Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to pro...
Alligation:
It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a
mixture of desired price.
Mean Price:
The cost of a unit quantity of the mixture is called the mean price.
Rule of Alligation:
If two ingredients are mixed, then
| |  | Quantity of cheaper |  | = | | C.P. of dearer - Mean Price | |
| Quantity of dearer | Mean price - C.P. of cheaper |
We present as under:
Suppose a container contains x of liquid from which y units are taken out and replaced by water.
| After n operations, the quantity of pure liquid = |  | x | | 1 - | y | | n |  | units. |
| x |
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 C
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 |
 5 x + 24 = 40 - 5 x
10 x = 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: |
| A. | Rs. 169.50 |
| B. | Rs. 170 |
| C. | Rs. 175.50 |
| D. | Rs. 180 |
Answer: Option C
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.
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 C
Suppose the can initially contains 7 x and 5 x 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 |
 |
| 7x - | 21 | |
| 4 |
| = | 7 |
| 5x - | 15 | + 9 |
| 4 |
| 9 |
| 28x - 21 | = | 7 |
| 20x + 21 | 9 |
252 x - 189 = 140 x + 147
112 x = 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?
|
| A. | 4 litres, 8 litres |
| B. | 6 litres, 6 litres |
| C. | 5 litres, 7 litres |
| D. | 7 litres, 5 litres |
Answer: Option B
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 |
| So, quantity of mixture taken from each can = | | 1 | x 12 | | = 6 litres. |
| 2 |
| Ratio of two mixtures = | 1 | : | 1 | = 1 : 1. |
| 8 | 8 |
Examples:- click here
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