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If 200 lb of a mixture contain 80% husk and 20% sand. Then

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tejal777
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If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   Updated on: 02 Dec 2013, 08:21
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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?

A. 1/4
B. 20/3
C. 1/2
D. 40
E. 60
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Official Answer
D

Originally posted by tejal777 on 13 Aug 2009, 03:27.
Last edited by Bunuel on 02 Dec 2013, 08:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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PareshGmat
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   22 Apr 2014, 02:46
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Mixture has 160 lbs husk & 40 lbs sand

After removal of husk, its percentage should be 75%, it means

percentage of sand should be 25%

25% of What?? = 40

What comes out to be 160

200 - 160 = 40

40lbs husk should be removed to have its share 75%

Answer = D

Algebraic Method:

Husk............ Sand........... Total
160............... 40.................. 200

Lets say quantity x husk is removed

160-x ............. 40 .................. 200-x

Setting up the equation

\(160-x = \frac{75}{100} (200-x)\)
x = 40

Answer = D
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Re: concentration of Husk? [#permalink] New post   13 Aug 2009, 05:10
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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] New post   26 Aug 2009, 03:00
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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] New post   26 Jan 2010, 22:50
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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] New post   27 Jan 2010, 02:11
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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] New post   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] New post   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] New post   17 May 2011, 07:50
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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] New post   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] New post   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: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   18 Apr 2014, 21:25
Total=200lb

Husk=160
Other=40

Consider X is taken out.

(160-X) = .75 (200-X) {Since the new solution has 75% conc.}

X=40
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   28 May 2014, 19:13
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Easy approach:

From the question you know that
Husk = 160lbs (80% of 200)
Sand = 40lbs (20% of 200)

After removing some Husk the concentration will be 75%. We need to find how much Husk to remove

Remember that the quantity of sand doesn't change, we still have 40lbs of sand, but now they represent 25% of the new mix.
If 25% = 40lbs, 75% equals 120lbs (simple math)

So we need to remove: 160-120 lbs = 40!!
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   01 Jun 2014, 21:51
PareshGmat wrote:
Mixture has 160 lbs husk & 40 lbs sand

After removal of husk, its percentage should be 75%, it means

percentage of sand should be 25%

25% of What?? = 40

What comes out to be 160

200 - 160 = 40

40lbs husk should be removed to have its share 75%

Answer = D

Algebraic Method:

Husk............ Sand........... Total
160............... 40.................. 200

Lets say quantity x husk is removed

160-x ............. 40 .................. 200-x

Setting up the equation

\(160-x = \frac{75}{100} (200-x)\)
x = 40

Answer = D



One more method to setup the equation:

\(40 = \frac{25}{100} * (200-x)\)

160 = 200 - x

x = 40
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   03 Aug 2015, 03:40
(160-x/200-x)*100=75

1000=25x => x=40

OR

0.8*(200)-x=0.75*(200-x)

10=0.25x => x=40

D
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   25 Sep 2016, 09:28
200(80/100) - x = (200-x)(75/100)

160 - x = 150 - 3/4 (x)

10 = -3/4 (x) + 4/4

10 = 1/4 (x) --> x = 40
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   04 May 2019, 07:34
Hi

I want to know, why can’t we apply the C1V1=C2V2 formula here to solve this question ? Thank you in advance !


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?

A. 1/4
B. 20/3
C. 1/2
D. 40
E. 60


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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   10 Jan 2022, 14:10
isn't 1/4 of the husk the same as 40lbs of husk?

husk = 160 - > 3/4*160 = 120
120 husk + 40 sand = 75%/25% split?
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   11 Jan 2022, 05:34
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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?


Amount of Husk in 200lb = 80% of 200 = 160 lb

Let the amount of husk to be extracted be X lb and new concentration of Husk is 75%.

75/100 = (160-x)/(200-x)

3/4 = (160-x)/(200-x)

600 - 3x = 640 - 4x

x = 40

Option D is the answer.

Thanks,
Clifin
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink] New post   09 Mar 2024, 23:05
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Re: If 200 lb of a mixture contain 80% husk and 20% sand. Then [#permalink]
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