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What percent of the solution is water?

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What percent of the solution is water? [#permalink]

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New post 27 Jan 2017, 09:38
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62% (01:31) correct 38% (01:03) wrong based on 208 sessions

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What percent of the solution is water?

(1) Adding 5 liters of water increases the percentage of water by 20%.
(2) There are 30 liters of solution before any additions or subtractions from the solution.

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Re: What percent of the solution is water? [#permalink]

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New post 28 Jan 2017, 14:44
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.
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Re: What percent of the solution is water? [#permalink]

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New post 20 Feb 2017, 02:09
Bunuel wrote:
What percent of the solution is water?

(1) Adding 5 liters of water increases the percentage of water by 20%.
(2) There are 30 liters of solution before any additions or subtractions from the solution.



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%.
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Re: What percent of the solution is water? [#permalink]

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New post 10 Apr 2018, 22:38
1
Bunuel wrote:
What percent of the solution is water?

(1) Adding 5 liters of water increases the percentage of water by 20%.
(2) There are 30 liters of solution before any additions or subtractions from the solution.



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)
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What percent of the solution is water? [#permalink]

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New post Updated on: 04 Jul 2018, 21:46
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

Originally posted by push12345 on 22 Apr 2018, 05:00.
Last edited by push12345 on 04 Jul 2018, 21:46, edited 1 time in total.
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Re: What percent of the solution is water? [#permalink]

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New post 03 Jun 2018, 04:56
Bunuel wrote:
What percent of the solution is water?

(1) Adding 5 liters of water increases the percentage of water by 20%.
(2) There are 30 liters of solution before any additions or subtractions from the solution.



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
Re: What percent of the solution is water?   [#permalink] 03 Jun 2018, 04:56
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