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# Solved Examples(Set 5) - Mixture and Alligation

 21. How many kg of rice at Rs.6.60 per kg be mixed with 56 kg of rice at Rs.9.60 per kg to get a mixture worth Rs.8.20 per kg? A. 56 kg B. 52 kg C. 44 kg D. 49 kg

Explanation:

By rule of alligation,

 Cost of 1 kg of 1st kind rice Cost of 1 kg of 2nd kind rice 6.6 9.6 Cost of 1 kg of the mixture 8.2 9.6 - 8.2 = 1.4 8.2 - 6.6 = 1.6

Quantity of 1st kind rice : Quantity of 2nd kind rice = 1.4 : 1.6 = 7 : 8
=> Quantity of 1st kind rice : 56 = 7 : 8
=> Quantity of 1st kind rice $=56 × \dfrac{7}{8} = 49$

 22. How many litres of water must be added to 16 liters of milk and water containing 10% water to make it 20% water in it? A. 3 litre B. 2 litre C. 4 litre D. 1 litre

Explanation:

By rule of alligation,

 % Concentrationof water in pure water (100) % Concentrationof water in the given mixture (10) Mean % concentration(20) 20 - 10 = 10 100 - 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here quantity of the mixture = 16 litres
=> Quantity of water : 16 = 1 : 8
=> Quantity of water $=16 × \dfrac{1}{8} = 2$ litre

 23. We have a 630 ml mixture of milk and water in the ratio 7:2. How much water must be added to make the ratio 7:3? A. 60 ml B. 70 ml C. 50 ml D. 80 ml

Explanation:

concentration of water in mixture1 $=\dfrac{2}{9}$ (since the ratio of milk and water = 7:2) ...(1)

concentration of water in pure water= 1 ...(2)

Now the above mentioned items are mixed to form mixture2 where milk and water ratio = 7 : 3
=> concentration of water in mixture2 $=\dfrac{3}{10}$

By rule of alligation,

 concentration of water in mixture1 $\left(\dfrac{2}{9}\right)$ concentration of water in pure water (1) Mean concentration $\left(\dfrac{3}{10}\right)$ $1-\dfrac{3}{10}=\dfrac{7}{10}$ $\dfrac{3}{10}-\dfrac{2}{9}=\dfrac{7}{90}$

=> Quantity of mixture1 : Quantity of water
$=\dfrac{7}{10}:\dfrac{7}{90}=\dfrac{1}{10}:\dfrac{1}{90}=1:\dfrac{1}{9}$

Given that Quantity of mixture1 = 630 ml
=> 630 : Quantity of water $=1:\dfrac{1}{9}$
=> Quantity of water $=630 ×\dfrac{1}{9}=70$ ml

 24. 3 litre of water is added to 11 litre of a solution containing 42% of alcohol in the water. The percentage of alcohol in the new mixture is A. 20% B. 33% C. 30% D. 25%

Explanation:

Solution 1

We have a 11 litre solution containing 42% of alcohol in the water.
=> Quantity of alcohol in the solution $=\dfrac{11 × 42}{100}$

Now 3 litre of water is added to the solution.
=> Total quantity of the new solution = 11 + 3 = 14

Percentage of alcohol in the new solution $=\dfrac{\dfrac{11 × 42}{100}}{14}× 100$
$=\dfrac{11 × 3}{100} = 33\%$

Solution 2

%Concentration of alcohol in pure water = 0
%Concentration of alcohol in mixture = 42
Quantity of water : Quantity of mixture = 3 : 11
Let the %concentration of alcohol in the new mixture $=x$

By rule of alligation,

 %Concentration of alcoholin pure water (0) %Concentration of alcoholin mixture(42) Mean %concentration $(x)$ $42-x$ $x-0=x$

But $(42 - x) : x = 3 : 11$
$\Rightarrow 11(42-x)=3x\\\Rightarrow 42 × 11 - 11x = 3x\\\Rightarrow 14x = 42 × 11 \\\Rightarrow x = 3× 11 = 33$

i.e., Percentage of alcohol in the new mixture is 33%

 25. Rs.460 was divided among 41 boys and girls such that each boy got Rs.12 and each girl got Rs.8. What is the number of boys? A. 30 B. 28 C. 33 D. 36

Explanation:

Solution 1

Assume that the number of boys = b and number of girls = g

number of boys + number of girls = 41
=> b + g = 41 ...(1)

Given that each boy got Rs.12 and each girl got Rs.8. Then the total amount is Rs.460
=> 12b + 8g = 460 ...(2)

Now we need to solve these equations to get b and g.

(1) × 8 => 8b + 8g = 8 × 41 = 328 ...(3)
(2) - (3) => 4b = 460 - 328 = 132
=> b $=\dfrac{132}{4}=33$

Solution 2

Given that amount received by a boy = Rs.12,
Amount received by a girl =Rs.8

Total amount = 460

Given that number of boys + number of girls = 41
Hence, mean amount $=\dfrac{460}{41}$

 By rule of alligation, Amount receivedby a boy (12) Amount receivedby a girl(8) Mean amount $\dfrac{460}{41}$ $\dfrac{460}{41}-8=\dfrac{132}{41}$ $12-\dfrac{460}{41}=\dfrac{32}{41}$

Number of boys : Number of girls
=$\dfrac{132}{41}:\dfrac{32}{41}=132:32\\=66:16=33:8$

Given that number of boys + number of girls = 41

Hence number of boys $=41 × \dfrac{33}{41} = 33$