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# GMAT Diagnostic Test Question 33

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Re: GMAT Diagnostic Test Question 33 [#permalink]

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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. Intern Joined: 02 Apr 2013 Posts: 4 Location: Russian Federation Followers: 0 Kudos [?]: 5 [0], given: 0 Re: GMAT Diagnostic Test Question 33 [#permalink] ### Show Tags 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! Intern Joined: 13 Jul 2012 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: GMAT Diagnostic Test Question 33 [#permalink] ### Show Tags 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. Manager Status: Looking to improve Joined: 15 Jan 2013 Posts: 177 GMAT 1: 530 Q43 V20 GMAT 2: 560 Q42 V25 GMAT 3: 650 Q48 V31 Followers: 1 Kudos [?]: 62 [0], given: 65 Re: GMAT Diagnostic Test Question 33 [#permalink] ### Show Tags 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.

Any comments on fallacy of this approach...
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Re: GMAT Diagnostic Test Question 33 [#permalink]

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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]

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17 Sep 2013, 00:51
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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# GMAT Diagnostic Test Question 33

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