Tough problem from tests : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 20 Feb 2017, 11:14

### 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

# Tough problem from tests

Author Message
TAGS:

### Hide Tags

Manager
Joined: 17 Aug 2009
Posts: 235
Followers: 5

Kudos [?]: 240 [0], given: 25

### Show Tags

16 Dec 2009, 04:08
00:00

Difficulty:

(N/A)

Question Stats:

29% (00:00) correct 71% (02:41) wrong based on 9 sessions

### HideShow timer Statistics

Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for$1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
A. 100p / (125 – p)
B. 150p / (250 – p)
C. 300p / (375 – p)
D. 400p / (500 – p)
E. 500p / (625 – p)

What is the simplest way to solve this??
Senior Manager
Joined: 30 Aug 2009
Posts: 286
Location: India
Concentration: General Management
Followers: 3

Kudos [?]: 164 [0], given: 5

Re: Tough problem from tests [#permalink]

### Show Tags

16 Dec 2009, 05:09
zaarathelab wrote:
Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for$1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
A. 100p / (125 – p)
B. 150p / (250 – p)
C. 300p / (375 – p)
D. 400p / (500 – p)
E. 500p / (625 – p)

What is the simplest way to solve this??

by plugging in numbers....will go with D. 400p / (500 – p)
Manager
Joined: 25 Aug 2009
Posts: 175
Location: Streamwood IL
Schools: Kellogg(Evening),Booth (Evening)
WE 1: 5 Years
Followers: 12

Kudos [?]: 178 [2] , given: 3

Re: Tough problem from tests [#permalink]

### Show Tags

16 Dec 2009, 14:26
2
KUDOS
Assume 1 copy of newspaper A was sold and 0 copies of newspaper B was sold
P = Percentage of Copies A sold = 100
R = Percentage of Revenue from A = 100
By substituting the value of P we easily know that D is the only answer choice to equal 100.

Hence D.

I feel bad about disrespecting such a fabulous question so I will provide the complete solution as well, if it helps anyone -

number of copies of Newspaper A sold = a.
number of copies of Newspaper B sold = b.

P/100=a/(a+b)
100/P = (a+b)/a
100/P = 1+b/a
b/a = 100/P -1

R/100=a/(a+1.25b)
100/R = (a+1.25b)/a
100/R = 1+5b/4a
substituting value of b/a
100/R = 1+125/P - 5/4
100/R = 125/P-1/4
100/R = (500-P)/4P
R=400P/(500-P)
_________________

Rock On

Intern
Joined: 23 Dec 2009
Posts: 27
Followers: 0

Kudos [?]: 83 [0], given: 7

Re: Tough problem from tests [#permalink]

### Show Tags

29 Dec 2009, 09:31
atish wrote:
Assume 1 copy of newspaper A was sold and 0 copies of newspaper B was sold
P = Percentage of Copies A sold = 100
R = Percentage of Revenue from A = 100
By substituting the value of P we easily know that D is the only answer choice to equal 100.

Hence D.

I feel bad about disrespecting such a fabulous question so I will provide the complete solution as well, if it helps anyone -

number of copies of Newspaper A sold = a.
number of copies of Newspaper B sold = b.

P/100=a/(a+b)
100/P = (a+b)/a
100/P = 1+b/a
b/a = 100/P -1

R/100=a/(a+1.25b)
100/R = (a+1.25b)/a
100/R = 1+5b/4a
substituting value of b/a
100/R = 1+125/P - 5/4
100/R = 125/P-1/4
100/R = (500-P)/4P
R=400P/(500-P)

I think that a more intuitive way to solve the problem will be:

consider n=total copies sold
p=percentage of copy sold that were A
1-p=percentage of copy sold that were B
therefore
r= 1*p*n/(1*p*n+1.25*(1-p)*n)=p/(p+1.25-1.25p)=p/(1.25-0.25p)
multiply the solution for 400 and obtain
r=400p/(500-100p)
Intern
Joined: 12 Oct 2009
Posts: 16
Followers: 1

Kudos [?]: 3 [0], given: 1

Re: Tough problem from tests [#permalink]

### Show Tags

05 Jan 2010, 22:25
I solved the problem by assuming that the store sold 100 newspapers. 50 NewspaperA and 50 NewspaperB. This means that "P"= 50%.

