Last Sunday a certain store sold copies of Newspaper A for $1.00 each

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. \(\frac{100p}{(125 – p)}\)

B. \(\frac{150p}{(250 – p)}\)

C. \(\frac{300p}{(375 – p)}\)

D. \(\frac{400p}{(500 – p)}\)

E. \(\frac{500p}{(625 – p)}\)



OG 2019 PS03144

Hi JeffTargetTestPrep, when you multiply the step with 4/4, why is it only multipy with 1.25 but not (100-P) behind ? Could you explain? Thanks

When we multiply 1.25(100 - p) by 4, we obtain 4 * 1.25 * (100 - p). This is a product containing three factors, so we can write it as (4 * 1.25) * (100 - p) = 5 * (100 - p). If we were to multiply 100 - p by 4, we wouldn't be able to multiply 1.25 by 4; we would obtain 1.25 * 4(100 - p) = 1.25 * (400 - 4p), which does not help us in any way. I think you are thinking of distributing multiplication over addition where if we multiply a sum or a difference by a constant, we multiply each term of that sum or difference by the same constant, but we don't have it here. In this step, we are simply multiplying a product containing two factors with another constant to obtain 4 * (1.25 * (100 - p)). The associative property of multiplication allows us to calculate (4 * 1.25) * (100 - p) instead, which is how we obtain 5 * (100 - p).

Assume total 100 copies are sold

then P of newspaper A and 100-p or newspaper b are sold

rev generated by them P*1 and (100-p)*1.25

now % rev of A = (rev by A/total rev)*100

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

from here r = Option D

Let's use the INPUT-OUTPUT approach.

Let's say that Newspaper A accounted for 20% of all newspapers sold. In other words, p = 20
This means that Newspaper B accounted for 80% of all newspapers sold.

BrentGMATPrepNow
This is very helpful, thank you! I am just confused on why we have to

The question asks us to find the value of r, the percentage of newspaper revenue from Newspaper A.
To determine this, let's say that 100 newspapers we sold IN TOTAL.
This means that 20 Newspaper A's were sold and 80 Newspaper B's were sold.

REVENUE:
Newspaper A: 20 newspapers at $1 apiece = $20
Newspaper B: 80 newspapers at $1.25 apiece = $100
So, TOTAL revenue = $120

Since Newspaper A accounted for $20 of revenue, we can say that Newspaper A accounted for 16 2/3% of revenue. In other words, r = 16 2/3
Aside: We know this because $20/$120 = 1/6 = 16 2/3%

So, when we INPUT p = 20, the OUTPUT is r = 16 2/3.
We'll now plug p = 20 into each answer choice and see which one yields an output of = 16 2/3

A. 100(20)/(125 - 20) = 2000/105.
IMPORTANT: If we want, we can use long division to evaluate this fraction (and others), but we can save a lot of time by applying some number sense. Since 2000/100 = 20, we know that 2000/105 will be SLIGHTLY less than 20. So, we can be certain that 2000/105 does not equal 16 2/3. As such, we can ELIMINATE A.

B. 150(20)/(250 - 20) = 3000/230. We know that 3000/200 = 15, so 3000/230 will be less than 15. So, we can be certain that 3000/230 does not equal 16 2/3. As such, we can ELIMINATE B.

C. 300(20)/(375 - 20) = 6000/355. Hmmm, this one is a little harder to evaluate. So,we may need to resort to some long division (yuck!). Using long division, we get 6000/355 = 16.9.... ELIMINATE C.

D. 400(20)/(500 - 20) = 8000/480 = 800/48 = 100/6 = 50/3 = 16 2/3. perfect! KEEP

E. 500(20)/(625 - 20) = 10000/605 = 100/6.05. Notice that, above, we saw that 100/6 = 16 2/3. So, 100/6.05 will NOT equal 16 2/3. ELIMINATE E.

Answer:

BrentGMATPrepNow
This is very helpful, thank you!
I am confused on the wording of the question... "which of the following expresses r in terms of p"? Wouldn't we be looking for r=1/6 and not 16 2/3% because it is just asking for r? Overall, the "r" vs. "r percent" and "p" versus "p percent" is a bit confusing to me. Any clarification would be greatly appreciated. Thank you!

Given: r percent of the store’s revenues from newspaper sales was from Newspaper A

So, if r = 1/6, then you're saying 1/6 percent of the store’s revenues from newspaper sales was from Newspaper A.
In other words, 0.1666...% of the store’s revenues from newspaper sales was from Newspaper A.

Instead, we want to say that 16 2/3 percent of the store’s revenues from newspaper sales was from Newspaper A.

Does that help?

BrentGMATPrepNow
Thank you so much for your response. I still always struggle with picking numbers. To clarify, you wouldn't start with picking a number for "r" in this case because the question asks for "r" in terms of "p"? Also, how do you know not to test another number such as p=10 and solving for "r" again to make sure that choice D is correct. Do you have any best practices for dealing with this? Thank you again so much for your help.

Hi avigutman

Loved your approach to the question! This is my first time looking at a ratio multiplied by percent, therefore, I was wary at first. I tried using numbers for the variables p and r to see if this approach works or not and it does. On L.H.S. you did 4*p/ 500*(100-p) I know that we are calculating the ratio of the revenues of newspapers A and B.
If p were a number, then 4*p = Revenue, but p is percent so what does 4*p% or 500(100-p)% give us? Revenue in terms of percent?

Let the total copies of newspaper(A+B) sold be 100
so the number of copies of A sold is p
number of copies of B sold is 100-p
thus revenue from A = p*1$ = p$
revenue from B = (100-p)5/4; because 1.25 = 5/4
percent of revenue from A = r = p/p+[(100-p)5/4)]= 400p / (500 – p)