Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 16 Apr 2014, 17:18

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# 20 / 2^5 = (1/ 2^m ) + ( 1 / 2^n ) What is m+n ? A) 2 b) 3

Author Message
TAGS:
Director
Joined: 01 Jan 2008
Posts: 513
Followers: 3

Kudos [?]: 43 [0], given: 0

20 / 2^5 = (1/ 2^m ) + ( 1 / 2^n ) What is m+n ? A) 2 b) 3 [#permalink]  22 Jun 2008, 22:53
20 /2^5 = (1/2^m) + ( 1 /2^n)

What is m+n ?

A) 2
b) 3
c)4
d) 5
e) Cannot be determined
Manager
Joined: 24 Apr 2008
Posts: 162
Followers: 1

Kudos [?]: 34 [0], given: 0

Re: Exponents [#permalink]  22 Jun 2008, 23:10
20/2^5 = 1/2^m + 1/2^n

or can be written as 5/2^3 = 1/2^m + 1/2^n
now 5 can be broken into 1+4 (other ways are not possible - 2+3 or 5+1 does not allow this eq.)

Hence m+n = 3+1 = 4
SVP
Joined: 17 Jun 2008
Posts: 1580
Followers: 10

Kudos [?]: 167 [0], given: 0

Re: Exponents [#permalink]  23 Jun 2008, 00:11
I think the answer should be (E) - cannot be determined.

If I take any of the values in the other four options and try to satisfy the equation, it does not hold true.
Senior Manager
Joined: 26 Jan 2008
Posts: 267
Followers: 2

Kudos [?]: 68 [0], given: 1

Re: Exponents [#permalink]  23 Jun 2008, 00:15
bhatiagp wrote:
20 /2^5 = (1/2^m) + ( 1 /2^n)

What is m+n ?

A) 2
b) 3
c)4
d) 5
e) Cannot be determined

20 / (2^5)
= 5/8
= 1/2 + 1/8
= 1/(2^1) + 1/(2^3)
=> m=1 & n=3

Hence (C)
_________________
Senior Manager
Joined: 07 Jan 2008
Posts: 419
Followers: 3

Kudos [?]: 55 [0], given: 0

Re: Exponents [#permalink]  23 Jun 2008, 00:17
Choose E.

20/2^5 = 13/2^5+7/2^5=5/2^3 = 1/2^3+1/2^1=1/2+3/2^3 = etc
Director
Joined: 01 Jan 2008
Posts: 513
Followers: 3

Kudos [?]: 43 [0], given: 0

Re: Exponents [#permalink]  23 Jun 2008, 00:42
Thanks .. The Answer is C= 4
Senior Manager
Joined: 07 Jan 2008
Posts: 419
Followers: 3

Kudos [?]: 55 [0], given: 0

Re: Exponents [#permalink]  23 Jun 2008, 01:58
CAN YOU POST THE OE? AS I MENTIONED, WE CAN EXTRACT INTIMAL FRACTION INTO ANY OTHER FRACTIONS.
VP
Joined: 18 May 2008
Posts: 1302
Followers: 10

Kudos [?]: 78 [0], given: 0

Re: Exponents [#permalink]  23 Jun 2008, 02:22
20/2^5=1/2^m + 1/2^n
now 20=4*5 can be written as 2^2*5
Thus 5/2^3
again 5 can be written as 4+1=2^2+1
Thus 2^2/2^3 +1/2^3=1/2 +1/2^3
Comparing m=1 and n=3
Thus m+n=1+3=4
Hence C
Senior Manager
Joined: 07 Jan 2008
Posts: 419
Followers: 3

Kudos [?]: 55 [0], given: 0

Re: Exponents [#permalink]  23 Jun 2008, 08:49
ritula wrote:
20/2^5=1/2^m + 1/2^n
now 20=4*5 can be written as 2^2*5
Thus 5/2^3
again 5 can be written as 4+1=2^2+1
Thus 2^2/2^3 +1/2^3=1/2 +1/2^3
Comparing m=1 and n=3
Thus m+n=1+3=4
Hence C

4+1=2^2+1 = 2+3
Thus 2/2^3 +3/2^3=1/2^2 +3/2^3 ==> m+n#4
Current Student
Joined: 12 Jun 2008
Posts: 288
Schools: INSEAD Class of July '10
Followers: 5

Kudos [?]: 33 [0], given: 0

Re: Exponents [#permalink]  23 Jun 2008, 09:04
lexis wrote:
4+1=2^2+1 = 2+3
Thus 2/2^3 +3/2^3=1/2^2 +3/2^3 ==> m+n#4

You don't satisfy here the form given in the question, which is \frac{1}{2^m}+\frac{1}{2^n}

(it is ONE over 2^n, not THREE over 2^n)
SVP
Joined: 30 Apr 2008
Posts: 1893
Location: Oklahoma City
Schools: Hard Knocks
Followers: 27

Kudos [?]: 398 [0], given: 32

Re: Exponents [#permalink]  23 Jun 2008, 09:32
I approched it like this: \frac{1}{?} + \frac{1}{?} = \frac{5}{8}

\frac{20}{2^5} = \frac{20}{32} = \frac{5}{8}

When adding fractions, the denominator needs to be the same, so we need to figure up all combinations that add up to 5 for the numerator

1+4
2+3
(the inverse of each doesn't matter as it will be the same fractions adding together in a different order).

So

\frac{1}{8} + \frac{4}{8} = 5/8 AND \frac{2}{8} + \frac{3}{8} = 5/8

We have to determine which of these fractions will allow us to reduce down to each having 1 as a numerator. That is \frac{1}{8} + \frac{4}{8} = 5/8 becomes \frac{1}{8} + \frac{1}{2} = 5/8.

Now we have to break the numerator down into base of 2 with exponents.

\frac{1}{2^1} + \frac{1}{2^3} = \frac{5}{8}

This gives us 1 and 3 so 1 + 3 for m + n = 4, or C

Much of this should be done in your head rather than on paper. THe explanation is long just to make sure everyone understands it. Speed it up for the actual test or you'll be too far behind!!

bhatiagp wrote:
\frac{20}{2^5} = \frac{1}{2^m} + \frac{1}{2^n}
What is m+n ?

A) 2
b) 3
c)4
d) 5
e) Cannot be determined

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

Last edited by jallenmorris on 23 Jun 2008, 09:36, edited 2 times in total.
Re: Exponents   [#permalink] 23 Jun 2008, 09:32
Similar topics Replies Last post
Similar
Topics:
2^(2m+1) = 2^(n+2) what is m+n ? a) if 2^(3n-1)=256 b) if 4 26 Jan 2005, 20:14
Is 3^m(1/2)^n>1 1) n=2m 2) n=2 5 03 Feb 2005, 13:22
20/2^5=1/2^m+1/2^n,where m and n are integers, mn= 2 10 Mar 2008, 01:03
1 20/2^5=1/2^m+1/2^n,where m and n are integers, mn= 8 15 Aug 2008, 19:26
1 m^2+n^2=5k+b ; b=0? I. m-n=5z+1 II. m+n=5t+3 10 19 Nov 2008, 12:54
Display posts from previous: Sort by