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QUESTION: 1

A mixture contains milk and water in the ratio 2 : 3 and the other contains them in the ratio 3 : 1 respectively. What weight of 2nd mixture must be taken so as to make a third mixture 7 litres in weight with 70% milk?

Solution:

C) 6 litres Explanation: 70/100 = 7/10

Milk in first = 2/(2+3) = 2/5, milk in second = 3/(3+1) = 3/4 By method of allegation: 2/5 3/4

. 7/10

3/4 – 7/10 7/10 – 2/5

1/20 : 3/10

1 : 6

So in 3rd mixture, 2nd mixture is [6/(1+6)] * 7 = 6 litres

QUESTION: 2

How many kg of wheat costing Rs 8 per kg must be mixed with 8 Kg of wheat costing Rs 12 per Kg, so that there may be gain of 20% by selling the mixture of Rs 11.52 per Kg?

Solution:

D) 12 kg Explanation: With 20% gain and SP = 11.52, CP = (100/120)*11.52 = 9.6 By method of allegation: 1st wheat (x kg) 2nd wheat (8 kg) 8 12

. 9.6

12-9.6 9.6-8

2.4 1.6

3 : 2

So x/8 = 3/2

QUESTION: 3

15 litres of the milk is drawn out of a jar and filled with water. This operation is performed 1 more times. If the ratio of the quantity of milk left in jar to that of water in jar is 16 : 9, what was the initial quantity of milk present in the jar?

Solution:

E) 75 litres Explanation: Let initial quantity of milk = x litres After two times, quantity of milk left in jar = x [1 – 15/x] So x [1 – 15/x] / x = 16/16+9 [1 – 15/x] = 16/25 Square root both sides, so [1 – 15/x] = 4/5 Solve, x = 75

QUESTION: 4

Wheat worth Rs 30 per kg and Rs 42 per kg are mixed with a third variety of wheat in the ratio 1 : 1 : 2 respectively. The mixture is worth Rs 42 per kg. Find the price (per kg) of the third variety of wheat.

Solution:

B) Rs 48 Explanation: Since 1st and 2nd wheat mixed in equal ratio, their average price = (30+42)/2 = 72/2 = Rs 36 Let 3rd variety of wheat be Rs x per kg So they are mixed as : (1+1) : 2 = 1 : 1 So 36 x . 42

x-42 42-36=6 so x-42/6 = 1/1 solve, x = 48

QUESTION: 5

A mixture contains A and B in the ratio 5 : 9. 14 litres of this mixture is taken out and 14 litres of B is poured in. Now the ratio of A to B becomes 2 : 5. Find the amount of B originally present in the mixture.

Solution:

B) 45 litres Explanation: Total = 5x+9x+14 = 14x+14 So 5x/9x+14 = 2/5 Solve, x = 4 So total = 14*4 + 14 = 70 litres So B = 9/(5+9) * 70 = 45

QUESTION: 6

A mixture contains A and B in the ratio 5 : 7. 24 litres of this mixture is taken out and 15 litres of A is poured in. Now the ratio of A to B becomes 10 : 7. Find the amount of B originally present in the mixture.

Solution:

E) 35 litres Explanation: Total = 5x+7x+24 = 12x+24 So 5x+15/7x = 10/7 Solve, x = 3 So total = 12*3 + 24 = 60 litres So B = 7/(5+7) * 60 = 35

QUESTION: 7

Milk contains 20% water. What quantity of pure milk should be added to 75 litres of this milk to reduce the quantity of water to 15%?

Solution:

A) 25 litres Explanation: To have final mixture with water 15% means, milk = (100-15) = 85% Pure milk is 100% milk, and in 75 litres of milk there is 80% milk so Milk (75 litres) Pure milk (x litres) 80 100

. 85

15 5

3 : 1

So 75/x = 3/1 Solve, x = 25

QUESTION: 8

An article is bought for Rs 560. Some of the part is sold at 20% profit and remaining at 15% loss giving a total of 10% profit. Find the cost price of part sold at 20% profit.

Solution:

C) Rs 400 Explanation: 20 -15

. 10

10-(-15) 20-10

25 : 10

5 : 2

So at 20% profit = 5/(5+2) * 560

QUESTION: 9

A man travelled a distance of 75 km in 5 hours partly on foot at the rate of 6 km/hour and partly by scooter at 18 km/hour. Find the distance travelled by scooter?

Solution:

D) 67.5 km Explanation: Average speed = 75/5 = 15 km/hr By method of allegation: 6 18

. 15

3 9

1 : 3

1 : 3 is the ratio of times.

So time by scooter = 3/(1+3) * 5 = 3.75 hrs So distance by scooter = 3.75 * 18

QUESTION: 10

A 50 litres mixture of milk and water contains 10% water. 10 litres of this mixture is replaced by 10 litres of milk. What is the percentage of water in the final mixture?

Solution:

B) 8%

Explanation: When 10 litres taken out, quantity of mixture left =50-10=40, so milk left = 90/100 *40 = 36 And water left = (10/100)*40 = 4 now 10 litres of milk poured in, so there is no change in quantity of water, and total mixture again becomes 50l so water now is 4/50 * 100 = 8%

QUESTION: 11

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?

Solution:

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

QUESTION: 12

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?

Solution:

A) 125/3 litres Explanation: Remaining milk = 60 [1 – (10/60)]^{2}

QUESTION: 13

A can contains 30liter of juice, from this container 5litres of juice was taken out and replaced by water. This process was repeated further two times. How much juice is now contained by the container?

Solution:

Answer – D.17 Solution: Quantity of milk in the final mixture = 30[1-(5/30)] = 30*(5/6) = 30*125/216 = 17.36 liter = 17

QUESTION: 14

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.

Solution:

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

QUESTION: 15

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.

Solution:

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

QUESTION: 16

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?

Solution:

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

QUESTION: 17

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%?

Solution:

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

QUESTION: 18

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?

Solution:

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

QUESTION: 19

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?

Solution:

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.

QUESTION: 20

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?

Solution:

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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