BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Exponents and Algebra questions

Expert replies
by bkk_marc » Sun Nov 21, 2010 1:58 am
Can anyone help me out with good explanation? Thanks in advance.

1. (1/5)^m (1/4)^18 = 1/2(10)^35

What is the value of m?

- 17
-18
-34
-35
-36

35



2. XY + Z = X(Y+Z), which of the following must be true?

-X=0, Z=0
-X=1, Y=1
-Y=1, Z=0
-X=1,Y=0
-X=1,Z=0

[spoiler] X=1,Z=0[/spoiler]
Dum vivimus, vivamus....
Join the discussion
Source: — Problem Solving |

by shovan85 » Sun Nov 21, 2010 2:15 am
bkk_marc wrote:Can anyone help me out with good explanation? Thanks in advance.

1. (1/5)^m (1/4)^18 = 1/2(10)^35

What is the value of m?

- 17
-18
-34
-35
-36

35
(1/5)^m (1/4)^18 = 1/2(10)^35

Take Left hand side of the equation and try to get all the base in primary digits

[( 5 ^ -1 ) ^ m ] * [(4 ^ -1) ^ 18]

= (5 ^ -m) * [(2 ^ -2) ^ 18 ]

= (5 ^ -m) * (2 ^ -36)

Now take Right hand side and factorize up to primary digits

1/2(10)^35

= 1 / [2 * (2*5) ^ 35]

= 1 / [(2 ^ 36) * (5 ^ 35)]

= (2 ^ -36) * (5 ^ -35)

Now compare LHS = RHS

=> (5 ^ -m) * (2 ^ -36) = (2 ^ -36) * (5 ^ -35)

=> -m = -35

=> m = 35
If the problem is Easy Respect it, if the problem is tough Attack it
Join the discussion

by shovan85 » Sun Nov 21, 2010 2:19 am
bkk_marc wrote:
2. XY + Z = X(Y+Z), which of the following must be true?

-X=0, Z=0
-X=1, Y=1
-Y=1, Z=0
-X=1,Y=0
-X=1,Z=0

[spoiler] X=1,Z=0[/spoiler]
XY + Z = X(Y+Z)

=> XY + Z = XY + XZ

=> (XY + Z) - (XY + XZ) = 0

=> Z - XZ = 0

=> Z (1 - X) = 0

Thus, Z = 0 or (1 - X) = 0

=> Z = 0 or X = 1
If the problem is Easy Respect it, if the problem is tough Attack it
Join the discussion

by bkk_marc » Sun Nov 21, 2010 5:06 am
First off, thank you for the clear explanation. I was just too narrow minded on plugging in and didn't consider doing it algebraically..

However, I have a question on the first question.

How did you factorized 1 / [2 * (2*5) ^ 35] to 1 / [(2 ^ 36) * (5 ^ 35)] ?

I understand everything else before and after this step.. i just dont see how you can make 2^36 from [2*(2*5)^35]

Thanks again for the clear steps
Dum vivimus, vivamus....
Join the discussion

by shovan85 » Sun Nov 21, 2010 7:49 am
bkk_marc wrote: How did you factorized 1 / [2 * (2*5) ^ 35] to 1 / [(2 ^ 36) * (5 ^ 35)] ?
1 / [2 * (2*5) ^ 35]

= 1 / [ (2 ^ 1) * { (2*5) ^ 35 } ]

= 1 / [ (2 ^ 1) * (2 ^35) * (5 ^ 35)]

= 1 / [ (2 ^ { 1 + 35 }) * (5 ^ 35)]

= 1 / [ (2 ^ 36) * (5 ^ 35)]
If the problem is Easy Respect it, if the problem is tough Attack it
Join the discussion

by bkk_marc » Sun Nov 21, 2010 7:52 am
AHh HA!

I understand :) Thanks!!!

I am a little on the slow side :(
Dum vivimus, vivamus....
Join the discussion

by Brian@VeritasPrep » Mon Nov 22, 2010 11:56 am
Hey guys,

Nice work on this problem! Just in response to the title of this post, I want to emphasize that there are three guiding principles for working with Exponents and Algebra on this test:

1) Find common bases, which generally means that you need to factor out bases into prime factors. That was a big key here.

2) Multiply, which means that a problem that involves addition or subtraction will almost always require you to factor out common terms to turn it into a multiplication problem.

3) Look for patterns, which you'll often see when the digits or values themselves are in play.


If you keep these three principles in mind, there shouldn't be any GMAT exponent-based problems that you can't solve.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

Looking for GMAT practice questions? Try out the Veritas Prep Question Bank. Learn More.
Join the discussion