Marta bought several pencils.

[Guest]

Source : GMATprep 2. DS Question

Marta bought several pencils. If each pencil was either 23 cents or 21 cent, how many 23 cents pencil did Marta buy?
1) Marta bought a total of 6 pencils
2) The total value of pencil Marta bought was 130 cents.

A
B
C
D
E

The OA is B. Can someone explain this? I got C, as we need both the information to arrive to solution

[Rustic Myth.]

There is no logical answer except trial n error method. Cause logic says two variable two equations, but gmat also wants us to hit and try.

there are 2= 23c and 4= 21c. pencil.

[Guest]

There is no logical answer except trial n error method. Cause logic says two variable two equations, but gmat also wants us to hit and try.

there are 2= 23c and 4= 21c. pencil.

Thanks Rustic.

But on the first hand, i figured that you have 2 equation and 2 unknown so both information are needed. How will i come to know that i have to give hit and trial priority rather than formulating equation and try to solve it?

[RonPurewal]

But on the first hand, i figured that you have 2 equation and 2 unknown so both information are needed. How will i come to know that i have to give hit and trial priority rather than formulating equation and try to solve it?

that's EXACTLY the trap that the test writers are trying to get you to fall into on this one. you took the bait.

here are a couple of things to know:
(1) one of the prime directives of standardized-test authors is to create a test on which test takers will not be able to succeed by the blind application of rules-of-thumb learned in school.
this is actually a really important dictum for test writers, because the idea is that standardized tests are supposed to measure some sort of reasoning ability that's independent of formal schooling.
therefore, there will be lots of problems that are written on purpose in a way such that conventional rules-of-thumb will lead to WRONG answers.
the "2 equation 2 variable rule" is violated so often in this regard that we actually talk about ditching it in our class slides!

(2) this problem is different because the solutions have to be WHOLE NUMBERS.
yes, that's right: you can't buy, say, 3.8 pencils.
if you have a linear equation whose solutions MUST BE WHOLE NUMBERS (0, 1, 2, ...), then you MUST TEST CASES in some sort of organized fashion.
unfortunately, there's really no better way to do problems like this.
these equations - on which the solutions are required to be whole numbers - are called "diophantine equations". entire ph.d. mathematics theses have been crafted on the topic of solving diophantine equations, but no one has really come up with simple, reliable ways to solve them at all - much less in the space of two minutes and with the relatively modest mathematical background required for the gmat.
there's just nothing you can do, besides testing cases.

one more thing:
if a reasonably difficult algebra problem requires whole-number solutions, you can be darned sure that the answer will turn out differently than it would if the solutions didn't have to be whole numbers.
this is actually the whole point of the problem: they want you to pick up on the ways in which the problem changes BECAUSE the solutions are required to be whole numbers.
therefore, ironically, you should trust that the solution to this problem won't be (c), because that's the solution you would get if this were a pure algebra problem (wihout the whole-number requirement).

[Guest]

In order to help you set up test cases, you can look at (2) and assume that the person only bought 23 cent pencils. 130/23 = 5.xxx So therefore you just have to test 0<x<6, where x is an integer. x<0 because the prompt tells you the person bought pencils at both prices.

[djjamal]

In order to help you set up test cases, you can look at (2) and assume that the person only bought 23 cent pencils. 130/23 = 5.xxx So therefore you just have to test 0<x<6, where x is an integer. x<0 because the prompt tells you the person bought pencils at both prices.

typo in last line, i meant x>0.

[Joey Z.]

1) Marta bought a total of 6 pencils -- obviously INSUFFICIENT

2) The total value of pencil Marta bought was 130 cents, which means 21m+23n = 130, where m and n are integers. Let's try solve this equation.

130 is not divisible by either 21 or 23, so m >= 1, n >= 1, and we have:
21(m-1) + 23(n-1) = 130-21-23 = 86, which again is not divisible by either 21 or 23, so m >= 2, n >= 2, and we have:
21(m-2) + 23(n-2) = 86-21-23 = 42, which is divisible by 21 but not 23. Hence, we have m = 4, n = 2. -- SUFFICIENT

[RonPurewal]

1) Marta bought a total of 6 pencils -- obviously INSUFFICIENT

2) The total value of pencil Marta bought was 130 cents, which means 21m+23n = 130, where m and n are integers. Let's try solve this equation.

