water(in grams)

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 185
Joined: Thu Feb 04, 2010 3:06 am
Thanked: 6 times

water(in grams)

by gmatnmein2010 » Fri Feb 19, 2010 8:09 pm
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
Source: — Data Sufficiency |

User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 am
Thanked: 137 times
Followed by:5 members

by thephoenix » Fri Feb 19, 2010 8:54 pm
gmatnmein2010 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.

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

User avatar
Legendary Member
Posts: 1022
Joined: Mon Jul 20, 2009 11:49 pm
Location: Gandhinagar
Thanked: 41 times
Followed by:2 members

by shashank.ism » Sun Feb 21, 2010 2:19 am
gmatnmein2010 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.

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
My Websites:
www.mba.webmaggu.com - India's social Network for MBA Aspirants

www.deal.webmaggu.com -India's online discount, coupon, free stuff informer.

www.dictionary.webmaggu.com - A compact free online dictionary with images.

Nothing is Impossible, even Impossible says I'm possible.