Next I calculated the Revenue.

Newspaper A 50x$1.00 =$50
Newspaper B 50x$1.25 =$62.50
Total Revenue $112.50 After that I found "r" r =$50/\$112.50 = .44444 x 100 = 44.4444%

Once I found "p" and "r", I just substituted "p" until one of the answers gave me the correct "r".

If you try plugging the answer into "A" first you will notice that 500/75 is too small to equal "r". So you can skip down to "E" or "D". Be careful with "D" because "D" says that "r" would equal 43.478. When you do the division you should get exactly 44.444, therefore the answer is "D".
Intern
Joined: 20 Dec 2009
Posts: 14
Followers: 1

Kudos [?]: 30 [0], given: 5

Re: Tough problem from tests [#permalink]

### Show Tags

06 Jan 2010, 02:08
It's a straight forward question which I think should be dealt straight to arrive at the solution without any complexity.

Because the question deals with percent and only involves newspaper A, assuming the total number of papers sold i.e., A+B = 100 will make it very easy to understand as well as solve.

So, let total number of papers be 100.
Hence, no. of newspaper A sold = p% of 100 = p
and no. of newspaper B sold = 100 - p.

Now r% of total revenue from papers comes from selling A.
Here total revenue from papers = p * 1 + (100 - p) * 1.25 = 125 - 0.25p
[revenue from A = p * 1 and from B = (100 - p) * 1.25]

Now %age of total revenue coming from A = revenue from A /total revenue * 100
= p/(125 - 0.25p) * 100
But it's already given that p/ (125-0.25p) *100 = r
Since all the answer options have p in the denominator, we'll have to remove 0.25 from the denominator. Simple observation leads us that multiplying by 400 will remove 0.25.
Hence, r = 400p/(500 - p). Ans (d)

P.S. I have elaborated the solving method here else this could be solved in 4 lines. Also, if you are expert in maths then as Atish has mentioned in his shortcut solution assuming 0 papers for B will give you the answer in seconds.
Manager
Joined: 17 Jan 2010
Posts: 149
Concentration: General Management, Strategy
GPA: 3.78
WE: Engineering (Manufacturing)
Followers: 1

Kudos [?]: 76 [0], given: 11

Re: Tough problem from tests [#permalink]

### Show Tags

21 Jan 2010, 14:05
atish wrote:
Assume 1 copy of newspaper A was sold and 0 copies of newspaper B was sold
P = Percentage of Copies A sold = 100
R = Percentage of Revenue from A = 100
By substituting the value of P we easily know that D is the only answer choice to equal 100.

Hence D.

I feel bad about disrespecting such a fabulous question so I will provide the complete solution as well, if it helps anyone -

number of copies of Newspaper A sold = a.
number of copies of Newspaper B sold = b.

P/100=a/(a+b)
100/P = (a+b)/a
100/P = 1+b/a
b/a = 100/P -1

R/100=a/(a+1.25b)
100/R = (a+1.25b)/a
100/R = 1+5b/4a
substituting value of b/a
100/R = 1+125/P - 5/4
100/R = 125/P-1/4
100/R = (500-P)/4P
R=400P/(500-P)

If the first approach is used (B) would also lead to the r=100...
Re: Tough problem from tests   [#permalink] 21 Jan 2010, 14:05
Similar topics Replies Last post
Similar
Topics:
1 average problem tough 1 18 May 2011, 14:12
13 Tough Coordinate Geometry problem from the gmac test 5 10 Jul 2010, 18:17
3 Really Tough Number Theory Problem 11 19 Apr 2010, 06:38
441 TOUGH & TRICKY SET Of PROBLEMS 165 12 Oct 2009, 08:22
9 Tough and tricky 4: addition problem 9 11 Oct 2009, 16:42
Display posts from previous: Sort by