Study the following data carefully and answer the questions:

Read the data given below carefully and answer the following questions:

There are five containers P, Q, R, S and T that contains the mixture of milk and water. Bar graph given below shows the average quantity of milk and water in the containers and it also shows the difference between quantity of milk and water in those containers.

Note: All the quantities of milk and water are multiple of ‘5’ and all the values are in litres.

Quantity of milk is more than quantity of water in every container.

1) If mixtures from containers P and Q are mixed in the ratio 2: 5 to form mixture A and difference between milk and water in the mixture A is 24.2 litres, then what is the total selling price of mixture A when sold at Rs.25 per litres?

a) Rs.3125

b) Rs.2695

c) Rs.2785

d) Rs.4225

e) None of these

2) Mixtures Q and R are mixed in the 28: X to form mixture B and when mixture B is sold at the price of pure milk, then profit percent earned is 58(1/3)%, then what is the value of ‘X’?

a) 15

b) 5

c) 10

d) 20

e) None of these

3) When 20 litres and 10 litres respectively is taken out from mixture R and mixture T and replaced with same amount of water. If this process is repeated one more time, then what will be the difference between amount of pure milk left in mixture R and that in mixture T?

a) 7 litres

b) 6 litres

c) 4.5 litres

d) 5.5 litres

e) None of these

4) If mixture R and S are mixed in the ratio 5: 9 to form mixture C and mixture C is sold at the price equal to 85% of the price of pure milk, then what will be the profit percent earned by the seller after selling the mixture C?

a) 21%

b) 19%

c) 23%

d) 27%

e) None of these

5) Mixtures R and T are mixed in certain ratio to form mixture D. From 280 litres of mixture D, 56 litres are taken out and replaced with same amount of water. This process is repeated one more time and now, quantity of water in the mixture D becomes 134.4 litres, then in what ratio mixture R and T are mixed to form mixture D?

a) 1: 1

b) 3: 2

c) 4: 3

d) 5: 4

e) None of these

Directions (1-5) :

Total quantity of milk and water in container P = 35 * 2 = 70 litres

Amount of milk in container P = (70 + 30)/2 = 50 litres

Amount of water in container P = (70 – 30)/2 = 20 litres

Total quantity of milk and water in container Q = 35 * 2 = 70 litres

Amount of milk in container Q = (70 + 10)/2 = 40 litres

Amount of water in container Q = (70 – 10)/2 = 30 litres

Total quantity of milk and water in container R = 50 * 2 = 100 litres

Amount of milk in container R = (100 + 60)/2 = 80 litres

Amount of water in container R = (100 – 60)/2 = 20 litres

Total quantity of milk and water in container S = 45 * 2 = 90 litres

Amount of milk in container S = (90 + 30)/2 = 60 litres

Amount of water in container S = (90 – 30)/2 = 30 litres

Total quantity of milk and water in container T = 50 * 2 = 100 litres

Amount of milk in container T = (100 + 40)/2 = 70 litres

Amount of water in container T = (100 – 40)/2 = 30 litres

1) Answer: b)

Let part of milk in mixture A = A

By the rule of alligation:

[(4/7) – A]: [A – (5/7)] = 2: 5

(20/7) – 5A = 2A – (10/7)

7A = (30/7)

A = 30/49

Ratio of milk to water in mixture A = 30: (49 – 30) = 30: 19

Total quantity of mixture A = 24.2 * [(30 + 19)/(30 – 19)] = 107.8 litres

Total selling price of mixture A = 107.8 * 25 = Rs.2695

2) Answer: c)

Since, profit percent earned after selling mixture B at the cost of pure milk is 58(1/3)%.

Hence, ratio of milk to water in mixture B = 100: 58(1/3) = 100: 58(1/3) = (300/3): (175/3) = 12: 7

By the rule of alligation:

Ratio of milk to water = [(4/5) – (12/19)]: [(12/19) – (4/7)]

= [(76 – 60)/95]: [(84 – 76)/133]

= (16/95): (8/133)

= 14: 5

= 28: 10

Hence, X = 10

3) Answer: d)

When 20 litres of mixture is taken out and replaced with same amount of water twice from mixture R, then amount of pure milk left in mixture R:

= 80(1 – 20/100)²

= 51.2 litres

When 10 litres of mixture is taken out and replaced with same amount of water twice from mixture T, then amount of pure milk left in mixture T:

= 70(1 – 10/100)²

= 56.7 litres

Required difference = 56.7 – 51.2 = 5.5 litres

4) Answer: b)

Let part of milk in mixture C = C

By the rule of alligation:

[(2/3) – C]: [C – (4/5)] = 5: 9

6 – 9C = 5C – 4

14C = 10

C = 5/7

Ratio of milk to water in mixture C = 5: (7 – 5) = 5: 2

Let amount of milk and water in mixture C is ‘5x’ and ‘2x’ respectively.

Cost price of mixture C = 5x * 1 = 5x

Selling price of mixture C = (5x + 2x) * 0.85 = 5.95x

Profit amount = 5.95x – 5x = 0.95x

Required percent = (0.95x/5) * 100 = 19%

5) Answer: a)

Let initial amount of milk in mixture D is ‘x’.

Amount of milk left in the mixture D = x[1 – 56/280]² = 134.4

0.64x = 134.4

x = 210

Ratio of milk to water in mixture D = 210: (280 – 210) = 210: 70 = 3: 1

By the rule of alligation:

Ratio in which mixtures R and T are mixed = [(7/10) – (3/4)]: [(3/4) – (4/5)]

= (1/20): (1/20)

= 1: 1