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ratio problem

by hutch27 » Sun Feb 10, 2013 2:48 pm
hi, i'm having trouble identifying a way to solve this problem. I'm better with more straightforawd ration questions but these more complex ones give me trouble.

its from the OG 13th ed.

at a certain school, the ratio of the number of second graders to the number of fourth graders is 8 to 5, and the ratio of the number of first graders to the number of second graders is 3 to 4. If the ratio of the number of third graders to the number of fourth graders is 3 to 2, what is the ratio of the number of first graders to the number of third grader?

a.) 16 to 15
b.) 9 to 5
c) 5 to 16
d) 5 to 4
e) 4 to 5

oa is e
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by Ian Stewart » Sun Feb 10, 2013 3:12 pm
hutch27 wrote:hi, i'm having trouble identifying a way to solve this problem. I'm better with more straightforawd ration questions but these more complex ones give me trouble.

its from the OG 13th ed.

at a certain school, the ratio of the number of second graders to the number of fourth graders is 8 to 5, and the ratio of the number of first graders to the number of second graders is 3 to 4. If the ratio of the number of third graders to the number of fourth graders is 3 to 2, what is the ratio of the number of first graders to the number of third grader?

a.) 16 to 15
b.) 9 to 5
c) 5 to 16
d) 5 to 4
e) 4 to 5

oa is e
The conventional way to do these problems is to get a common value in each ratio. We know the ratio of 2nd to 4th graders is 8 to 5, and the ratio of 1st to 2nd graders is 3 to 4. If we multiply this second ratio by 2 to get a ratio of 6 to 8, we will have 8 second graders in each ratio, so we know the three-part ratio of 1st to 2nd to 4th graders is 6 to 8 to 5. Finally, we also know the ratio of 3rd to 4th graders is 3 to 2. If we multiply this ratio by 5, and the three-part ratio by 2, we will have 10 fourth graders in each ratio, so we get the four part ratio of 1st to 2nd to 3rd to 4th graders is 12 to 16 to 15 to 10. From here you can see the answer is 12 to 15, or 4 to 5.

I'd always organize the information above in a table, which makes it easy to see how to proceed, but I haven't worked out how to present tables in this forum unfortunately.

You can also do these questions algebraically, using the fact that ratios are just fractions, and working out how to cancel what you don't want. So here, if a is the number of 1st graders, b is the number of 2nd graders, c is the number of 3rd graders, and d is the number of 4th graders, the information in the question tells us:

b/d = 8/5
d/c = 2/3
a/b = 3/4

We want to find a/c. Notice if we just multiply the three fractions on the left sides of those three equations, we will get a/c exactly: (b/d)(d/c)(a/b) = a/c. So a/c must be equal to the product of the three fractions on the right sides: a/c = (8/5)(2/3)(3/4) = 4/5.
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by GMATGuruNY » Sun Feb 10, 2013 3:12 pm
At a certain school, the ratio of the number of second graders to the number of fourth graders is 8 to 5, and the ratio of the number of first graders to the number of second graders is 3 to 4. if the ratio of the number of third graders to the number of fourth graders is 3 to 2, what is the ratio of the number of first graders to the number of third graders?

A. 16 to 15
B. 9 to 5
C. 5 to 16
D. 5 to 4
E. 4 to 5
1st/3rd = 1st/2nd * 2nd/4th * 4th/3rd.

In the equation above, all of the values in red cancel out.
Thus:
1st/3rd = 3/4 * 8/5 * 2/3 = 4/5.

The correct answer is E.
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