19
Q:

# How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?

 A) 36 Kg B) 42 Kg C) 54 Kg D) 63 Kg

Answer:   D) 63 Kg

Explanation:

By the rule of alligation:
C.P. of 1 kg sugar of 1st kind      C.P. of 1 kg sugar of 2nd kind

${\color{Blue}&space;\therefore&space;}$ Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.

Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind.
Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.

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

Answer & Explanation Answer: A) 6 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.

2 14
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.

3 31
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 %

Answer & Explanation Answer: A) 27 %

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 80
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

Answer & Explanation Answer: B) 29.63 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.

6 145
Q:

A Container has a capacity of 48 litres and is full of milk. 4 liters of milk is taken out of it and replaced by same quantity of water. Again 4 liters of the mixture is taken out and replaced by the same quantity of water . The process is repeated five times. How much milk is left in the mixture ?

$\inline \dpi{100} \fn_cm \frac{Milk\: Left\: in\: the\: container\: after\: the\: operations}{Original\: quantity\: of\: milk\: in\: the\: container\: }$ = $\inline \dpi{100} \fn_cm (\frac{Total\: quantity\: of\: milk - quantity\: drawn\: each\: time}{Total\: mixture })^n$

$\inline \dpi{100} \fn_cm \frac{Milk\: left\: in\: the\: container}{48 }=[\frac{48-4}{5}]^5$

Milk left in the container = $\inline \dpi{100} \fn_cm [\frac{44}{48}]^5\times 48$

$\inline \dpi{100} \fn_cm \frac{11}{12}\times \frac{11}{12}\times \frac{11}{12}\times \frac{11}{12}\times \frac{11}{12}\times 48$

$\inline \dpi{100} \fn_cm \frac{161051}{5184}$ = $\inline \dpi{100} \fn_cm 31\frac{347}{5184}$ liters