130 is not divisible by either 21 or 23, so m >= 1, n >= 1, and we have:
21(m-1) + 23(n-1) = 130-21-23 = 86, which again is not divisible by either 21 or 23, so m >= 2, n >= 2, and we have:
21(m-2) + 23(n-2) = 86-21-23 = 42, which is divisible by 21 but not 23. Hence, we have m = 4, n = 2. -- SUFFICIENT

interesting.

i think i may have an idea of the general principle behind this approach, but would you mind elaborating a bit for the benefit of other forum readers?

[mohitkant]

Hi,

I have a doubt on a similar GMATprep Question, i'd start a another thread but it would defeat the purpose. So here it goes.

At the bakery, Lew spent a total of $6.00 for one kind of cupcake ad one kind of doughnut. How many doughnuts did he buy?
(1) The price of 2 donuts was $0.10 leass than the price of 3 cupcakes
(2) The average price of 1 doughnut and 1 cupcake was $0.35

Here's my approach.

Let Price of Doughnut be D , number of Doughnuts - n1 , Price of Cupcake be C, number of Cupcakes - n2

So.. We have n1D + n2C = 6 (the total amount spent on Doughnuts and Cupcakes)

1. 2D =3C - 0.10 - Not Sufficient.

2. D+C/2 = 0.35 - Not Sufficient.

Taking both the equations together, we can find the value of D and C, but we don't have n1 and n2.

The answer should be E, however, i chose C, because..n1 and n2 can only be positive integers and there could only be one possible combination of n1 and n2 that would result in a total of 6 dollars. Hence C should have been the answer.

So, is the criterion to decide between C and E on such questions is to check where you are on the Test, i.e. if it is in top 10, the answer is more likely to be C, if the question is 15+ then it is E.

[mschwrtz]

That's a very interesting problem mohitkant. I'm going to skip to the point where you have solved for the prices, since your question really starts there.

A doughnut cost 40 cents and a cupcake costs 30 cents. What number of doughnuts (and cupcakes) might Marta have purchased? She might have purchased 3 doughnuts and 16 cupcakes, or 6 doughnuts and 12 cupcakes, or 9 doughnuts and 8 cupcakes, or 12 doughnuts and 4 cupcakes. So the answer is E.

There's quite a lot going on here that's not revealed by trying values, but if you encounter a question like this in practice or on the actual test, get the answer first, then muse on the other stuff later.

Here's some of that other stuff: Notice that the cost of 3 doughnuts is the same as the cost of 4 cupcakes. This suggests that if we come up with one answer we'll be able to swap out 3 doughnuts for 4 cupcakes to produce another answer.

By the way, this problem might seem to violate Ron's rule of thumb above, because the requirement that the numbers be integers doesn't change the answer; it's still C. Ron's rule is generally very reliable. What makes this question exceptional is that most test takers will treat the question as solved once they have determined the costs of the doughnut and the cupcake, the values represented by d and c in your equations. If C were in fact the right answer, most people who chose it would do so for the wrong reason.

You did a little extra work in the question stem, which work very nearly got you the right answer. You noted that you were solving not for d but for n1. That habit--of taking the trouble to rephrase the question--will stand you in good stead.

[nirmal.merchant]

I think the difference between the pencils problem and the donuts and cupcakes problem is that in the first problem you cannot buy only 23 cent pencils or only 21 cent pencils and not have any change left. Whereas in the latter problem it is possible to buy only donuts or only cup cakes which gives you two solutions right there... Would love to hear if that's a good test to evaluate whether it is worth spending the effort in finding the solution.

