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%.

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

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

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 .

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.

The amount of petrol left after 4 operations =  200 × 1 - 40 200 4 =81.92 

Hence the amount of kerosene = 200 - 81.92 = 118.08 liters

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

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 Here amount being replaced denotes the quantity which is to be withdrawn in each time.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

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.

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.