Mixtures2

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shrey2287: Senior | Next Rank: 100 Posts; Posts: 33; Joined: Thu Apr 15, 2010 9:36 pm; Thanked: 1 times

Mixtures2

by shrey2287 » Mon Oct 17, 2011 4:44 am

7. A certain quantity of 40% solution is replaced with 25% solution such that
the new concentration is 35%. What is the fraction of the solution that was
replaced?
(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4

OA - B

PL xplain

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Source: Beat The GMAT — Problem Solving | Mon Oct 17, 2011 4:44 am

shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 am; Thanked: 230 times; Followed by:21 members

by shankar.ashwin » Mon Oct 17, 2011 4:51 am

Ratio of 25% / Ratio of 40% = 40-35/35-25 = 1/2

So 1 part of 25% solution is added to 2 parts of 40% soln to from a 35% solution.

Hence 1/3 of the 40% must have been removed

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enjoylife1788: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Sat Aug 15, 2009 10:31 pm; Location: India

by enjoylife1788 » Mon Oct 17, 2011 5:08 am

I still don't get it. Can someone please provide a detailed explanation.

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enjoylife1788: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Sat Aug 15, 2009 10:31 pm; Location: India

by enjoylife1788 » Mon Oct 17, 2011 5:31 am

I was able to solve it by way of backtracking. Not ideal but at times gets the work done.

I started by choosing C. So if both the new solution and the old one are 50 percent each. Then the concentration is 20ml ( from the non substituded solution, 50ml * 0.40 ) + 12.5 ml ( from the substitued, 50ml * 0.25 ) .

Final concentration is (20+12.5)/100 which is 32.5. So its slightly less than final intended result. So we should substitute a bit less. Hence, Answer Choice B.

Even though I could get the answer this way, I would like to know the exact math involved.