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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

Therefore the answer is A
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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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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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Thanks a lot, now I get it.
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statement 2) is useless. Its Truism analogous to "sun rises in the east". Hence A.
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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

So answer - option A
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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

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

Bunuel
tania
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.

Answer: A.

Hope it's clear.
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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
Bunuel, in these questions of mixture is there no need to specify that there is no other substances in the solution?

Bunuel
tania
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.

Answer: A.

Hope it's clear.
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tania
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.

The prompt indicates that a 35% acid solution must be mixed with a 0% acid solution -- the pure water -- to yield a 10% acid solution.
Using alligation, we can determine the required ratio of 35% solution to 0% solution.

Statement 1:
Since the prompt enables us to determine the required RATIO of 35% solution to 0% solution, and Statement 1 gives us the AMOUNT of 35% solution, we can calculate the required amount of 0% solution -- in other words, the required amount of water.
SUFFICIENT.

Statement 2:
\(\frac{acid}{total} = \frac{7}{7+13} = \frac{7}{20} = \frac{35}{100} =\) 35%
Statement 2 indicates that the 35% solution is composed solely of acid and water in a 7:13 ratio, yielding a solution that is 35% acid.
Since the amoutn of 35% solution is unknown, the required amount of water cannot be determined.
INSUFFICIENT.

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