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# GMAT Diagnostic Test Question 26

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Founder
Joined: 04 Dec 2002
Posts: 15143
Location: United States (WA)
GMAT 1: 750 Q49 V42
GMAT Diagnostic Test Question 26 [#permalink]

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06 Jun 2009, 23:08
Expert's post
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GMAT Diagnostic Test Question 26
Field: word problems
Difficulty: 750

new price per dozen = ($12/[E+2])*12 now, the equation is; (old price per dozen) - (new price per dozen) = 1 i.e. {($12/E)*12} - {($12/[E+2])*12} = 1 solve for E, 144/E - 144/(E+2) = 1 144E + 288 - 144E = E(E+2) 288=E^2 +2E E^2 + 2E - 288 = 0 By factoring we get E^2 + 18E - 16E -288 = 0 E(E+18)-16(E+18)=0 (E+18)(E-16)=0 E= - 18, or 16 rejecting negative value we get E=16 (the original no of eggs purchased) no. of eggs brought home = E+2 or 16 + 2 = 18 Therefore, answer is E Actually, I used the same approach, but did not dare to do the factorization on the timed question, so I used the quadratics formula. E^2 + 2E - 288 = 0 E = {-2 +/- sqrt[4 - 4*(-288)]}/2 => {-2 +/- sqrt[1156]}/2 => {-2 +/- 34}/2 => E =16 => E+2 = 18 Of course, I wasted precious time finding the square root of 1156. I find both methods (quadratics vs factorization) equally cumbersome for this equation. _________________ Please kudos if my post helps. Manager Joined: 22 Jul 2009 Posts: 191 Re: GMAT Diagnostic Test Question 27 [#permalink] ### Show Tags 23 Aug 2009, 12:50 6 This post received KUDOS Flyingbunny's method is the fastest, as by working by dozens instead of units it results in smaller numbers and simpler calculations. Thumbs up for him. Allow me to recap the 3 methods herein presented using the same nomenclature. A-Target760's Uses equation: {($12/E)*12} - {($12/[E+2])*12} = 1 ...it results in E^2+2E-288=0 B-dzyubam's Uses equations: 1. EP=12 2. (E+2)(P-(1/12))=12 ...it results in E^2+2E-288=0 C-flyingbunny's Uses equations: 1. EP=12 2. (E+(1/6))(P-1)=12 .....it resutls in 6E^2+E-12=0. Easier to find the root than with other methods. _________________ Please kudos if my post helps. Manager Joined: 27 May 2009 Posts: 214 Re: GMAT Diagnostic Test Question 27 [#permalink] ### Show Tags 22 Oct 2009, 12:29 4 This post received KUDOS 1 This post was BOOKMARKED my approach..... Let him buy m number of eggs.... so when the prize came down by 1$ per 12 eggs that means per egg it came down by 1/12.
so equation becomes...
12/m = 12/(m+2) + 1/12

....... 12/m ---- original price per egg
12/(m+2) --- new price per egg
1/12 -- the amount by which the new price per egg came down.

U can now subsitute the values given in option and come at the ans.
_________________

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Manager
Joined: 19 Nov 2007
Posts: 222
Re: GMAT Diagnostic Test Question 27 [#permalink]

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09 Nov 2009, 23:47
3
KUDOS
N= Number of dozen;
X= price per dozen;

NX=12
(N+1/6)(X-1)=12

Solving for N we get N=4/3
The total number of eggs = 4/3*12+2=18
Manager
Joined: 22 Sep 2009
Posts: 215
Location: Tokyo, Japan
Re: GMAT Diagnostic Test Question 27 [#permalink]

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21 Nov 2009, 05:13
I solved it all the way till X^2 + 2X +288 =0; then gave up and guess
Intern
Joined: 06 Nov 2009
Posts: 4
Re: GMAT Diagnostic Test Question 27 [#permalink]

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26 Nov 2009, 21:26
Quote:
suppose before added two more eggs, the price per dozen is X and there are Y dozen
so we have XY=12
and (X-1)(Y+1/6)=12

from above we can get Y=16/12, there are 16 eggs.
in the end, the cook brings 16+2=18 eggs home.

Hello,
I feel that flyingbunny's way is the simplest but there's one part I don't get it.

