# Aptitude Questions: Mixtures And Alligations Set 8

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 !!

1. 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               35
.        29
6                4
3 : 2

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)             milk (10 litres)
0                                      50
.                    44
6                                      44
6 : 44
3 : 22
So x/10 = 3/22

3. 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               65
.        58
7                 16
And
54               65
.        58
7                  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

4. 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 – (1/(1+4))*x = 5 – x/5
Milk left = 20 – (4/(1+4))*x = 20 – 4x/5
Now x litres of water is added too, so water becomes = 5 – x/5 + x = 5 + 4x/5
So [5 + 4x/5] / [20 – 4x/5] = 1/3
75x+12x = 100x-4x
16x = 25
x=25/16 = 1.5625

5. 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 – 4/60]2

6. 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              2nd wheat
8.70                                 x
.                   9
2                                    3
So (x – 9)/(9 – 8.70) = 2/3
Solve, x = 9.20

7. 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 – (4/10)*x + x = 4 + (3x/5)
Copper is now = 6 – (6/10)*x = 6 – (3x/5)
Now [4 + (3x/5)] / [6 – (3x/5)] = 3/2
Solve, x = 10/3
So part of mixture drawn out is (10/3)*10 = 1/3

8. 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                       x
.           61
x-61              61-52=9
so x-61/9 = 4/3
solve, x = 73

9. 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

10. 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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