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Q:

In a mixture of milk and water, there is only 26% water. After replacing the mixture with 7 liters of pure milk , the percentage of milk in the mixture become 76%. The quantity of mixture is:

A) 65 liters	B) 91 liters
C) 38 liters	D) None of these

Answer & Explanation Answer: B) 91 liters

Explanation:

Milk Water

74% 26% (initially)

76% 24% ( after replacement)

Left amount = Initial amount $(1 - \frac{r e p l a c e d a m o u n t}{t o t a l a m o u n t})$

24 = 26 $(1 - \frac{7}{k})$

=> k = 91

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Filed Under: Alligation or Mixture

Q:

The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture ?

A) 4	B) 5
C) 8	D) None of these

Answer & Explanation Answer: B) 5

Explanation:

Wine Water

8L 32L

1 : 4

20 % 80% (original ratio)

30 % 70% (required ratio)

In ths case, the percentage of water being reduced when the mixture is being replaced with wine.

so the ratio of left quantity to the initial quantity is 7:8

Therefore , $\frac{7}{8} = [1 - \frac{K}{40}]$

=> K = 5 Lit

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Q:

A jar was full with honey. A person used to draw out 20% of the honey from the jar and replaced it with sugar solution. He has repeated the same process 4 times and thus there was only 512 gm of honey left in the jar, the rest part of the jar was filled with the sugar solution. The initial amount of honey in the jar was filled with the sugar solution. The initial amount of honey in the jar was:

A) 1.25 kg	B) 1 kg
C) 1.5 kg	D) None of these

Answer & Explanation Answer: A) 1.25 kg

Explanation:

Let the initial amount of honey in the jar was K, then

$512 = K {(1 - \frac{1}{5})}^{4}$ $[∵ 20 % = \frac{20}{100} = \frac{1}{5}]$

or

$512 = K {(\frac{4}{5})}^{4}$

Therefore, K = 1250

Hence initially the honey in the jar= 1.25 kg

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Q:

From a container of wine, a thief has stolen 15 liters of wine and replaced it with same quantity of water.He again repeated the same process. Thus in three attempts the ratio of wine and water became 343:169. The initial amount of wine in the container was:

A) 75 liters	B) 100 liters
C) 150 liters	D) 120 liters

Answer & Explanation Answer: D) 120 liters

Explanation:

$\frac{W i n e (l e f t)}{W a t e r (a d d e d)} = \frac{343}{169}$

It means $\frac{W i n e (l e f t)}{W i n e (i n i t i a l a m o u n t)} = \frac{343}{512}$ $[∵ 343 + 169 = 512]$

Thus ,

$343 x = 512 x {(1 - \frac{15}{K})}^{3}$

$\Rightarrow \frac{343}{512} = {(\frac{7}{8})}^{3} = {(1 - \frac{15}{k})}^{3}$

=> K = 120

Thus the initial amount of wine was 120 liters.

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Q:

From a tank of petrol , which contains 200 liters of petrol, the seller replaces each time with kerosene when he sells 40 liters of petrol(or mixture). Everytime he sells out only 40 liters of petrol (pure or impure). After replacing the petrol with kerosene 4th time, the total amount of kerosene in the mixture is

A) 81.92L	B) 96L
C) 118.08L	D) None of these

Answer & Explanation Answer: C) 118.08L

Explanation:

The amount of petrol left after 4 operations = $200 \times {(1 - \frac{40}{200})}^{4} =$ 81.92

Hence the amount of kerosene = 200 - 81.92 = 118.08 liters

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Q:

From the 50 liters of milk, 5 liters of milk is taken out and after it 5 liters of water is added to the rest amount of milk. Again 5 liters of milk and water is drawn out and it was replaced by 5 liters of water. If this process is continued similarly for the third time, the amount of milk left after the third replacement:

A) 45L	B) 36.45L
C) 40.5L	D) 42.5L

Answer & Explanation Answer: B) 36.45L

Explanation:

General Formula:

Final or reduced concentration = initial concentration x ${(1 - \frac{a m o u n t b e i n g r e p l a c e d i n e a c h o p e r a t i o n}{t o t a l a m o u n t})}^{n}$

where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time.

Therefore, $50 \times {(1 - \frac{5}{50})}^{3}$

= 36.45 L

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Q:

In what ratio should water and wine be mixed so that after selling the mixture at the cost price a profit of 33.33% is made?

A) 1:4	B) 1:3
C) 2:3	D) 3:4

Answer & Explanation Answer: B) 1:3

Explanation:

33.33% profit means there is one part water and 3 part is pure wine. so the required ratio of water and wine in the mixture is 1:3

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Filed Under: Profit and Loss

Q:

A milk man has 20 liters of milk. If he mixes 5 liters of water, which is freely available, in 20 liters of pure milk.If the cost of pure milk is Rs.18 per liter,then the profit of the milkman, when he sells all the mixture at cost price is:

A) 20%	B) 25%
C) 33.33%	D) 18%

Answer & Explanation Answer: B) 25%

Explanation:

when the water is freely available and all the water is sold at the price of the milk, then the water gives the profit on the cost of 20 liters of milk.

Therefore, profit percentage = (5/20) x 100 =25 % [ Profit %= (Profit/C.P)*100]

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Filed Under: Profit and Loss

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