2
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

Explanation:

General Formula:

Final or reduced concentration = initial concentration $\inline \dpi{100} \fn_cm \left ( 1-\frac{amount\: being\: replaced \: in \: each\: operation}{total\: amount} \right )^{n}$

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 Here amount being replaced denotes the quantity which is to be withdrawn in each time.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,    $\inline \dpi{100} \fn_cm 50\times \left ( 1-\frac{5}{50} \right )\times \left ( 1-\frac{5}{50} \right )\times \left ( 1-\frac{5}{50} \right )$

$\inline \dpi{100} \fn_cm 50\times \frac{45}{50}\times \frac{45}{50}\times \frac{45}{50}=50\times \left ( \frac{45}{50} \right )^{3}=50\times \left ( \frac{9}{10} \right )^{3}$

= 36.45 L

Q:

In a mixture of milk and water the proportion of water by weight was 75%. If in 60 gm of mixture 15 gm water was added, what would be the percentage of water ? (Weight in gm)

 A) 80% B) 70% C) 75% D) 62%

Explanation:

Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm.

After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and

weight of a mixture = 60 + 15 = 75 gm.

So % of water = 100 x 60/75 = 80%.

1 7
Q:

The amount of water (in ml) that should be added to reduce 9 ml lotion, containing 50% alcohol, to a lotion containing 30% alcohol is ?

 A) 6 ml B) 11 ml C) 15 ml D) 9 ml

Explanation:

Let us assume that the lotion has 50% alcohol and 50% water.
ratio = 1:1
As the total solution is 9ml
alcohol = water = 4.5ml
Now if we want the quantity of alcohol = 30%
The quantity of water = 70%
The new ratio = 3:7
Let x ml of water be added
We get,
$\inline \fn_jvn \small \frac{4.5}{4.5+x}=\frac{3}{7}$
=> x=6
Hence 6ml of water is added.

3 15
Q:

If a man buys 1 lt of milk for Rs.12 and mixes it with 20% water and sells it for Rs.15, then what is the percentage of gain ?

 A) 25% B) 30% C) 17% D) 19%

Explanation:

He has gain = 15 - 12 = 3,
Gain% = (3/12) x 100 = (100/4) = 25.
He has 25% gain.

4 32
Q:

One type of liquid contains 25 % of benzene, the other contains 30% of benzene. A can is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of benzene in the new mixture.

 A) 27 % B) 26 % C) 29 % D) 21 %

Explanation:

Let the percentage of benzene = X
(30 - X)/(X- 25) = 6/4 = 3/2
=> 5X = 135
X = 27

So, required percentage of benzene = 27 %

5 82
Q:

A container contains 50 litres of milk. From that 8 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container ?

 A) 24.52 litres B) 29.63 litres C) 28.21 litres D) 25.14 litres

Explanation:

Given that container has 50 litres of milk.

After replacing 8 litres of milk with water for three times, milk contained in the container is:

$\inline&space;\fn_jvn&space;\small&space;\Rightarrow&space;\left&space;[&space;50\left&space;(&space;1-\frac{8}{50}&space;\right&space;)^{3}&space;\right&space;]$

$\inline&space;\fn_jvn&space;\small&space;\Rightarrow&space;\left&space;(&space;50\times&space;\frac{42}{50}&space;\times&space;\frac{42}{50}\times&space;\frac{42}{50}\right&space;)$ = 29.63 litres.