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m25,#30

Author Message
Intern
Joined: 14 Feb 2013
Posts: 3
GMAT 1: Q V
GPA: 3.12
Followers: 1

Kudos [?]: 10 [0], given: 8

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01 Oct 2013, 10:35
is there a way to solve this problem using allegations?
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099

Kudos [?]: 93560 [1] , given: 10578

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02 Oct 2013, 01:39
1
KUDOS
Expert's post
aemarroquin wrote:
is there a way to solve this problem using allegations?

On page 1: m25-75372.html#p1122182
_________________
Intern
Joined: 14 Feb 2013
Posts: 3
GMAT 1: Q V
GPA: 3.12
Followers: 1

Kudos [?]: 10 [0], given: 8

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02 Oct 2013, 07:41
oh I see the explanation now. Thanks for the help on this problem. There are so many ways to solve these types of problems and I want to develop a strategy to always approach them in the most efficient manner.

Kudos(?) to both of you.
Intern
Joined: 21 Jul 2014
Posts: 23
Followers: 0

Kudos [?]: 2 [0], given: 13

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01 Sep 2014, 08:52
ritula wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

(A) $$\frac{1}{5}$$
(B) $$\frac{1}{4}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{3}{4}$$
(E) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Can sum1 give a simpler explanation?

Solved it by plugging in the option C first :

If 1/2 of the original solution is replaced, that means 50 out of 100 was replaced.
so the replaced part will not have 30% of 50= 15units of acid
the remaining 50 units of the original solution (which was not replaced) has 50% = 25 units of acid
total we have 25+15=40% which is equal to the goal for plugging in ! bingo.. C is the answer !!

Bunuel, please comment if this is an OK approach to solve the question?
Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7099

Kudos [?]: 93560 [0], given: 10578

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01 Sep 2014, 09:04
apsForGmat wrote:
ritula wrote:
Some part of a 50% solution of acid was replaced with an equal amount of 30% solution of acid. If, as a result, 40% solution of acid was obtained, what part of the original solution was replaced?

(A) $$\frac{1}{5}$$
(B) $$\frac{1}{4}$$
(C) $$\frac{1}{2}$$
(D) $$\frac{3}{4}$$
(E) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Can sum1 give a simpler explanation?

Solved it by plugging in the option C first :

If 1/2 of the original solution is replaced, that means 50 out of 100 was replaced.
so the replaced part will not have 30% of 50= 15units of acid
the remaining 50 units of the original solution (which was not replaced) has 50% = 25 units of acid
total we have 25+15=40% which is equal to the goal for plugging in ! bingo.. C is the answer !!

Bunuel, please comment if this is an OK approach to solve the question?

Yes, that's correct.
_________________
Re: m25,#30   [#permalink] 01 Sep 2014, 09:04

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