Percentage
- In what ratio must a mixture of 30% alcohol strength be mixed with that of 50% alcohol strength so as to get a mixture of 45% alcohol strength ?
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Let x litres of first mixture is mixed with y litres of the second mixture.
According to the question,
⇒ 0.3x + 0.5y = 9 0.7x + 0.5y 11
Correct Option: B
Let x litres of first mixture is mixed with y litres of the second mixture.
According to the question,⇒ 0.3x + 0.5y = 9 0.7x + 0.5y 11
⇒ 6.3x + 4.5y = 3.3x + 5.5y
⇒ 6.3x – 3.3x = 5.5y – 4.5y
⇒ 3x = y⇒ x = 1 : 3 y
- 200 litres of a mixture contains 15% water and the rest is milk. The amount of milk that must be added so that the resulting mixture contains 87.5% milk is
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In 200 litres of mixture,
Quantity of milk = 85 × 200 = 170 litres 100
Quantity of water = 200 - 170 = 30 litres
Let the quantity of additional milk added be m litres.
According to the question,170 + m × 100 = 87.5 200 + m
Correct Option: C
In 200 litres of mixture,
Quantity of milk = 85 × 200 = 170 litres 100
Quantity of water = 200 - 170 = 30 litres
Let the quantity of additional milk added be m litres.
According to the question,170 + m × 100 = 87.5 200 + m
⇒ (170 + m) × 100 = 17500 + 87.5m
⇒ 100m – 87.5m = 17500 – 17000
⇒ 12.5m = 500⇒ m = 500 = 40 litres 12.5
- In 50 gm alloy of gold and silver, the gold is 80% by weight. How much gold should be mixed to this alloy so that the weight of gold would become 95% ?
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Initial quantity of gold = 50 × 80 = 40 gm 100
Let 'm ' gm be mixed.(40 + m) = (50 + m) × 95 100
Correct Option: B
Initial quantity of gold = 50 × 80 = 40 gm 100
Let 'm ' gm be mixed.(40 + m) = (50 + m) × 95 100 ⇒ 40 + m = (50 + m) × 19 20
⇒ 800 + 20m = 950 + 19m
⇒ m = 150 gm
- How much pure alcohol has to be added to 400 ml of a solution containing 15% of alcohol to change the concentration of alcohol in the mixture to 32% ?
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According to question ,
Alcohol = 
15 × 400 
ml = 60 ml. 100
Water = 400 - 60 = 340 ml.
Let m ml of alcohol be added.Then, 60 + m × 100 = 32 400 + m ⇒ 60 + m = 32 = 8 400 + m 100 25
Correct Option: B
According to question ,
Alcohol = 15 × 400 ml = 60 ml. 100
Water = 400 - 60 = 340 ml.
Let m ml of alcohol be added.Then, 60 + m × 100 = 32 400 + m ⇒ 60 + m = 32 = 8 400 + m 100 25
⇒ 1500 + 25m = 3200 + 8m
⇒ 17m = 1700
⇒ m = 100 ml
- 40 litres of a mixture of milk and water contains 10% of water, the water to be added, to make the water content 20% in the new mixture is :
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Water content in 40 litres of mixture
= 4 × 10 = 4 litres 100
∴ Milk content = 40 – 4 = 36 litres
Let m litres of water is mixed.Then, 4 + m = 20 40 + m 100
Correct Option: D
Water content in 40 litres of mixture
= 4 × 10 = 4 litres 100
∴ Milk content = 40 – 4 = 36 litres
Let m litres of water is mixed.Then, 4 + m = 20 40 + m 100 ⇒ 4 + m = 1 40 + m 5
⇒ 20 + 5m = 40 + m
⇒ 4m = 20 ⇒ m = 5 litres
