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Ratio

by RobertP » Tue Jan 26, 2016 4:33 pm
Hello everyone!

Ok, this is exactly a kind of question that I have a lot of dificulites answering, especially because it involves more of a math logic that math skills per say.

Anyone have any suggestions how I can improve this skill?

Thank you!

Certain store sold copies of the newspaper A for USD 1 each and copies of newspaper B for USD 1.25 each, and the store did not sell any other newspaper that day. If r% of the stores revenue is related to newspaper A and p% of the newspapers sold were copies of newspaper A, which of the following expression indicates 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
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by Brent@GMATPrepNow » Tue Jan 26, 2016 4:51 pm
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 (aka plugging in numbers).

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 » Tue Jan 26, 2016 4:51 pm
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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by ceilidh.erickson » Sat Jan 30, 2016 3:22 pm
First of all, I want to point out that this is a notoriously VERY difficult problem (it's #205 in OG2016, #198 in OG2015).

For most of my students, I'd recommend Brent's 2nd method: picking 100 to get to a quick 50/50 chance. This is the sort of question that can really bog students down with timing if they try to approach it algebraically. For most people, this should probably be a skip/guess.

But, I think it's worthwhile to understand the algebra mechanics on the crazy-hard ones, so you can apply that logic on more doable ones. Take time to go through the algebraic explanation posed in the OG, repeat the steps yourself, and see if it makes sense.

But don't worry if you have difficulty answering this one! Probably every expert on this forum would have picked numbers instead of solving algebraically.
Ceilidh Erickson
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Harvard Graduate School of Education
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by Matt@VeritasPrep » Sun Jan 31, 2016 7:50 pm
Plugging in numbers is certainly a good idea, but we can use algebra too.

Suppose we sold x copies of A and y copies of B. We know that

x + y = # of copies
x + 1.25y = revenue

Let's start by removing a variable: finding y in terms of x. Since x = (p/100)*(x + y), we know that

y = 100x/p - x

So the total # of copies is really

x + 100x/p - x,

or 100x/p.

Likewise, the revenue is x + 1.25y, or
x + 1.25*(100x/p - x), or
x + 125x/p - 1.25x, or
125x/p - .25x.

Now we want p in terms of r.

We know (from earlier) that x = (p/100)*(# of copies) and that x = (r/100)*(revenue), so we can set these equal:

(p/100)*(# of copies) = (r/100)*(revenue)

or

p * (# of copies) = r * (revenue)

Now let's sub in our # of copies and our revenue:

p * (100x/p) = r * (125x/p - .25x)

100x = 125xr/p - .25xr

Dividing by x gives

100 = 125r/p - .25r

and we rearrange to

100p = 125r - .25pr

400p = 500r - pr

400p/(500 - p) = r

Woof!
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