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only cherry pies and apple pies

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by sanju09 » Tue Jul 23, 2013 10:26 pm
A certain bakery sells only cherry pies and apple pies, which cost 20% more than cherry pies. If the bakery's pie sales on Wednesday totaled $111, how many cherry pies were sold?

(1) Cherry pies cost $15 each.

(2) The bakery sold a total of 7 pies on Wednesday.


Source: The Princeton Review
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Source: — Data Sufficiency |
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by GMATGuruNY » Tue Jul 23, 2013 11:10 pm
sanju09 wrote:A certain bakery sells only cherry pies and apple pies, which cost 20% more than cherry pies. If the bakery's pie sales on Wednesday totaled $111, how many cherry pies were sold?

(1) Cherry pies cost $15 each.

(2) The bakery sold a total of 7 pies on Wednesday.
Let c = the number of cherry pies sold and a = the number of apple pies sold.

Statement 1: Cherry pies cost $15 each.
Since apple pies cost 20% more, the cost of each apple pie = 15 + .2(15) = 18.

Since the total cost of all of the pies is $111, we get:
15c + 18a = 111
5c + 6a = 37.

Since c and a must be INTEGER VALUES, the resulting equation above might have ONLY ONE SOLUTION.
5c represents a MULTIPLE OF 5 less than 37:
5, 10, 15, 20, 25, 30, 35.
6a represents a MULTIPLE OF 6 less than 37:
6, 12, 18, 24, 30, 36.
Only the two values in red will yield a sum of 37.
Thus, 5c = 25 and 6a = 12, implying that c=5 and a=2.
SUFFICIENT.

Statement 2: The bakery sold a total of 7 pies on Wednesday.
c + a = 7.
It's possible that c=1 and a=6 or that c=2 and a=5.
Since c can be different values, INSUFFICIENT.

The correct answer is A.
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by Matt@VeritasPrep » Thu Jul 25, 2013 9:38 am
Interesting note about this problem: if you assume that the price of a cherry pie is an integer (which you obviously can't), Statement 2 is sufficient.

c = # of cherry pies
(7 - c) = # of apple pies
x = cost of a cherry pie
1.2x = cost of an apple pie

cx + (7-c)(1.2x) = 111 over the natural numbers gives c = 5, x = 15

What's even more interesting, though, is that statement 2 SHOULD be sufficient, because any value of c from 0 to 7 (other than 5) gives an impossible price!

For example, if c = 1, x = $555/41, a non-terminating decimal. How would you possibly give change for a pie like that ... I suppose you'd have to sell it in batches of 41. :D
Set prices really need to be decimals that terminate before the thousandths place.

Anyway, this seems like a sloppy one, I'd toss it out.
Last edited by Matt@VeritasPrep on Sat Jul 27, 2013 8:23 am, edited 1 time in total.
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by sanju09 » Fri Jul 26, 2013 10:14 pm
$555/41 is not irrational, and price could be any real number, and definitely any rational number.
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by Matt@VeritasPrep » Sat Jul 27, 2013 8:20 am
sanju09 wrote:$555/41 is not irrational, and price could be any real number, and definitely any rational number.
Whoops, you're right - I meant to say "non-terminating decimal".

But as 555/41 is a non-terminating decimal, you absolutely cannot charge $555/41 for a single pie using any accepted modern currency unless you (1) do not accept cash or (2) only sell the pies in batches of 41. Otherwise you'll be forced to alter the price when you give change: if I pay $1 for a pie that costs $(2/3), for instance, the change will be either $.33 (in which case the price was $.67) or $.34 (in which case the price was $.66). In neither case is the price $(2/3).

You could argue that a currency might have (a) coin(s) worth $(1/3), but that currency is not dollars, the currency stated in the problem.

One last note: if you don't accept the assumption that a price in dollars must terminate before the thousandths place, why would you accept the assumption that pies must be sold whole (that the # of each pie sold must be an integer)? Both are equally reasonable, yet neither is explicitly stipulated.
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by sanju09 » Sat Jul 27, 2013 11:46 pm
Matt@VeritasPrep wrote:
sanju09 wrote:$555/41 is not irrational, and price could be any real number, and definitely any rational number.
Whoops, you're right - I meant to say "non-terminating decimal".

But as 555/41 is a non-terminating decimal, you absolutely cannot charge $555/41 for a single pie using any accepted modern currency unless you (1) do not accept cash or (2) only sell the pies in batches of 41. Otherwise you'll be forced to alter the price when you give change: if I pay $1 for a pie that costs $(2/3), for instance, the change will be either $.33 (in which case the price was $.67) or $.34 (in which case the price was $.66). In neither case is the price $(2/3).

You could argue that a currency might have (a) coin(s) worth $(1/3), but that currency is not dollars, the currency stated in the problem.

One last note: if you don't accept the assumption that a price in dollars must terminate before the thousandths place, why would you accept the assumption that pies must be sold whole (that the # of each pie sold must be an integer)? Both are equally reasonable, yet neither is explicitly stipulated.
Dear Matt, I never wanted you to edit over and again, but, let's say, if a candy store sells 3 candies for $2, are we going to throw them out of the real world, simply because if we buy two candies, the change currency doesn't exist?

Even in this problem, as per the correct calculation you shown here, if 1 cherry pie costs $555/41 (a non-terminating, or recurring decimal), then at the same time 6 apple pies also cost $3996/41 (yet another decimal of the same kind), making the whole bill equal to $111, the currency you are most happy to accept.

That is said, that kind of discussion would take us far away from the point of the question with no take away in the end of the day. We at The Princeton Review take full care of all those little and big things before making a question available for the test takers at large.

