water(in grams)

[gmatnmein2010]

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.

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

[thephoenix]

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.

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

Let x be the original volume, so original volume is 0.35x/x

so 0.35x/(x+w) = 0.1

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.
not suff

hence A

[shashank.ism]

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.

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

Let the soln be x gm.
so wt. of water present = 0.65x; acid = 0.35 x
let y gm water be added
so (0.65x )/(x+y) = 0.10 --> 0.65x = 0.1x+0.10 y --> 0.55x =0.10 y --> 5.5x =y
St. 1 ) x= 50 , y= 5.5x50 = 275 gm suff.
St.2 ) 0.35x/0.65x = 7/13 .... which is already known.. not suff.
Ans A