Last visit was: 24 Apr 2026, 07:27 It is currently 24 Apr 2026, 07:27
Home
Messages
Tests
Schools
Events
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,814
Own Kudos:
811,002
 [4]
Given Kudos: 105,871
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,814
Kudos: 811,002
 [4]
15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}} Updated on: 25 Oct 2025, 02:13
Originally posted by Bunuel on 02 Sep 2015, 22:28.
Last edited by Bunuel on 25 Oct 2025, 02:13, edited 1 time in total.
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

80% (01:44) correct 20% (02:23) wrong based on 260 sessions

History

Date
Time
Result
Not Attempted Yet
\(\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}\)

(A) \(2^{7}\)
(B) \(2^{8}\)
(C) \(2^{9}\)
(D) \(15(2^{7})\)
(E) \(15 (2^{8})\)
ShowHide Answer
Official Answer
B
User avatar
KS15
User avatar
Joined: 21 May 2013
Last visit: 25 Jul 2019
Posts: 531
Own Kudos:
259
 [1]
Given Kudos: 608
Posts: 531
Kudos: 259
 [1]
15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}} 02 Sep 2015, 22:58
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
\(\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}\)

(A) \(2^{7}\)
(B) \(2^{8}\)
(C) \(2^{9}\)
(D) \(15(2^{7})\)
(E) \(15 (2^{8})\)


Kudos for a correct solution.
Show more

Denominator=1/2^5+2^6+2^7+2^8=1/2^3+2^2+2+1/2^8
Fraction becomes=15*2^8/15=2^8
Answer B
User avatar
AbdurRakib
User avatar
Joined: 11 May 2014
Last visit: 03 Mar 2026
Posts: 464
Own Kudos:
43,752
 [3]
Given Kudos: 220
Status:I don't stop when I'm Tired,I stop when I'm done
Location: Bangladesh
Concentration: Finance, Leadership
Schools: Wharton '20 Yale '20 Tepper '20
GPA: 2.81
WE:Business Development (Real Estate)
Schools: Wharton '20 Yale '20 Tepper '20
Posts: 464
Kudos: 43,752
 [3]
15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}} 03 Sep 2015, 00:02
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Simplifying the denominator of this fraction in necessary to calculate,as below

\(\frac{15}{2^-5+2^-6+2^-7+4^-4}\)

=\(\frac{15}{2^-5+2^-6+2^-7+2^-8}\)
=\(\frac{15}{2^-5(1+2^-1+2^-2+2^-3)}\)
=\(\frac{15}{2^-5(1+1/2+1/4+1/8)}\)
=\(\frac{15}{2^-5(15/8)}\)
=\(\frac{15}{2^-5(15/2^3)}\)
=\(\frac{15}{2^-8(15)}\)
=\(\frac{1}{2^-8}\)
=2^8

So,the Correct answer is B
User avatar
kunal555
User avatar
Joined: 29 Jul 2015
Last visit: 17 Jun 2019
Posts: 143
Own Kudos:
781
 [2]
Given Kudos: 59
Posts: 143
Kudos: 781
 [2]
15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}} 03 Sep 2015, 08:27
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
\(\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}\)

(A) \(2^{7}\)
(B) \(2^{8}\)
(C) \(2^{9}\)
(D) \(15(2^{7})\)
(E) \(15 (2^{8})\)


Kudos for a correct solution.
Show more

\(\frac{15}{({2^{-5}+2^{-6}+2^{-7}+4^{-4}})}\)

= \(\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+2^{-8}}}}\)

= \(\frac{15}{{2^{-5}({{1+2^{-1}+2^{-2}+2^{-3}})}\)

= \(\frac{(2^5)(15)}{(1 + 1/2 + 1/4 + 1/8)}\)

=\(\frac{(2^5)(15)}{(15/8)}\)

= \(8(2^5)\)

= \((2^3)(2^5)\)

= \(2^8\)

Answer:- B
User avatar
peachfuzz
User avatar
Joined: 28 Feb 2014
Last visit: 27 Jan 2018
Posts: 268
Own Kudos:
369
 [1]
Given Kudos: 132
Location: United States
Concentration: Strategy, General Management
Products:
My Reviews
Posts: 268
Kudos: 369
 [1]
15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}} 04 Sep 2015, 14:30
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
\(\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}\) =?

The denominator becomes 2^-5(15/8)

15*(8/15)*2^-5
=2^8

Answer:
(B) \(2^{8}\)
User avatar
anudeep133
User avatar
Joined: 10 Aug 2015
Last visit: 14 Dec 2018
Posts: 94
Own Kudos:
282
 [1]
Given Kudos: 20
Posts: 94
Kudos: 282
 [1]
15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}} 06 Sep 2015, 07:58
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Solution:

Denominator = \(2^{-5} + 2^{-6} + 2^{-7} + 2^{-8}\) = \(2^{-8}[2^{3} + 2^{2} + 2 + 1]\) = \(2^{-8}(8+4+2+1)\) = \(2^{-8}(15)\)

Therefore, ans = \(\frac{(15)}{(2^{-8}(15))}\) = \(2^{8}\)

Option B
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,814
Own Kudos:
811,002
Given Kudos: 105,871
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,814
Kudos: 811,002
15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}} 07 Sep 2015, 03:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
\(\frac{15}{{{2^{-5}+2^{-6}+2^{-7}+4^{-4}}}}\)

(A) \(2^{7}\)
(B) \(2^{8}\)
(C) \(2^{9}\)
(D) \(15(2^{7})\)
(E) \(15 (2^{8})\)


Kudos for a correct solution.
Show more

VERITAS PREP OFFICIAL SOLUTION:

When you face exponents and addition, you’ll almost always need to factor out common exponential terms so that you can multiply, which is the operation that lends itself to more flexibility with exponent rules. Here, you can first take \(4^{-4}\) and express it as \((2^{2})^{-4}\), which equals \(2^{-8}\), so that you have all common bases in the denominator.

Then, in order to multiply, factor out a \(2^{-5}\) from each term to get: \(2^{-5} (1 + 2^{-1} + 2^{-2} + 2^{-3})\). The \(2^{-5}\) is now a multiplicative term, which you can express also as \(\frac{1}{(2^{5})}\), allowing you to flip it to the numerator, which becomes \(2^{5}* 15\).

Back to the denominator, you can express \((1 + 2^{-1} + 2^{-2} + 2^{-3})\) as \(1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8}\), which equals \(1 + \frac{7}{8} = \frac{15}{8}\).

Putting the entire fraction together, the fraction is \(\frac{15(2^{5})}{(\frac{15}{8})}\). The 8 can be expressed as \(2^{3}\), again providing a common exponential base, making the fraction \(\frac{15(2^{5})}{\frac{15}{2^{3}}}\). Dividing by a fraction is the same as multiplying by its reciprocal, leaving \(\frac{15(2^{5}) * 2^{3}}{15}\). The 15s cancel, leaving \(2^{5} * 2^{3}\) which equals \(2^{8}\), answer choice B.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,972
Own Kudos:
1,117
Posts: 38,972
Kudos: 1,117
15 / {{2^{-5}+2^{-6}+2^{-7}+4^{-4}}} 25 Oct 2025, 02:10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109814 posts
Krunaal
Tuck School Moderator
853 posts