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# If p1 and p2 are the populations and r1 and r2 are

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If p1 and p2 are the populations and r1 and r2 are [#permalink]

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02 Mar 2005, 23:54
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If $$p_1$$ and $$p_2$$ are the populations and $$r_1$$ and $$r_2$$ 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) $$p_1 > p_2$$
(2) $$r_2 > r_1$$

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[Reveal] Spoiler: OA

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03 Mar 2005, 03:35
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B/w C and E
p1,p2,r1,r2 are all >0
so multiplying
if it is true that if a>b
,multiplying for c>0, ac>bc
and it becomes stronger if c>d
so that ac>>bd
then a/d>>b/c
so C
please don't kill me if i'm wrong because i'll do it by myself

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03 Mar 2005, 03:55
E as well.
i and ii does not give us any clue on size of distribution to finalize a ratio

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03 Mar 2005, 10:33
C for me.

beause once we have both , the only possibility when the Population is higher and at the same time the representatives a fewer, obviously the rate would be smaller.

looking forward for OA.
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03 Mar 2005, 10:53
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C it is
we want Rn/Pn
n=[1,2]
Both combined, if R2 is biggest and P2 is smallest, we know immediately that the ratio Rn/Pn is larger for City2
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03 Mar 2005, 11:11
I am looking forward to the OA here because i am crossed between C and E.
Given population vs. reps, is it okay to always assume population will be higher than reps so in each district?
Yes it makes sense that population should be larger than reps but then i have read GMAT guidelines that suggests test takers should only work with the information given to them as opposed to inferring extra data from a given information.

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03 Mar 2005, 12:08
"C"

We need to see if P1/R1 > P2/R2

state 1....insuff.....as R1 can be > R2 making P1/R1 small.

state 2....insuff...again don't know P1 and P2

combine.....we know P1 > P2.....we know that R1 < R2 then overall....P1?R1 will be even greater than P2/R2....suff

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03 Mar 2005, 12:11
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Basic rules for inequalities:
(in the example: a>b, c>d)

You need to flip directions when both side are multiplied by a negative number:
-a<-b, -c<-d

You need to flip directions when 1 is divided by both side:
1/a<1/b, 1/c<1/d

You can only add or multiply them when their signs are the same direction:
a+c>b+d
ac>bd

You can only apply substractions and divisions when their signs are the opposite direction:
a>b, d<c
a-d>b-c
a/d>b/c
(You can't say a/c>b/c. It is WRONG)

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03 Mar 2005, 12:13
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(1) p1 > p2
(2) r1 < r2

Combine: p1/r1>p2/r2
Sufficient.
(C)

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02 Sep 2008, 12:47
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rao_1857 wrote:
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

1) We need r1 and r2 values
insuffcient
2) we need p1 and p2 values.
insuffcient

combine

p1 > p2
r2 > r1

p1/r1(always) >p2/r2

Suffcient
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Last edited by x2suresh on 02 Sep 2008, 18:37, edited 1 time in total.

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02 Sep 2008, 15:04
Not sure I follow. I would have said D.

If P1=10 P2=9 and R1=5 R2=1

Then P1/R1 = 2
and P2/R2 = 9

In this case, p1/r1 < p2/r2

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02 Sep 2008, 15:43
C

We need to know the population size AND number of reps to get a ratio

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02 Sep 2008, 20:48
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Is P1/R1 greater of P2/R2 greater

Combining both info, It is clear that P1/R1 is greater because the numerator is greater and denominator is lesser when compared to the other fraction. C

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03 Sep 2008, 11:43
We want to know if p1/r1 > p2/r2

Since all quantities are positive, we can cross multiply to get the equivalent inequality

p1r2>p2r1

If both positive factors on the left are greater than both positive factors on the right
then the inequality is true.

Note: We could have started with the inequality p1/r1<p2/r2 and then the equivalent inequality
would be false but the statements would still be sufficient together.

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27 Aug 2010, 16:17
metallicafan wrote:
Which way is better to solve this question? Using algebra or testing with some values or numbers? Why? I used numbers, but OG used algebra.

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

My initial attempt would be to use algebra because it is fool proof. But I fall back to using numbers only when the algebraic method fails or is too cumbersome.

Ratio of population to representatives -- P1/R1 v\$ P2/R2

Statement 1: Without the relationship between the population and representatives this is insufficient.

Statement 2: Insufficient because of the above reasons.

However combining them we release P1/R1 would be higher since P1 is higher (than P2) and R1 is smaller than R2.

Hence a higher value (P1) is divided by a lower value (R1) -- this makes P1/R1 value higher.

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Re: If p1 and p2 are the populations and r1 and r2 are [#permalink]

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08 Apr 2012, 23:58
I didn't understand this. Could someone please explain? thanks.
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Re: If p1 and p2 are the populations and r1 and r2 are [#permalink]

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09 Apr 2012, 01:28
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I didn't understand this. Could someone please explain? thanks.

If $$p_1$$ and $$p_2$$ are the populations and $$r_1$$ and $$r_2$$ 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?

Question: is $$\frac{p_1}{r_1}>\frac{p_2}{r_2}$$? Since $$p_1$$, $$p_2$$, $$r_1$$ and $$r_2$$ are positive integers then we can cross-multiply and rephrase the question: is $$p_1*r_2>p_2*r_1$$?

(1) $$p_1 > p_2$$. Not sufficient by itself since no info about $$r_1$$ and $$r_2$$.
(2) $$r_2 > r_1$$. Not sufficient by itself since no info about $$p_1$$ and $$p_2$$.

(1)+(2) Now, since each multiple ($$p_1$$ and $$r_2$$) on the left hand side is greater than the respective multiple ($$p_2$$ and $$r_1$$) on the right hand side then $$p_1*r_2>p_2*r_1$$. Sufficient.

Hope it's clear.
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Re: If p1 and p2 are the populations and r1 and r2 are [#permalink]

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30 Oct 2012, 02:17
Bunuel wrote:
I didn't understand this. Could someone please explain? thanks.

If $$p_1$$ and $$p_2$$ are the populations and $$r_1$$ and $$r_2$$ 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?

Question: is $$\frac{p_1}{r_1}>\frac{p_2}{r_2}$$? Since $$p_1$$, $$p_2$$, $$r_1$$ and $$r_2$$ are positive integers then we can cross-multiply and rephrase the question: is $$p_1*r_2>p_2*r_1$$?

(1) $$p_1 > p_2$$. Not sufficient by itself since no info about $$r_1$$ and $$r_2$$.
(2) $$r_2 > r_1$$. Not sufficient by itself since no info about $$p_1$$ and $$p_2$$.

(1)+(2) Now, since each multiple ($$p_1$$ and $$r_2$$) on the left hand side is greater than the respective multiple ($$p_2$$ and $$r_1$$) on the right hand side then $$p_1*r_2>p_2*r_1$$. Sufficient.

Hope it's clear.

had 2 been r1>r2, ans would have been E. .Bunuel ,pls correct me if I am wrong
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Re: If p1 and p2 are the populations and r1 and r2 are [#permalink]

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Re: If p1 and p2 are the populations and r1 and r2 are   [#permalink] 17 Dec 2015, 15:07

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