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Steve has 4 pens. The values of the pens are \$2, \$4, \$7, and

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Manager
Joined: 21 Sep 2008
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Steve has 4 pens. The values of the pens are \$2, \$4, \$7, and [#permalink]

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15 Nov 2008, 14:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Steve has 4 pens. The values of the pens are \$2, \$4, \$7, and \$12. Which of the following could NOT be the total value of any collection of these pens?

\$14
\$16
\$17
\$21
\$23
I just can't understand the answer. My answer was D because with four given prices, you can't make 21 dollars.

hoice (A): \$14 = \$12 + \$2. Discard choice (A).

Choice (B): \$16 = \$12 + \$4. Discard choice (B).

Choice (C): \$17. There does not appear any way to obtain a total of \$17. Let's look at the remaining choices to be sure that this choice is the value that cannot be obtained.

Choice (D): \$21 = \$12 + \$7 + \$2. Discard choice (D).

Choice (E): \$23 = \$12 + \$7 + \$4. Discard choice (E).Now that all 4 incorrect answer choices have been eliminated, we know that choice (C) must be correct.

Am I just not thinking hard enough?

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Joined: 05 Jul 2008
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Re: Quantative: Kaplan practice question [#permalink]

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15 Nov 2008, 14:57
hwiya320 wrote:
Steve has 4 pens. The values of the pens are \$2, \$4, \$7, and \$12. Which of the following could NOT be the total value of any collection of these pens?

\$14
\$16
\$17
\$21
\$23
I just can't understand the answer. My answer was D because with four given prices, you can't make 21 dollars.

hoice (A): \$14 = \$12 + \$2. Discard choice (A).

Choice (B): \$16 = \$12 + \$4. Discard choice (B).

Choice (C): \$17. There does not appear any way to obtain a total of \$17. Let's look at the remaining choices to be sure that this choice is the value that cannot be obtained.

Choice (D): \$21 = \$12 + \$7 + \$2. Discard choice (D).

Choice (E): \$23 = \$12 + \$7 + \$4. Discard choice (E).Now that all 4 incorrect answer choices have been eliminated, we know that choice (C) must be correct.

Am I just not thinking hard enough?

Please do not post answers and your doubts as they cut into the thoughts of others. You can start asking Q's once others show their approach. Most of the times, such doubts will be addressed in the responses. If not you are more than welcome to have a discussion.

17 can be obtained if steve had 7 + 2 + 4 +4 So discard C

B T W, the way I understood the Q is Steve has to have 4 pens of some value. He just cannot have all the value in 1 pen or 2 pens with value of 3rd and 4th equal to 0.

16 = 4 X 4

12 = 3 X4 + 2

17 = 2 X4 + 2 + 7

21

23 = 12 + 7 + 2 +2

21 remains

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Re: Quantative: Kaplan practice question [#permalink]

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15 Nov 2008, 18:11
kaplan got it wrong it has to be 21..otherwise the question doesnt make sense if he has 4 pens..

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SVP
Joined: 29 Aug 2007
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Re: Quantative: Kaplan practice question [#permalink]

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15 Nov 2008, 18:37
icandy wrote:
hwiya320 wrote:
Steve has 4 pens. The values of the pens are \$2, \$4, \$7, and \$12. Which of the following could NOT be the total value of any collection of these pens?

\$14
\$16
\$17
\$21
\$23
I just can't understand the answer. My answer was D because with four given prices, you can't make 21 dollars.

hoice (A): \$14 = \$12 + \$2. Discard choice (A).

Choice (B): \$16 = \$12 + \$4. Discard choice (B).

Choice (C): \$17. There does not appear any way to obtain a total of \$17. Let's look at the remaining choices to be sure that this choice is the value that cannot be obtained.

Choice (D): \$21 = \$12 + \$7 + \$2. Discard choice (D).

Choice (E): \$23 = \$12 + \$7 + \$4. Discard choice (E).Now that all 4 incorrect answer choices have been eliminated, we know that choice (C) must be correct.

Am I just not thinking hard enough?

Please do not post answers and your doubts as they cut into the thoughts of others. You can start asking Q's once others show their approach. Most of the times, such doubts will be addressed in the responses. If not you are more than welcome to have a discussion.

17 can be obtained if steve had 7 + 2 + 4 +4 So discard C

B T W, the way I understood the Q is Steve has to have 4 pens of some value. He just cannot have all the value in 1 pen or 2 pens with value of 3rd and 4th equal to 0.

16 = 4 X 4

12 = 3 X4 + 2

17 = 2 X4 + 2 + 7

21

23 = 12 + 7 + 2 +2

21 remains

D. 21 is convincing...

I was understanding the problem differently as the question is ambigious........
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Kudos [?]: 843 [0], given: 19

Manager
Joined: 21 Sep 2008
Posts: 78

Kudos [?]: 3 [0], given: 0

Location: Atlanta, Georgia
Schools: Georgetown, UNC, Emory, GeorgiaTech
WE 1: Product Engineering
WE 2: Logistics & Inventory
WE 3: Product Information & Pricing Manager
Re: Quantative: Kaplan practice question [#permalink]

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16 Nov 2008, 19:48
icandy wrote:

Please do not post answers and your doubts as they cut into the thoughts of others. You can start asking Q's once others show their approach. Most of the times, such doubts will be addressed in the responses. If not you are more than welcome to have a discussion.

I'll remember to do that next time. I was just so frustrated after getting problems wrong.

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Manager
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Re: Quantative: Kaplan practice question [#permalink]

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17 Nov 2008, 02:40
Choice C is correct and the explanation is very clear

17 = 7 + 2 + 4 +4

Steve has 4 pens and these pens have different prices, 2, 4, 7, 12. There are no 2 pens which have the same price \$4.

21 = 12 + 7 + 2 <==correct

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Re: Quantative: Kaplan practice question   [#permalink] 17 Nov 2008, 02:40
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