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 !!

Questions Penned by Yogit

**An alloy contains iron, copper and zinc in the ratio of 3:4:2. Another alloy contains copper, zinc and tin in the ratio of 10:5:3. If equal quantities of both alloys are melted, then weight of tin per kg in the new alloy**

a) 1/8 kg

b) 1/10 kg

c) 1/12 kg

d) 1/14 kg

e) None of theseAnswer –**c) 1/12 kg**

**Explanation :**

I:C:Z = 3:4:2 (in first alloy) and C:Z:T = 10:5:3

Equal quantities is taken. So, I:C:Z = 6:8:4 in first alloy and C:Z:T = 10:5:3

I = 6

C = 8 + 10 = 18

Z = 4+5 = 9

T = 3

So weight of tin = 3/36 = 1/12**8 litres are drawn from a flask containing milk and then filled with water. The operation is performed 3 more times. The ratio of the quantity of milk left and total solution is 81/625. How much milk the flask initially holds?**

a) 10ltr

b) 20ltr

c) 30ltr

d) 40ltr

e) None of theseAnswer –**b) 20ltr**

**Explanation :**

let initial quantity be Q, and final quantity be F

F = Q*(1 – 8/Q)^4

81/625 = (1-8/Q)^4

3/5 = 1 – 8/Q

Q = 20**A 40 litre mixture contains milk and water in the ratio of 3:2. 20 litres of the mixture is drawn of and filled with pure milk. This operation is repeated one more time. At the end what is the ratio of milk and water in the resulting mixture?**

a) 5:1

b) 6:1

c) 8:1

d) 9:1

e) None of theseAnswer –**d) 9:1**

**Explanation :**

milk = 40*3/5 = 24 and water = 16 litres initially

milk = 24 – 20*3/5 + 20 = 32 – 20*4/5 + 20 = 36

water = 16 – 20*2/5 = 8 – 20*1/5 = 4**Two vessels contain milk and water in the ratio of 7:3 and 2:3 respectively. Find the ratio in which the contents of both the vessels must be mixed to get a new mixture containing milk and water in the ratio 3:2.**

a) 2:1

b) 2:3

c) 3:1

d) 3:5

e) None of theseAnswer –**a) 2:1**

**Explanation :**

Let the ratio be k:1

then in first mixture, milk = 7k/10 and water = 3k/10

and in second mixture, milk = 2/5 and water = 3/5

[7k/10 + 2/5]/[3k/10 3/5] = 3/2

K = 2, so ratio will be 2:1**How many Kgs of rice A costing rupees 20 per kg must be mixed with 20 kg of rice B costing rupees 32 per kg, so that after selling them at 35 rupees per kg, he gets a profit of 25%.**

a) 10 kg

b) 40 kg

c) 24 kg

d) 25 kg

e) None of theseAnswer –**a) 10 kg**

**Explanation :**

by rule of alligation,

20 32

…….28………..

4 8

So, x = 10**How many litres of water must be added to 60 litre mixture that contains milk and water in the 7:3 such that the resulting mixture has 50% water in it?**

a) 12

b) 16

c) 24

d) 28

e) None of theseAnswer –**c) 24**

**Explanation :**

milk = (7/10)*60 = 42 and water = 18

so water must be added = 42 – 18 = 24**A sample of x litre is replaced from a container containing milk and water in the ratio 2:3 by pure milk. If the container hold 30 litres of the mixture, and after the operation proportion of milk and water is same. Find the value of X?**

a) 4

b) 5

c) 6

d) 7

e) None of theseAnswer –**b) 5**

**Explanation :**

milk = 30*2/5 = 12 and water = 30*3/5 = 18

milk = 12 – x*2/5 + x and water = 18 – x*3/5

equate both equation and we get x = 5**Two cans P and Q contains milk and water in the ratio of 3:2 and 7:3 respectively. The ratio in which these two cans be mixed so as to get a new mixture containing milk and water in the ratio 7:4.**

a) 4:7

b) 7:3

c) 7:4

d) 7:5

e) None of theseAnswer –**c) 7:4**

**Explanation :**

Milk in 1^{st}can = 3/5 and water = 2/5. Similarly in second can milk = 7/10 and water = 3/10.

Take the ratio = K:1

(3k/5 + 7/10)/(2k/5 + 3/10) = 7/4

Solve for k, we get k = 7/4. So the ratio is 7:4**A trader mixes 6ltr of milk costing 5000 rupees with 7ltr of milk costing 6000 rupees per litre. The trader also mixes some quantity of water to the mixture so as to bring the price to 4800 per litre. How many litres of water is added**

a) 1ltr

b) 2ltr

c) 3ltr

d) 4ltr

e) None of theseAnswer –**b) 2ltr**

**Explanation :**

(6*5000 + 7*6000)/(13 + w) = 4800 (w is the amount of water added)**How many kilograms of rice costing Rs. 9 per kg must be mixed with 27kg of rice costing Rs.7 per kg so that there may be gain of 10% by selling the mixture at Rs.9.24 per kg?**

a) 63

b) 56

c) 49

d) 35

e) None of theseAnswer –**a) 63**

**Explanation :**

By rule of allegation: –

9 7

….8.4…..

1.4 0.6

So, x/27 = 1.4/0.6, we get x = 63 kg

**Note:** Dear Readers if you have any doubt in any chapter in Quants you can ask here. We will clear your doubts

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