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Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct

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ccax
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Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 13:35
Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct each.

How many Pencils did he buy?

(1) He paid 278 ct for all
(2) A+B = 18



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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 13:43
17A + 13B = 278
A + B = 18

A & B can be found from the two equations.

C.
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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 17:54
1) 17p + 13r = 278
Only possible set (p,r) = (11,7)

Sufficient.


2) Does not provide any useful information.

Ans A
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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 19:01
ccax wrote:
Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct each.
How many Pencils did he buy?
(1) He paid 278 ct for all
(2) A+B = 18


A is sufficient. there must be only one value for each of A and B to make their total values 278 ct.
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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 19:21
A. i agree with above

this looks like a kaplan question :) they pull this trick all the time
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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 19:25
Yeah.. agree with A.
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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 19:49
I understand this but is there a shortcut to find the matching set? as here how did we determine 17r+13q=278 is only possible for (11,7). So every time i encounter question like this I have to spend 5-6 mins trying all combinations. I understood at first glace that this is sometime kinda trap but still do I have to waste time everytime?
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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 19:54
ywilfred wrote:
1) 17p + 13r = 278
Only possible set (p,r) = (11,7)

Sufficient.


2) Does not provide any useful information.

Ans A
what's the fastest way to solve for (11,7)
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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   04 Apr 2006, 20:22
Not sure if it's a good method for you, but at least it works for me

17p + 13r = 278
13(p+r) + 4p = 278
13(p+r) = 278-4p

Can quickly work out which multiple of 4 results in a multiple of 13.

So for p = 11,
So p+r = 18
r = 18-11 = 7.
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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink] New post   05 Apr 2006, 08:05
NTLancer wrote:
this looks like a kaplan question :)


Well it's not a Kaplan question, but I didn't pick prime
numbers (13 and 17) for the prices by accident :-D

In the case of primes, the first match with two possibilities
could come at A*B. In the case above, this would be at
13*17 = 221. It's easy to see that the difference of 57
(278-221) can't be split in k*13+n*17. So (1) is sufficient.

Always pay a bit more attention if you see primes or numbers
with a big prime factor in questions similar to this one.



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Re: Fred bought A Pencils for 17 ct each and B Rubbers for 13 ct [#permalink]
05 Apr 2006, 08:05
Moderator:
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

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