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 5 of 8.
Dhanwin said:
1 decade ago
Different approach:
==============
Can 1 ---- 25% water + 75% milk.
Ratio ---- 1:3 (25/75 = 1/3).
Can 2 ---- 50% water + 50% milk.
Ratio ----- 1:1 (50/50 = 1/1).
If I take p liters from can 1 and q liters from can 2.
So that total should be 12 liters with water : milk = 3 : 5.
So I get following equation.
P+q/3p+q = 3/5 -------- (equation 1).
Now try checking the given options in questions and the answer should be the "option" which satisfies the (equation 1).
Ex:
Let p = 6 and q = 6.
Then 6+6/(3*6+6) = 3/5.
So the equation 1 is satisfied. So the answer is 6 liters from can 1 and 6 liters from can 2.
Other options given won't satisfy the (equation 1).
==============
Can 1 ---- 25% water + 75% milk.
Ratio ---- 1:3 (25/75 = 1/3).
Can 2 ---- 50% water + 50% milk.
Ratio ----- 1:1 (50/50 = 1/1).
If I take p liters from can 1 and q liters from can 2.
So that total should be 12 liters with water : milk = 3 : 5.
So I get following equation.
P+q/3p+q = 3/5 -------- (equation 1).
Now try checking the given options in questions and the answer should be the "option" which satisfies the (equation 1).
Ex:
Let p = 6 and q = 6.
Then 6+6/(3*6+6) = 3/5.
So the equation 1 is satisfied. So the answer is 6 liters from can 1 and 6 liters from can 2.
Other options given won't satisfy the (equation 1).
Sri said:
1 decade ago
Can anybody tell me how 5/8 arrived here?
Sri Divya said:
1 decade ago
Hi friends, Please explain how 5/8 became the mean price?
How to calculate mean price?
How to calculate mean price?
Gokulpriya said:
1 decade ago
How did 5/8 become mean prize what is this value?
Vaishnavi said:
1 decade ago
Please explain me in a simple way guys?
I can't get the logic?
I can't get the logic?
ANAND PATEL said:
1 decade ago
From the first can mixture is X liter.
From the first can mixture is Y liter.
NOW X + Y = 12.
IN 12 LITER portion of water is 4.5 and milk IS 7.5 LITER.
X( (1/4)+ (1/2))+ Y ((3/4)+(1/2))= 4.5 + 7.5.
SO X( (1/4)+ (1/2))= 4.5 => X= 6 LITER.
Y ((3/4)+(1/2)) = 7.5 +> Y = 6 LITER.
From the first can mixture is Y liter.
NOW X + Y = 12.
IN 12 LITER portion of water is 4.5 and milk IS 7.5 LITER.
X( (1/4)+ (1/2))+ Y ((3/4)+(1/2))= 4.5 + 7.5.
SO X( (1/4)+ (1/2))= 4.5 => X= 6 LITER.
Y ((3/4)+(1/2)) = 7.5 +> Y = 6 LITER.
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.
Vikram Ojha said:
1 decade ago
1st can Water:milk = 1:3.
In 2nd can water:milk = 1:1.
In 12 liters = water:milk = 3:5.
So in 12 liters we have 3(12)/8 = 4.5 liter of water.
Lets us consider we have x liters of water in new mix.
So our equation will be,
1(x)/4 + 1(x)/2 = 4.5.
x = 6 liters.
So amount of water in new mixture = 6 liters.
And that of milk will be = 12-x = 6 liters.
In 2nd can water:milk = 1:1.
In 12 liters = water:milk = 3:5.
So in 12 liters we have 3(12)/8 = 4.5 liter of water.
Lets us consider we have x liters of water in new mix.
So our equation will be,
1(x)/4 + 1(x)/2 = 4.5.
x = 6 liters.
So amount of water in new mixture = 6 liters.
And that of milk will be = 12-x = 6 liters.
Dhruv Sahni said:
1 decade ago
Container A has 25% water, Container B has 50% of water and after mixing these two final mixture has 3/8 -> 37.5% of water.
So by allegation method the ratio of A and B in final mixture will be 25% 50% 37.5%.
50%-37.5% : 37.5%-25%.
= 12.5% = 12.5% = 1:1.
So Ratio of A:B in final mixture is 1:1, hence in 12 liter of mixture there will be 6L of A and 6L of B.
So by allegation method the ratio of A and B in final mixture will be 25% 50% 37.5%.
50%-37.5% : 37.5%-25%.
= 12.5% = 12.5% = 1:1.
So Ratio of A:B in final mixture is 1:1, hence in 12 liter of mixture there will be 6L of A and 6L of B.
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.
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