123 + Interview Questions in QUANTITATIVE APTITUDE in MIXTURE PROBLEMS Page 3 InterviewSolution

101.	There are two alloys of copper and aluminium, alloy A has 40% aluminium and alloy B has 10% copper. In what ratio should these alloys be mixed so that the cost of total copper becomes equal to the cost of total aluminium in the mixture? (The cost price of aluminium is Rs. 200 per kg and cost price of copper is Rs. 500 per kg)1). 11 ∶ 242). 13 ∶ 223). 21 ∶ 114). 16 ∶ 19
Answer» Suppose ‘x’ kg of alloy A and ‘y’ kg of alloy B is MIXED; ∴ Amount of copper in new mixture = (0.6x + 0.1y) kg Amount of aluminium in new mixture = (0.4x + 0.9y) kg Since COST PRICE of aluminium is Rs. 200 per kg and cost price of copper is Rs. 500 per kg; ∴ (0.6x + 0.1y) × 500 = (0.4x + 0.9y) × 200 ⇒ 300x + 50Y = 80x + 180y ⇒ 220x = 130y ⇒ x ? y = 13 ? 22

101.

There are two alloys of copper and aluminium, alloy A has 40% aluminium and alloy B has 10% copper. In what ratio should these alloys be mixed so that the cost of total copper becomes equal to the cost of total aluminium in the mixture? (The cost price of aluminium is Rs. 200 per kg and cost price of copper is Rs. 500 per kg)1). 11 ∶ 242). 13 ∶ 223). 21 ∶ 114). 16 ∶ 19

Answer»

Suppose ‘x’ kg of alloy A and ‘y’ kg of alloy B is MIXED;

∴ Amount of copper in new mixture = (0.6x + 0.1y) kg

Amount of aluminium in new mixture = (0.4x + 0.9y) kg

Since COST PRICE of aluminium is Rs. 200 per kg and cost price of copper is Rs. 500 per kg;

∴ (0.6x + 0.1y) × 500 = (0.4x + 0.9y) × 200

⇒ 300x + 50Y = 80x + 180y

⇒ 220x = 130y

⇒ x ? y = 13 ? 22

Discussion

102.	A milk seller and a fruit seller are two enemies, jealous from each other’s progress. One day the fruit seller found that the milk seller had mixed 20% water in the milk. Out of fear he replaced one litre of mixture with one litre of pure milk which was worth Rs. 80. Milk seller had 20 litre of mixture initially, but still after fruit seller’s work on his mixture, he was in profit. Find his profit percentage.1). 25%2). 29.45%3). 23.46%4). 31.57%
Answer» Initially for 20 litre mixture, ⇒ Milk : Water = 16 : 4 For 19 litre mixture, ⇒ Milk = 19 × 80/100 = 15.2 ⇒ Water = 19 – 15.2 = 3.8 After replacement, ⇒ Milk = 15.2 + 1 = 16.2 ⇒ Water = 3.8 ⇒ 1 litre milk = Rs. 80 ⇒ 15.2 litre milk = 80 × 16.2 = Rs. 1296 ⇒ Cost Price = Rs. 1296 ⇒ Selling Price = 80 × 20 = 1600 ⇒ PROFIT % = (1600 – 1296)/1296 × 100 = 23.46%

102.

A milk seller and a fruit seller are two enemies, jealous from each other’s progress. One day the fruit seller found that the milk seller had mixed 20% water in the milk. Out of fear he replaced one litre of mixture with one litre of pure milk which was worth Rs. 80. Milk seller had 20 litre of mixture initially, but still after fruit seller’s work on his mixture, he was in profit. Find his profit percentage.1). 25%2). 29.45%3). 23.46%4). 31.57%

Answer»

Initially for 20 litre mixture,

⇒ Milk : Water = 16 : 4

For 19 litre mixture,

⇒ Milk = 19 × 80/100 = 15.2

⇒ Water = 19 – 15.2 = 3.8

After replacement,

⇒ Milk = 15.2 + 1 = 16.2

⇒ Water = 3.8

⇒ 1 litre milk = Rs. 80

⇒ 15.2 litre milk = 80 × 16.2 = Rs. 1296

⇒ Cost Price = Rs. 1296

⇒ Selling Price = 80 × 20 = 1600

⇒ PROFIT % = (1600 – 1296)/1296 × 100 = 23.46%

Discussion

103.	The amount of alcohol in two varieties of whiskey are 27% volume/volume and 42% volume/volume. In what ratio should they be mixed so that the mixture has 33% volume/volume alcohol?1). 8 : 72). 3 : 23). 2 : 14). 1 : 1
Answer» Let the whiskeys are mixed in AMOUNTS M litres and N litres. ⇒ Amount of alcohol in first VARIETY = 0.27M ⇒ Amount of alcohol in second variety = 0.42N Total amount of mixture = (M + N) litres Amount of alcohol in mixture = 33% volume/volume = 0.33(M + N) ⇒ 0.27M + 0.42N = 0.33(M + N) ⇒ 0.09N = 0.06M ⇒ M/N = 3/2 ∴ RATIO in which they should be mixed is 3 : 2.

