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# Marta bought several pencils. If each pencil was either a

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SVP
Joined: 21 Jul 2006
Posts: 1551
Followers: 8

Kudos [?]: 296 [0], given: 1

Marta bought several pencils. If each pencil was either a [#permalink]  11 Dec 2007, 03:43
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Marta bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, how many 23-cent pencils did Marta buy?

(1) Marta bought a total of 6 pencils

(2) The total value of the pencils Marta bought was 130 cents

Director
Joined: 12 Jul 2007
Posts: 865
Followers: 12

Kudos [?]: 214 [0], given: 0

Let x = 23 cent pencil
Let y = 21 cent pencil
Both x and y are >= 0 (she can't buy negative pencils)

1. x + y = 6

this could be any combination of x and y to get 6.

(0,6)(1,5)(2,4)(3,3)(4,2)(5,1)(6,0)

INSUFFICIENT

2. 23x +21y = 130

Plug in numbers to see which combinations turns out right

23x = 130 - 21y

4(21) + 2(23) = 130

is the only combination that works out. It just happened that x + y = 6 in this equation, we didn't use the information from statement 1 at all.

SUFFICIENT

SVP
Joined: 21 Jul 2006
Posts: 1551
Followers: 8

Kudos [?]: 296 [0], given: 1

the OA is B. But i honestly thought that this question was mean because we always understood that an equation with 2 variables can never be solvable, now i'm seeing the contrary here. is there any other way to solve this problem without manually coming up with different numbers to experiement with? i don't usually like coming up with numbers because it can waste time. are there any direct steps that one can take? when can we say that 2 variables with only 1 equation is solvable and when not???
Senior Manager
Joined: 18 Oct 2007
Posts: 449
Location: USA
Schools: Tepper '11
Followers: 5

Kudos [?]: 56 [0], given: 2

I selected B and got it right on my practice exam, but I too was curious about the theoretics behind this. Is something considered solvable even if you have to simply try out every combination? Conversely, if there is only one solution, does it by definition mean there must be a mathematical way of solving it?
Director
Joined: 12 Jul 2007
Posts: 865
Followers: 12

Kudos [?]: 214 [0], given: 0

I think if given the proper constraints there is only one correct answer possible you count it, even if it takes the guess and check method.
Intern
Joined: 12 Nov 2007
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 0

My method [#permalink]  15 Dec 2007, 14:18
I found answer following this reasoning:
- we have 23*a + 21*b=130
- a and b are both integers (this is actually a second constraint)
In order to restrict the possiblities, we can look at the unit digits of 23 and 21 (3 and 1) and note that we need to have 3*a + 1*b= "number with 0 as unit digit"
Thus we have the folloiwing possiblities:
- a=3, b=1
- a=2, b=4
- a=1. b=9

The only one which gives the initial formula equal to 130 ia for a=2 and b=4.
CEO
Joined: 29 Mar 2007
Posts: 2591
Followers: 16

Kudos [?]: 232 [0], given: 0

B. Just by a few trials.

I like gindro79's approach.
SVP
Joined: 21 Jul 2006
Posts: 1551
Followers: 8

Kudos [?]: 296 [0], given: 1

yeah me too. this approach is indeed much easier.
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