CMAT Mixtures and Solutions Questions

Initially Milkman has 90 liters of milk
Now amount of water added = 10 liters 
So total quantity becomes : 100 liters 
He sold 1/5 of quantity 
so he sold 2 liters of water and 18 liters of milk (A total of 20 liters is sold )
Now he is left with 10-2 = 8 liters of water  and 90-18 = 72 liters of milk.
Now he adds 20 liters of water
so the amount of milk and water with him will be : 72 and 28 respectively.

We have formula :
F = $$I\left(1-\frac{K}{\left(V\right)}\right)^n$$
Here n =3 V =500 k =50 
substituting we get
F/I = 729/1000
Therefore in percentage F = 72.9 percent

Initial amount of Kerosene(I) = 5% of 5 kg mixture of petrol and kerosene = $$\frac{5}{100}\times 5000 = 250$$ grams 

Let say 'x' gm of Kerosene is added to the mixture.

The final amount of Kerosene in the mixture after the addition is given as 10%.

$$\Rightarrow \frac{250 + x}{5000 + x} = \frac{10}{100}$$

$$\Rightarrow 2500 + 10x = 5000 + x$$

$$\Rightarrow 9x = 2500$$

$$\Rightarrow x = 275$$ gm approximately.

Therefore additionally 275 gm of kerosene is to be added to the mixture to make it 10%.

From 32 litres, 4 litres is drawn out and is replaced with water.

Proportion of wine drawn = 4/32 = 1/8th of the total volume of the wine in the mixture.

When wine is drawn out for the second time, $$(1/8)\times (28) = 3.5$$ litres of wine will be drawn out.

Wine remaining in the jar = 28-3.5 = 24.5 litres

When wine is drawn out for the third time, $$(1/8)\times (24.5) =3.0625$$ litres of wine will be drawn out.

Wine remaining in the jar = 24.5 - 3.0625 = 21.4375 litres

Water in the jar = 10.5625 litres

Ratio of wine remaining in the jar to water remaining in the jar = 343:169