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# Mixtures

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Senior Manager
Joined: 08 Jan 2009
Posts: 329
Followers: 2

Kudos [?]: 91 [0], given: 5

Mixtures [#permalink]  07 Sep 2009, 23:01
00:00

Difficulty:

(N/A)

Question Stats:

67% (03:39) correct 33% (03:24) wrong based on 3 sessions
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

Manager
Joined: 28 Jul 2009
Posts: 124
Location: India
Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business
Followers: 6

Kudos [?]: 62 [3] , given: 12

Re: Mixtures [#permalink]  07 Sep 2009, 23:11
3
KUDOS
tkarthi4u wrote:
If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

Well, there are no Answer options, eh! Anyways, according to me, the answer is 80% or 4/5th of the original mix was replaced.
Using amalgamation,

50--------------25
--------30--------
5---------------20

ie : 5:20 => 1:4.

But I am not sure what "percentage of Original alcohol was replaced" means. If I did get it right, the answer for alcohol automatically comes out to be the same ie 80% or 4/5th.

What is the OA?
_________________

GMAT offended me. Now, its my turn!
Will do anything for Kudos! Please feel free to give one.

Senior Manager
Joined: 08 Jan 2009
Posts: 329
Followers: 2

Kudos [?]: 91 [0], given: 5

Re: Mixtures [#permalink]  08 Sep 2009, 17:35
Thank you. i get it. The answer is correct.

i needed to clarify one thing.

The final ratio is [ ( Concentartion of 50% alcohol mix) / ( Concentartion of 25% alcohol mix)] in the new mixture. ie 30% alcohol mixture

so to find the part replaced we are finding the amount of 25% alcohol mix in the new mixture = 4/5.

Am i correct.
Manager
Joined: 10 Jul 2009
Posts: 169
Followers: 1

Kudos [?]: 44 [0], given: 8

Re: Mixtures [#permalink]  08 Sep 2009, 19:25
As bhanushalinikhil (phew lengthy name ) mentioned the ratio 1:4 is the ratio of 50% solution to 25% solution. So four parts of 25% solution should be mixed with 1 part of 50% solution. So the replaced amount = 4/5.
Senior Manager
Joined: 08 Jan 2009
Posts: 329
Followers: 2

Kudos [?]: 91 [0], given: 5

Re: Mixtures [#permalink]  08 Sep 2009, 22:40
Thank you. Let us see the below question. I am just checking if i am getting my mixture concepts correctly.

How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?There are 50 grams of the 35%-solution.

50 * 35 /100 = 17.5

17.5 / ( 50 + x) = 10 /100 1750 - 500 = 10 x so x = 125.

Is my method correct? How can do it via amalgamation?
Senior Manager
Joined: 08 Jan 2009
Posts: 329
Followers: 2

Kudos [?]: 91 [0], given: 5

Re: Mixtures [#permalink]  10 Sep 2009, 18:35
Can some help on the below question
Senior Manager
Joined: 31 Aug 2009
Posts: 420
Location: Sydney, Australia
Followers: 6

Kudos [?]: 149 [0], given: 20

Re: Mixtures [#permalink]  10 Sep 2009, 20:31
If you put all the % into the following format:

35--------------0 (since water is 0%)
--------10--------
10---------------25

That means Ratio of 35% mixture:water = 10: 25 = 2:5
If there's 50 grams of 35% mixture you need 5x25 grams of water = 125grams
Re: Mixtures   [#permalink] 10 Sep 2009, 20:31
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