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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, 09:02
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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
[Reveal] Spoiler: OA
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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 27 Feb 2012, 12:17
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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, 01:23
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, 01:31
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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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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 17 Jan 2013, 01:40
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] New post 11 Jun 2013, 11:29
Bunuel, could you please explain this a little more wordy.
I don't get the step from this
Bunuel wrote:
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};

to this
Bunuel wrote:
\frac{A}{C}=\frac{12}{15}=\frac{4}{5}.


Thanks in advance.
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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 11 Jun 2013, 12:04
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Marcoson wrote:
Bunuel, could you please explain this a little more wordy.
I don't get the step from this
Bunuel wrote:
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};

to this
Bunuel wrote:
\frac{A}{C}=\frac{12}{15}=\frac{4}{5}.


Thanks in advance.


We have B:D=16:10, C:D=15:10 --> B:C=16:15. Since A:B=12:16, then A:B:C=12:16:15 --> A:C=12:15=4:5.

Hope it's clear.
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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 12 Jun 2013, 00:36
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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


Second / Four = 8/5
First / Second = 3/4
Third / Four = 3/2

==> First = 3/4 Second = 3/4 (8/5 four) = 3/4 * 8/5 * (2/3 Third) = 4//5 Third
==> First / Third = 4/5

E is correct
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Re: At a certain school, the ratio of the number of second grade [#permalink] New post 21 Feb 2014, 12:20
I actually plugged in the number.

I took 130. (8+5=13)

So
(1)2nd to 4th - 8:5
2nd grades (130/13=10) 10*8 and 4th graders 10*5

(2) 1st to 2nd - 3:4
2nd is 80 80/4 =20 so 1st 20*3=60

(3) 3rd to 4th 3:2
I know 4th is 50 so 3rd is (50/2=25) 25*3=75

(4)1st graders to 3rd graders?
60/75=4/5

But my method was more time consuming I think rather than the ones I saw here. I'll have to check.

Update: Actually it took me almost same time using the unknown multiplier method as well.
Re: At a certain school, the ratio of the number of second grade   [#permalink] 21 Feb 2014, 12:20
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