# 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:

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.

176 48183
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 =$\left[x{\left(1-\frac{8}{x}\right)}^{4}\right]$litres

$⇒{\left[1-\frac{8}{x}\right]}^{4}={\left(\frac{2}{3}\right)}^{4}$

$⇒x=24$

115 25061
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 =$\left(7x-\frac{7}{12}×9\right)=\left(7x-\frac{21}{4}\right)$ litres.

Quantity of B in mixture left =  $\left(5x-\frac{5}{12}×9\right)=\left(5x-\frac{15}{4}\right)$litres.

$\frac{\left(7x-\frac{21}{4}\right)}{\left(5x-\frac{15}{4}\right)+9}=\frac{7}{9}$

$⇒\frac{28x-21}{20x+21}=\frac{7}{9}$

$⇒x=3$

So, the can contained 21 litres of A.

112 23728
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.

82 16476
Q:

A vessel is filled with liquid, 3 parts of which are water and 5 parts of 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?

 A) 1/3 B) 1/4 C) 1/5 D) 1/7

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 =  $\left(3-\frac{3x}{8}+x\right)$litres.

Quantity of syrup in new mixture =  $\left(5-\frac{5x}{8}\right)$ litres.

$\left(3-\frac{3x}{8}+x\right)=\left(5-\frac{5x}{8}\right)$

=>   5x + 24 = 40 - 5x

=>   10x = 16    => x = 8/5

So, part of the mixture replaced = $\left(\frac{8}{5}×\frac{1}{8}\right)$ = 1/5.

102 13657
Q:

A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

 A) 10 liters B) 20 liters C) 30 liters D) 40 liters

Explanation:

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters

Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)

Thus, (30 + P) = 25% of (150 + P)

Solving, we get P = 10 liters

35 11984
Q:

The ratio of expenditure and savings is 3 : 2 . If the income increases by 15% and the savings increases by 6% , then by how much percent should his expenditure increases?

 A) 25 B) 21 C) 12 D) 24

Explanation:

Therefore x = 21%

53 11658
Q:

The ratio of petrol and kerosene in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the kerosene, the ratio become 2:3. Then total quantity of the mixture in the container is:

 A) 25 B) 30 C) 45 D) cannot be determined

Explanation:

pool : kerosene

3  :  2(initially)

2  :  3(after replacement)

(for petrol)   $\frac{2}{3}=\left(1-\frac{10}{k}\right)$

=> K = 30

Therefore the total quantity of the mixture in the container is 30 liters.