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

Correct Answer: 11:9

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

 

=>  3 4 - x x - 1 4 = 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 .

Correct Answer: 25%

Explanation:

As water costs free, water sold at cost price of milk gives the profit.

Required profit % = 5/20 x 100 = 5 x 5 = 25%.

Correct Answer: 2:1

Explanation:

Given rate of wheat at cheap = Rs. 2.90/kg

Rate of wheat at cost = Rs. 3.20/kg

Mixture rate = Rs. 3/kg

Ratio of mixture = 

          2.90                              3.20

                              3

(3.20 - 3 = 0.20)            (3 - 2.90 = 0.10)

 

0.20 : 0.10 = 2:1

 

Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.

Correct Answer: 50 gm

Explanation:

Initial quantity of copper = 80 100 x 50 = 40 g 

And that of Bronze = 50 - 40 = 10 g

Let 'p' gm of copper is added to the mixture

=> 50 + p x 90 100 = 40 + p

=> 45 + 0.9p = 40 + p

=> p = 50 g

Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.

Correct Answer: 7:5

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

Correct Answer: 3:2

Explanation:

Let he mixes the oils in the ratio = x : y

Then, the cost price of the oils = 60x + 65y

Given selling price = Rs. 68.20

=> Selling price = 68.20(x+y)

Given profit = 10% = SP - CP

=> 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y)

=> 6x + 6.5y = 8.20x + 3.20y

=>2.2x = 3.3y

=> x : y = 3 : 2

Correct Answer: 36.45L

Explanation:

General Formula:

Final or reduced concentration = initial concentration x  1 - 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 × 1 - 5 50 3

= 36.45 L

Correct Answer: 60 lit

Correct Answer: 87.48 lit

Explanation:

For these type of problems,

Quantity of Diesel remained =  q 1 - p q n   

 

Here p = 12 , q = 120 

 

=>  120 1 - 12 120 3 = 120 x 0.9 x 0.9 x 0.9 = 87.48 lit.

Correct Answer: 105/103

Explanation:

As given equal amounts of alloys are melted, let it be 1 kg.

Required ratio of gold and silver =  5 13 + 5 8 8 13 + 3 8 = 105 103 .

 

Hence, ratio of gold and silver in the resulting alloy = 105/103.