Marla buys two types of pencils only, which costs 21 cents o

[rschunti]

What is the best way to approach below problem?

Marla buys two types of pencils only, which costs 21 cents or 23 cents. How many 23 cent pencils did she buy?

(A) She bought 6 pencils in total
(B) The total cost of her purchase was 130 cents

[RonPurewal]

i'm assuming this is a paraphrase of the actual problem (which is fine), but next time please try to copy the problem verbatim. this problem doesn't seem to have any subtleties, but some problems feature subtleties that might be lost if you don't copy the problem word for word.

--

statement (1) is the easier statement: if all you know is that she bought 6 pencils, you of course have no idea how many of those 6 were of each type. this statement is therefore insufficient.

statement (2):
you have 21x + 23y = 130. because that's one equation in two variables, your first instinct is probably to say 'insufficient!!'
the problem here, though, is that this equation must have solutions that are nonnegative integers (and is therefore called a 'diophantine equation', if you like mathematical terms). because of that restriction, it's quite possible that there's only one feasible solution; the only way to find out within a reasonable amount of time is to exhaust the possibilities:
the total cost was 130 cents. so, you can just find all possible values of the total cost of the 23-cent pencils, and then subtract these from 130 and find whether the resulting differences are possible.

so, here we go:

POSSIBLE MULTIPLES OF 23 CENTS ... REMAINING CENTS
0 ...................................................... 130
23 .................................................... 107
46 .................................................... 84
69 .................................................... 61
92 .................................................... 38
115 .................................................. 15


red lights don't work (because the 'remaining' quantities aren't multiples of 21 cents). green light works.
sufficient.

THERE IS NO FASTER WAY TO HANDLE STATEMENTS / EQUATIONS LIKE THIS ONE WHEN THE SOLUTIONS MUST BE WHOLE NUMBERS.
YOU MUST TEST CASES.
to all the theory types out there: sorry. :(

answer = b

[shottayardie]

how do you recognize diphantine equations?

[RonPurewal]

how do you recognize diphantine equations?

a diophantine equation is just an equation whose solutions have to be whole numbers (0, 1, 2, ..., and yes, 0 is considered a whole number).

it's all based on context: either
(1) the problem statement will directly state that the solutions must be integers, or, more commonly,
(2) the context of the word problems will dictate that the solutions must be integers. i.e., if a problem calls for a school class of x boys and y girls, then obviously x and y must be whole numbers (barring the possibility that you're watching some gory school horror movie).

[kevinmarmstrong]

You could also deduce (1) from (2), since 130/21 < 7 and 130/23 > 5, implying that x + y could only be 6.

[adiagr]

Dear all,

Continuing with the discussion, I want to ask, how to be sure whether a given equation will have unique solution?


For e.g. consider the following GMAT prep question

At a certain bakery, each roll costs r cents and each doughnut costs d cents. if Alfredo bought rolls and doughnuts at the bakery, how many cents did he pay for each roll?
1) Alfredo paid $5 for 8 rolls and 6 doughnuts.
2) Alfredo would have paid $10 if he had bought 16 rolls and 12 doughnuts.

In above question

From (1)

8r + 6d = 500

From (2)

16 r + 12 d = 1000

It can be seen that (2) is essentially same as (1)

so no new information is obtained from (2).

But here main doubt is about (1). How do i know whether (1) is sufficient or not?

(Highlight to see OA) OA is E

[akhp77]

Usually, two variables linear equation required two equations to get unique solution.

But, here, we have to find only +ve integral solution. This is the extra boundary condition.

21x + 23y = 130 = 84 + 46 = 21*4 + 23*2
x = 4, y = 2
No other +ve integral solution possible.

BUT

8r + 6d = 500
4r + 3d = 250
220 + 30 = 55*4 + 10*3; r = 55, d = 10
160 + 90 = 40*4 + 30*3; r = 40, d = 30

We have multiple solutions. It might be two or more solutions.
I have not tested it.

[adiagr]

You are quite right. But is hit & trial method the only way to go about in these cases.

Aditya

[akhp77]

You are quite right. But is hit & trial method the only way to go about in these cases.

Aditya

Usually, YES but idea is that one has to attack on DS problem.

Above both problems have very limited possibilities. So, you can find out easily.