103.

The amount of alcohol in two varieties of whiskey are 27% volume/volume and 42% volume/volume. In what ratio should they be mixed so that the mixture has 33% volume/volume alcohol?1). 8 : 72). 3 : 23). 2 : 14). 1 : 1

Answer»

Let the whiskeys are mixed in AMOUNTS M litres and N litres.

⇒ Amount of alcohol in first VARIETY = 0.27M

⇒ Amount of alcohol in second variety = 0.42N

Total amount of mixture = (M + N) litres

Amount of alcohol in mixture = 33% volume/volume = 0.33(M + N)

⇒ 0.27M + 0.42N = 0.33(M + N)

⇒ 0.09N = 0.06M

⇒ M/N = 3/2

∴ RATIO in which they should be mixed is 3 : 2.

Discussion

104.	A vessel is full of a mixture of kerosene and petrol in which there is 20% kerosene. 5 litres are drawn off and then the vessel is filled with petrol. If the kerosene is now 15%, how much does the vessel hold?1). 30 litres2). 35 litres3). 26 litres4). 32 litres
Answer» Let the total CAPACITY of VESSEL be x litres Kerosene in the vessel = 20x/100 Kerosene after 5 litres are drawn = 15x/100 Kerosene in 5 litres of mixture = (20 × 5)/100 ⇒ 20x/100 – (20 × 5)/100 = 15x/100 ⇒ 20x – 100 = 15x ⇒ 5x = 100 ⇒ x = 20 Thus, Vessel holds 20 litres of kerosene

104.

A vessel is full of a mixture of kerosene and petrol in which there is 20% kerosene. 5 litres are drawn off and then the vessel is filled with petrol. If the kerosene is now 15%, how much does the vessel hold?1). 30 litres2). 35 litres3). 26 litres4). 32 litres

Answer»

Let the total CAPACITY of VESSEL be x litres

Kerosene in the vessel = 20x/100

Kerosene after 5 litres are drawn = 15x/100

Kerosene in 5 litres of mixture = (20 × 5)/100

⇒ 20x/100 – (20 × 5)/100 = 15x/100

⇒ 20x – 100 = 15x

⇒ 5x = 100

⇒ x = 20

Thus, Vessel holds 20 litres of kerosene

Discussion

105.	44 litres of milk-water mixture of ratio 7 ∶ 4 is mixed with 40 litres of another milk-water mixture of ratio 5 ∶ 3. If 2 litres of water and 7 litres of milk were further added, what is the ratio of milk and water in the final mixture?1). 8 ∶ 52). 12 ∶ 73). 16 ∶ 94). 20 ∶ 11
Answer» ⇒ Ratio of first MIXTURE = 7 : 4 ⇒ Quantity of MILK in 44 litres of first mixture = (7/11) × 44 = 28 litres ⇒ Quantity of water in 44 litres of first mixture = (4/11) × 44 = 16 litres Similarly, ⇒ Ratio of second mixture = 5 : 3 ⇒ Quantity of milk in 40 litres of second mixture = (5/8) × 40 = 25 litres ⇒ Quantity of water in 40 litres of second mixture = (3/8) × 40 = 15 litres Further, 2 litres of water and 7 litres of milk was ADDED, ⇒ Total quantity of milk in new mixture = 28 + 25 + 7 = 60 litres ⇒ Total quantity of water in new mixture = 16 + 15 + 2 = 33 litres ∴ Ratio of milk and water in FINAL mixture = 60 : 33 = 20 : 11

105.

44 litres of milk-water mixture of ratio 7 ∶ 4 is mixed with 40 litres of another milk-water mixture of ratio 5 ∶ 3. If 2 litres of water and 7 litres of milk were further added, what is the ratio of milk and water in the final mixture?1). 8 ∶ 52). 12 ∶ 73). 16 ∶ 94). 20 ∶ 11

Answer»

