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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]

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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
[Reveal] Spoiler: OA
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Re: At a certain school, the ratio of the number of second grade [#permalink]

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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: $$\frac{A}{B}=\frac{3}{4}=\frac{12}{16}$$;

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

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Re: At a certain school, the ratio of the number of second grade [#permalink]

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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]

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22 Apr 2012, 02:31
Expert's post
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]

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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]

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11 Jun 2013, 12:29
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:
$$\frac{A}{C}=\frac{12}{15}=\frac{4}{5}$$.

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Re: At a certain school, the ratio of the number of second grade [#permalink]

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11 Jun 2013, 13:04
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Expert's post
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:
$$\frac{A}{C}=\frac{12}{15}=\frac{4}{5}$$.

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]

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12 Jun 2013, 01:36
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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

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]

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21 Feb 2014, 13:20
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

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.
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Re: At a certain school, the ratio of the number of second grade [#permalink]

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16 Aug 2014, 13:47
Let S= Second, P = Fourth , F = First, T = Third

Given: S/P = 8/5
F/S = 3/4
T/P = 3/2

F/T = (S/P)(F/S)(P/T) = (8/5)(3/4)(2/3) = 4/5
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Re: At a certain school, the ratio of the number of second grade [#permalink]

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11 Sep 2014, 01:42
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

:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

B:D -- 8:5
A:B --3:4
C:D - 3:2

A:B:D -- 6:8:5

12:16:10

apply c:d here
then
12:16:15:10

c:d -- 12:15
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Re: At a certain school, the ratio of the number of second grade [#permalink]

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11 Sep 2014, 04:07
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Table approach would suit to the problem quick and accurate. factor all the ratios such that two ratio have common number to match up.

Attachment:

table.jpg [ 18.75 KiB | Viewed 4792 times ]

based on table, 1st: 3rd = 12:15 = 4:5
Attachments

table.jpg [ 18.75 KiB | Viewed 4784 times ]

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Re: At a certain school, the ratio of the number of second grade [#permalink]

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01 Apr 2015, 03:58
Bunuel
Bunuel wrote:
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.

Hi there,
I ddn't get why he multiplied 3/4*8x.

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Re: At a certain school, the ratio of the number of second grade [#permalink]

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01 Apr 2015, 04:04
Expert's post
elisabettaportioli wrote:
Bunuel
Bunuel wrote:
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.

Hi there,
I ddn't get why he multiplied 3/4*8x.

The ratio of the number of first graders to the number of second graders is 3 to 4: a/b = 3/4 = a/(8x) = 3/4 --> a = 3/4*8x.
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Re: At a certain school, the ratio of the number of second grade [#permalink]

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15 Apr 2016, 21:58
is it possible to equate A/B=3/4 and B/D=8/5 to A/B=6/8 and B/D=8/5 since you have 2 numbers for the 2nd grade or do both numbers have to be in the numerator/denominator?
Re: At a certain school, the ratio of the number of second grade   [#permalink] 15 Apr 2016, 21:58
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# At a certain school, the ratio of the number of second grade

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