The value of B ?

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The value of B ?

by avik.ch » Mon Sep 17, 2012 12:04 am
If 3^a4^b = c, what is the value of b?

(1) 5a = 25

(2) c = 36

OA = C

I think the second statement is sufficient.

Source : Kaplan

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by everything's eventual » Mon Sep 17, 2012 2:25 am
hmmm..I am not sure how you did it...but this was my method :

Statement 1) is def not sufficient.

Statement 2) says c = 36 = 3^2 * 2^2

Equating both sides we get (5^4)^b = 2

Then we get b = 0 and 2) should be sufficient.

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by vk_vinayak » Mon Sep 17, 2012 2:47 am
avik.ch wrote:If 3^a4^b = c, what is the value of b?

(1) 5a = 25

(2) c = 36

OA = C

I think the second statement is sufficient.

Source : Kaplan
1) Tells us about only a. INSUF

2) (3^a) * (4^b) = 36 = (3^2) * (4^1) => b=1. SUF.
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by avik.ch » Mon Sep 17, 2012 2:57 am
So we can reasonably conclude that the OA is wrong.

Thanks !!

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by avik.ch » Mon Sep 17, 2012 2:59 am
everything's eventual wrote:
Statement 2) says c = 36 = 3^2 * 2^2

Equating both sides we get (5^4)^b = 2

Then we get b = 0 and 2) should be sufficient.
I am not sure how you are getting b=0 ?? Can you please elaborate !!

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by vk_vinayak » Mon Sep 17, 2012 3:20 am
avik.ch wrote:So we can reasonably conclude that the OA is wrong.

Thanks !!
I searched for this question, and found that OA is technically correct, but the question is incomplete. Refer Stewart's explanation at: https://www.beatthegmat.com/a-data-from- ... 81048.html

B is the logical answer.
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by imskpwr » Mon Sep 17, 2012 6:05 am
avik.ch wrote:If 3^a4^b = c, what is the value of b?

(1) 5a = 25

(2) c = 36

OA = C

I think the second statement is sufficient.

Source : Kaplan
At first, I was a bit skeptical as we don't know anything about the RANGE of a and b, but now i am sure now that it's a trick q.

I mean a and b could be integers or non integers; rationals or irrationals;etc.

But once we know that a=2(integer), we can say Definitely that b has to be an integer to satisfy the equation(3^a4^b = c) for c=36.

Hence, BOTH STATEMENTS ARE REQUIRED TO ANSWER.
From 2, we cannot deduce that a and b are only "whole numbers". There may be other values which can satisfy the equation....this would be easier if you think of a solution where a and b are not whole numbers.

This must be a VERY HIGH LEVEL Q.....probably when one is in 50s....and when you are at that level you don't get peanuts so beware of such qs at that level.
I don't think GMAT ask such qs, But Indian CAT do throw such q at you.

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by everything's eventual » Mon Sep 17, 2012 5:44 pm

everything's eventual wrote:
Statement 2) says c = 36 = 3^2 * 2^2

Equating both sides we get (5^4)^b = 2

Then we get b = 0 and 2) should be sufficient.

I am not sure how you are getting b=0 ?? Can you please elaborate !!


Hello, I messed up the whole question. Sorry. I had a brain freeze.

First of all I understood the question as follows :

[3 ^ (a^4)]^ b = c

Statement 2) says c = 36 = 3^2 * 2^2

So comparing original equation and value of C, we get (a^4)^b = 2

Now we cannot solve the above equation unless we know the value of a. From statement 1) we get a= 5.

Therefore, (5^4)^b = 2 and so b = 0.

So yes, according to my flawed explanation the answer is C.