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If p1 and p2 are the populations and r1 and r2 are [#permalink]
02 Mar 2005, 22:54

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00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

76% (01:43) correct
24% (00:27) wrong based on 106 sessions

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?

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

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.

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 _________________

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.

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)

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 _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Last edited by x2suresh on 02 Sep 2008, 17:37, edited 1 time in total.

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

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.

Re: Populations and Representatives [#permalink]
27 Aug 2010, 15:17

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.

Answer to the question:

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.

Answer: C _________________

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Re: If p1 and p2 are the populations and r1 and r2 are [#permalink]
09 Apr 2012, 00:28

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Expert's post

mymbadreamz 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.

Re: If p1 and p2 are the populations and r1 and r2 are [#permalink]
30 Oct 2012, 01:17

Bunuel wrote:

mymbadreamz 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.

Answer: C.

Hope it's clear.

had 2 been r1>r2, ans would have been E. .Bunuel ,pls correct me if I am wrong _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.