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OG #38

by Reft » Sun Jul 11, 2010 4:47 pm
If p1 and p2 are the populations and r1 and r2 are the numbers of representatives of District 1 and District 2 , respectively, the ratio of the population to the number of representatives is greater for which of the two districts?

(1) p1 > p2

(2) r2 > r1

OA: C


I understand why they don't work individually, but do not understand how they work together.
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Source: — Data Sufficiency |

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by Rahul@gurome » Sun Jul 11, 2010 5:04 pm
Ratio for district D1 = p1/r1
Ratio for district D2 = p2/r2

(1) p1 > p2 but we don't know r1 and r2, so (1) is NOT SUFFICIENT.

(2) r2 > r1 but we don't know p1 and p2, so (2) is NOT SUFFICIENT.

Combining (1) and (2), since p1 > p2 and r2 > r1, so when we divide p1/r1, we will get a larger number than the number we get by dividing p2/r2.
Hence p1/r1 > p2/r2

[spoiler]The correct answer is (C).
[/spoiler]
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by barcebal » Sun Jul 11, 2010 9:32 pm
Another way to think about it with some verbal/numbers

If the price of computer in State A is greater than state B the ONLY way for the TOTAL computer cost in state B to be greater than State A is if State B charges a higher tax rate than State A, which takes you back to understanding how Statement II is NS.

But since the tax rate in state B is less than A it is impossible for the cost of the computer in State B to get more expensive.

Now with some numbers.

Assume the pre-tax cost in State A is 100 and has a tax rate of 10%. Total cost of the computer is $110. If the computer in State B cost $99. Even if the tax rate in State B=the tax rate in State A, the total cost of the computer in State B is 108.90. But the tax rate has to be LOWER. There's no way to make up the ground if State A has a higher base price and a higher sales tax rate.
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by anoopprasad » Sun Jul 11, 2010 9:42 pm
Is P1/R1 > P2/R2 - that is the question
Since R1 and R2 cannot be negative we can cross multiply
So the question becomes

is P1R2 > P2R1
Since P1>P2 and R2>R1 we can conclude P1R2 > P2R1

So C
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by GMATGuruNY » Mon Jul 12, 2010 3:22 am
Question: Is p1/r1 > p2/r2?

Let's take an example with real numbers:

Is 2/3 > 3/5?

One way to compare fractions is to give them a common denominator:

2/3 = 10/15
3/5 = 9/15

So the question becomes:

Is 10/15 > 9/15?

Since the two fractions now have the same denominator, we just have to compare the numerators: 10 > 9, so 2/3 > 3/5.

How did we get the 10 in 10/15? (denominator in 3/5) * (numerator in 2/3) = 5 * 2 = 10.
How did we get the 9 in 9/15? (denominator in 2/3) * (numerator in 3/5) = 3 * 3 = 9.

To compare fractions, we multiply the denominator in each fraction by the numerator in the other and compare the products.

Using this logic, Is p1/r1 > p2/r2? can be rewritten as:

Is r2 * p1 > r1 * p2?

Now let's examine statements 1 and 2 together:

Statement 1: p1 > p2
Statement 2: r2 > r1
So r2 * p1 > r1 * p2.
SUFFICIENT.

The correct answer is C.

Hope this helps!
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by funshy » Mon Jul 12, 2010 8:44 am
let:
p1/r1 =y and
p2/r2 =x
objective: can you establish y>x or y<x. OR put in another way can you establish (y/x) greater than 1 or (y/x) less than 1. That is (y/x) > 1 or (y/x) < 1 obtainable from the information given?

y/x=(p1/r1)/(p2/r2)=(p1/r1)*(r2/p2) = (p1/p2)*(r2/r1)

No single statement provide the information to solve the equation above, but combining the two statements solve the problem hence answer is C
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