Friends see can also solve the problem..
initially water and milk in first can are in 1/4 and 3/4
in the second can 1/2 and 1/2
therefore total (1/4)+(1/2) of water and (3/4)+(1/2)
i.e (3/4) of water and (5/4) of milk
If we compare it already in the ratio of 3:5
So we have add equal amount of water and milk in order to maintain the ratio same..
initially water and milk in first can are in 1/4 and 3/4
in the second can 1/2 and 1/2
therefore total (1/4)+(1/2) of water and (3/4)+(1/2)
i.e (3/4) of water and (5/4) of milk
If we compare it already in the ratio of 3:5
So we have add equal amount of water and milk in order to maintain the ratio same..
Charan said:
1 decade ago
Lets take total in terms of mixture in x & y
new water mixture = x/4+x/2
milk = 3y/4 + y/2
ratio comes out as x:y = 3:5
thats its self gives the new mixture
2x=12 or 2y=12
6ltr n 6ltr
new water mixture = x/4+x/2
milk = 3y/4 + y/2
ratio comes out as x:y = 3:5
thats its self gives the new mixture
2x=12 or 2y=12
6ltr n 6ltr
Kunal said:
1 decade ago
CAN 1 -water 1/4, milk 3/4 x.
CAN 2 - water 1/2, milk 1/2 y.
Required " 3/8, " 5/8.
x/4+y/2=3/8 3x/4+y/2=5/8.
Solving x=1/2, y=1/2.
x:y = 1:1.
So 6, 6.
CAN 2 - water 1/2, milk 1/2 y.
Required " 3/8, " 5/8.
x/4+y/2=3/8 3x/4+y/2=5/8.
Solving x=1/2, y=1/2.
x:y = 1:1.
So 6, 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.
Ajinx999 said:
1 decade ago
Let x L of milk be removed from can A (25% water). Since, 12 L of milk is finally required, (12-x) L of milk is removed from can B (50% water).
Considering following ratio
(water from A) + (water from B) 3
--------------------------------------- = ---
(pure_milk from A) + (pure_milk from B) 5
(1/4)x + (1/2)(12-x) 3
-------------------- = ---
(3/4)x + (1/2)(12-x) 5
Solving for x, x = 6
So 6 L is removed from can A and (12-6) = 6 L is removed from can B.
Considering following ratio
(water from A) + (water from B) 3
--------------------------------------- = ---
(pure_milk from A) + (pure_milk from B) 5
(1/4)x + (1/2)(12-x) 3
-------------------- = ---
(3/4)x + (1/2)(12-x) 5
Solving for x, x = 6
So 6 L is removed from can A and (12-6) = 6 L is removed from can B.
Amisha said:
1 decade ago
Can1 75% milk Can2 50% milk
mean 62.5%(3:5,so milk = 5/8*100 = 62.5)
12.5 12.5
so, ratio is = 1:1 so, in 12 lit mixture 6lit from 1can and 6 lit from 2 can
mean 62.5%(3:5,so milk = 5/8*100 = 62.5)
12.5 12.5
so, ratio is = 1:1 so, in 12 lit mixture 6lit from 1can and 6 lit from 2 can
ANIL KUMAR JAT said:
1 decade ago
With the help of option
M W
6 3 : 1
6 1 : 1
M=3+4.5=7.5
W=3+1.5=4.5
THERFORE W:M=4.5:7.5=3:5 IS GIVEN SO ANSWER 6,6 IS CORRECT.
M W
6 3 : 1
6 1 : 1
M=3+4.5=7.5
W=3+1.5=4.5
THERFORE W:M=4.5:7.5=3:5 IS GIVEN SO ANSWER 6,6 IS CORRECT.
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