Marta bought several pencils.

[tim]

:)

[bradlynn33]

An equation of the form 21x+23 y = 130 can be solved mathematically if you now how to find simultaneous integral solution or if you have heard of Diophantine equation. Joey.z has provided the method .for the benefit of the reader the method is :
21x = 130-23y ...(A)
you can either substitute 1, 2 ,3 .. for y or for x in (A) and see what happens .. let us substitute Y =1 in (A)
then 21x = 130-23(1) = 107
21x= 107 ;x = 5.09 (ruled out as Martha is not buying pencil in decimals.. the pencils must be whole numbers giving us a hint that we are looking at simultaneous integral solution or Diophantine equation).
we now put y= 2 in equation (A)
we get ; 21 x= 130-23(2)
21 x= 130-46
21x=84 ; x=4 . so we get the solution y= 2 and x=4 and therefore the answer (B).
point to note : in equation (A) when you have to substitute ; you substitute for y instead of x . Ron can u tell why so ???others can reason out too.

[RonPurewal]

bradlynn33,
two things

1/
it seems that you are just showing the method of testing cases (as mentioned earlier in the thread) here.
it's somewhat hard for me to understand what's going on in your post -- mostly because the spacing, organization, etc. are sort of weird -- but it seems that you're just testing integer values of y and seeing whether you get an integer value of x.

if so, that's the same thing that's already suggested above.

if not, please explain, thanks.

2/
this is not how DS works:

we now put y= 2 in equation (A)
we get ; 21 x= 130-23(2)
21 x= 130-46
21x=84 ; x=4 . so we get the solution y= 2 and x=4 and therefore the answer (B).

here, you seem to be implying that you are finished as soon as you get the pair (4, 2).
that's not how DS works. in fact, there will always be at least one valid solution for each statement, so finding one valid solution, in and of itself, doesn't mean anything.
the real issue in DS is whether you can find a second solution!

so, here, after you find x = 4 and y = 2, you still have to test the rest of the cases.
you haven't actually established that the statement is sufficient until you get through all of the remaining cases without finding an additional solution.

point to note : in equation (A) when you have to substitute ; you substitute for y instead of x

the process is the same either way, so it doesn't really matter.
it's ever so slightly more efficient to substitute for y, because there are fewer possibilities -- the upper limit is 130, so you only have to try y = 0/1/2/3/4/5, whereas you'd have to try x = 0/1/2/3/4/5/6.
but, realistically speaking, we're talking about a difference of, what, two or three seconds. so it doesn't matter at all which one you choose to substitute for.

[Jbv626]

I think a good way to simplify the testing is the following:

In statement 2, it is easy to realize that 6 are the pencils that marta bought. 6x20=120. With such costs, no other solution is possible. Then, we have yet 10 cents left until 130. With the 3s of 23 and the 1s of 21, we can see that only two 3s and four 1s work.

Another way to solve the commented statement is a sort of application of the concept of wheighted average (that ron has taught in the forum):

130/22=5.9090. Here we can see that the average cost of 22 (3 pencils of each kind) is higher than possible. So, first thing is to check two 23s and four 21s in order to make the average lower. As far as we have the number of 130, it is easy to realize that no other combination is possible.

Neither of these ways is elegant, but both are fast.

[bradlynn33]

Dear Ron ,

Finding one solution is good enough to stop finding solution to the question. Why this so? This is because of restrictions given in the problem. Here goes my explanation:

Given the equation 21x+ 23y =130
Let me put this mathematical equation into words part by part so it is easy to understand.If x is a whole number then when we write 21x : this means in words x is a multiple of 21.

If we make a set that contains multiples of 21 for this question and denote it by A then

Set A =(21,42,63,84,105,126..)

Stopping at 126 because the total money spent by Martha is 130 cents. The elements in set A is the money that martha can spend when she buys pencil worth 21 cents. Similarly for 23y, the set be denoted by B.

So set B =(23,46,69,92,115).

If martha buys only pencil worth 21 cents she cannot spend 130 cents .This means Martha has also bought pencil worth 23 cents. Now if you were to take elements of Set A (21,42,63,84,105,126) and Set B = (23,46,69,92,115) in a way that they add up to 130 there is ONE AND ONLY ONE WAY THEY CAN ADD UPTO 130 . That is 46+ 84. Mr.Ron that is exactly the reason I stopped after finding only one solution in my earlier post. The restriction given in this particular is such that only one solution is possible.

On to the question that in my earlier post I have given only a way of how to substitute - I would like to say that "I want the readers on manhattan forum to realise that an equation of the form 21x+23y= 130 can be solved by method of substitution .The moment people see equation with two variable and only one equation being given- people think it cannot be solved .This is exactly what Ron you have said earlier in this thread.

One more thing I would like to add: Mathematical equation when put into words or put in the form of question a lot of us get confused. I would also like to thank Ron for a quick reply earlier.

[jlucero]

Dear Ron ,

Finding one solution is good enough to stop finding solution to the question. Why this so? This is because of restrictions given in the problem. Here goes my explanation:

Given the equation 21x+ 23y =130
Let me put this mathematical equation into words part by part so it is easy to understand.If x is a whole number then when we write 21x : this means in words x is a multiple of 21.

If we make a set that contains multiples of 21 for this question and denote it by A then

Set A =(21,42,63,84,105,126..)

Stopping at 126 because the total money spent by Martha is 130 cents. The elements in set A is the money that martha can spend when she buys pencil worth 21 cents. Similarly for 23y, the set be denoted by B.

So set B =(23,46,69,92,115).

If martha buys only pencil worth 21 cents she cannot spend 130 cents .This means Martha has also bought pencil worth 23 cents. Now if you were to take elements of Set A (21,42,63,84,105,126) and Set B = (23,46,69,92,115) in a way that they add up to 130 there is ONE AND ONLY ONE WAY THEY CAN ADD UPTO 130 . That is 46+ 84. Mr.Ron that is exactly the reason I stopped after finding only one solution in my earlier post. The restriction given in this particular is such that only one solution is possible.

Your method here is much clearer than from your description above, but it's hard to express algebra sometimes in text. It looks good, and the main thing is to make sure that there's only one possible answer. That was what Ron mentioned, and I agree. In every DS question there will be at least one solution. The goal is to see if there is a second possible solution.

On to the question that in my earlier post I have given only a way of how to substitute - I would like to say that "I want the readers on manhattan forum to realise that an equation of the form 21x+23y= 130 can be solved by method of substitution .The moment people see equation with two variable and only one equation being given- people think it cannot be solved .This is exactly what Ron you have said earlier in this thread.

I agree with this. If you're referring to your earlier question of why this is the case- it's because there are restrictions in the real world that don't happen in algebra. In the real world, there's an infinite number of numbers that add up to 10, but in a game that only counts by 3s and 7s, there's only one way.

One more thing I would like to add: Mathematical equation when put into words or put in the form of question a lot of us get confused. I would also like to thank Ron for a quick reply earlier.

They can be confusing, but if it didn't confuse someone, it wouldn't be on the GMAT. Learning how to get better at translating is the goal.

[NinaP494]

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.

Ron:
1. It is such a neat rule. Why do you think it will never have a chance to use?
2. More importantly, is the converse true?
thks

[RonPurewal]

1. It is such a neat rule. Why do you think it will never have a chance to use?

well... i mean, just look at it.

More importantly, is the converse true?

what are you asking?
the "converse" of If A, then B is If B, then A. if i try to do that here, i get a nonsense statement. so, "converse" probably isn't the word you are looking for.

can you just write it out explicitly, please? what's the theorem/statement that you are proposing?