Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are given some** Mixture and Alligations** problems that is important for competitive exams. We have included Some questions that are repeatedly asked in exams !!!

**A 30 litres milk-water mixture contains 50% milk. How much pure milk need to be added to this mixture to make the mixture 30% water?**

A) 10 litres

B) 15 litres

C) 18 litres

D) 20 litres

E) 25 litres**D) 20 litres**

Explanation:

30 litres mixture contains 15 litres of water. When milk added to this, quantity of water will remain same (i.e. 15 l)

Let x l of pure milk to be added, then

30% of (30+x) = 15

Solve, x = 20**A can contains 60 litres of milk. 10 litres of this milk is taken out and replaced with water. This process is repeated again. Find the amount of remaining milk in the mixture?**

A) 125/3 litres

B) 26 2/3 litres

C) 40 litres

D) 45 litres

E) 25 litres**A) 125/3 litres**

Explanation:

Remaining milk = 60 [1 – (10/60)]^{2}**A mixture contains A and B in the ratio 2 : 5. 7 litres of mixture is replaced by 18 litres of A and the new ratio becomes 8 : 11 respectively. What is the amount of A present in the original mixture?**

A) 20 litres

B) 18 litres

C) 44 litres

D) 24 litres

E) None of these**D) 24 litres**

Explanation:

A = 2x, B = 5x

After 7 litres draw out:

A = 2x – (2/7)*7 = 2x – 2 , B = 5x – (5/7)*7 = 5x – 5

After 18 l of A put back, A = (2x – 2) + 18 = 2x +16

So (2x+16)/(5x-5) = 8/11

Solve, x = 12

So A = 2x = 2*12**Rs 2000 is lent in 2 parts, 1 part at 8% per annum and 2nd part at 15% per annum. At the end of a year Rs 216 is received as simple interest. Find the part lent at 15% p.a.**

A) Rs 1200

B) Rs 800

C) Rs 900

D) Rs 750

E) Rs 1500**B) Rs 800**

Explanation:

216 = 2000*r*1/100

r = 10.8%

By the method of allegation:

1st part 2nd part

8 15

. 10.8

4.2 2.8

4.2 : 2.8 = 3 : 2

At 15% = (2/5) * 2000**1st mixture contains 30% zinc and rest copper and a 2nd mixture contains 20% zinc and rest copper. Some quantity is taken out of 1st mixture and twice this quantity is taken from 2nd mixture and mixed in a bottle. Find the ratio of copper to zinc in the bottle.**

A) 31 : 12

B) 23 : 7

C) 22 : 13

D) 25 : 9

E) None of these**B) 23 : 7**

Explanation:

Let x litres taken from 1st mixture, then 2x litres from 2nd mixture. So ratio of copper to zinc =

70% of x + 80% of 2x : 30% of x + 20% of 2x**A 132 litres of mixture contains milk and water in the ratio 5 : 7. How much milk need to be added to this mixture so that the new ratio is 13 : 11 respectively?**

A) 36 litres

B) 40 litres

C) 28 litres

D) 32 litres

E) 30 litres**A) 36 litres**

Explanation:

Milk in original = (5/12) * 132 = 55 l, so water = 132 – 55 = 77 l

Let x l of milk to be added, so

(55+x)/77 = 13/11

Solve, x = 36**How many kilograms of wheat at Rs 42 per kg be mixed with 25 kg of wheat at Rs 24 per kg so that on selling the mixture at Rs 40 per kg, there is a gain of 25%?**

A) 24 kg

B) 20 kg

C) 42 kg

D) 36 kg

E) 22 kg**B) 20 kg**

Explanation:

Let x kg of wheat at Rs 42 per kg be mixed

SP = 40, gain% = 25%, so CP = (100/125) * 40 = Rs 32

So

1st wheat(x kg) 2nd wheat(25 kg)

42 24

. 32

8 10

So 8/10 = x/25

Solve, x = 20**150 kg of wheat is at Rs 7 per kg. 50 kg is sold at 10% profit. At what rate per kg the remaining need to be sold so that there is a profit of 20% on the total price?**

A) Rs 9.75

B) Rs 8.25

C) Rs 8.75

D) Rs 10.25

E) Rs 10**C) Rs 8.75**

Explanation:

Let remaining 100 kg at x%. so,

50 kg 100 kg

10% x%

. 20%

(x-20) 10

So (x-20)/10 = 50 kg/100 kg

Solve, x = 25%

100 kg costs = 100*7 = Rs 700

So at 25% profit SP of 100 kg is (125/100) * 700 = Rs 875

So SP of 1 kg = 875/100 = 8.75**A can contains milk and water in the ratio 3:1. A part of this mixture is replaced with milk, and now the new ratio of milk to water is 15:4.What proportion of original mixture had been replaced by milk?**

A) 9/19

B) 8/19

C) 6/17

D) 3/19

E) 4/19**D) 3/19**

Explanation:

Let total original quantity = x litres, Let y litres replaced.

After y litres of mixture drawn out,

Milk = [3/(3+1)] * x – [3/(3+1)] * y

Water = [1/(3+1)] * x – [1/(3+1)] * y

Now y litres of milk poured in can. Milk becomes (3/4)*x – (3/4)*y +y = (3/4)*x +(1/4)*y

Now [(3/4)*x +(1/4)*y] / [(1/4)*x – (1/4)*y] = 15/4

Solve, y = (3/19)* x

So 3/19 of original mixture removed.**A can contains 40 litres of milk at Rs 3.5 per litre. How much water must be added to this can so that the cost of milk reduces to Rs 2 per litre?**

A) 20 litres

B) 30 litres

C) 24 litres

D) 38 litres

E) None of these**B) 30 litres**

Explanation:

Milk(40 litres) Water(x litres)

3.5 0

. 2

2 1.5

Ratio = 2 : 1.5 = 4 : 3

so 4/3 = 40/x

solve, x = 30

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