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Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product

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Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product  [#permalink]

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New post 16 Jan 2019, 17:32
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Question Stats:

50% (01:59) correct 50% (01:46) wrong based on 90 sessions

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Set A = {2, 3, 4, 5}, and set B = {4, 5, 6, 7, 8}. If P = the product of one number chosen from set A and one number chosen from set B, how many DIFFERENT values of P are possible?

A) 14
B) 15
C) 16
D) 18
E) 20

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Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product  [#permalink]

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New post Updated on: 14 Feb 2019, 04:24
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IMO, D (16)

I solved it by multiplying each 2 numbers from different sets and excluding repeated products:
2|| 8,10,12,14,16
3|| 12,15,18,21,24
4|| 16,20,24,28,32
5|| 20,25,30,35,40

but is there a faster way?
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Originally posted by MahmoudFawzy on 16 Jan 2019, 18:35.
Last edited by MahmoudFawzy on 14 Feb 2019, 04:24, edited 1 time in total.
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Re: Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product  [#permalink]

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New post 17 Jan 2019, 12:05
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Mahmoudfawzy83 wrote:
but is there a faster way?


Nice work - kudos to you!!

I don't believe there's a faster solution. In fact, I created the question to highlight the fact that, when it comes to counting questions, many students are reluctant to simply list and count the possible outcomes.

In this case, all 5 answer choices are relatively small, so we can be sure that it won't take long to list and count the outcomes.

That said, even when the answer choices are large, listing possible outcomes can often lead to useful insights regarding the correct answer.

I talk about all of this (and more) in the following video:

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Re: Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product  [#permalink]

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New post 17 Jan 2019, 12:10
GMATPrepNow wrote:
I talk about all of this (and more) in the following video


Big Fan for your channel, it helped me much (Kudos for your channel :-D )
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Re: Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product  [#permalink]

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New post 13 Feb 2019, 11:57
GMATPrepNow Shouldn't the answer be 15?

Mahmoudfawzy83 wrote:
IMO, D (16)

I solved it by multiplying each 2 numbers from different sets and excluding repeated products:
2|| 8,10,12,14,18 2*8=16
3|| 12,15,18,21,24
4|| 16,20,24,28,32
5|| 20,25,30,35,40

but is there a faster way?
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Re: Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product  [#permalink]

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New post 13 Feb 2019, 12:24
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energetics wrote:
GMATPrepNow Shouldn't the answer be 15?

Mahmoudfawzy83 wrote:
IMO, D (16)

I solved it by multiplying each 2 numbers from different sets and excluding repeated products:
2|| 8,10,12,14,18 2*8=16
3|| 12,15,18,21,24
4|| 16,20,24,28,32
5|| 20,25,30,35,40

but is there a faster way?


Good question.

Mahmoudfawzy83 accidentally wrote 18 (instead of 16) in the first row.
This means the 16 in the third row should be a duplicate.
However, at the same time if we change 18 to 16 in the first row, then the 18 in the second row is no longer a duplicate.

Here's the revised table:

2|| 8,10,12,14,16
3|| 12,15,18,21,24
4|| 16,20,24,28,32
5|| 20,25,30,35,40

Cheers,
Brent
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Re: Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product  [#permalink]

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New post 13 Feb 2019, 12:25
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Mahmoudfawzy83 wrote:
GMATPrepNow wrote:
I talk about all of this (and more) in the following video


Big Fan for your channel, it helped me much (Kudos for your channel :-D )


Thanks Mahmoudfawzy83!!!
I'm glad you like it.

Cheers,
Brent
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Re: Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product  [#permalink]

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New post 13 Feb 2019, 12:40
I solved by simply multiplication. I think we need to look at the numbers in this type of que if they are easy to multiple then we should do it .

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Re: Set A = {2, 3, 4, 5} and set B = {4, 5, 6, 7, 8}. If P = the product   [#permalink] 13 Feb 2019, 12:40
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