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Last Sunday a certain store sold copies of Newspaper A for $
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15 May 2020, 10:47
15 May 2020, 10:47
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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. \(\frac{500p}{(625 – p)}\)
A. \(\frac{100p}{(125 – p)}\)
B. \(\frac{150p}{(250 – p)}\)
C. \(\frac{300p}{(375 – p)}\)
D. \(\frac{400p}{(500 – p)}\)
E. \(\frac{500p}{(625 – p)}\)
Re: Last Sunday a certain store sold copies of Newspaper A for $
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15 May 2020, 10:47
15 May 2020, 10:47
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Re: Last Sunday a certain store sold copies of Newspaper A for $
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17 May 2020, 09:22
17 May 2020, 09:22
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Carcass 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. \(\frac{100p}{(125 – p)}\)
B. \(\frac{150p}{(250 – p)}\)
C. \(\frac{300p}{(375 – p)}\)
D. \(\frac{400p}{(500 – p)}\)
E. \(\frac{500p}{(625 – p)}\)
A. \(\frac{100p}{(125 – p)}\)
B. \(\frac{150p}{(250 – p)}\)
C. \(\frac{300p}{(375 – p)}\)
D. \(\frac{400p}{(500 – p)}\)
E. \(\frac{500p}{(625 – p)}\)
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