⇒ Ratio of first MIXTURE = 7 : 4

⇒ Quantity of MILK in 44 litres of first mixture = (7/11) × 44 = 28 litres

⇒ Quantity of water in 44 litres of first mixture = (4/11) × 44 = 16 litres

Similarly,

⇒ Ratio of second mixture = 5 : 3

⇒ Quantity of milk in 40 litres of second mixture = (5/8) × 40 = 25 litres

⇒ Quantity of water in 40 litres of second mixture = (3/8) × 40 = 15 litres

Further, 2 litres of water and 7 litres of milk was ADDED,

⇒ Total quantity of milk in new mixture = 28 + 25 + 7 = 60 litres

⇒ Total quantity of water in new mixture = 16 + 15 + 2 = 33 litres

∴ Ratio of milk and water in FINAL mixture = 60 : 33 = 20 : 11

Discussion

106.	In a chemistry lab two beakers A and B contains 36% and 40% of spirit respectively. If two liters from A is mixed with 4 liters of B. The ratio of sprit and water in the resulting mixture is:1). 19 : 462). 29 : 463). 29 : 554). 29 : 45
Answer» A has 36% spirit. Amount of spirit in 2 liters of A = 0.36 × 2 = 0.72 liters B has 40% spirit. Amount of spirit in 4 liters of B = 0.4 × 4 = 1.6 liters Total amount of mixture = 2 + 4 = 6 liters Amount of spirit in the FINAL mixture = 0.72 + 1.6 = 2.32 liters Amount of water in the final mixture = 6 – 2.32 = 3.68 Required RATIO = 2.32 : 3.68 = 29 : 46

106.

In a chemistry lab two beakers A and B contains 36% and 40% of spirit respectively. If two liters from A is mixed with 4 liters of B. The ratio of sprit and water in the resulting mixture is:1). 19 : 462). 29 : 463). 29 : 554). 29 : 45

Answer»

A has 36% spirit.

Amount of spirit in 2 liters of A = 0.36 × 2 = 0.72 liters

B has 40% spirit.

Amount of spirit in 4 liters of B = 0.4 × 4 = 1.6 liters

Total amount of mixture = 2 + 4 = 6 liters

Amount of spirit in the FINAL mixture = 0.72 + 1.6 = 2.32 liters

Amount of water in the final mixture = 6 – 2.32 = 3.68

Required RATIO = 2.32 : 3.68 = 29 : 46

Discussion

107.	Two solutions of 90% and 97% purity are mixed, resulting in 21 L of mixture of 94% purity. The quantity of the second solution in the resulting mixture, (in L) is1). 152). 123). 94). 6
Answer» advice - use alligation let see-AB 9097 94 (97-94):(94-90) 3:4=7 = 21ltrs (GIVEN) \|\|\|\| \|\|1 = 3ltrs *\|43ltrs = 12ltrs (SECOND SOLUTION) = answer 33 = 9ltrs (first solution)*

107.

Two solutions of 90% and 97% purity are mixed, resulting in 21 L of mixture of 94% purity. The quantity of the second solution in the resulting mixture, (in L) is1). 152). 123). 94). 6

Answer»

advice - use alligation

let see-AB

9097

94

(97-94):(94-90)

3:4=7 = 21ltrs (GIVEN)

||||

||1 = 3ltrs

|4*3ltrs = 12ltrs (SECOND SOLUTION) = answer

3*3 = 9ltrs (first solution)

Discussion

108.	1). 1 : 22). 2 : 33). 2 : 54). 1 : 6
Answer» From the given data, Selling price of MIXTURE is Rs. 408 per kg Profit% = (selling price/Cost price – 1) × 100 ⇒ 20 = (408/Cost price – 1) × 100 ⇒ 0.2 + 1 = 408/Cost price ⇒ Cost price of mixture = 408/1.2 = Rs.340 per kg ∴ Ratio of Assam Tea to the Darjeeling Tea = (340 - 300) : (400 - 340) = 40 : 60 = 2 : 3 so, answer is2 : 3

108.

1). 1 : 22). 2 : 33). 2 : 54). 1 : 6

Answer»

From the given data,

Selling price of MIXTURE is Rs. 408 per kg

Profit% = (selling price/Cost price – 1) × 100

⇒ 20 = (408/Cost price – 1) × 100

⇒ 0.2 + 1 = 408/Cost price

⇒ Cost price of mixture = 408/1.2 = Rs.340 per kg

∴ Ratio of Assam Tea to the Darjeeling Tea = (340 - 300) : (400 - 340) = 40 : 60 = 2 : 3

so, answer is2 : 3

Discussion

109.	100 litres of milk is added with 40 litres of water. What will be the difference between proportion of milk and water? 1). 4/72). 5/73). 3/74). Cannot be determined
Answer» Total QUANTITY of MIXTURE = 100 + 40 = 140 litres ⇒ Proportion of MILK in mixture = 100/140 = 5/7 ⇒ Proportion of WATER in mixture = 40/140 = 2/7 ∴ Difference in proportion = 5/7 – 2/7 = 3/7

109.

