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# Solved Examples(Set 3) - Mixture and Alligation

 11. In what ratio must tea worth Rs. 60 per kg be mixed with tea worth Rs. 65 a kg such that by selling the mixture at Rs. 68.20 a kg ,there can be a gain 10%? A. 2 : 3 B. 3 : 4 C. 3 : 2 D. 4 : 3

Explanation:

Cost Price(CP) of 1 kg mixture = Rs. 68.20

Profit = 10%

Cost Price(CP) of 1 kg mixture $=\dfrac{100}{(100+\text{Profit}\%)}× \text{SP}$
$=\dfrac{100}{(100+10)}× 68.20\\\\= \dfrac{100}{110}× 68.20=\dfrac{682}{11}= \text{Rs. }62$

By rule of alligation

 CP of 1 kg teaof 1st kind CP of 1 kg teaof 2nd kind 60 65 Mean Price 62 65 - 62 = 3 62 - 60 = 2

Hence required ratio = 3 : 2

 12. A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially? A. 21 B. 19 C. 17 D. 23

Explanation:

Let initial quantity of P in the container be $7x$
and initial quantity of Q in the container be $5x$

Now 9 litres of mixture is drawn off from the container.
Quantity of P in 9 litres of the mixture drawn off
$=9× \dfrac{7}{12} = \dfrac{63}{12}=\dfrac{21}{4}$
Quantity of Q in 9 litres of the mixture drawn off
$=9× \dfrac{5}{12} = \dfrac{45}{12}=\dfrac{15}{4}$

Hence,
Quantity of P remaining in the mixture after 9 litres is drawn off
$=7x-\dfrac{21}{4}$
Quantity of Q remaining in the mixture after 9 litres is drawn off
$=5x-\dfrac{15}{4}$

Since the container is filled with Q after 9 litres of mixture is drawn off, quantity of Q in the mixture
$= 5x-\dfrac{15}{4}+9=5x+\dfrac{21}{4}$

Given that the ratio of P and Q becomes 7 : 9

$\Rightarrow \left(7x-\dfrac{21}{4}\right) : \left(5x+\dfrac{21}{4}\right) = 7 : 9\\\Rightarrow 9\left(7x-\dfrac{21}{4}\right)=7\left(5x+\dfrac{21}{4}\right)\\\Rightarrow 63x-\left(\dfrac{9× 21}{4}\right)=35x+\left(\dfrac{7× 21}{4}\right)\\\Rightarrow 28x=\left(\dfrac{16× 21}{4}\right)\\\Rightarrow x=\left(\dfrac{16× 21}{4× 28}\right)$

Litres of P contained in the container initially
$=7x = \left(\dfrac{7 ×16× 21}{4× 28}\right)\\= \dfrac{16× 21}{4× 4}= 21$

 13. A vessel is filled with liquid, 3 parts of which are water and 5 parts are 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. $\dfrac{1}{3}$ B. $\dfrac{1}{4}$ C. $\dfrac{1}{5}$ D. $\dfrac{1}{6}$

Explanation:

Let the quantity of the liquid in the vessel = 8 litre. Then,
quantity of water in the liquid = 3 litre,
and quantity of syrup in the liquid = 5 litre.

Suppose $x$ litre of the mixture is drawn off and replaced with water. Then,
Quantity of water in the new mixture
$=3-\dfrac{3x}{8}+x$
Quantity of syrup in the new mixture
$=5-\dfrac{5x}{8}$

Given that in the new mixture, quantity of water = quantity of syrup
$\Rightarrow 3 - \dfrac{3x}{8} + x = 5 - \dfrac{5x}{8}\\\Rightarrow \dfrac{10x}{8} = 2 \\\Rightarrow \dfrac{5x}{4} = 2\\\Rightarrow x = \dfrac{8}{5}$

i.e., if the quantity of the liquid is 8 litre, $\dfrac{8}{5}$ litre of the mixture needs to be drawn off and replaced with water so that the mixture may be half water and half syrup.

It means $\dfrac{1}{5}$ of the mixture needs to be drawn off and replaced with water so that the mixture may be half water and half syrup.

 14. In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre? A. 1 : 2 B. 2 : 2 C. 3 : 1 D. 1 : 3

Explanation:

By rule of alligation,

 Cost Price of1 litre water Cost Price of1 litre milk 0 12 Mean Price 8 12-8=4 8-0=8

Required Ratio = 4 : 8 = 1 : 2

 15. In what ratio must tea at Rs.62 per kg be mixed with tea at Rs. 72 per kg so that the mixture must be worth Rs. 64.50 per kg? A. 3 : 1 B. 2 : 1 C. 1 : 2 D. 1 : 3

Explanation:

By rule of alligation,

 Cost of 1 kg of 1st kind tea Cost of 1 kg of 2nd kind tea 62 72 Mean Price 64.5 72-64.5=7.5 64.5-62=2.5

Required Ratio = 7.5 : 2.5 = 3 : 1