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GMAT Diagnostic Test Question 33

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Re: GMAT Diagnostic Test Question 33 [#permalink] New post 03 Oct 2012, 02:08
bb wrote:
GMAT Diagnostic Test Question 33
Field: word problems (mixture)
Difficulty: 650
Rating:


A kilogram of nut mixture contains X% chestnuts and Y% walnuts and sells for $7.00/kg. If the ratio of chestnuts is increased by 50% so that the new mixture is sold for $8.00/kg, what is the price of a kg of walnuts?

A. $1.00
B. $2.50
C. $5.00
D. $7.50
E. $10.00


I hate to bump an old thread but isn't there an issue re. what it means to "increase a ratio by 50%"? "Increasing a ratio" by 50% would e.g. involve increasing a ratio of 2:3 to 1:1 (e.g. we go from 0.66*1.5 = 1). THAT is a 50% increase in a ratio. Using a plugging in numbers approach, a movement from a 40% / 60% mixture ratio to a 50% / 50% ratio would represent an increase in the ratio by 50%. Yet this cannot satisfy any of the answer choices. The (1.5X) to (1-1.5X) answer used by gmat tiger is potentially troublesome as it can lead to negative quantities for the Walnut side.
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Re: GMAT Diagnostic Test Question 33 [#permalink] New post 06 Apr 2013, 13:43
(a1) X/100=c/(w+c)
(a2) Y/100=w/(w+c)=(100-X)/100

(b1) A price of c
(b2) B price of w

(c1) Ac+Bw=7(c+w)
(c2) Ac/(c+w)+Bw/(c+w)=7

(d1) AX/100+B(100-X)/100=7
(d2) A1.5X/100+B(100-1.5X)/100=8
*substitute c/(c+w) with X/100, and w/(w+c) with (100-X)/100

(e1) A0.5X/100-B0.5/100=1
*subtract d1 from d2
(e2) 200=AX-BX
(e3) 200+BX=AX

(f1) (200+BX)/100+B(100-X)/100=7
*substitute AX in d1 with e3
(f2) 700=200+BX+100B-BX
(f3) 500=100B

(g1) B=5
*the price of a kg of w

To avoid ambiguity stem should have STATED that the "proportion" of c in the mixture OR that the ratio of c to the whole mixture IS increased by 50%.

Hope this helps!
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Re: GMAT Diagnostic Test Question 33 [#permalink] New post 06 Apr 2013, 20:27
Survival wrote:
(a1) X/100=c/(w+c)
(a2) Y/100=w/(w+c)=(100-X)/100

(b1) A price of c
(b2) B price of w

(c1) Ac+Bw=7(c+w)
(c2) Ac/(c+w)+Bw/(c+w)=7

(d1) AX/100+B(100-X)/100=7
(d2) A1.5X/100+B(100-1.5X)/100=8
*substitute c/(c+w) with X/100, and w/(w+c) with (100-X)/100

(e1) A0.5X/100-B0.5/100=1
*subtract d1 from d2
(e2) 200=AX-BX
(e3) 200+BX=AX

(f1) (200+BX)/100+B(100-X)/100=7
*substitute AX in d1 with e3
(f2) 700=200+BX+100B-BX
(f3) 500=100B

(g1) B=5
*the price of a kg of w

To avoid ambiguity stem should have STATED that the "proportion" of c in the mixture OR that the ratio of c to the whole mixture IS increased by 50%.

Hope this helps!


Gotcha. Although it's worth pointing out that the text in the question isn't ambiguous - it's just plain wrong. Your amendments would correct it, however.
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Re: GMAT Diagnostic Test Question 33 [#permalink] New post 07 Apr 2013, 11:33
How about using some logical reasoning rather than pure algebra.

E.g. X% and Y% mixture costs $7 and X% is increased by 50% => 1.5X% in the mixture increases the price by 1$.

A 50% increase in X results in a $1 increase in price. This implies the cost X% is $2 since 50% of $2 is $1. Hence the cost for Y% is $7 - $2 = $5.

Answer is C.

Any comments on fallacy of this approach...
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Re: GMAT Diagnostic Test Question 33 [#permalink] New post 17 Sep 2013, 00:46
I wonder if Bunuel or Karishma can provide their ways to this problem...:) Admittedly, GMAT TIGER's way is fairly clear...just wonder if there is a more concise way. Thanks.
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Re: GMAT Diagnostic Test Question 33 [#permalink] New post 17 Sep 2013, 00:51
Expert's post
obs23 wrote:
I wonder if Bunuel or Karishma can provide their ways to this problem...:) Admittedly, GMAT TIGER's way is fairly clear...just wonder if there is a more concise way. Thanks.


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Re: GMAT Diagnostic Test Question 33   [#permalink] 17 Sep 2013, 00:51
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