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QUESTION: 1

In two alloys copper and zinc are in the ratio of 1:3 and 4:1 respectively. 20 kg of first alloy and 35 kg of second alloy and some quantity of pure zinc is melted together. The final alloy has copper and zinc in the ratio of 5:4. Find the amount of pure zinc melted.

Solution:

Answer – b) 4.4 Explanation : In 1 alloy copper = (1/4)*20 = 5kg and zinc = (3/4)*20 = 15kg in 2 alloy copper = (4/5)*35 = 28kg and zinc = (1/5)*35 = 7kg So, 33/(22+x) = 5/4 (X is the amount of pure zinc added)

QUESTION: 2

In what ratio three kinds of rice costing 1.45rs, 1.54rs and 1.70rs must be mixed so that the mixture can be sold at 1.65rs per kg.

Solution:

Answer – c) 11:20:44 Explanation : By the rule of allegation, 145 154

…………165…………….

11 20

154 170

………165…………….

5 11

Final ratio = 11:20:44

QUESTION: 3

A container filled with liquid containing 4 parts of water and 6 parts of milk.How much of mixture must be drawn off and filled with water so that the mixture contains half milk and half water.

Solution:

Answer – c) 1/6 Explanation : Let water = 40ltr and milk is 60ltr.

Water = 40 – x*(2/5) + x and milk = 60 – x*(3/5) [x is the amount of mixture taken out] Equate both the equation, we get x = 50/3.

Now, mixture drawn off = (50/3)/100 = 1/6

QUESTION: 4

A trader has 1500 kg of wheat. One part of it is sold at 10 percent profit and other part at 18 percent profit. He gains a total of 16 percent on the whole lot.The quantity sold at 10% is

Solution:

Answer – b) 375 Explanation : Ratio => 1:3. So quantity sold at 10% = (1/4)*1500 = 375

QUESTION: 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.

Solution:

Answer – c) 7:4 Explanation : Milk in 1 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

QUESTION: 6

A dishonest seller professes to sell his milk at cost price but he mixes water with milk and gains 25 percent, then find the percentage of milk in the mixture.

Solution:

Answer – c) 80% Explanation : Suppose initially there is 100ltr of milk costing 100 rupees.

Now he gains 25% means in 100ltr of milk he add 25ltr water, so percentage of milk in the mixture = (100/125)*100 = 80%

QUESTION: 7

Fresh fruit contains 75 percent water and dry fruit contains 20 percent water.How much dry fruit can be obtained from 150 kg of fresh fruit.

Solution:

Answer – c) 47 Explanation : Dry fruit obtained from 150kg of fresh fruit = (25/100)*150 = (80/100)*x.

Solve for x x=47

QUESTION: 8

How much water must be added to 50 litre of milk costing 10 rupees per litre so as to bring the cost of milk to 8 rupees per litre.

Solution:

Answer – b) 12.5 Explanation : By using the allegation rule Water: milk = 1:4 = x:50

QUESTION: 9

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

Solution:

Answer – b) 2ltr Explanation : (6*5000 + 7*6000)/(13 + w) = 4800 (w is the amount of water added)

QUESTION: 10

There are three vessels each of 20 litre capacity is filled with the mixture of milk and water. The ratio of milk and water are 2:3, 3:4 and 4:5 respectively. All the vessels are emptied into fourth vessel, then find the ratio of milk and water in the final mixture.

Solution:

Answer – b) 401/544 Explanation : Milk = 2/5 + 3/7 + 4/9 and water = 3/5 + 4/7 + 5/9 so ratio will be 401/544

QUESTION: 11

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?

Solution:

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

QUESTION: 12

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?

Solution:

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

QUESTION: 13

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%?

Solution:

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

QUESTION: 14

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?

Solution:

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

QUESTION: 15

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?

Solution:

B) 52.26 litres Explanation: Performed 2 times, so Milk left = 60 [1 – 4/60]^{2}

QUESTION: 16

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?

Solution:

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

QUESTION: 17

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?

Solution:

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

QUESTION: 18

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.

Solution:

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

QUESTION: 19

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?

Solution:

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

QUESTION: 20

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?

Solution:

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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