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GMAT Diagnostic Test Question 26

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GMAT Diagnostic Test Question 26 [#permalink] New post 06 Jun 2009, 22:08
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A
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Question Stats:

28% (03:58) correct 72% (02:51) wrong based on 141 sessions
GMAT Diagnostic Test Question 26
Field: word problems
Difficulty: 750


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
[Reveal] Spoiler: OA

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Last edited by bb on 29 Sep 2013, 14:27, edited 2 times in total.
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 06 Jul 2009, 05:48
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Explanation:
Official Answer: E

Say the # of eggs the cook originally got was x;
The price per egg then would be \frac{12}{x} and the price per dozen would be 12*\frac{12}{x}.

Now, since the cook talked the seller into adding two more eggs then he finally got x+2 eggs (notice that x+2 is exactly what we should find);
So, the price per egg became \frac{12}{x+2} and the price per dozen became 12*\frac{12}{x+2}.

As after this the price per dozen went down by a dollar then 12*\frac{12}{x}-12*\frac{12}{x+2}=1 --> \frac{144}{x}-\frac{144}{x+2}=1. At this point it's better to substitute the values from answer choices rather than to solve for x. Answer choices E fits: if x+2=18 then \frac{144}{16}-\frac{144}{18}=9-8=1.

Answer: E.

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Last edited by bb on 29 Sep 2013, 14:27, edited 2 times in total.
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 24 Jul 2009, 04:37
P indicates the price per dozen?
In that case shouldnt you take (P-1) to multiply with (e+2)?

How did you come up with (P- 1/12)?
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 25 Jul 2009, 09:48
You could plug in numbers......just start with E or D first....
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 26 Jul 2009, 09:36
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here is my solution:

let's say no of eggs purchased = E

price per dozen of eggs in first case = ($12/E)*12 i.e. price per egg multiply by 12, to get price per dozen
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

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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 31 Jul 2009, 15:12
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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.
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 20 Aug 2009, 07:10
Target760 wrote:
here is my solution:

let's say no of eggs purchased = E

price per dozen of eggs in first case = ($12/E)*12 i.e. price per egg multiply by 12, to get price per dozen
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


Hey man, could you pls explain how did you come to your factorization. I do not get the way you're going from :
E^2 + 2E - 288 = 0
to
E(E+18)-16(E+18)=0

Thx very much
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 21 Aug 2009, 22:17
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my way:

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.

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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 22 Aug 2009, 14:44
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defoue wrote:

Hey man, could you pls explain how did you come to your factorization. I do not get the way you're going from :
E^2 + 2E - 288 = 0
to
E(E+18)-16(E+18)=0

Thx very much


Need to factors of 288 that when subtracted render 2.
Prime factorization of 288 gives: 2x2x2x2x2x3x3
Try combining primes to get two numbers whose difference is 2.
There are 7 primes, so one number would have 3 primes and the other 4 primes.
First combinations to try would be 3x3x2 and 2x2x2x2. That's 18 and 16.
18-16=2. Bingo, we got the factors we were looking for.
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 22 Aug 2009, 14:51
Target760 wrote:
here is my solution:

let's say no of eggs purchased = E

price per dozen of eggs in first case = ($12/E)*12 i.e. price per egg multiply by 12, to get price per dozen
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.

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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 23 Aug 2009, 11:50
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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.

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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 22 Oct 2009, 11:29
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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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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 09 Nov 2009, 22:47
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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
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 21 Nov 2009, 04:13
I solved it all the way till X^2 + 2X +288 =0; then gave up and guess
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 26 Nov 2009, 20:26
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 !
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 30 Nov 2009, 21:44
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!
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 04 Jan 2010, 20:57
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.
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Re: GMAT Diagnostic Test Question 27 [#permalink] New post 10 Jan 2010, 23:21
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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A cook [#permalink] New post 21 Jul 2010, 09:57
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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
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Re: A cook [#permalink] New post 21 Jul 2010, 10:02
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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, 10:02
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