100 litres of milk is added with 40 litres of water. What will be the difference between proportion of milk and water? 1). 4/72). 5/73). 3/74). Cannot be determined

Answer»

Total QUANTITY of MIXTURE = 100 + 40 = 140 litres

⇒ Proportion of MILK in mixture = 100/140 = 5/7

⇒ Proportion of WATER in mixture = 40/140 = 2/7

∴ Difference in proportion = 5/7 – 2/7 = 3/7

Discussion

110.	In a 5 litre solution of water and HCl, content of HCl is 20%. In order to attain a desirable result, strong HCl is required. Hence, 10 litres of 100% pure HCl is added. What is the strength of HCl in the mixture now?1). 73%2). 73.33%3). 73.66%4). 80%
Answer» TOTAL content = 5 litres Content of HCl = 20% of 5 litres ⇒ 20/100 × 5 = 1 litre After 10 litres of 100% HCl is added, Total content = 5 + 10 = 15 litres STRENGTH of HCl = HCl content/Total content × 100 ⇒ (1 + 10) /15 × 100 ⇒ 11/15 × 100 = 73.33% ∴ Strength of HCl = 73.33%

110.

In a 5 litre solution of water and HCl, content of HCl is 20%. In order to attain a desirable result, strong HCl is required. Hence, 10 litres of 100% pure HCl is added. What is the strength of HCl in the mixture now?1). 73%2). 73.33%3). 73.66%4). 80%

Answer»

TOTAL content = 5 litres

Content of HCl = 20% of 5 litres

⇒ 20/100 × 5 = 1 litre

After 10 litres of 100% HCl is added,

Total content = 5 + 10 = 15 litres

STRENGTH of HCl = HCl content/Total content × 100

⇒ (1 + 10) /15 × 100

⇒ 11/15 × 100 = 73.33%

∴ Strength of HCl = 73.33%

Discussion

111.	A 24 liters of cough syrup and water mixture contains cough syrup and water in the ratio 3 : 5. What liters of the mixture should be taken out and replaced with pure cough syrup so that the final mixture contains syrup and water in equal proportions?1). 20/3 liters2). 3 liters3). 32/5 liters4). 24/5 liters
Answer» In 24 liters of MIXTURE, cough syrup = 3/8 × 24 = 9 liters and WATER = 15 liters 24 liters mixture must have 12 liters water and 12 liters cough syrup, 3 liters of water should be taken out, since we are only adding cough syrup. Let y liters of mixture taken out such that 5/8 × y = 3, On SOLVING, y = 24/5 liters

111.

A 24 liters of cough syrup and water mixture contains cough syrup and water in the ratio 3 : 5. What liters of the mixture should be taken out and replaced with pure cough syrup so that the final mixture contains syrup and water in equal proportions?1). 20/3 liters2). 3 liters3). 32/5 liters4). 24/5 liters

Answer»

In 24 liters of MIXTURE, cough syrup = 3/8 × 24 = 9 liters and WATER = 15 liters

24 liters mixture must have 12 liters water and 12 liters cough syrup, 3 liters of water should be taken out, since we are only adding cough syrup.

Let y liters of mixture taken out such that 5/8 × y = 3,

On SOLVING, y = 24/5 liters

Discussion

112.	1). 6 ∶ 3 ∶ 72). 6 ∶ 3 ∶ 83). 6 ∶ 3 ∶ 94). 6 ∶ 3 ∶ 10
Answer» The bucket CONTAINS 5 litres of milk. Rajesh removes 20% of the milk that is 1 litre of milk and adds 2 litres of water The solution PRESENT now in the bucket is 4 litres of milk and 2 litres of water Mahesh removes 40% of this solution, that is 1.6 litres of milk and 0.8 litres of water and he adds 4 litres of soya SAUCE The solution present now in the bucket is 2.4 litres of milk and 1.2 litres of water and 4 litres of soya sauce and their ratio will be 6 ? 3 ? 10 respectively Now Ramesh removes 30% of this solution but does not replace it with any other solution Thus, the FINAL ratio of this LIQUIDS won’t change after this operation ∴ Required Ratio of milk, soya sauce and water in the bucket = 6 ? 10 ? 3

112.

