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# How much water (in grams) should be added to a 35%-solution

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Intern
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How much water (in grams) should be added to a 35%-solution  [#permalink]

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27 Dec 2009, 14:59
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61% (01:36) correct 39% (01:34) wrong based on 589 sessions

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How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?

(1) There are 50 grams of the 35%-solution.
(2) In the 35%-solution the ratio of acid to water is 7:13.
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27 Dec 2009, 17:37
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tania wrote:
How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?

There are 50 grams of the 35%-solution.
In the 35%-solution the ratio of acid to water is 7:13.

I know the above one is easy, can someone please explain to me in detail, how the above problem should be solved?

Set the equation 0.35S=0.1(S+W), W=2.5S --> W=? You can see that only thing we need to know to calculate the value of W is S (initial amount of solution in grams).

The equation basically is saying that acid in grams in initial 35% solution (0.35S), equals to the acid in grams in 10% solution after water is added (0.1(S+W)), as amount of acid is not changed.

(1) Directly gives the value of S=50 --> W=2.5*50=125gr. Sufficient.

(2) A/W=7/13, this information is the same as given in the stem: there is 35% of acid in the solution. Out of 20 shares (7+13=20) 7 is acid or 7/20=35%. Not sufficient.

Hope it's clear.
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28 Dec 2009, 09:12
The answer is A because from the statement 1, you have enough information to answer the question.
if you know the total is 50 gram, you have 35% of acid and 65% of water. by knowing the total, you can determine the amount in gram of acid 50*35/100 and the amount in gram of water 50*65/100, so you don't need to pursue the calculation because you know that you have enough information to answer the question.

The second statement is irrelevant because this information is already stated on the question stem, if you know the amount of acid is 35%, so the amount of water is 65%, so the ratio is 3.5:6.5 which is the same as 7:13

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25 Feb 2011, 12:50
3.How much water (in grams) should be added to the 35%-solution of acid to obtain the 10%-solution?
1. There are 50 grams of the 35%-solution
2. In the 35%-solution the ratio of acid to water is 7:13
what is the best way to approach mixture problems ?

1) I tried to solve this problem using the mixture composition:
x-original solution in grams
w-water to be added in grams

0.35x + w = 0.1(x + w)

but obviously this is the wrong equation. Can some one point out why this eqn is wrong.
Thanks,

-Mike
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25 Feb 2011, 14:39
1
First of all, the second statement tells us nothing as 35% means 7:13 or 35%:65% ratio.
From the first statement we can find how much water and acid we have in 50g of 35% solution and it's sufficient. If it's not obvious, we can use following calculations:

acid = 0.35x
water in the solution: 0.65x
% of new solution = 0.1 = 0.35x / (w + x)

0.1*(w + x) = 0.35x - meaning: acid in new and old solutions (it should be the same number)

so, here is your mistake: 0.35x + w = 0.1(x + w)
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Joined: 28 Nov 2010
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25 Feb 2011, 15:17
Thanks a lot, now I get it.
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26 Feb 2011, 20:53
statement 2) is useless. Its Truism analogous to "sun rises in the east". Hence A.
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23 Apr 2011, 08:34
Statement 2 gives us the information that we already have from the question
Statement 1 - We can calculate the answer by using this statement...
If 50 is the solution, then 17.5 is the acid
Solve
17.5/(50+x)=10/100

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Re: How much water (in grams) should be added to a 35%-solution  [#permalink]

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13 Jan 2015, 15:12
A quick way to solve such problems:

(old - new)/(new) = dilution
(35-20)/(25) = 2.5 = dilution

From 1: 2.5*50 = 125 gm ---- SUFFICIENT
From 2: (7/20)*100 = 35% ------ No new info, same as the question ------- INSUFFICIENT

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Re: How much water (in grams) should be added to a 35%-solution  [#permalink]

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20 Dec 2015, 03:45
We need to find the mass and not the ratio.

Statement gives the mass. Sufficient

Statement 2 gives the ratio. THere might be 1000 gms of 35% solutions or 10,000 gms of it. While the ratio remains same. the grams of water would vary depending on the mass of the solutions.

Hence A
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Re: How much water (in grams) should be added to a 35%-solution  [#permalink]

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11 Mar 2016, 01:42
3
tania wrote:
How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?

(1) There are 50 grams of the 35%-solution.
(2) In the 35%-solution the ratio of acid to water is 7:13.

Responding to a pm:
Here is how you will use scale method here:
You want to add pure water (0% acid solution) to 35% acid solution to get 10% acid solution.

w1/w2 = (35 - 10)/(10 - 0) = 25/10 = 5/2

So you will need to add 5 parts water to 2 parts 35% solution.

(1) There are 50 grams of the 35%-solution.
5/2 = Water/50
Water = 125 gms

Sufficient.

(2) In the 35%-solution the ratio of acid to water is 7:13.
This just tells us that 35% solution has 7 parts acid in every 20 parts of solution. This is same as saying that there are 35 parts acid in 100 parts of solution. No new information.
Not sufficient.

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Re: How much water (in grams) should be added to a 35%-solution  [#permalink]

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20 Jun 2018, 06:21
Bunuel, in these questions of mixture is there no need to specify that there is no other substances in the solution?

Bunuel wrote:
tania wrote:
How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?

There are 50 grams of the 35%-solution.
In the 35%-solution the ratio of acid to water is 7:13.

I know the above one is easy, can someone please explain to me in detail, how the above problem should be solved?

Set the equation 0.35S=0.1(S+W), W=2.5S --> W=? You can see that only thing we need to know to calculate the value of W is S (initial amount of solution in grams).

The equation basically is saying that acid in grams in initial 35% solution (0.35S), equals to the acid in grams in 10% solution after water is added (0.1(S+W)), as amount of acid is not changed.

(1) Directly gives the value of S=50 --> W=2.5*50=125gr. Sufficient.

(2) A/W=7/13, this information is the same as given in the stem: there is 35% of acid in the solution. Out of 20 shares (7+13=20) 7 is acid or 7/20=35%. Not sufficient.

Hope it's clear.
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Re: How much water (in grams) should be added to a 35%-solution  [#permalink]

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20 Jun 2018, 20:23
Hello

Generally speaking, when we say 35% acid solution, it means it has 35% acid and rest 65% is water only, no other material present.
Here we have to mix 35% acid solution with water (which is 0% acid) to obtain a 10% acid solution (which should have 10% acid and rest 90% water).

devlin wrote:
Bunuel, in these questions of mixture is there no need to specify that there is no other substances in the solution?

Bunuel wrote:
tania wrote:
How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?

There are 50 grams of the 35%-solution.
In the 35%-solution the ratio of acid to water is 7:13.

I know the above one is easy, can someone please explain to me in detail, how the above problem should be solved?

Set the equation 0.35S=0.1(S+W), W=2.5S --> W=? You can see that only thing we need to know to calculate the value of W is S (initial amount of solution in grams).

The equation basically is saying that acid in grams in initial 35% solution (0.35S), equals to the acid in grams in 10% solution after water is added (0.1(S+W)), as amount of acid is not changed.

(1) Directly gives the value of S=50 --> W=2.5*50=125gr. Sufficient.

(2) A/W=7/13, this information is the same as given in the stem: there is 35% of acid in the solution. Out of 20 shares (7+13=20) 7 is acid or 7/20=35%. Not sufficient.

Hope it's clear.
Re: How much water (in grams) should be added to a 35%-solution &nbs [#permalink] 20 Jun 2018, 20:23
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