p percent of newspapers were A

Let T be the total newspapers

No of A newspapers = percent of A newspapers(p percent given) * T
= p/100*T = pT/100
Revenue from all A newspapers
= No of A newspapers * cost of each A newspaper
= pT/100 * 1
= pT/100 (1)

Therefore 100-p percent is the B newspapers

No of B newspapers = percent of B newspapers(p percent given) * T
= 100-p /100*T

Revenue from all B newspapers
= No of B newspapers * cost of each B newspaper
= 100-p/100*T * 1.25 (2)

Total Revenue = pT/100 + (100-p) /100 * T * 1.25
(Adding (1) and (2))

r is the percentage of revenue obtained from A newspapers

= Revenue from all A newspapers / Total Revenue * 100

= [pT/100] / [pT/100 + (100-p) /100 * T * 1.25 ] * 100


Take T/100 OUT AS COMMON FACTOR FROM THE NUMERATOR AND DENOMINATOR

T/100 ( P ) DIVIDED BY T/100 (P+ ((100-P)*1.25)) * 100

T/100 cancels out

so we are left with P / P+125-1.25P *100
= 100P / 125-.25P

Multiplying the numerator and denominator by 4 we get
D) 400P / 500-P

Did anyone try this by substituting numbers?

I did and got the D as ans. i felt my method was easier

what is your method?

i took 10 newspaper of A = 10$
4 newspaper of B = 15$
so, 14 newspaper =15$

r =revenue from A = 10/15*100 =66.6%
p= 10/14*100 =71%(Approx)
So. I putted the values of P in the eqn and tried to find out the r(Both in %)

d. ( 400*71)/429 ~ (i took it as 430)
caclualte it and comes out 66.4
As both the figures are in % . so this is correct

this is a really tough question..was this in not official gmat material like
mgmat or kaplan?

Let No of papers of A type = A
revenue from A type = A $

Let No of papers of B type = B
revenue from B type = 1.25B $


r (A + 1.25B) / 100 = A

p (A + B) / 100 = A

Solve both and you will have your ans

rA + 1.25rB = 100A

(100-r)A = 1.25rB

B/A = (100-r)/1.25r ---------------1


pA + pB = 100A
pB = (100-p)A

B/A = (100-p)/p ---------------2


equate both B/A
(100-r)/5r = (100-p)/4p

20/r - 1/5 = (100-p)/4p

20/r = (100-p)/4p + 1/5

20/r = (500-p)/20p

r = 400p/(500-p)

D ans

Hope this helps

I have a question for the last solution...

When the equations are set equal each other... why is the denominator r multiplied by 5 and the p by 4?

Thank you

A and B is the opnly new paper the shop keeper sold

so i plugin numbers ! (percentages any numebr wil do ) but i always take multiles of 10

total revenue is 50

r =50% and r =25% that is 25 newspapers

that is 25$ he earned by newpaper A

and and rest 50% that is p=50% from B that is numeber of B news papers is 20

so subsitute back p=20 and u sghould get r to be equal to 25


co incidence i always start with D (Manhatan)

400*20/500-20)

400*20/480

-->(20*20)/5
(5*20)/6
which satisfies the coindtion !!

this took more than 3 minutes !!

wats the source ???

Vishu

650 or bust wrote:I have a question for the last solution...

When the equations are set equal each other... why is the denominator r multiplied by 5 and the p by 4?

Thank you

1.25 is actually 5/4

(100-r)/1.25r
will become

4*(100-r)/5r

nwo take this 4 to denom of other expr

PS: I actually divided both equations by 4, which is same as we did above

Hope this helps!!

gmat009 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)

OA is [spoiler]"D". [/spoiler]Plz. explain

LETS KEEP IT SIMPLE! 8)

[ P x 1 / ( 100 -P) 1.25 ] = R /(100-R)

R = 400 P / (500 – P)

Hint: This is a GMAT test, not a MATH midterm. Keep things simple! Keep things under 2 minutes!