GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Oct 2019, 14:42

### 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

# If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 22 Jun 2014
Posts: 1093
Location: India
Concentration: General Management, Technology
GMAT 1: 540 Q45 V20
GPA: 2.49
WE: Information Technology (Computer Software)
If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

### Show Tags

18 Dec 2018, 10:26
2
2
00:00

Difficulty:

15% (low)

Question Stats:

80% (01:59) correct 20% (02:11) wrong based on 166 sessions

### HideShow timer Statistics

If $$r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$ and $$s = 1 + \frac{1}{3}r$$, then $$s$$ exceeds $$r$$ by

(A) $$\frac{1}{3}$$
(B) $$\frac{1}{6}$$
(C) $$\frac{1}{9}$$
(D) $$\frac{1}{27}$$
(E) $$\frac{1}{81}$$

Project PS Butler : Question #85

Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS

_________________
Director
Status: Manager
Joined: 27 Oct 2018
Posts: 671
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)
If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

### Show Tags

Updated on: 31 Jul 2019, 17:40
1
we can rewrite R and S as the following:
$$R = 3^0 + 3^{-1} + 3^{-2} + 3^{-3}$$

$$S = 3^0 + 3^{-1}R$$
$$S = 3^0 + 3^{-1}(3^0 + 3^{-1} + 3^{-2} + 3^{-3}) = 3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}$$

upon subtraction,
$$S-R$$ = $$(3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4})$$ $$-(3^0 - 3^{-1} - 3^{-2} - 3^{-3})$$ $$= 3^{-4}$$

which is E
_________________
Thanks for Kudos

Originally posted by MahmoudFawzy on 18 Dec 2018, 12:57.
Last edited by MahmoudFawzy on 31 Jul 2019, 17:40, edited 1 time in total.
Intern
Joined: 14 Jan 2018
Posts: 2
Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

### Show Tags

19 Dec 2018, 08:52
1
r= 1 + 1/3 + 1/9 + 1/27
Apply G.P summation formula, sum1 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum1 = 13/9
s= 1 + 1/3(r)
i.e s = 1 + 1/3 + 1/9 + 1/27 + 1/81
Apply G.P summation formula, sum2 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum2 = 40/27
Do sum1 - sum2 = 1/27

Option (D)
Director
Status: Manager
Joined: 27 Oct 2018
Posts: 671
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)
If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

### Show Tags

19 Dec 2018, 09:10
AmanAngrish wrote:
r= 1 + 1/3 + 1/9 + 1/27
Apply G.P summation formula, sum1 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum1 = 13/9
s= 1 + 1/3(r)
i.e s = 1 + 1/3 + 1/9 + 1/27 + 1/81
Apply G.P summation formula, sum2 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum2 = 40/27
Do sum1 - sum2 = 1/27

Option (D)

your concept is very interesting and correct,
however, I think there is a small mistake in the application.

I think n=4 for sum1 (as 4 elements are added),
and n=5 for sum 2 (as 5 elements are added).

sum1 would be $$40/27$$
sum2 would be $$121/81$$
and by subtraction, the result would be $$1/81$$
_________________
Thanks for Kudos
VP
Joined: 09 Mar 2016
Posts: 1230
If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

### Show Tags

19 Dec 2018, 12:01
1
$$r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$ =$$\frac{40}{27}$$

$$s = 1 + (\frac{1}{3} * \frac{40}{27})$$ = $$\frac{121}{81}$$

$$\frac{121}{81}$$ - $$\frac{40}{27}$$ = $$\frac{1}{81}$$

E
Director
Joined: 09 Mar 2018
Posts: 997
Location: India
Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

### Show Tags

12 Feb 2019, 08:08
HKD1710 wrote:
If $$r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$ and $$s = 1 + \frac{1}{3}r$$, then $$s$$ exceeds $$r$$ by

(A) $$\frac{1}{3}$$
(B) $$\frac{1}{6}$$
(C) $$\frac{1}{9}$$
(D) $$\frac{1}{27}$$
(E) $$\frac{1}{81}$$

Project PS Butler : Question #85

Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS

So i thought of some other method, it did not work

$$r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$

$$s = 1 + \frac{1}{3}r$$

Just substitute r in the second expression

s = 1 + 1/3 * $$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$

So it will give us 1/81

E
_________________
If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
Intern
Status: Active
Affiliations: NA
Joined: 27 Feb 2019
Posts: 25
Location: India
Schools: ISB '21, NUS '22, NTU '21
GPA: 3.8
WE: Business Development (Manufacturing)
Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by  [#permalink]

### Show Tags

31 Jul 2019, 07:21
[quote="HKD1710"]If $$r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$ and $$s = 1 + \frac{1}{3}r$$, then $$s$$ exceeds $$r$$ by

(A) $$\frac{1}{3}$$
(B) $$\frac{1}{6}$$
(C) $$\frac{1}{9}$$
(D) $$\frac{1}{27}$$
(E) $$\frac{1}{81}$$

$$s = 1 + \frac{1}{3}r$$ = $$1+ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}$$

Now, s-r = $$1+ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}$$ - ($$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}$$) = $$\frac{1}{81}$$
Re: If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by   [#permalink] 31 Jul 2019, 07:21
Display posts from previous: Sort by

# If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne