SNAP Mixtures and Solutions Questions

Let the percentage of salt in the solution be $$x$$%. The amount of salt will be constant. Upon equating

$$5\%of\ 6l\ =\ x\%\ of\ 5l$$

It gives $$x=6$$

Let the volume be V of 30% alcohol solution
Now amount of alcohol = 0.3V
Amount of alcohol in 40 liters solution = 24
Now total volume = V+40
Total amount of alcohol = (V+40)(0.5) = 0.5V +20
Now Equating amount of Alcohol :
we get 0.3V +24 = 0.5V +20
we get V = 20

300 gms of solution has 40% salt 
so the amount of salt = 300(0.4) = 120 
Now let salt added be x 
So we get 120+x/300+x = 1/2
we get 240+2x =300+x
we get x= 60

The fraction of milk in A = 4/(4+3) = 4/7

The fraction of milk in B = 2/(2+3) = 2/5

Let them be mixed in the ratio x:y such that their fraction is 1/2
It can be written as 

$$\frac{4x}{7}$$ + $$\frac{2y}{5}$$ = $$\frac{x+y}{2}$$

Upon solving,

$$\frac{4x}{7}$$ - $$\frac{x}{2}$$ = $$\frac{y}{2}$$ - $$\frac{2y}{5}$$

$$\frac{x}{14}$$ = $$\frac{y}{10}$$

$$\frac{x}{y}$$ = $$\frac{7}{5}$$

Required ratio is 7:5

Initially, A has  10 L wine and B has 10L water

Process 1 : When 1 L of water is poured from B to A

A now has 11L ( 10 L wine and 1L water) and B has 9L water

Process 2:  1L from A is put into B

1L of liquid from A will have the composition of 10/11L Wine and 1/11L Water

Composition of A(10L)  = $$\frac{100}{11}$$ L wine and $$\frac{10}{11}$$ L water

Composition of B(10L) = $$\frac{10}{11}$$ L wine and $$\frac{100}{11}$$ L water