Have a nice time.
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by snigdha1605 » Sun Jul 28, 2013 3:38 am
Hi,

Can anyone tell me the Problems #s in OG which are similar to this question?

Regards,
Snigdha
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by sanju09 » Sun Jul 28, 2013 4:36 am
snigdha1605 wrote:Hi,

Can anyone tell me the Problems #s in OG which are similar to this question?

Regards,
Snigdha
See # 17, 46, 67 of DS in OG, hope that works.
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by Matt@VeritasPrep » Sun Jul 28, 2013 11:42 am
but, let's say, if a candy store sells 3 candies for $2, are we going to throw them out of the real world, simply because if we buy two candies, the change currency doesn't exist?
Of course you can sell things in batches (3 for $2) -- in fact, I mentioned this already in an earlier post -- but that isn't the same thing as selling them for $(2/3) each: you're assuming that I buy the candies in multiples of 3.

What you've given is a non-response: if the store chooses to sell me two candies at such a price, US currency ($) will force that price to be rounded. I cannot be charged $(4/3) in hard currency.
Even in this problem, as per the correct calculation you shown here, if 1 cherry pie costs $555/41 (a non-terminating, or recurring decimal), then at the same time 6 apple pies also cost $3996/41 (yet another decimal of the same kind), making the whole bill equal to $111, the currency you are most happy to accept.
That's because the prices depend on each other in an equation that's equal to $111: cx + (7-c)(1.2x) = $111. Of course it will cost $111 for all seven: the cost equation presupposes that! The point is that you ought to then discard solutions that give "unchargeable" prices, which I'd say any natural value of c other than 5 will do.

I'm also still waiting to hear your justification for why I have to buy a whole pie, given the text of this question - pies are easily (and frequently) sold in halves, eighths, and quarters.

But I'm happy to be done with this: I've made my case.
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by Matt@VeritasPrep » Sun Jul 28, 2013 11:52 am
snigdha1605 wrote:Hi,

Can anyone tell me the Problems #s in OG which are similar to this question?

Regards,
Snigdha
Algebraic equations for which only integer solutions are accepted are called Diophantine equations, and mathematicians have been puzzling over them for thousands of years: there are all sorts of excellent practice questions out there (beyond the ones in the OG). The GMAT seems fond of these questions, as they come up fairly often in the official guides and CAT's.

Here's a great introduction to GMAT-level Diophantine equations: well worth reading!
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by maneshbabutm » Mon Jul 03, 2017 5:05 am
Matt@VeritasPrep wrote:Interesting note about this problem: if you assume that the price of a cherry pie is an integer (which you obviously can't), Statement 2 is sufficient.

c = # of cherry pies
(7 - c) = # of apple pies
x = cost of a cherry pie
1.2x = cost of an apple pie

cx + (7-c)(1.2x) = 111 over the natural numbers gives c = 5, x = 15

What's even more interesting, though, is that statement 2 SHOULD be sufficient, because any value of c from 0 to 7 (other than 5) gives an impossible price!
It's possible to solve this.

Your equation is cx + (7-c)(1.2x) = 111

=> cx + 8.4x - 1.2cx = 111

=> 84x - 2cx = 1110
=> 42x - cx = 555
=> x(42-c) = 555

the factors of 555 are 3,5 and 37

Removing roots that make 42-c negative (since x can't be negative), the possible values for x and 42-c are (15, 37, 111, 185) and (37, 15, 5, 3) respectively.

Therefore, the values for c are 5, 27, 37 and 39. But the total number of pies sold is 7. So the only possible value of c is 5.

So the answer is D.

I don't see any logical issues with this approach. Please feel free to comment.
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by Jay@ManhattanReview » Mon Jul 03, 2017 9:21 pm
sanju09 wrote:A certain bakery sells only cherry pies and apple pies, which cost 20% more than cherry pies. If the bakery's pie sales on Wednesday totaled $111, how many cherry pies were sold?

(1) Cherry pies cost $15 each.

(2) The bakery sold a total of 7 pies on Wednesday.

Source: The Princeton Review
Say x = the number of cherry pies sold and y = the number of apple pies sold.

Statement 1: Cherry pies cost $15 each.

Since an apple pie cost 20% more, the cost of an apple pie = 1.2*15 = 18.

Thus,

15x + 18y = 111
=> 5x + 6y = 37

=> x = (37-6y)/5 = 7 + (2-6y)/5;

Since x and y both are positive integers, (2-6y) must be a multiple of "-5."

Say 2-6y = -5 => y = a fraction, not a possible value;
Again, say 2-6y = -10 => y = 2, a possible value; this gives x = 7 + (-10/5) = 7-2 = 5, a possible value

So one of the possible values of x = 5.

Let's try if x can take other values too.

Since x = 7 + (2-6y)/5, we must ensure that -6 ≤ (2-6y)/5 ≤ 0 or -30 ≤ 2-6y ≤ 0.

=> -32 ≤ -6y ≤ -2

=> 5.33 ≥ y ≥ 0.33; note that the sign of inequality is reversed!

=> y may be 5, 4, 3, 2, or 1.

However, @ y = 5, 4, 3, and 1, x is a fraction (not a positive integer), thus, we have x = 2 and y = 5. Sufficient.

Statement 2: The bakery sold a total of 7 pies on Wednesday.

=> x + y = 7

There can be may possible values of x; for example, x = 1, y = 6; x = 2, y = 5; x = 3, y = 4, etc.
No unique value of x. Insufficient.

The correct answer: A

Hope this helps!

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