[RonPurewal]

I think the difference between the pencils problem and the donuts and cupcakes problem is that in the first problem you cannot buy only 23 cent pencils or only 21 cent pencils and not have any change left. Whereas in the latter problem it is possible to buy only donuts or only cup cakes which gives you two solutions right there... Would love to hear if that's a good test to evaluate whether it is worth spending the effort in finding the solution.

this may or may not be true, but what i'm seeing here is that this poster is willing to exert 10X amount of effort in hypothesizing and memorizing dubious "rules", in order to avoid exerting 1X amount of effort in simply testing cases.
in either of the two problems discussed in this thread, it really doesn't take a whole lot of time or effort to test the different possible cases.
especially in the cupcake/doughnut problem, in which there are several possible combinations (listed thoughtfully by michael schwartz, above), of which any two will lead quickly to the correct answer (e).

really, guys -- if you see a problem like these ones, in which the answers must be whole numbers and in which the numbers are not terribly large, TESTING CASES is still the best, easiest, and most reliable way to go.
i'm still consistently surprised at the degree to which people seem to be opposed at all costs to testing cases.

[jp.jprasanna]

1) Marta bought a total of 6 pencils -- obviously INSUFFICIENT

2) The total value of pencil Marta bought was 130 cents, which means 21m+23n = 130, where m and n are integers. Let's try solve this equation.

130 is not divisible by either 21 or 23, so m >= 1, n >= 1, and we have:
21(m-1) + 23(n-1) = 130-21-23 = 86, which again is not divisible by either 21 or 23, so m >= 2, n >= 2, and we have:
21(m-2) + 23(n-2) = 86-21-23 = 42, which is divisible by 21 but not 23. Hence, we have m = 4, n = 2. -- SUFFICIENT

interesting.

i think i may have an idea of the general principle behind this approach, but would you mind elaborating a bit for the benefit of other forum readers?

Hi - Ron Totally agree with testing cases.

but if one is under time pressure, can we use the below logic

1/ In the pencil question we have 23 and 21 Cent pencil each 23 and 21 share no common factors so x23 + y21 =130 then there must be a unique value for x and y so that x23 + y21 equals 130 otherwise the problem would not exist (also we have to take each stat as true - Martha DID BUY 130 cents value of pencil using 23 and 21 cent pencils and there is also a integer constrain on x and y)

2/ Another set of nos x2 + y3 = 10 now since 2 and 3 share no common factors (except for 1) we can say there must be unique value for x and y for the equation x2 + y3 = 10 to exist

But the converse is not true for example
3/ x6 +y10 = 100 (x and y share factors) values for x and y can be 10,4 or 5,7

4/ x4 + y6 = 20 (x and y share factors) value for x and y can ONLY be 2,2

So for the pencil problem can we directly choose B instead of testing nos because they DO NOT share common factors? Or would there be exceptions?

See also problem no 10 in Advanced quant guide page 124

[RonPurewal]

but if one is under time pressure, can we use the below logic
...
no common factors
...

"no common factors" doesn't really help you here.
for instance:
5x + 7y = 47 --> there are two different pairs (x, y) that solve this equation: (8, 1) and (1, 6).
5x + 7y = 48 --> this time, there's only one such pair: (4, 4).
both of these equations have "no common factors", so there goes that theory.
also, it's quite easy to make equations with no common factors that will have tons and tons of solutions, just by picking a really big number for the constant term. for instance, 7x + 3y = 1,000,000,000 doesn't have any common factors among its coefficients, but there will obviously be lots of different pairs of integers that solve it.

if you really, really, really are dying to learn obscure rules that you'll probably never have the chance to use, there's always this: if you have ax + by = c and c > 2ab, then there will always be more than one pair (x, y) that solves the equation.

--

more importantly... in response to this:

Hi - Ron Totally agree with testing cases.

but if one is under time pressure...

on a problem like this one, you should definitely be able to test cases in under a minute. probably under thirty seconds. so this is really a non-issue.

i'm not trying to be a mind reader here, but, in my experience, in 99% of all cases when people say "...but what if it takes too much time?", what the person is really thinking is, "... but i'm not used to that strategy! that makes me uncomfortable!"
i bet the same thing is happening here. you've got to expand your repertoire to succeed on this exam.

[nitestr]

For I was taking more time than I should by testing numbers. Joey Z's method is pretty cool