Quote:
we have XY=12

OK

Quote:
(X-1)

means in regular english "the price per dozen was reduced by 1$" Quote: (X-1)(Y+1/6)=12 Where does this 1/6 comes from ? Thanks ! Manager Joined: 18 Nov 2009 Posts: 57 Re: GMAT Diagnostic Test Question 27 [#permalink] ### Show Tags 30 Nov 2009, 22:44 suppose before added two more eggs, the price per dozen is X and there are Y dozen so we have XY=12 and (X-1)(Y+1/6)=12 hi flying bunny, how did you get XY=12? i get X=price/dozen, and there are Y dozen, but how does that equate to 12? thanks! Manager Joined: 08 Jul 2009 Posts: 170 Re: GMAT Diagnostic Test Question 27 [#permalink] ### Show Tags 04 Jan 2010, 21:57 djxilo wrote: my approach: Worked backwards plugging in answer choices. We basically need to find the answer choice that gets us a saving of$1.00/12 eggs (~8 cents savings per egg) If we plug-in answer choice E, we get a total price paid per egg of $0.67. Subtracting two eggs for the same purchase price of$12 implies an original price of $0.75 per egg.$0.75 - $0.66 = ~8 cents in savings. This is a good approach, solving it the other direction. Intern Joined: 09 Jan 2010 Posts: 1 Re: GMAT Diagnostic Test Question 27 [#permalink] ### Show Tags 11 Jan 2010, 00:21 marcos4 wrote: Quote: suppose before added two more eggs, the price per dozen is X and there are Y dozen so we have XY=12 and (X-1)(Y+1/6)=12 from above we can get Y=16/12, there are 16 eggs. in the end, the cook brings 16+2=18 eggs home. Hello, I feel that flyingbunny's way is the simplest but there's one part I don't get it. Quote: we have XY=12 OK Quote: (X-1) means in regular english "the price per dozen was reduced by 1$"

Quote:
(X-1)(Y+1/6)=12

Where does this 1/6 comes from ?
Thanks !

1/6 came from the 2 free eggs. 12 * 1/6 = 2.
_________________

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Albert Pike

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Intern
Joined: 13 May 2010
Posts: 5

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21 Jul 2010, 10:57
1
KUDOS
A cook went to a market to buy some eggs and paid $12. But since the eggs were quite small, he talked the seller into adding two more eggs, free of charge. As the two eggs were added, the price per dozen went down by a dollar. How many eggs did the cook bring home from the market? A.8 B.12 C.15 D.16 E.18 Ms. Big Fat Panda Status: Three Down. Joined: 09 Jun 2010 Posts: 1918 Concentration: General Management, Nonprofit Re: A cook [#permalink] ### Show Tags 21 Jul 2010, 11:02 5 This post received KUDOS Let us assume he buys n eggs and it costs him$12. So the cost per egg = $$\frac{12}{n}$$.

Cost per dozen eggs at the original price = $$\frac{12}{n}*12$$ = $$\frac{144}{n}$$. (Since there are 12 eggs in a dozen and we are multiplying the cost per egg calculated above with 12)

Original Cost per dozen = $$\frac{144}{n}$$

Now, the number of eggs he buys becomes n+2, and the cost remains the same. So the cost per egg is now: $$\frac{12}{n+2}$$.

Using the same logic, cost per dozen = Cost per egg * 12 = $$\frac{12}{n+2}*12 = \frac{144}{n+2}$$.

New Cost per dozen = $$\frac{144}{n+2}$$

It's given that the new cost per dozen = original cost per dozen - 1

$$\frac{144}{n+2}=\frac{144}{n} - 1$$

So you get: $$\frac{144}{n} - \frac{144}{n+2} = 1$$

Cross multiplying: $$144( \frac{1}{n} - \frac{1}{n+2}) = 1$$

$$( \frac{1}{n} - \frac{1}{n+2}) = \frac{1}{144}$$

$$\frac{n+2 - n}{(n)(n+2)} = \frac{1}{144}$$

$$288 = n(n+2)$$

Solving this you get 18.

Alternatively plugging numbers will work faster. Just find out cost per dozen for each of the numbers and compare them to get answer.
Re: A cook   [#permalink] 21 Jul 2010, 11:02

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# GMAT Diagnostic Test Question 26

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