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

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

GMATPrepNow Shouldn't the answer be 15?

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

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

Cheers,
Brent

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 .

Posted from my mobile device

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 think that there is one quick way. Correct me if I'm wrong! So, there are 4 and 5 numbers in a set! So... maximum possible values of P is 4C1 x 5C1 = 20. Now, we need to delete the double counted values. Numbers 4 and 5 are common in both sets. If these were different numbers and we count possible of values of P... with only 2 in each set... then a total of 8 different values of P is possible. But, these numbers .. 4 and 5 of first set and 4 and 5 of second set are the same. So, products will be repeated. So, among 8 possible P values... 4 P value are genuine and 4 are repeated. So, answer is (Total Possible - Repeat) = 20 - 4 = 16.

Good idea, but I think you're still performing a version of listing and counting, except you're counting the duplicates only (and subtracting them from 20).
How do you handle the duplication of 4 x 6 (4 from set A and 6 from set B) and 3 x 8 (3 from set A and 8 from set B), since they both have products of 24?

----------------------
Yes, you are right! In some ways, it is lite version of counting! We might as well just do it systematically... to even catch 4x6 and 3x8 duplication. I get it now... you're right. Whenever possible, it's best to simply list and count! Thank you for the thoughts!

split the set A into 2 subset A1={2,3} and A2={4,5), then calculate the A1B and A2B separatedly.
choose one from A1 and then choose one from B1={4,5,6,7,8}, the total different numbers should be 2 * 5= 10
then choose one from A2, and choose one from B2={6,7,8}, in order to remove repeat with 4 and 5,
then the result shold be 2 * 3 = 6
so add up those two step's result 10+6 is the final answer.

­Total number of choices = 4C1 x 5C1
Total numbers repeating = 12  16 24 32  = 4 
Different sol = 20-4 =16

Given: Set A = {2, 3, 4, 5}, and set B = {4, 5, 6, 7, 8}.

Asked: 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?

Different values of P = 4*5 =20

IMO E

Posted from my mobile device