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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?

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.

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.

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

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 \(\frac{144}{16}-\frac{144}{18}=9-8=1\).

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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I do not suffer from insanity. I enjoy every minute of it.

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.

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 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

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 \(\frac{144}{16}-\frac{144}{18}=9-8=1\).

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

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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