1). 6 ∶ 3 ∶ 72). 6 ∶ 3 ∶ 83). 6 ∶ 3 ∶ 94). 6 ∶ 3 ∶ 10

Answer»

The bucket CONTAINS 5 litres of milk. Rajesh removes 20% of the milk that is 1 litre of milk and adds 2 litres of water

The solution PRESENT now in the bucket is 4 litres of milk and 2 litres of water

Mahesh removes 40% of this solution, that is 1.6 litres of milk and 0.8 litres of water and he adds 4 litres of soya SAUCE

The solution present now in the bucket is 2.4 litres of milk and 1.2 litres of water and 4 litres of soya sauce and their ratio will be 6 ? 3 ? 10 respectively

Now Ramesh removes 30% of this solution but does not replace it with any other solution

Thus, the FINAL ratio of this LIQUIDS won’t change after this operation

∴ Required Ratio of milk, soya sauce and water in the bucket = 6 ? 10 ? 3

Discussion

113.	1). 64 litres2). 69 litres3). 78 litres4). None of the above
Answer» Juice ? Milk = 3 ? 14 AMOUNT = 51 LITRES Amount of juice = 9 litres Amount of milk = 51 - 9 = 42 litres Let x amount of juice be added, such that RATIO becomes 9 ? 14 (9 + x) ? 42 = 9 ? 14 ⇒ (9 + x)(14) = (42)(9) ⇒ x = 18 New amount = 51 + 18 = 69 litres

113.

1). 64 litres2). 69 litres3). 78 litres4). None of the above

Answer»

Juice ? Milk = 3 ? 14

AMOUNT = 51 LITRES

Amount of juice = 9 litres

Amount of milk = 51 - 9 = 42 litres

Let x amount of juice be added, such that RATIO becomes 9 ? 14

(9 + x) ? 42 = 9 ? 14

⇒ (9 + x)(14) = (42)(9)

⇒ x = 18

New amount = 51 + 18 = 69 litres

Discussion

114.	The glasses of capacity 3 L, 5 L and 7 L contain mixture of milk and water with milk concentrations 60%, 50% and 20% respectively. The contents of three glasses are emptied into a large vessel. Find the ratio of milk to water in the resultant mixture.1). 52 : 492). 11 : 233). 127 : 634). 19 : 31
Answer» In the 3L glass, Concentration of MILK = 60% ∴ Volume of milk = 60% of 3L = 3 × 60/100 = (9/5) = 1.8L---(1) ⇒ Concentration of water = 40% ∴ Volume of water = 40% of 3L = 3 × 40/100 = 6/5 = 1.2L---(2) SIMILARLY, in the 5L glass, Volume of milk = 50% of 5 = 2.5L---(3) Volume of water = 5L – 2.5L = 2.5L---(4) In the 7L glass, Volume of milk = 20% of 7L = (7/5) = 1.4 L---(5) Volume of water = 7 – (7/5) = (28/5) = 5.6L ---(6) ∴ From 1, 2, 3, 4, 5 and 6, In the TOTAL mix, Total volume of milk = 1.8 + 2.5 + 1.4 = 5.7L Total volume of water = 1.2 + 2.5 + 5.6 = 9.3L ∴ ratio of milk: water = 5.7: 9.3 = 19: 31 Thus, ratio of Milk : Water = 19 : 31.

114.

The glasses of capacity 3 L, 5 L and 7 L contain mixture of milk and water with milk concentrations 60%, 50% and 20% respectively. The contents of three glasses are emptied into a large vessel. Find the ratio of milk to water in the resultant mixture.1). 52 : 492). 11 : 233). 127 : 634). 19 : 31

Answer»

In the 3L glass,

Concentration of MILK = 60%

∴ Volume of milk

= 60% of 3L

= 3 × 60/100

= (9/5) = 1.8L---(1)

⇒ Concentration of water = 40%

∴ Volume of water

= 40% of 3L

= 3 × 40/100

= 6/5 = 1.2L---(2)

SIMILARLY, in the 5L glass,

Volume of milk

= 50% of 5

= 2.5L---(3)

Volume of water

= 5L – 2.5L

= 2.5L---(4)

In the 7L glass,

Volume of milk

= 20% of 7L

= (7/5) = 1.4 L---(5)

Volume of water = 7 – (7/5) = (28/5) = 5.6L ---(6)

∴ From 1, 2, 3, 4, 5 and 6,

In the TOTAL mix,

Total volume of milk = 1.8 + 2.5 + 1.4 = 5.7L

Total volume of water = 1.2 + 2.5 + 5.6 = 9.3L

∴ ratio of milk: water = 5.7: 9.3 = 19: 31

Thus, ratio of Milk : Water = 19 : 31.

Discussion

115.	A drum contains forty liters of whisky. Four liters of whisky is taken out and replaced by soda. This process is carried out twice further. How much whisky is now contained by the container?1). 202). 29.163). 33.24). 37.11
Answer» Suppose a solution contains x units of a WHISKY from which y units are taken out and replaced by soda After n repeated OPERATIONS, QUANTITY of pure liquid REMAINING in solution ⇒ x(1 − y/x)ⁿ = x(1 − y/x)ⁿ units Where n = NUMBER of replacements So, whisky in the drum now: ⇒ 40 (1 − 4/40)³= 40(1 − 1/10)³ ⇒ 29.16 ∴ 29.16 units of whisky is in the container

115.

