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vik123
 
 

GMAT Prep CAT-1 Question

by vik123 Sun Aug 03, 2008 9:30 am

On a certain sightseeing tour the ratio of the number of children to the number of women was 2 to 5. What is the number of men on the tour?

1. On the tour, the ratio of number of children to number of men was 5 to 11
2. The number of women on the tour was less than 30

Each statement is definitely not enough in itself to solve the problem.

Even when taken together, I think we cannot determine the number of men, because the second statement just says that W < 30 but not equal to. Hence my answer was E.

However GMAT preps answer was C. That is it claims that both statements together are sufficient. Am I missing anything or is the GMAT prep wrong?
Smeet
 
 

Re: GMAT Prep CAT-1 Question

by Smeet Wed Aug 06, 2008 9:52 pm

vik123 Wrote:On a certain sightseeing tour the ratio of the number of children to the number of women was 2 to 5. What is the number of men on the tour?

1. On the tour, the ratio of number of children to number of men was 5 to 11
2. The number of women on the tour was less than 30

Each statement is definitely not enough in itself to solve the problem.

Even when taken together, I think we cannot determine the number of men, because the second statement just says that W < 30 but not equal to. Hence my answer was E.

However GMAT preps answer was C. That is it claims that both statements together are sufficient. Am I missing anything or is the GMAT prep wrong?



Since you have been given that the ratio of children to women is 2:5, you know for a fact that the no of women has to be a multiple of 5. Therefor women can be 5,10,15,.........

From i, you also know that the no of children has to be a multiple of 5.

From 2, you know women is less than 30.

From both 1 and 2 you know children and women are multiple of 5 and women less than 30. The only possible value of children is 10 ( since it has to be a multiple of 5, not it cannot be 20 , in that case women will be 50 which is more than 30), so no of women women is 25. From this we can find out the no of men.

Ans is c.
RonPurewal
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by RonPurewal Thu Aug 14, 2008 3:35 am

to attack statement (1), it helps to have a protocol for COMBINING RATIOS. fortunately, this is pretty easy to do.

here's what you do:
* find the term or thing that's common to the 2 ratios ("children", in this case)
* MULTIPLY the ratios by appropriate factors (in exactly the same way you'd multiply the numerator and denominator of a fraction - because ratios, after all, are just glorified fractions) to generate that least common multiple in both
* combine the ratios

for statement (1) here, then:
the prompt tells us that C : W = 2 : 5, and statement (1) tells us that C : M = 5 : 11.
the two C terms are 2 and 5, so the least common multiple is 10.
multiply by 5's to give C : W = 10 : 25, and multiply by 2's to give C : M = 10 : 22.
therefore, C : W : M = 10 : 25 : 22.
from this ratio, it's easy to see that the # of children must be a multiple of 10, the # of women must be a multiple of 25, and the # of men must be a multiple of 22.
combined with statement (2), this means that the numbers must actually be 10, 25, and 22, since any multiple thereof would be way too big to satisfy the "under 30" criterion.
rustom.hakimiyan
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Re:

by rustom.hakimiyan Sat Jun 14, 2014 2:30 am

RonPurewal Wrote:to attack statement (1), it helps to have a protocol for COMBINING RATIOS. fortunately, this is pretty easy to do.

here's what you do:
* find the term or thing that's common to the 2 ratios ("children", in this case)
* MULTIPLY the ratios by appropriate factors (in exactly the same way you'd multiply the numerator and denominator of a fraction - because ratios, after all, are just glorified fractions) to generate that least common multiple in both
* combine the ratios

for statement (1) here, then:
the prompt tells us that C : W = 2 : 5, and statement (1) tells us that C : M = 5 : 11.
the two C terms are 2 and 5, so the least common multiple is 10.
multiply by 5's to give C : W = 10 : 25, and multiply by 2's to give C : M = 10 : 22.
therefore, C : W : M = 10 : 25 : 22.
from this ratio, it's easy to see that the # of children must be a multiple of 10, the # of women must be a multiple of 25, and the # of men must be a multiple of 22.
combined with statement (2), this means that the numbers must actually be 10, 25, and 22, since any multiple thereof would be way too big to satisfy the "under 30" criterion.


Hi Ron,

I realize that in this problem, since we are dealing with people, we are restricted to whole numbers. That being said, are there cases when the unknown multiplier x could be a fraction?

I actually did an extra step here and tried to reduce 25:10:22 further but realized that this ratio was the GCF of the group of ratios.

Thanks.
RonPurewal
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Re: Re:

by RonPurewal Mon Jun 16, 2014 1:50 pm

If you've properly reduced the ratio to lowest-terms (i.e., you can't "pull anything else out" of all the numbers), then the multiplier can't be a fraction without making at least one of the actual numbers into a fraction.

This is actually the whole point of "lowest terms" ratios, by the way. If you can do that, then your ratio isn't in "lowest terms".

E.g., if you take the ratio 8:6:10 and multiply it by 5/2, then you get 20:15:25. But that only happens because 8:6:10 wasn't properly reduced"”you could have "pulled out" a 2, which is, non-coincidentally, the same as the bottom of the fraction you were able to multiply by.