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Hi I have a doubt in this with my approach. Let us assume price per dozen of eggs is x, and he bought total y eggs. So he bought y/12 dozens of eggs. Now we can have 2 equations : y/12 * x = 12 and (y+2)/12 * (x-1) = 12 But this does not give y = 18 , could someone please help me ? Regards, Subhash
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subhashghosh wrote: Hi
I have a doubt in this with my approach.
Let us assume price per dozen of eggs is x, and he bought total y eggs. So he bought y/12 dozens of eggs.
Now we can have 2 equations :
y/12 * x = 12
and
(y+2)/12 * (x-1) = 12
But this does not give y = 18 , could someone please help me ?
Regards,
Subhash Your interpretation is correct. and y will not be equal to 18. It will be 16. y/12 * x = 12 x = 144/y (y+2)/12 * ((144/y)-1)=12 This gets transformed into same equation; y^2+2y-288=0Solve for roots; y=16 or y=-18 y can't be negative. y=16. y- number of eggs initially purchases = 16 After adding 2 eggs to this 16+2=18.
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Oops.. I just made an oversight error. Thanks for clarifying it. Regards, Subhash
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Re: A question from diagnostic test [#permalink]
 07 Feb 2011, 20:51
sara933 wrote: 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?
Answer is 18, but I did not get the explanation. Another approach: Let us say the price per dozen is p. Since he paid a total of $12, he must have bought 12/p dozens. Total number of eggs = (\frac{12}{p})*12(because a dozen has 12 eggs) But, he got 2 extra eggs so now total number of eggs = (\frac{12}{p})*12 + 2The price per dozen went down by $1 so new price per dozen = (p - 1) New total number of eggs = \frac{12}{(p-1)} * 12(\frac{12}{p})*12 + 2 = \frac{12}{(p-1)} * 12We see that we have numerator 144 and denominator p on one side and (p-1) on the other side. Since these are number of eggs, they must be integral values. 144 = 2^4*3^2We need two consecutive integers both of which divide 144. You can see that 8 and 9 both divide 144. Let's confirm: (\frac{12}{9})*12 + 2 = \frac{12}{8} * 12 = 18So the cook brought 18 eggs home.
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Re: GMAT Diagnostic Test Question 27 [#permalink]
09 Feb 2011, 13:11
wilbase wrote: 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. thats what I did as well. worked for me.
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Re: GMAT Diagnostic Test Question 27 [#permalink]
17 Apr 2011, 19:53
price per dozen difference is 1
144/x - 144(x+2) = 1
solving we get x = 16
=> x+2 = 18
Answer is E.
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Re: GMAT Diagnostic Test Question 27 [#permalink]
18 Apr 2011, 00:11
im having trouble understanding how multiplying \frac{1}{12} reduces the price per dozen by a dollar?
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I stopped at 16 without adding 2 more eggs. Now i know my mistake. Good Question
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n/12 = no of dozens thus 12/(n/12) - 12/[(n+2)/12] = 1 144/n - 144/n+2 = > n = 16 thus 18
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Re: GMAT Diagnostic Test Question 27 [#permalink]
04 Sep 2011, 06:28
As long as we understand that the cook gets the 2 additional eggs at the same price of 12, we can setup an equation
N- Number of eggs
P - Price per dozen
12 eggs cost $P
N eggs cost NP/12 and this = 12 --- 1st equation
Now
12 eggs cost $P-1
N+2 eggs cost (N+2)(P-1)/12 and this = 12 ---2nd equation. Here,we know that the cook got the 2 eggs for the same amount of $12
Combining both equations, we have
NP=(N+2)(P-1)
NP = 144
Finally, we get N^2+2N-288=0. Solving this equation, we get N=16. Therefore, the number the cook took back home is 18
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Re: GMAT Diagnostic Test Question 27 [#permalink]
25 Oct 2011, 16:54
urchin wrote: my approach.....
Let him buy m number of eggs....
so when the prize came down by 1$ 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. Hi , I do find your approach to fit more my way of thinking and approaching this problem but... and I apologize if this is very stupid, but I'm substituting 18 on the final formula, and can't get prove that it is correct.... is 12/18 equal to 12/20 + 1/12 ?? .. .. what am I doing wrong?
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Re: GMAT Diagnostic Test Question 27 [#permalink]
15 Nov 2011, 19:16
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 clear as crystal explanation :D thanksssss
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whiplash2411 wrote: 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. Easiest way to solve is algebra
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Re: GMAT Diagnostic Test Question 27 [#permalink]
05 Jan 2012, 06:19
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
Can anyone solve it???? I do not get N=4/3.
Thanks!
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Re: GMAT Diagnostic Test Question 27
[#permalink]
05 Jan 2012, 06:19
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