11
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

Explanation:

$\frac{wine(left)}{water(added)}=\frac{343}{169}$

It means    $\frac{wine(left)}{wine(initial\:&space;amount)}=\frac{343}{512}$       $\left&space;(&space;\because&space;\:&space;\:&space;343+169=512&space;\right&space;)$

Thus ,       $343x=512x\left&space;(&space;1-\frac{15}{k}&space;\right&space;)^{3}$

$\Rightarrow&space;\:&space;\:&space;\frac{343}{512}=\left&space;(&space;\frac{7}{8}&space;\right&space;)^{3}=\left&space;(&space;1-\frac{15}{k}&space;\right&space;)^{3}$

$\Rightarrow\:&space;\:&space;\left&space;(&space;1-\frac{15}{k}&space;\right&space;)=\frac{7}{8}=\left&space;(&space;1-\frac{1}{8}&space;\right&space;)$

$\Rightarrow&space;\:&space;\:&space;K=120$

Thus the initial amount of wine was 120 liters.

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:

1 25
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.

11 405
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 = $\inline \fn_jvn \left [ q\left ( 1-\frac{p}{q} \right )^{n} \right ]$

Here p = 12 , q = 120

=> $\inline \fn_jvn \left [ 120\left ( 1-\frac{12}{120} \right ) ^{3}\right ]$

=> 120 x 0.9 x 0.9 x 0.9

=> 120 x 0.729

= 87.48 lit.

7 356
Q:

Manideep purchases 30kg of barley at the rate of 11.50/kg and 20kg at the rate of 14.25/kg. He mixed the two and sold the mixture in the shop. At what price per kg should he sell the mixture to make 30% profit to him ?

 A) 15.84 B) 14.92 C) 13.98 D) 16.38

Explanation:

Given, Manideep purchases 30kg of barley at the rate of 11.50/kg nad 20kg at the rate of 14.25/kg.

Total cost of the mixture of barley = (30 x 11.50) + (20 x 14.25)

=> Total cost of the mixture = Rs. 630

Total kgs of the mixture = 30 + 20 = 50kg

Cost of mixture/kg = 630/50 = 12.6/kg

To make 30% of profit

=> Selling price for manideep = 12.6 + 30% x 12.6

=> Selling price for manideep = 12.6 + 3.78 = 16.38/kg.

6 389
Q:

There are two mixtures of honey and water in which the ratio of honey and water are as 1:3 and 3:1 respectively. Two litres are drawn from first mixture and 3 litres from second mixture, are mixed to form another mixture. What is the ratio of honey and water in it ?

 A) 111:108 B) 11:9 C) 103:72 D) None

Explanation:

From the given data,

The part of honey in the first mixture = 1/4

The part of honey in the second mixture = 3/4

Let the part of honey in the third mixture = x

Then,

1/4             3/4

x

(3/4)-x     x-(1/4)

Given from mixtures 1 & 2 the ratio of mixture taken out is 2 : 3

=> $\inline \fn_jvn \frac{\frac{3}{4}-x}{x-\frac{1}{4}}=\frac{2}{3}$

=> Solving we get the part of honey in the third mixture as 11/20

=> the remaining part of the mixture is water = 9/20

Hence, the ratio of the mixture of honey and water in the third mixture is 11 : 9 .