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Marta bought several pencils. If each pencil was either a

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Marta bought several pencils. If each pencil was either a [#permalink] New post 13 Jun 2007, 07:36
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A
B
C
D
E

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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
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 [#permalink] New post 13 Jun 2007, 08:05
B

1) insuff
2) if you brute force it, you can see that 4,2 is the only possible combination that'll add up to 130 cents.
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 [#permalink] New post 13 Jun 2007, 08:13
Agree, B it is. However, this is a trick question, and in the actual GMAT with the entire time pressure one can fall into the trap.

I got B using trial and error.

Hey Caas, r u a Barcelona or Real Madrid Fan by the way? :-D
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 [#permalink] New post 13 Jun 2007, 08:17
X & Y wrote:
Agree, B it is. However, this is a trick question, and in the actual GMAT with the entire time pressure one can fall into the trap.

I got B using trial and error.

Hey Caas, r u a Barcelona or Real Madrid Fan by the way? :-D


No :) I am more fan of Italian football.
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 [#permalink] New post 13 Jun 2007, 11:53
possible options for the 23 cent pencils:

one = 23
two = 46
three = 69
four = 92
five = 115
six = 138 (too big !)

possible options for the 21 cent pencils:

(130-23)/21 = not an integer.
(130-46)/21 = 4
(130-69)/21 = not an integer.
(130-115)/21 = not an integer.

the answer is (B)

:-D
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 [#permalink] New post 13 Jun 2007, 22:54
Hey guys,

It would take me very long to count all the possible ways on the exam.
Does anyone know any shortcut?
How to quickly understand that there is only one couple of solutions: 2 and 4???
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 [#permalink] New post 13 Jun 2007, 23:24
The process of a quick solution as I see it
Consider (2)

21X + 23Y = 130
1) Check the max quantity of pencils.
130/21 = 6

2) Options of quantities: (total must be 6 the biggest)
21 cents 23 cents
1 5
2 4
3 3
4 2
5 1

3) Check the last digit of multiplication of all the options:
It must be 0
Ex: 21*1 + 23*5 = ...6
21*2+23*4 = ... 2+....2 = 4
.....
21*4+23*2=130

Is it OK or can I do it quicklier?
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 [#permalink] New post 14 Jun 2007, 00:14
St1:
We don't have total amount spent to set up and equation. Insufficient.

St2:
23n + 21p = 130
Only one set works, n = 2, and p = 4. Sufficient.

Ans B

Recognizw 23 is a prime, so just find the multiples and subtract off from 130 and see which one cna be divided evenly by 21
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 [#permalink] New post 14 Jun 2007, 00:21
shortcut:

assume 23 pencils and 21 pencils has the same number of pencils:

(23+21)*2 = 88 (too small)

(23+21)*3 = 132 (too small)

(23+21)*4 = 176 and 176-130 = 46

meaning we have too many 23 pencils. since 46/23 = 2 so we are left with four 21 penciles and two 23 pencils.

:-D
  [#permalink] 14 Jun 2007, 00:21
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