Last Sunday a certain store sold copies of Newspaper A of $1.00 each and copies of ewspaper B for $1.25 each, and the store sold no other newspapers that day. If r percent of the store's revenue from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold weere 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
Answer: D
PLease breakdown and explain
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- sanju09
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A problem based on percents should be dealt with 100 as total. Let's take total number of newspapers sold to be 100. Hence the number of copies of newspaper A is p, and 100 - p are that of newspaper B. But we hate variables when in hurry, so keeping the given prices to mind, let's assign a suitable number for p. How about p = 20? This will make 1.25 to deal with 80, how easy!oquiella wrote:Last Sunday a certain store sold copies of Newspaper A of $1.00 each and copies of ewspaper B for $1.25 each, and the store sold no other newspapers that day. If r percent of the store's revenue from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold weere 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
Answer: D
PLease breakdown and explain
Revenue from newspaper A is $1 (20) = $20, and the revenue from newspaper B is $1.25 (80) = $100. Thus the total revenue is $120. Now, as we need to answer r, which is the percent of revenue from newspaper A; we know that, r = ($20/$120) × 100 = 50/3%. All options are fractions and so is our target answer 50/3. Hence, the denominator of a possible answer must be divisible by 3, when we plug in p = 20 there. Let's check, one by one:
A. 100p/125-p: 125 - 20 = 105 is divisible by 3, so keep it!
B. 150p/250-p: 250 - 20 = 230 is not divisible by 3, so eliminate it!
C. 300p/375-p: 375 - 20 = 355 is not divisible by 3, so eliminate it!
D. 400p/500-p: 500 - 20 = 480 is divisible by 3, so keep it!
E. 500p/625-p: 625 - 20 = 605 is not divisible by 3, so eliminate it!
Left with A and D, turn to check the numerators now:
A. 100p/125-p: 125 - 20 = 105 and 100×20/105 is NOT 50/3.
[spoiler]Hence (D)[/spoiler]
We can check:
D. 400p/500-p: 500 - 20 = 480 and 400×20/480 = [spoiler]50/3 BINGO!![/spoiler]
For algebraic approach
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- Brent@GMATPrepNow
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Let's use the INPUT-OUTPUT approach.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 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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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.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)
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