## 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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