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confused about this problem..Power problem

Expert replies
by Mo2men » Fri Nov 11, 2016 1:47 am
If 10^a * 3^b * 5^C =450^n, what is the value of c?

(1) a is 1.
(2) b is 2.

Source: Veritas

How come the OA is E???

Thanks
Last edited by Mo2men on Thu May 18, 2017 3:15 am, edited 1 time in total.
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Source: — Data Sufficiency |

by GMATGuruNY » Fri Nov 11, 2016 5:00 am
Mo2men wrote:If 10^a * 3^b * 5^C =450^n, what is the value of c?

(1) a is 1.
(2) b is 2.
Statements combined:
Case 1: n=0, with the result that (10^a)(3^b)(5^c) = 1
Substituting a=1 and b=2 into (10^a)(3^b)(5^c) = 1, we get:
(10¹)(3²)(5^c) = 1
5^c = 1/90
c = a very ugly number.

Case 2: n=1, with the result that (10^a)(3^b)(5^c) = 450
Substituting a=1 and b=2 into (10^a)(3^b)(5^c) = 450, we get:
(10¹)(3²)(5^c) = 450
5^c = 5
c = 1.

Since c can be different values, the two statements combined are INSUFFICIENT.

The correct answer is E.
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by Jay@ManhattanReview » Thu Jan 05, 2017 5:20 am
Mo2men wrote:If 10^a * 3^b * 5^C =450^n, what is the value of c?

(1) a is 1.
(2) b is 2.

Source: Veritas

How come the OA is E???

Thanks
The trap in this question is your thinking that a, b, c, and n are integers. This is bolstered more by the information given in the statements: a = 1 and b = 2. However, we must not assume that a, b, c, and n all are integers.

We are given that 10^a * 3^b * 5^C = 450^n;

Doing prime factorization, we get 2^a * 3^b * 5^(a+c) = 2^n * 3^(2n) * 5^(2n)

The second trap is the value of n would only be determined by a, b, and/or c, i.e. n itself cannot have any value.

Let us discuss each statement one by one.

S1: a = 1

By 2^1 * 3^b * 5^(1+c) = 2^n * 3^(2n) * 5^(2n)

We see that the exponent (a=1) of 2 on the LHS should be equal to the exponent (n) of 2 on the RHS, thus n = 1, and 1 + c = 2n = 2*1 = 2 => c = 1. However, we cannot conclude that the unique value of c =1.

What is n = 0?

In that case, 2^1 * 3^b * 5^(1+c) = 1. The value of c would is indeterminable. Insufficient.

S2: We need not discuss S2. Its fate is the same as that of S1.

S1 and S2: We already have c = 1 from S1.

Let us find out its value if n = 0.

If a = 1 and b = 2, 2^1 * 3^2 * 5^(1+c) = 1

=> 2*9*5*5^c = 1

=> 90 * 5^c = 1 => 5^c = 1/90

=> c < 1; there is no need to calculate the value of c, we need to be assured that its other than 1.

=> c = 1 or c < 1. Insufficient.

Answer: E

Hope this helps!

-Jay
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by crackverbal » Thu Jan 05, 2017 10:21 pm
Hi Mo2men,

Here it is easy to jump into conclusion saying both statements together sufficient.

But that's the trick is, here they have not talked about whether a , b and c are integers.

So obviously each statements alone are not sufficient.

Considering together,

Just ask yourself there are 4 unknowns and you know the value of 2 unknowns from the statements.

But still nothing about c and n.

So just be careful, since no condition whether they are integers are not,

we can have n = 1, c = 1

or we can have n = 0, where we have a different c value.

So the answer is E.

Remember if nothing mentioned about the values, then the values could be any real number.

Hope this clear.
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by crackverbal » Thu Jan 05, 2017 10:22 pm
Hi Mo2men,

Here it is easy to jump into conclusion saying both statements together sufficient.

But that's the trick is, here they have not talked about whether a , b and c are integers.

So obviously each statements alone are not sufficient.

Considering together,

Just ask yourself there are 4 unknowns and you know the value of 2 unknowns from the statements.

But still nothing about c and n.

So just be careful, since no condition whether they are integers are not,

we can have n = 1, c = 1

or we can have n = 0, where we have a different c value.

So the answer is E.

Remember if nothing mentioned about the values, then the values could be any real number.

Hope this clear.
Join Free 4 part MBA Through GMAT Video Training Series here -
https://gmat.crackverbal.com/mba-throug ... video-2018

Enroll for our GMAT Trial Course here -
https://gmatonline.crackverbal.com/

For more info on GMAT and MBA, follow us on @AskCrackVerbal
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