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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

You can alway solve through discriminant, but it takes you to 4 + 288*4 = 289*4 =1156 - which makes a bit problem if you dont know the square root of 1156.

My approach:

- as we have -288 one root of the equation must be negative and one positive - as we have 2E, the yE and xE must sum in 2E ( y+x = 2 - the positive must be greater by 2 in absolute value)

now: 288 is pretty close to 289 which is 17^2. Try 17+1 and 17-1 which makes 16*18 -> now 16 must be negative. then: (x-16)*(x+18) = 0 _________________

I used the same approach..and got my answer.This is a little tricky and took me just about 2 mins..the point is you need to be a little quick on quadratic equations in the calculation part.Nevertheless a good one.

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)

The way I did it: Let the total number of eggs bought originally be x,

setting up the eqn 12*(12/x - 12/(x+2))=1 ............... I

Now during the test I used the brute force method where I solved the eqn and then tried to use the quadratic formula to calculate roots of the eqn given above which solves down to i.e x^2+2x-288=0 where I got stuck trying to find the square root of 1156 (its 34 btw). In retrospect I would set up the eqn and start plugging in values to solve it.

when you plug D it satisfies the eqn which boils down to 12*(3/4-2/3) = 1 = RHS of I

The only trick in this is that it gives the value of x which is the total number of eggs before the cook bargained his way to getting more. The answer that the question is looking for is really x+2=18 since it says how many did the cook go home with? (hands up if you missed that part!) hence E _________________

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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?

Yeah, this one is a bit tricky. Our equation we want will look like

Price per dozen before - Price per dozen after (we got the two eggs) = 1 (the difference was 1$)

How do we find the price per dozen?

One way is taking the unit price and multiply by 12

(total price/no of eggs) * 12 = price per dozen

let x denote no of eggs and total price was 12$

(12/x)*12 = dozen price before we got the two eggs for free (12/x+2)*12 = dozen price after we got the two eggs for free, and the difference should be 1 so.

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 = \((\frac{12}{p})*12 + 2\)

The price per dozen went down by $1 so new price per dozen = (p - 1) New total number of eggs = \(\frac{12}{(p-1)} * 12\) \((\frac{12}{p})*12 + 2 = \frac{12}{(p-1)} * 12\)

We 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. \(144 = 2^4*3^2\) We need two consecutive integers both of which divide 144. You can see that 8 and 9 both divide 144. Let's confirm: \((\frac{12}{9})*12 + 2 = \frac{12}{8} * 12 = 18\) So the cook brought 18 eggs home. _________________

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....

Of course, I wasted precious time finding the square root of 1156. I find both methods (quadratics vs factorization) equally cumbersome for this equation.

You spent some extra time getting to 1156 and then sqrting it. Would have been more efficient to do 4+4(288) in the radical, which leads to 4(1+288) = (2^2)(17^2). Much quicker.

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

This is a hard equation to come up with during the test. The price of eggs was $12. Let \(E\) denote the number of eggs and \(P\) denote the price.

\((E+2)\times(P-\frac{1}{12})=12\)

From this equation we get that \(E = 16\).

Answer: \(E+2=16+2=18\).

This equation is much more easier than the rest. But definately a 750 level ques.

Honestly, I do not understand \((P-\frac{1}{12})\)...

Secondly, how do you solve from \((E+2)\times(P-\frac{1}{12})=12\) -->\(E = 16\) - with this "shortcut" it is pretty much useless. Please explain to me how it could be solved, with two variables...Thanks. _________________

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