# 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 = $\inline [x(1-\frac{y}{x})^n]$ 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.$\inline \fn_jvn \small \left ( \frac{126+135}{2} \right )$ = 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.

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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 =$\inline (7x-\frac{7}{12}\times 9)=(7x-\frac{21}{4})$ litres.

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

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

$\inline \Rightarrow \frac{28x-21}{20x+21}=\frac{7}{9}$

$\inline \Rightarrow 252x-189=140x+147$

$\inline \Rightarrow x=3$

So, the can contained 21 litres of A.

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

$\inline \fn_cm \therefore \left [ \frac{x(1-\frac{8}{x})^4}{x} \right ]=\frac{16}{81}$

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

$\inline \fn_cm \Rightarrow \left ( \frac{x-8}{x} \right )=\frac{2}{3}$

${\color{Black}&space;\Rightarrow&space;3x-24=2x}$

${\color{Black}&space;\Rightarrow&space;x=24}$

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

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

${\color{Blue}&space;\therefore&space;}$ $\inline (3-\frac{3x}{8}+x)=(5-\frac{5x}{8})$

=>   5x + 24 = 40 - 5x

=>   10x = 16    => x = 8/5

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

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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.

${\color{Blue}&space;\therefore&space;}$ Quantity of 2nd kind = (3/5 x 1000)kg = 600 kg.