# Alligation or Mixture Questions

FACTS  AND  FORMULAE  FOR  ALLIGATION  OR  MIXTURE  QUESTIONS

I. 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 a desired price.

II. Mean Price : The cost price of a unit quantity of the mixture is called the mean price.

III. Rule of Alligation : Suppose Rs.x per unit be the price of first ingradient mixed with another ingradient (cheaper) of price Rs.y per unit to form a mixture whose mean price is Rs. z per unit, then

Quantity of cheaper : Quantity of dearer

= ( C.P of dearer - Mean Price ) : ( Mean Price - C. P of cheaper )

= ( x- z ) : ( z - y )

IV. Suppose a container contains x units of liquid from which y units are taken out and replaced by water. After  operations , the quantity of pure liquid = $x{\left(1-\frac{y}{x}\right)}^{n}$ units.

Q:

8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?

 A) 18 litres B) 24 litres C) 32 litres D) 42 litres

Explanation:

Let the quantity of the wine in the cask originally be x litres

Then, quantity of wine left in cask after 4 operations =$x1-8x4$litres

$⇒1-8x4=234$

$⇒x=24$

698 185180
Q:

A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is

 A) 400 kg B) 560 kg C) 600 kg D) 640 kg

Explanation:

By the rule of alligation:

Profit of first part                         Profit of second part

So, ratio of 1st and 2nd parts = 4 : 6 = 2 : 3.

Therefore, Quantity of 2nd kind = (3/5 x 1000)kg = 600 kg.

390 108742
Q:

Tea worth of Rs. 135/kg & Rs. 126/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.1700 C) Rs. 175.50 D) Rs. 180

Explanation:

Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50

So, Now 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'.

Cost of 1 kg tea of 1st kind         Cost of 1 kg tea of 2nd kind

x-153/22.50 = 1  => x - 153 = 22.50  => x=175.50.

Hence, price of the third variety = Rs.175.50 per kg.

591 89104
Q:

How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?

 A) 36 Kg B) 42 Kg C) 54 Kg D) 63 Kg

Explanation:

By the rule of alligation:

C.P. of 1 kg sugar of 1st kind      C.P. of 1 kg sugar of 2nd kind

Therefore, Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.

Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind.

Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.

178 64283
Q:

The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is

 A) Rs. 19.50 B) Rs. 19 C) Rs. 18 D) Rs. 18.50

Explanation:

Let the price of the mixed variety be Rs. x per kg. By the rule of alligation, we have :

Cost of 1 kg of type 1 rice           Cost of 1 kg of type 2 rice

$∴$(20-x)/(x-15) = 2/3

$⇒$ 60 - 3x = 2x - 30

$⇒$ x = 18.

150 55142
Q:

In what proportion water must be added to spirit to gain 20% by selling it at the cost price ?

Let the C.P of spirit be = Rs.10 per litre

Then S.P l litre of the mixture = Rs. 11

Profit = 20%

Therefore, C.P of the mixture = Rs. $100120×10$ = Rs. 25/3 per litre

Therefore, Ratio of water and spirit = 1 : 5

52487
Q:

In a pot, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk, the pot would be full and ratio of milk and water would become 6 : 5. Find the capacity of the pot ?

 A) 11 lit B) 22 lit C) 33 lit D) 44 lit

Explanation:

Let the capacity of the pot be 'P' litres.
Quantity of milk in the mixture before adding milk = 4/9 (P - 8)
After adding milk, quantity of milk in the mixture = 6/11 P.
6P/11 - 8 = 4/9(P - 8)
10P = 792 - 352 => P = 44.

The capacity of the pot is 44 liters.

61 51495
Q:

A can contains a mixture of two liquids A and B in 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?

 A) 10 B) 20 C) 21 D) 25

Explanation:

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

Quantity of A in mixture left =$7x-712×9=7x-214$ litres.

Quantity of B in mixture left =  $5x-512×9=5x-154$litres.

$7x-2145x-154+9=79$

$⇒28x-2120x+21=79$

$⇒x=3$

So, the can contained 21 litres of A.