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Last Sunday a certain store sold copies of newspaper a for..

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Last Sunday a certain store sold copies of newspaper a for..

by AAPL » Fri Mar 16, 2018 5:41 am
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 % of the stores revenue for newspaper sales was for. Newspaper a and if p % 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

The OA is D.

I solved this PS question as follow,

Let the total number of newspaper sold be 100

Then,
No. Of A newspaper sold =p
No. Of B newspaper sold = 100-p

Rev generated by A = p dollars
Rev generated by B = (100-p)1.25
Total rev = 125-0.25p
Now,
R% of (125-0.25p) =p

Solving the above will give the solution. Option D.

Is there another strategic approach to solve this question? Can any expert help, please? Thanks!
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by GMATGuruNY » Fri Mar 16, 2018 5:52 am
AAPL 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 % of the stores revenue for newspaper sales was for. Newspaper a and if p % 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
Plug in values that make the math easy.

Let the number of copies of B sold = 4, so that the revenue from B = 4(1.25) = 5.
Let the number of copies of A sold = 4, so that the revenue from A = 4*1 = 4.
Since (copies of A)/(total copies) = 4/8 = 1/2, p = (1/2)100 = 50.
Since (revenue from A)/(total revenue) = 4/9, r = (4/9)100 = 400/9. This is our target.

Now we plug p=50 into the answers to see which yields our target of 400/9.

Answer choice D looks like a good bet, since it includes 400:
(400*50)/(500-50) = (400*50)/450 = 400/9.

The correct answer is D.
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by Brent@GMATPrepNow » Fri Mar 16, 2018 6:16 am
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)
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.

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: D

Cheers,
Brent
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by Brent@GMATPrepNow » Fri Mar 16, 2018 6:16 am
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)
If you're not sure how to proceed with this question, or if you're behind on time and you want to catch up, you can give yourself a 50-50 chance in about 10 seconds.

To do so, we'll see what happens when we use an EXTREME value for p.
Say p = 100
In other words, 100% of the newspapers sold were Newspaper A.
This means that 100% of the revenue is from Newspaper A.
In other words, when p = 100, then r = 100

At this point, we'll plug in 100 for p and see which one yields a value of 100.
Only answer choices B and D work.
B) 150(100)/(250-100) = 100 PERFECT
D) 400(100)/(500-100) = 100 PERFECT

Now take a guess (B or D) and move on.

Cheers,
Brent
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