GMAT Question of the Day

A store sells newspaper A at price of $1 and B at price of $1.25. In a certain day, if r percent of the total income is from A, and p percent of the total copies sold are A, in terms of r, p=?

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Take GMAT Team tagged this post with: GMAT Quantitative Descrete Questions, GMAT Question of the Day Read 1234 articles by Take GMAT Team

1. daveparthiv says:

   August 10, 2007 at 12:22 pm

   My answer is:

   r = 100p/(p +123.75)

   My method was:
   i) Let x be the total copies
   ii) copies of A = px/100
   iii) copies of B would be: (100-p)x/100

   iv) Total income from A = px/100 $
   v) Total income from B = (125x-1.25px)/100
   vi) Adding iv) and v) gives total income, which is: (125x+(p-1.25)x)/100

   Therefore, r = 100*px/(125x+px-1.25x)…….. since r is % of total income from A.
   r = px/(123.75x+px)
   = p/(p+123.75)
2. Jess says:

   August 10, 2007 at 2:44 pm

   Hi,

   I think best way wud be using the figures.
   Say there are 60 copies of each paper.
   By selling paper A, income received = 60$
   By selling paper B, income received = 75$
   Total income = 135$
   Now r% = 60/135 and p% = 75/135 = 5/4*r%..which is the ans reqd..
   You can verify the same with any number of papers.
3. Raj says:

   August 10, 2007 at 3:10 pm

   Suppose total 100 Copies.
   A – > P Copies
   B -> 100 – p Copies

   Total Income = P + (100 – P)1.25

   A’s Income i.e r % = (P+(100-P)1.25 )r / 100
4. DiD says:

   August 11, 2007 at 1:00 am

   say r= 50% then p = r * 1.25/(1+1.25)

   thus p = 5/9 % (cross checked)
5. DiD says:

   August 11, 2007 at 1:16 am

   sorry

   1.25r / (0.25r + 100)

   say 50% revenues from A then p % = 1.25(50) / (.25*50 + 100) = 62.5/112.5 = 5/9
6. AP says:

   August 11, 2007 at 7:57 pm

   Answer should be:

   p = 1.25r / (0.0025r + 1)
7. tapas says:

   September 20, 2007 at 2:39 pm

   I got the ans: 500r / (400+r)
8. Krishna says:

   September 10, 2009 at 6:38 pm

   My answer:

   Considering the ratio of incomes for B and A

   (100-p) * 1.25 / p = (100-r)/r

   => p = 500 r/(r+400)
9. Ankit says:

   September 10, 2009 at 10:43 pm

   is anybody able to see the options??
10. Binita says:

    September 11, 2009 at 5:22 pm

    GMAT TEAM plz give true answer
11. Binita says:

    September 11, 2009 at 7:37 pm

    Total Income of A = r * x (where x=Total Income)
    Total Income of B = (100 – r) * x

    Total Copies of A = p * y (where y = Total Copies)
    Total Copies of B = (100 – p) * y

    Now ,

    Total Income Of A = Total Copies of A * Amount of each copy (viz is 1)
    So , r*x = p * y * 1
    Hence , x/y = p/r —–(i)

    Similarly , Total Income of B = Total Copies of B * Amount of each Copy (viz 1.25)
    So , (100 – r) * x = (100 – p) * y
    Solve n u will get , x/y = (100 – p)(1.25) / (100 – r) —–(ii)

    Equate i n ii and solve u will get the answer as,

    p= (125 * r) / (100 – 0.25r)
12. Binita says:

    September 11, 2009 at 7:38 pm

    is it rt ?
13. Mahadev Suvarna says:

    September 11, 2009 at 10:14 pm

    Let “X” be the no. of A newspaper copies sold.
    Let “Y” be the no. of B newspaper copies sold.
    From the given statements:
    Eq.1:
    r(X+1.25Y)/100 = X
    Eq. 2
    p(X+Y)/100 = X

    SIMPLIFICATION GIVES p = 1.125 r
14. WT says:

    September 12, 2009 at 2:19 am

    p = 1.25r / (1 + 0.25r) if p and r are expressed as decimals from 0 to 1

    p = 125r / (100 + 0.25r) if p and r are expressed as percentages from 0 to 100
15. Binita says:

    September 14, 2009 at 4:28 pm

    whats the true answer????????????????
16. krishna74 says:

    September 18, 2009 at 8:01 pm

    p=500r/(400+r)
17. vd says:

    March 8, 2010 at 9:08 pm

    lets understand the solution:
    Assume,total no of copies sold=C
    copies sold for A=C*p
    copies sold for B=C*(100-p)
    Now,

    r = [(C*p*1$)/{C*p*1$+C*(100-p)*1.25$}]*100
    r= 500r/(400+r)