A drum contains forty liters of whisky. Four liters of whisky is taken out and replaced by soda. This process is carried out twice further. How much whisky is now contained by the container?1). 202). 29.163). 33.24). 37.11

Answer»

Suppose a solution contains x units of a WHISKY from which y units are taken out and replaced by soda

After n repeated OPERATIONS, QUANTITY of pure liquid REMAINING in solution

⇒ x(1 − y/x)ⁿ = x(1 − y/x)ⁿ units

Where n = NUMBER of replacements

So, whisky in the drum now:

⇒ 40 (1 − 4/40)³= 40(1 − 1/10)³

⇒ 29.16

∴ 29.16 units of whisky is in the container

Discussion

116.	There are two containers of equal capacity. The ratio of milk to water in the first container is 3 : 1, in the second container 5 : 2. If they are mixed up, the ratio of milk to water in the mixture will be1). 28 : 412). 41 : 283). 15 : 414). 41 : 15
Answer» Let the volume of the two containers be V. Now, ? Milk to water ratio in the first container = 3 ? 1 ⇒ Volume covered by milk = 3V/4 and Volume covered by water = V/4 In the second container, milk to water ratio = 5 ? 2 ⇒ Amount of milk = 5V/7 and amount of water = 2V/7 When both the container’s content are MIXED, amount of milk = $(\frac{{3V}}{4} + \frac{{5V}}{7} = \frac{{21V + 20V}}{{28}} = \frac{{41V}}{{28}})$ And amount of water = $(\frac{V}{4} + \frac{{2V}}{7} = \frac{{7V + 8V}}{{28}} = \frac{{15V}}{{28}})$ Now, Ratio of milk to water in the new mixture = $(\frac{{\frac{{41V}}{{28}}}}{{\frac{{15V}}{{28}}}} = \frac{{41}}{{15}} = 41\;:15)$

116.

There are two containers of equal capacity. The ratio of milk to water in the first container is 3 : 1, in the second container 5 : 2. If they are mixed up, the ratio of milk to water in the mixture will be1). 28 : 412). 41 : 283). 15 : 414). 41 : 15

Answer»

Let the volume of the two containers be V.

Now,

? Milk to water ratio in the first container = 3 ? 1

⇒ Volume covered by milk = 3V/4 and Volume covered by water = V/4

In the second container, milk to water ratio = 5 ? 2

⇒ Amount of milk = 5V/7 and amount of water = 2V/7

When both the container’s content are MIXED, amount of milk = $(\frac{{3V}}{4} + \frac{{5V}}{7} = \frac{{21V + 20V}}{{28}} = \frac{{41V}}{{28}})$

And amount of water = $(\frac{V}{4} + \frac{{2V}}{7} = \frac{{7V + 8V}}{{28}} = \frac{{15V}}{{28}})$

Now, Ratio of milk to water in the new mixture = $(\frac{{\frac{{41V}}{{28}}}}{{\frac{{15V}}{{28}}}} = \frac{{41}}{{15}} = 41\;:15)$

Discussion

117.	A vessel contains a mixture of P and Q in the ratio of 5 : 3. 16 liters of this mixture is taken out and 5 liters of P is poured in. The new mixture has ratio of P to Q as 11 : 6. Find the total original quantity of mixture.1). 80 liters2). 96 liters3). 98 liters4). 84 liters
Answer» P = 5x, Q = 3x The quantity of P and Q in 16 liters of the mixture. Quantity of P = (16 × 5x)/8x = 10 Quantity of Q = (16 × 3x)/8x = 6 Now, 5 liters of P POURED in and then RATIO becomes 11 ? 6 (5x - 10 + 5)/3x - 6 = 11/6 (5x - 5)/(3x - 6) = 11/6 Solve, x = 12 So total mixture originally = 8x= 8 × 12 = 96 liters

117.

A vessel contains a mixture of P and Q in the ratio of 5 : 3. 16 liters of this mixture is taken out and 5 liters of P is poured in. The new mixture has ratio of P to Q as 11 : 6. Find the total original quantity of mixture.1). 80 liters2). 96 liters3). 98 liters4). 84 liters

Answer»

P = 5x, Q = 3x

The quantity of P and Q in 16 liters of the mixture.

