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concentration of Husk? [#permalink]
 13 Aug 2009, 03:27
Question Stats:
52% (02:11) correct
47% (00:58) wrong based on 0 sessions
if 200 lb of a mixture contain 80% husk and 20%sand. Then how much husk needs to be extracted in order to have 75% concentration of Husk? 1/4 20/3 1/2 40 60 DETAILED SOLUTION PLEASE>>>
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Re: concentration of Husk? [#permalink]
13 Aug 2009, 05:10
tejal777 wrote: if 200 lb of a mixture contain 80% husk and 20%sand. Then how much husk needs to be extracted in order to have 75% concentration of Husk?
1/4
20/3
1/2
40
60
DETAILED SOLUTION PLEASE>>> We need to extract 40 lb of husk. 80 % of 200 = 160 20 % of 200 = 40 Now we have t remove husk to have 75 % of Husk ( key point sand is not supposed to be removed) Let X be the new quantity after removing husk So, 40 = 25 % of X X = 160 Husk will 120 lbs , hence we need to remove 1/4 of the total husk or 40 lbs
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Re: concentration of Husk? [#permalink]
13 Aug 2009, 05:53
nitishmahajan wrote: tejal777 wrote: if 200 lb of a mixture contain 80% husk and 20%sand. Then how much husk needs to be extracted in order to have 75% concentration of Husk?
1/4
20/3
1/2
40
60
DETAILED SOLUTION PLEASE>>> We need to extract 40 lb of husk. 80 % of 200 = 160 20 % of 200 = 40 Now we have t remove husk to have 75 % of Husk ( key point sand is not supposed to be removed) Let X be the new quantity after removing husk So, 40 = 25 % of X X = 160 Husk will 120 lbs , hence we need to remove 1/4 of the total husk or 40 lbs Yeah.....perfect solution!
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Re: concentration of Husk? [#permalink]
13 Aug 2009, 15:15
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Re: concentration of Husk? [#permalink]
26 Aug 2009, 03:00
1
This post received
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40 it is...
ONE MORE WAY TO APPROACH ..ALTHOUGH CONCEPT IS SIMILAR....
(160-x) = (200-x)75/100
Solve for x...
x = 40
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Re: concentration of Husk? [#permalink]
25 Jan 2010, 02:05
working with the answers helps to solve the question quickly
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Re: concentration of Husk? [#permalink]
26 Jan 2010, 22:50
Good question. When I first tried it, I kind of overlooked the word extracted. Once I realized the answer i came up with was not an answer choice...I decided to look at the question again. I need to figure out how to overcome this problem I sometimes have with reading a question and not catching certain key words.
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Re: concentration of Husk? [#permalink]
27 Jan 2010, 02:11
I would suggest a generalized strategy:
1) reading the question in parts and not the whole in one go.
2) writing down the necessary numbers to make it simpler (i.e. converting the English into Numbers).
In the above example, the key is to understand that 'sand' should not be extracted, which means that the 'sand' quantity in the mixture will not change. Only the proportion(% in terms of the mixture) will change.
First 'sand' was 20% of the mixture i.e. 40lbs. Later-on this 40lbs 'sand' becomes 25% of the mixture. Now the rest 75% of the mixture will be 120lbs of husk. Which is a reduction of 40lbs of 'husk', which is your answer.
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Re: concentration of Husk? [#permalink]
26 Jan 2011, 10:40
one more way to approach 200*80/100-x=75/100*(200-x) solve for x=40
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Re: concentration of Husk? [#permalink]
11 Mar 2011, 05:13
80% of 200 - X = 75% of (200-X)
Solve for X
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Re: concentration of Husk? [#permalink]
12 Mar 2011, 02:49
160-x/200-x = 3/4 So x = 40 and the answer is D.
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Re: concentration of Husk? [#permalink]
17 May 2011, 07:50
80% of 200 = 160 lb of husk ---> Total Husk in 200 lb. 75% concentration of total husk = 75% of 160 lb = (3/4) * 160 = 120 Total amount to be extracted = Initial Amount of Husk - 75% concentration of Husk Total amount to be extracted = 160 lb - 120 lb = 40 lb
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Re: concentration of Husk? [#permalink]
17 May 2011, 11:55
(160-x)/(200-x) = 75/100=3/4 x=640-600 = 40
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Re: concentration of Husk? [#permalink]
17 May 2011, 16:53
let x be the removed husk quantity
=> (80/100)(200)-x = (75/100)(200-x)
=> x = 40
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Re: concentration of Husk? [#permalink]
18 Dec 2011, 02:39
macsin wrote: I would suggest a generalized strategy:
1) reading the question in parts and not the whole in one go.
2) writing down the necessary numbers to make it simpler (i.e. converting the English into Numbers).
In the above example, the key is to understand that 'sand' should not be extracted, which means that the 'sand' quantity in the mixture will not change. Only the proportion(% in terms of the mixture) will change.
First 'sand' was 20% of the mixture i.e. 40lbs. Later-on this 40lbs 'sand' becomes 25% of the mixture. Now the rest 75% of the mixture will be 120lbs of husk. Which is a reduction of 40lbs of 'husk', which is your answer. Quality explanation!!!! Thanks! 
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Re: concentration of Husk?
[#permalink]
18 Dec 2011, 02:39
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