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

Question

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: how-much-water-in-grams-should-be-added-to-a-35-solution-88479.html

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Official Answer

A

alwynjoseph · Accepted Answer

Statement 1 alone is sufficient

if the total weight of the solution is 50 grams which has got 17.5 g ( 35%) of acid and 32.5 g (65% water)

to make it as a solution of 10% acid, then 10% of x should be 17.5 g => x =175 g.
Current weight of water is 32.5, hence we need to add (157.5-32.5) = 125 g of water.

Statement 2: gives you the ratio of acid:water = 7:13 which is already given in the stem of the question. so no use

Bunuel · Answer

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. OPEN DISCUSSION OF THIS QUESTION IS HERE: how-much-water-in-grams-should-be-added-to-a-35-solution-88479.html

crejoc · Answer

Agree with  alwynjoseph .. it is A

bipolarbear · Answer

Write out the equations:

Let x be the original volume, so original volume is .35x/x

so .35x/(x+w) = .1

We're looking for w, so we just need to know x to solve the equation

1. There are 50 grams of the 35%-solution
=> x = 50, so we can solve for w => SUFFICIENT

2. In the 35%-solution the ratio of acid to water is 7:13
we know the proportion of x but not x itself, as x can still be 1g, 2g...etc.

Hence answer is (A)

dolly12 · Answer

'A'

(1) - 50gms -> 17.5 gms acid and 32.5 gms water - needs to get to total 175 gms for 10%. Needs 125gm more water

(2) - solution can be 10gms, 50, 100 gms..... will change the amount of water required to get to 10%

kamalkicks · Answer

AGREE WITH A

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

statement 1 sufficent 
 let x be water added
 17.5/50+x= 1/10

statement 2 in sufficient
 still gives us ratio 35 :65

hence a

MHIKER · Answer

1. Water in 50 grams = 50*75% = 37.5 grams
so (37.5/35%)*10%
Statement 1 is sufficient
2. it gives ratio. Insufficient 
Ans. A

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

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

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