# Problems on Mixtures and Alligations For IBPS PO & CLERK ## Problems on Mixtures and Alligations For IBPS PO & CLERK TYPE 1: USING WEIGHTED AVERAGE

The concept of weighted average is useful when a different part of the group have the same property but the proportion of each part is different. In such a case, the proportion of each part to the whole has to be factored in while finding out the average value of the attribute or property. For instance, if a milkman mixes the 2 litres of milk costing Rs. 6 per litre and 1 litre of milk costing rs. 9 per litre, then the average cost the mixture to be calculated using weighted averages.

Here the attribute is the cost and weight is the quantity of milk.

Cost of mixture =  = Rs.  7 per litre.

Points to remember:

• If the equal weights of two equal types of milk are mixed, the price of the mixture will be Rs. 7.50. This can be calculated by finding the simple average.
• The attribute should all be of the same unit. Similarly, the weights should also be of the same

Question 1. A vessel is filled with liquid, which is 3 parts water and 5 parts milk. How much of the liquid should be drawn off and replaced by water to make it half water and half milk?

Solution: Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.

water in new mixture =(3-3x/8+x) syrup in new mixture =(5-5x/8)

Then (3-3x/8+x) = (5-5x/8)

5x + 24 = 40 – 5x

10x=16==>x=8/5

So part of mixture replaced is 8/5*1/8=1/5.

Question 2.  Milk and water are in a Can A as 4:1 and in Can B as 3:2. For Can C, if one takes equal quantities from A and B, find the ratio of milk to water in C.

Solution: Ratio of only milk in vessel A = 4: 5

The ratio of only milk in vessel B = 3: 5

Let ‘x’ be the quantity of milk in vessel C

(3/5-x)/(x-4/5)=1/1

X=7/10

Therefore, the quantity of milk in vessel C = 7

=> Water quantity = 10 – 7 = 3

Hence the ratio of milk & water in vessel 3 is 7 : 3

Question 3.  A container contains 50 litres of milk. From this container, 5 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Solution: Amount of milk left after 3 operations ={50 (1-5/50)3 }

=50 * 9/10 * 9/10 * 9/10

=36.45

Question 4.  4 kg of a metal contains 1/5 copper and rest in Zinc. Another 5 kg of metal contains 1/6 copper and rest in Zinc.The ratio of Copper and Zinc into the mixture of these two metals?

Solution: Copper in 4 kg = 4/5

Zinc in 4 kg = 4*4/5=16/5

Copper in 5 kg = 5/6 and

Zinc in 5 kg = 5*5/6=25/6

Therefore, Copper in mixture = 4/5 + 5/6=49/30

and Zinc in the mixture = 16/5 + 25/6=221/30

Therefore the required ratio = 49: 221

Question 5.  Rs. 69 were divided among 115 students so that each girl gets 50 paise less than a boy. Thus each boy received twice the paise as each girl received. The no. of girls in the class is

Solution.  Here each girl receives 50 paise and each boy receives 100 paise and the average receiving of each student.

=6900/115=60paise

50…………………….100

…………..60

40……………………..10

4:1

5 == 115

=> 4? == 92

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Question 6.  A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:

Solution: Let CP of 1 litre milk be Rs 1

Then SP of 1 litre of mixture =Rs 1

Gain 25% CP of 1 litre mixture =Rs (100/125*1)=4/5

Milk………………..Water

1………………………0

…………..4/5

4/5……………………1/5

Ratio 4:1

Hence percentage of water in the mixture =(1/5*100)

=20%

Question 8.  A trader sells total 315 TV sets. He sells black and white TV sets at a loss of 6% and color TV sets at a profit of 15% thus he gains 9% on the whole. What is the no. of black and white sets which he has sold?

Solution: -6………………..+15

…………..9

6………………………15

2……………………..5

7 == 315

2 ? == 90

=> x=90

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August 26, 2019

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