It is currently 20 Mar 2018, 23:52

### 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

# Last year the price per share of Stock X increased by k

Author Message
Director
Joined: 27 Jun 2008
Posts: 526
WE 1: Investment Banking - 6yrs
Last year the price per share of Stock X increased by k [#permalink]

### Show Tags

17 Sep 2008, 07:16
1
KUDOS
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Last year the price per share of Stock X increased by k percent and the earnings per share
of Stock X increased by m percent, where k is greater than m. By what percent did the
ratio of price per share to earnings per share increase, in terms of k and m?
A. k m %
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %

Whats the simplest way to work on such problems? Plugin numbers?
SVP
Joined: 07 Nov 2007
Posts: 1764
Location: New York

### Show Tags

17 Sep 2008, 08:26
1
KUDOS
pawan203 wrote:
Last year the price per share of Stock X increased by k percent and the earnings per share
of Stock X increased by m percent, where k is greater than m. By what percent did the
ratio of price per share to earnings per share increase, in terms of k and m?
A. k m %
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %

Whats the simplest way to work on such problems? Plugin numbers?

S2=S1(1+K/100)
E2=E1(1+M/100)

(S2/E2)= (S1/E1)*((1+K/100)/(1+M/100))
= (S1/E1)*(1+((K+100)/(M+100))-1)
=(S1/E1)*(1+ (K-M)/(M+100) )

100(k- m)] / (100 + m) %
_________________

Smiling wins more friends than frowning

Retired Moderator
Joined: 18 Jul 2008
Posts: 921

### Show Tags

17 Sep 2008, 16:23
(S2/E2)= (S1/E1)*((1+K/100)/(1+M/100))
= (S1/E1)*(1+((K+100)/(M+100))-1)
=(S1/E1)*(1+ (K-M)/(M+100) )

I do not understand how you went about it here (bold). Can you explain?

Thanks.

x2suresh wrote:
pawan203 wrote:
Last year the price per share of Stock X increased by k percent and the earnings per share
of Stock X increased by m percent, where k is greater than m. By what percent did the
ratio of price per share to earnings per share increase, in terms of k and m?
A. k m %
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %

Whats the simplest way to work on such problems? Plugin numbers?

S2=S1(1+K/100)
E2=E1(1+M/100)

(S2/E2)= (S1/E1)*((1+K/100)/(1+M/100))
= (S1/E1)*(1+((K+100)/(M+100))-1)
=(S1/E1)*(1+ (K-M)/(M+100) )

100(k- m)] / (100 + m) %
Director
Joined: 27 Jun 2008
Posts: 526
WE 1: Investment Banking - 6yrs

### Show Tags

18 Sep 2008, 14:07
Can you please elaborate? woudln't plugin in numbers be easier?
Manager
Joined: 11 Jan 2008
Posts: 54

### Show Tags

18 Sep 2008, 14:26
assume original price per share is 100 and earnings is also 100.
assume k=20 and m=10.

100/100 + X = 120/110 where X is the increase.

solving this we get X to be 1/11.

putting same values in D,

[100(k – m)] / (100 + m) %

100(10)/110 we also get 1/11.

so D.

Is this correct way of doing guys?
Current Student
Joined: 11 May 2008
Posts: 552

### Show Tags

18 Sep 2008, 19:36
jackychamp wrote:
assume original price per share is 100 and earnings is also 100.
assume k=20 and m=10.

100/100 + X = 120/110 where X is the increase.

solving this we get X to be 1/11.

putting same values in D,

[100(k – m)] / (100 + m) %

100(10)/110 we also get 1/11.

so D.

Is this correct way of doing guys?

any method which is fast and gives result is ok. UR method is also fine.
VP
Joined: 18 May 2008
Posts: 1204

### Show Tags

01 Oct 2008, 17:30
Actually It is % increase in S1/E1 = {(S1/E1)*((1+K/100)/(1+M/100))- (S1/E1)}/ (S1/E1)
NOw take (S1/E1) common and solve, u will get the answer

(S2/E2)= (S1/E1)*((1+K/100)/(1+M/100))
= (S1/E1)*(1+((K+100)/(M+100))-1)
=(S1/E1)*(1+ (K-M)/(M+100) )

I do not understand how you went about it here (bold). Can you explain?

Thanks.

x2suresh wrote:
pawan203 wrote:
Last year the price per share of Stock X increased by k percent and the earnings per share
of Stock X increased by m percent, where k is greater than m. By what percent did the
ratio of price per share to earnings per share increase, in terms of k and m?
A. k m %
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %

Whats the simplest way to work on such problems? Plugin numbers?

S2=S1(1+K/100)
E2=E1(1+M/100)

(S2/E2)= (S1/E1)*((1+K/100)/(1+M/100))
= (S1/E1)*(1+((K+100)/(M+100))-1)
=(S1/E1)*(1+ (K-M)/(M+100) )

100(k- m)] / (100 + m) %
VP
Joined: 17 Jun 2008
Posts: 1328

### Show Tags

01 Oct 2008, 22:31
pawan203 wrote:
Last year the price per share of Stock X increased by k percent and the earnings per share
of Stock X increased by m percent, where k is greater than m. By what percent did the
ratio of price per share to earnings per share increase, in terms of k and m?
A. k m %
B. (k – m) %
C. [100(k – m)] / (100 + k) %
D. [100(k – m)] / (100 + m) %
E. [100(k – m)] / (100 + k + m) %

Whats the simplest way to work on such problems? Plugin numbers?

1+k/100 /1+m/100 *100 is the answer !!!!
IMO D
_________________

cheers
Its Now Or Never

Re: PS - Q17   [#permalink] 01 Oct 2008, 22:31
Display posts from previous: Sort by