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 !!

**Two varieties of rice costing Rs 25 and Rs 35 respectively are mixed in a certain ration and the resulting mixture is sold at a profit of 20% at Rs 34.8. What is the respective ratio in which they are mixed?**

A) 5 : 9

B) 4 : 1

C) 3 : 2

D) 19 : 7

E) 16 : 9**C) 3 : 2**

Explanation:

SP = 34.8, profit = 20%, so CP = (100/120)*34.8 = 29

By method of allegation:

25 Â Â Â Â Â Â Â 35

. Â Â Â Â 29

6 Â Â Â Â Â Â Â Â 4

3 : 2**How much quantity of water should be mixed with 10 l of milk costing Rs 50 per litre so that the resultant mixture is to be sold at Rs 44 per kg?**

A) 9.5 litres

B) 2 litres

C) 1 7/11 litres

D) 1 4/11 litres

E) 12 litres**D) 1 4/11 litres**

Explanation:

CP of water = Rs 0

By method of allegation:

water (x kg) Â Â Â Â Â Â milk (10 litres)

0 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 50

. Â Â Â Â Â Â Â Â Â Â 44

6 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 44

6 : 44

3 : 22

So x/10 = 3/22**In what ratio must the 3 varieties of wheat costing Rs 42, Rs 54 and Rs 65 respectively be mixed so that the resulting mixture is sold for Rs 63.8 at a profit of 10%?**

A) 7 : 20 : 7

B) 4 : 8 : 13

C) 5 : 5 : 21

D) 5 : 9 : 21

E) 7 : 7 : 20**E) 7 : 7 : 20**

Explanation:

SP = 63.8, Profit=10%, so CP = (100/110)*63.8 = 58

Now 58 is greater than 42 and 54 and less than 65

So

42 Â Â Â Â Â Â Â 65

. Â Â Â Â 58

7 Â Â Â Â Â Â Â Â 16

And

54 Â Â Â Â Â Â Â 65

. Â Â Â Â 58

7 Â Â Â Â Â Â Â Â Â 4

**So 1 part of 1st wheat A, 1 part of 2nd wheat B and 2 parts of 3rd wheat C gives**

A : C = 7 : 16, and B : C = 7 : 4

So A : B : C = 7 : 7 : (16+4)

***we have taken 2 parts of C so it is added – when there are 3 varieties to be mixed**

it is**not**done like simple calculation of A : C and B : C**A can containing 25 litres of mixture of milk and water has 80% milk in it. How much quantity of the mixture be drawn out and replaced with water such that the new ratio of water to milk becomes 1 : 3?**

A) 1.5625 litres

B) 4.725 litres

C) 9.25 litres

D) 10 litres

E) 6.875 litres**E) 1.5625 litres**

Explanation:

Milk = (80/100)*25 = 20, so water = 5 l

So water : milk = 5 : 20 = 1 : 4

Let x litres drawn out

So water left = 5 â€“ (1/(1+4))*x = 5 â€“ x/5

Milk left = 20 â€“ (4/(1+4))*x = 20 â€“ 4x/5

Now x litres of water is added too, so water becomes = 5 â€“ x/5 + x = 5 + 4x/5

So [5 + 4x/5] / [20 â€“ 4x/5] = 1/3

75x+12x = 100x-4x

16x = 25

x=25/16 = 1.5625**A can contains 60 litres of milk. 4 litres of milk is drawn and replaced with water. This procedure is repeated once again. How much quantity of milk is remained in the can?**

A) 45 litres

B) 52.26 litres

C) 55 litres

D) 40.54 litres

E) 58.92 litres**B) 52.26 litres**

Explanation:

Performed 2 times, so

Milk left = 60 [1 â€“ 4/60]^{2}**A variety of wheat costing Rs 8.70 is mixed with another variety in a ratio 2 : 3. If the mixture is sold at Rs 8.10 making a loss of 10%, then what is the cost of 2nd variety of wheat?**

A) Rs 9420

B) Rs 9.20

C) Rs 10

D) Rs 9.50

E) Rs 10.20**A) Rs 9.20**

Explanation:

SP = 8.1, loss = 10%, so CP = (100/90)*8.1 = Rs 9

1st wheat Â Â Â Â Â Â Â 2nd wheat

8.70 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

. Â Â Â Â Â Â Â Â Â 9

2 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 3

So (x â€“ 9)/(9 â€“ 8.70) = 2/3

Solve, x = 9.20**An alloy contains 4 parts bronze and 6 parts copper. How much part of mixture should be drawn out and replaced with bronze so that the ratio of bronze to copper gets reversed in new mixture?**

A) 2/5

B) 1/3

C) 1/4

D) 10/3

E) 12/7**B) 1/3**

Explanation:

Total = 4+ 6 = 10

The old ratio is 4/6 = 2/3, so new ratio of bronze to copper should be 3/2

Let x kg of mixture is drawn out and then x kg of bronze added.

So

Bronze is now = 4 â€“ (4/10)*x + x = 4 + (3x/5)

Copper is now = 6 â€“ (6/10)*x = 6 â€“ (3x/5)

Now [4 + (3x/5)] / [6 â€“ (3x/5)] = 3/2

Solve, x = 10/3

So part of mixture drawn out is (10/3)*10 = 1/3**Wheat worth Rs 50 per kg and Rs 56 per kg are mixed with a third variety of wheat in the ratio 2 : 2 : 3 respectively. If the mixture obtained is worth Rs 61 per kg. Find the price (per kg) of the third variety of wheat.**

A) Rs 87

B) Rs 73

C) Rs 62

D) Rs 55

E) Rs 76**B) Rs 73**

Explanation:

Since 1st and 2nd wheat mixed in equal ratio, their average price = (50+56)/2 = Rs 52

Let 3rd variety of wheat be Rs x per kg

So they are mixed as : (2+2) : 3 = 4 : 3

So

52 Â Â Â Â Â Â Â Â Â Â Â x

. Â Â Â Â Â 61

x-61 Â Â Â Â Â Â Â 61-52=9

so x-61/9 = 4/3

solve, x = 73**A mixture of milk and water contains 25% water. 12 litres of this mixture is drawn out and replaced with 5 litres of water. If the new ratio of water to milk becomes 2 : 5, what is the amount of milk originally present in the mixture?**

A) 82 litres

B) 75 litres

C) 95 litres

D) 80 litres

E) 84 litres**E) 84 litres**

Explanation:

Milk : water = 75% : 25% = 3 : 1

Total = 3x+x+12 = 4x+12

So (x+5)/3x = 2/5

Solve, x = 25

So total = 4*25 + 12 = 112 litres

So originally milk = 3/(3+1) * 112 = 84**Mixture A contain water and milk in the ratio 2 : 5 and mixture B contain them in the ratio 3 : 4 respectively. Equal quantities from both the mixture are taken and mixed to form mixture C. What is the ratio of milk to water in mixture C?**

A) 17 : 3

B) 9 : 5

C) 13 : 6

D) 10 : 7

E) 12 : 5**B) 9 : 5**

Explanation:

Let x litres taken from both mixtures,

Then new ratio of milk to water is

[5/(5+2)] * x + [4/(3+4)] * x : [2/(5+2)] * x + [3/(3+4)] * x

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