10
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

Explanation:

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

$512=K{\left(1-\frac{1}{5}\right)}^{4}$

or

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

Therefore, K = 1250

Hence initially the honey in the jar= 1.25 kg

Q:

In a 100 litre of mixture the ratio of milk and water is 6:4. How much milk must be added to the mixture in order to make the ratio 3 : 1?

 A) 85 B) 60 C) 55 D) 45

Explanation:

Let M litres milk be added

=>

=> 60 + M = 120

=> M = 60 lit.

1 9
Q:

A mixture contains 25% milk and rest water. What percent of this mixture must taken out and replaced with milk so that in mixture milk and water may become equal.

 A) 31.8% B) 31% C) 33.33% D) 29.85%

Explanation:

Now, take percentage of milk and applying mixture rule

25          100

50

50            25  = 2 : 1

Hence required answer =  1/3 or 33.33%

8 120
Q:

The concentration of glucose in three different mixtures (glucose and alcohol) is  respectively. If 2 litres, 3 litres and 1 litre are taken from these three different vessels and mixed. What is the ratio of glucose and alcohol in the new mixture?

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

Explanation:

Concentration of glucose are in the ratio = $\frac{1}{2}:\frac{3}{5}:\frac{4}{5}$

Quantity of glucose taken from A = 1 liter out of 2

Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit

Quantity of glucose taken from C = 0.8 lit

So, total quantity of glucose taken from A,B and C = 3.6 lit

So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit

Ratio of glucose to alcohol = 3.6/2.4 = 3:2

7 265
Q:

A milkman claims to sell milk at its cost price, still, he is making a profit of 30% since he has mixed some amount of water in the milk. What is the % of milk in the mixture?

 A) 71.02% B) 76.92% C) 63.22% D) 86.42%

Explanation:

Let the milk he bought is 1000 ml

Let C.P of 1000 ml is Rs. 100

Here let he is mixing K ml of water

He is getting 30% profit

=> Now, the selling price is also Rs. 100 for 1000 ml

=> 100 : K%

= 100 : 30

10 : 3 is ratio of milk to water

=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92%

9 435
Q:

The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?

 A) 8:3 B) 6:7 C) 7:5 D) 11:7

Explanation:

Milk in 1-litre mixture of A = 4/7 litre.

Milk in 1-litre mixture of B = 2/5 litre.

Milk in 1-litre mixture of C = 1/2 litre.

By rule of alligation we have required ratio X:Y

X                  :                 Y

4/7                                2/5

\                      /

(Mean ratio)
(1/2)

/                      \

(1/2 – 2/5)     :       (4/7 – 1/2)

1/10                      1/1 4

So Required ratio = X : Y = 1/10 : 1/14 = 7:5

9 333
Q:

In a mixture of 240 lt. water is 20% and rest is Milk. What quantity of mixture should be taken out and replaced with water so that water becomes 40%?

 A) 60 lit B) 55 lit C) 45 lit D) 50 lit

Explanation:

7 455
Q:

In a 48 ltr mixture, the ratio of milk and water is 5:3. How much water should be added in the mixture so as the ratio will become 3:5 ?

 A) 24 lit B) 16 lit C) 32 lit D) 8 lit

Explanation:

Given mixture = 48 lit

Milk in it = 48 x 5/8 = 30 lit

=> Water in it = 48 - 30 = 18 lit

Let 'L' lit of water is added to make the ratio as 3:5

=> 30/(18+L) = 3/5

=> 150 = 54 + 3L

=> L = 32 lit.

12 755
Q:

A container contains 120 lit of Diesel. From this container, 12 lit of Diesel was taken out and replaced by kerosene. This process was further repeated for two times. How much diesel is now there in the container ?

 A) 88.01 lit B) 87.48 lit C) 87.51 lit D) 87.62 lit

Explanation:

For these type of problems,

Quantity of Diesel remained = $\left(q{\left(1-\frac{p}{q}\right)}^{n}\right)$

Here p = 12 , q = 120

=> $\left(120{\left(1-\frac{12}{120}\right)}^{3}\right)$ = 120 x 0.9 x 0.9 x 0.9 = 87.48 lit.