[RonPurewal]

Dear all,

Continuing with the discussion, I want to ask, how to be sure whether a given equation will have unique solution?

i'm going to copy and paste from my own post, in this same thread.

i said:

THERE IS NO FASTER WAY TO HANDLE STATEMENTS / EQUATIONS LIKE THIS ONE WHEN THE SOLUTIONS MUST BE WHOLE NUMBERS.
YOU MUST TEST CASES.
to all the theory types out there: sorry. :(

i wasn't kidding about that -- the best way to investigate problems like this is, really, no kidding, simply to test cases and see whether more than one of them works.

i mean, yeah, there are graduate-level techniques in number theory that could let you deal with the general case of this sort of thing, but those techniques are not worth the time to learn for this test.

in any case, the above is the answer to your question itself. what seems more important, however, is to address the underlying issue behind your question.

in particular (assuming that you read the posts appearing above yours), it seems as though you are doing literally everything in your power to avoid plugging numbers -- i.e., despite my explicit statement above that you can't use algebra to solve problems like this, you're still looking for an algebraic solution!

you MUST understand the following:
this test is designed to be hostile to people who try to attack EVERYTHING with routine algebra.
in other words, there will be problems that are explicitly designed so that they CAN'T be solved with routine algebraic operations -- instead, you will have to think about the boundaries/restrictions/special cases of some problem, or (as in the problems at hand here) you will just have to plug in various cases and watch what happens.
the point is that you're going to have to be flexible. algebra is certainly important, but you can't bank on algebra to do absolutely everything that is necessary on this exam.

[adiagr]

Ron,

Thanks for this insight. Will try solving by plugging numbers and not mere algebra.

Really appreciate your effort and also your straight forward approach.

Thanks

Aditya

[mschwrtz]

Glad that Ron was able to help. He is 100% right that there's nothing to be gained ON THE GMAT from trying to determine without checking all values whether there might be a couple of solutions. If there were scores of solutions to check, or if each took a long time to check, than it might be worthwhile to think more deeply about number theory, but those conditions don't obtain ON THE GMAT.

[vinversa]

Could it be that.... Nevertheless...it is worth mentioning.... Even if it is not true in all cases... it may be in most... and could turn out as a time saver: "If prime then equation alone is enough."


For a certain play performance, adults’ tickets were sold for $12 each and children’s tickets were sold for $8 each. How many children’s tickets were sold for the performance?

(1) The total revenue from the sale of adults’ and children’s tickets for the performance was $5,040.
(2) The number of adults’ tickets sold for the performance was 1/3 the total number of adults’ and children’s tickets sold for the performance.

(1) $12.A + $8.B = 5040
This is not enough. Since both 12 and 8 are non prime numbers. A & B can adorn several different solutions unlike X & Y in the Martha (Martha bought several pencils) problem $23.X + $21.Y = 130 (here 23 is a prime number, thus there is possibility for only one solution to this equation) - Same case with Joanna and her stamps [Joanna bought only $0.15 stamps and $0.29 stamps] ($0.15A + $0.29B = $4.40) - [29 is prime here]
In the case of Martha and Joanna - that one equation is sufficient.

(2) A=1/3(A+C)
2A = C
OA = C

[RonPurewal]

i'm a little bit worried by this response:

Could it be that.... Nevertheless...it is worth mentioning.... Even if it is not true in all cases... it may be in most...

if you think you can depend on a "rule" that is not true in all cases, and for which the best you can do is "may" and "most", then you really need to understand the purpose of this test better than you do.[/i]
one of the MAIN PURPOSES of this exam is to make sure that students get problems WRONG if they just follow rules all the time.
the problems on which this happens are not "trick questions" -- they are the reason why the exam is there in the first place. the purpose of this exam is to make sure that people who think critically about the mathematics -- and who therefore are acutely aware of boundaries, restrictions, and exceptions, and don't just follow rules all the time -- score higher than people who just try to memorize a bunch of rules.

in other words:
if you have a rule that is "true most of the time", then the EXCEPTIONS to that rule will be tested MORE OFTEN than will the rule itself.
count on it.
for instance, the infamous "n equations for n variables" rule -- the exceptions to this rule are tested at least 5 times for every 1 time that the rule itself is tested. (this problem is one such exception -- there are 2 variables and 2 equations, but you only need one of the equations.)

and could turn out as a time saver:

perhaps, but it really doesn't take that long to test the values!
check out my solution above -- you could easily do all that in well under a minute.

"If prime then equation alone is enough."

nope. this is actually unreliable both ways -- i can give you an equation that's sufficient even though the coefficients are not prime, and i can also give you an equation that's insufficient even though the coefficients are prime.

here are all 4 possibilities for whole number solutions:
Prime coefficients, sufficient: 5x + 7y = 51 (the only solution is x = 6 and y = 3)
Prime coefficients, insufficient: 5x + 7y = 47 (could be x = 1 and y = 6, or x = 8 and y = 1)
note how similar these are -- you aren't going to find a simple rule.

now, just multiply these equations by 2 to get the other two possibilities:
Nonprime coefficients, sufficient: 10x + 14y = 102 (6, 3 is still the only solution)
Nonprime coefficients, insufficient: 10x + 14y = 94 (same two solutions 1, 6 and 8, 1)

--

again, this post is an example of what i'm talking about up there -- people are excessively hostile to the idea of plugging in possibilities on these problems.
* please be more open-minded to the idea that you may have to plug in test values!

* rules are not everything!

* testing values really doesn't take that long!

* on problems like these, it's way, way easier than trying to formulate rules!