Quantity of P = (16 × 5x)/8x = 10

Quantity of Q = (16 × 3x)/8x = 6

Now, 5 liters of P POURED in and then RATIO becomes 11 ? 6

(5x - 10 + 5)/3x - 6 = 11/6

(5x - 5)/(3x - 6) = 11/6

Solve, x = 12

So total mixture originally = 8x= 8 × 12 = 96 liters

Discussion

118.	1). 2 : 5 : 32). 3 : 5 : 23). 1 : 5 : 64). 7 : 4 : 6
Answer» The DENSITY of Iron is twice that of Copper and four times that of Silver. Suppose density of Silver be D. ⇒ Density of Iron will be 4D, and that of Copper will be 2D. Let the volumes of Iron, Copper and Silver taken be P, Q and R, respectively. Weight = Volume × Density ⇒ Weights of Iron, Copper and Silver will be 4PD, 2QD and RD, respectively. ⇒ 4PD : 2QD : RD = 2 : 5 : 3 ⇒ 4P : 2Q : R = 2 : 5 : 3 ⇒ 4P : 4Q : 4R = 2 : 10 : 12 ⇒ P : Q : R = 1 : 5 : 6 ∴ Ratio of volumes should be 1 : 5 : 6.

118.

1). 2 : 5 : 32). 3 : 5 : 23). 1 : 5 : 64). 7 : 4 : 6

Answer»

The DENSITY of Iron is twice that of Copper and four times that of Silver.

Suppose density of Silver be D.

⇒ Density of Iron will be 4D, and that of Copper will be 2D.

Let the volumes of Iron, Copper and Silver taken be P, Q and R, respectively.

Weight = Volume × Density

⇒ Weights of Iron, Copper and Silver will be 4PD, 2QD and RD, respectively.

⇒ 4PD : 2QD : RD = 2 : 5 : 3

⇒ 4P : 2Q : R = 2 : 5 : 3

⇒ 4P : 4Q : 4R = 2 : 10 : 12

⇒ P : Q : R = 1 : 5 : 6

∴ Ratio of volumes should be 1 : 5 : 6.

Discussion

119.	A butler stole wine from a butt of Sherry which contained 18% of spirit and he replaced what he had stolen by wine containing only 8% spirit. The butt was then 14% strength only. How much of the butt did he steal?1). 1/42). 2/53). 3/54). 3/4
Answer»

Discussion

120.	A mixture contains liquid A and liquid B in the ratio of 5 : 4. If 2 litres of liquid B is added to it, then the ratio of liquid A and liquid B becomes 7 : 6. What is the quantity (in litres) of liquid A in the mixture?1). 322). 353). 404). 42
Answer» Let the quantity of the TWO liquids A and B be 5x and 4x respectively. According to the question (5x)/(4x + 2) = 7/6 ⇒ 30x = 28X + 14 ⇒ 2x = 14 ⇒ x = 7 ∴ Quantity of LIQUID A in MIXTURE = 7 × 5 = 35 liters

120.

A mixture contains liquid A and liquid B in the ratio of 5 : 4. If 2 litres of liquid B is added to it, then the ratio of liquid A and liquid B becomes 7 : 6. What is the quantity (in litres) of liquid A in the mixture?1). 322). 353). 404). 42

Answer»

Let the quantity of the TWO liquids A and B be 5x and 4x respectively.

According to the question

(5x)/(4x + 2) = 7/6

⇒ 30x = 28X + 14

⇒ 2x = 14

⇒ x = 7

∴ Quantity of LIQUID A in MIXTURE = 7 × 5 = 35 liters

Discussion

121.	Three bottles of equal capacity containing mixture of milk and water in ratio 2 : 5, 3 : 4 and 4 : 5 respectively. These three bottles are emptied into a large bottle. What will be the ratio of milk and water respectively in the large bottle?1). 73 : 1062). 73 : 1163). 73 : 1134). 73 : 189
Answer» Let the capacity be Z litres For bottle 1, 2 : 5 implies that milk is 2a and water is 5a ⇒ 2a + 5a = z ⇒ a = z/7…..(i) For bottle 2, 3 : 4 implies that milk is 3b and water is 4B ⇒ 3b + 4b = z ⇒ b = z/7…..(ii) For bottle 1, 4 : 5 implies that milk is 4c and water is 5c ⇒ 4c + 5c = z ⇒ c = z/9…..(III) Total milk is = 2a + 3b + 4c Using (i), (ii) and (iii) ⇒ milk = 2 × z/7 + 3 × z/7 + 4 × z/9 Total water is 5a + 4b + 5c ⇒ water = 5 × z/7 + 4 × z/7 + 5 × z/9 Ratio of milk and water is GIVEN as ⇒ (2 × z/7 + 3 × z/7 + 4 × z/9) / ( 5 × z/7 + 4 × z/7 + 5 × z/9) ⇒ 73/116 ∴ ratio is 73 : 116

121.

