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At a certain school, the ratio of the number of second grade

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At a certain school, the ratio of the number of second grade [#permalink] New post 27 Feb 2012, 10:02
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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
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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 27 Feb 2012, 13:17
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BANON wrote:
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


Let # of first, second, third and fourth graders be A, B, C and D respectively. Question: \frac{A}{C};

Given: \frac{B}{D}=\frac{8}{5}, \frac{A}{B}=\frac{3}{4}, and \frac{C}{D}=\frac{3}{2};

Equate D's: \frac{B}{D}=\frac{8}{5}=\frac{16}{10} --> \frac{C}{D}=\frac{3}{2}=\frac{15}{10}. Now, equate B's: \frac{A}{B}=\frac{3}{4}=\frac{12}{16};

\frac{A}{C}=\frac{12}{15}=\frac{4}{5}.

Answer: E.
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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 22 Apr 2012, 02:23
How can you solve this with an unknown multiplier?

let a=1st b=2nd c=3rd and d=4th graders

b/d=8/5 or b=8x and d=5x

so we get a=3/4*(8x)=6x and c=15/2*x

which obviously is not correct, why can't I apply this concept here?
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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 22 Apr 2012, 02:31
BN1989 wrote:
How can you solve this with an unknown multiplier?

let a=1st b=2nd c=3rd and d=4th graders

b/d=8/5 or b=8x and d=5x

so we get a=3/4*(8x)=6x and c=15/2*x

which obviously is not correct, why can't I apply this concept here?


Actually it's a perfectly valid approach. You've done everything right, there is just one final step missing: A/C=(6x)/(15x/2)=12/15=4/5.
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Comparing Ratios [#permalink] New post 13 Sep 2012, 08:34
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

The explanation in the OG is convoluted. What's the shortcut to solving this?
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Re: Comparing Ratios [#permalink] New post 13 Sep 2012, 08:37
praha317 wrote:
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

The explanation in the OG is convoluted. What's the shortcut to solving this?


Merging similar topics. Please ask if anything remains unclear.

Also, please read and follow: rules-for-posting-please-read-this-before-posting-133935.html
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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 17 Jan 2013, 02:40
1-2 is 3:4 ----(a)
2-4 is 8:5 ----(b)
which means (aX2)

1-4 is 6:5 ----(c)
4-3 is 2:3 ----(d)
which means (cX2 & dX5)

1-3 is 12:15 or 4:5
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Re: At a certain school, the ratio of the number of second grade   [#permalink] 17 Jan 2013, 02:40
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