What percent of the solution is water?

Let P litres be the quantity of Initial solution, containing x litres of water.

hence % of water in Initial solution = \({(x/P) * 100}\)

Statement 1 says Adding 5 litres of water increases the % of water by 20%

Therefore New Quantity of Solution = (P+5) litres, containing (x+5) litres of water.

% Increase = \({(x+5)/(P+5)} - {(x/P)} = 20/100\)

we get an equation in 2 unknowns, hence St 1 alone is Insufficient.

Statement 2 says Quantity of Initial Solution, P = 30 litres, since no other info is provided, we can safely say St 2 alone is Insufficient.

Combining St 1 & 2

We can solve the equation in St 1, by substituting P = 30 & find value of x.

St 1 & 2 together are Sufficient.

Answer C.

Thanks,
GyM
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assume there is W liter water in S liter of solution therefore (W/S)=?

1. assume there is W liter water in S liter solution the percentage is (W/S)*100 after adding a 5 liter of water there is S+5 liter of solution and W+5 of water. algebraically we can write (W+5)/(S+5)=(W/S)+.2* (20 percent increase is equal to .2) it is clear that there is no way to solve for w/s. insuff

2. S=30. clearly insufficient

1+2. having S=30 we can solve for w from * then W/S. Sufficient.

Initial conc. of water = W1 - To be found
Final conc. of water = W2
Initial volume of mixture = V1
Final volume of mixture = V2

Ci * Vi = Cf * Vf

St-1:
W1 = ?
W2 = 20%
V1 = ?
V2 = V1+5

W1 * V1 = 20 * (V1+5)

After simplification: W1 = 20 + (5/V1)

V1 can have multiple values. Hence, St-1 NS.

St-2:
W1 = ?
W2 = ?
V1 = 30
V2 = ?

W1 * 30 = W2 * V2
Hence, St-2 NS.

St-1 + St-2:
W1 = 20 + (5/V1)
V1 = 30

We get unique W1.

Therefore, answer is (C) (St-1 + St-2)

One correction:
Since the distance between 1.2x and x = 1.2x - x = 0.2x, the equation should include the red values shown above.
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Bunuel,

Need your help here. Is option A not clear-I can get a unique solution in which we have 25 litres of water in a 30 litre solution and adding 5 litres increases the % of water by 20%.

Let x litre for one substance and y litre for water

I)(y+5)*100/x+y+5= y*100/x+y
Insufficient two unknowns

II)x+y=30 insufficient

Combining both statements
We get answer
C

Another approach
As per statement 1
25 l of water addition would increase value by 100 percent
Means currently water is 25 but we don't know total volume of solution so insufficient

Statement 2 gives no info about water

Combining both we can get
25 l water is present in total 30 l of solution

Give kudos if it helps
Posted from my mobile device

thanks gymrat for the explanation

Let x be percentage of water in y liters of solution.

1) Using mixture formula, we get xy + 5 = 1.2x(y+5) -> here we have 2 variables, so need to find both x and y - Insufficient
2) y = 30 - clearly insufficient

Combining 1+2 - x can be found, Ans. C

Dear GMATGuruNY

Can the alligation method used as follows for statement 1?

(1) Adding 5 liters of water increases the percentage of water by 20%.

Let the water percent in original solution S= x

The water percentage after adding water = 1.2x

Pure water = 100%

Solution.................Mixture......................Water
x.........0.2............1.2x......100-1.2x.......100

S/W = (100-1.2x)/0.2

Is the equation above correct?

Thanks in advance

Thanks Mitch, However, after taking closer look. I think there is small mistake in red. isn't it?

Good catch!
I've amended my response accordingly.
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