Three bottles of equal capacity containing mixture of milk and water in ratio 2 : 5, 3 : 4 and 4 : 5 respectively. These three bottles are emptied into a large bottle. What will be the ratio of milk and water respectively in the large bottle?1). 73 : 1062). 73 : 1163). 73 : 1134). 73 : 189

Answer»

Let the capacity be Z litres

For bottle 1, 2 : 5 implies that milk is 2a and water is 5a

⇒ 2a + 5a = z

⇒ a = z/7…..(i)

For bottle 2, 3 : 4 implies that milk is 3b and water is 4B

⇒ 3b + 4b = z

⇒ b = z/7…..(ii)

For bottle 1, 4 : 5 implies that milk is 4c and water is 5c

⇒ 4c + 5c = z

⇒ c = z/9…..(III)

Total milk is = 2a + 3b + 4c Using (i), (ii) and (iii)

⇒ milk = 2 × z/7 + 3 × z/7 + 4 × z/9

Total water is 5a + 4b + 5c

⇒ water = 5 × z/7 + 4 × z/7 + 5 × z/9

Ratio of milk and water is GIVEN as

⇒ (2 × z/7 + 3 × z/7 + 4 × z/9) / ( 5 × z/7 + 4 × z/7 + 5 × z/9)

⇒ 73/116

∴ ratio is 73 : 116

Discussion

122.	In a vessel, a mixture of milk and coffee is in the ratio 9 ∶ 1. If 20 liters of the mixture is taken out and same quantity of coffee is poured into the vessel, the ratio of milk and coffee becomes 5 ∶ 2. Find the original quantity of Milk in the vessel?1). 79.3 liters2). 82.3 liters3). 87.3 liters4). 77.3 liters
Answer» LET the ORIGINAL quantity of milk be 9X and coffee be x According to the question, $(\frac{{\left( {9x - 20} \right) \times \frac{9}{{10}}\;}}{{\left( {x - 20} \right) \times \frac{1}{{10}} + 20}} = \frac{5}{2})$ $(\RIGHTARROW \frac{{9x - 18\;}}{{x + 18}} = \frac{5}{2})$ ⇒ (9x – 18) × 2 = (x + 18) × 5 ⇒ 18x – 36 = 5x + 90 ⇒ 13x = 126 ⇒ x = 9.69 or 9.7 ∴ Original quantity of milk = 9x = 9 × 9.7 = 87.3 liters

122.

In a vessel, a mixture of milk and coffee is in the ratio 9 ∶ 1. If 20 liters of the mixture is taken out and same quantity of coffee is poured into the vessel, the ratio of milk and coffee becomes 5 ∶ 2. Find the original quantity of Milk in the vessel?1). 79.3 liters2). 82.3 liters3). 87.3 liters4). 77.3 liters

Answer»

LET the ORIGINAL quantity of milk be 9X and coffee be x

According to the question,

$(\frac{{\left( {9x - 20} \right) \times \frac{9}{{10}}\;}}{{\left( {x - 20} \right) \times \frac{1}{{10}} + 20}} = \frac{5}{2})$

$(\RIGHTARROW \frac{{9x - 18\;}}{{x + 18}} = \frac{5}{2})$

⇒ (9x – 18) × 2 = (x + 18) × 5

⇒ 18x – 36 = 5x + 90

⇒ 13x = 126

⇒ x = 9.69 or 9.7

∴ Original quantity of milk = 9x = 9 × 9.7 = 87.3 liters

Discussion

123.	In a mixture, ink and water are in the ratio of 8 : 5. If 24 litres of water is added to the mixture, then the new ratio becomes 8 : 11. What is the total quantity (in litres) of water in the new mixture?1). 402). 423). 444). 48
Answer» LET the initial quantity of ink and water in the mixture be 8a and 5A respectively When 24 litres of water is added to the mixture, then the NEW ratio becomes 8 : 11 ⇒ [8a/(24 + 5a)] = 8/11 ⇒ (8a × 11) = 8 × (24 + 5a) ⇒ 88a = 192 + 40a ⇒ 88a - 40a = 192 ⇒ 48A = 192 ⇒ a = (192/48) = 4 ∴ Quantity of water in the new mixture = (5 × 4) + 24 = 44

123.

In a mixture, ink and water are in the ratio of 8 : 5. If 24 litres of water is added to the mixture, then the new ratio becomes 8 : 11. What is the total quantity (in litres) of water in the new mixture?1). 402). 423). 444). 48

Answer»

LET the initial quantity of ink and water in the mixture be 8a and 5A respectively

When 24 litres of water is added to the mixture, then the NEW ratio becomes 8 : 11

⇒ [8a/(24 + 5a)] = 8/11

⇒ (8a × 11) = 8 × (24 + 5a)

⇒ 88a = 192 + 40a

⇒ 88a - 40a = 192

⇒ 48A = 192

⇒ a = (192/48) = 4

∴ Quantity of water in the new mixture = (5 × 4) + 24 = 44

Discussion

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