Aptitude - Alligation or Mixture
Exercise : Alligation or Mixture - General Questions
- Alligation or Mixture - Formulas
- Alligation or Mixture - General Questions
1.
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Answer: Option
Explanation:
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
| Quantity of water in new mixture = |  |
3 - | 3x | + x |  |
litres |
| 8 |
| Quantity of syrup in new mixture = | |
5 - | 5x | |
litres |
| 8 |
 |
|
3 - | 3x | + x | |
= | |
5 - | 5x | |
| 8 | 8 |
5x + 24 = 40 - 5x
10x = 16
| x = |
8 | . |
| 5 |
| So, part of the mixture replaced = | |
8 | x | 1 | |
= | 1 | . |
| 5 | 8 | 5 |
2.
Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:
Answer: Option
Explanation:
Since first and second varieties are mixed in equal proportions.
| So, their average price = Rs. | |
126 + 135 | |
= Rs. 130.50 |
| 2 |
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.
By the rule of alligation, we have:
| Cost of 1 kg of 1st kind Cost of 1 kg tea of 2nd kind | ||
| Rs. 130.50 | Mean Price Rs. 153 | Rs. x |
| (x - 153) | 22.50 | |
| |
x - 153 | = 1 |
| 22.50 |
x - 153 = 22.50
x = 175.50
3.
A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
Answer: Option
Explanation:
Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
| Quantity of A in mixture left = | |
7x - | 7 | x 9 | |
litres = | |
7x - | 21 | litres. |
| 12 | 4 |
| Quantity of B in mixture left = | |
5x - | 5 | x 9 | |
litres = | |
5x - | 15 | litres. |
| 12 | 4 |
| |
|
= | 7 | |||||
|
9 |
|
28x - 21 | = | 7 |
| 20x + 21 | 9 |
252x - 189 = 140x + 147
112x = 336
x = 3.
So, the can contained 21 litres of A.
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 |
5.
In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?
Answer: Option
Explanation:
By the rule of alligation:
| Cost of 1 kg pulses of 1st kind Cost of 1 kg pulses of 2nd kind | ||
| Rs. 15 | Mean Price Rs. 16.50 | Rs. 20 |
| 3.50 | 1.50 | |
Required rate = 3.50 : 1.50 = 7 : 3.
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