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

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Intern
Joined: 23 Mar 2016
Posts: 24
Schools: Tulane '18 (M\$)
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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02 May 2016, 15:01
super simple, just a minute and you're done (if not quicker)

given

1) 1s:2n......... 2n:4th ........3r:4th
2) 3:4 .............8:5............3:2
3) 6:8:5 .... 3:2
(6:8:5)*2 ...(3:2)*5
(12:16:10) (15:10)
12:16:15:10
12:15 = 1st:3rd
(12:15)/3 = 4:5
Manager
Joined: 17 Jul 2016
Posts: 54
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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18 Dec 2017, 11:11
2nd to 4th= 8/5
1st to 2nd= 3/4 = 6/8
3rd to 4th= 3/2

let 1st=6
2nd=8
4th=5
3rd is 1.5 times 4th so 1.5 (5)=7.5

1st to 3rd= 6/7.5=12/15=4/5
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Joined: 07 Dec 2014
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At a certain school, the ratio of the number of second grade  [#permalink]

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18 Dec 2017, 21:44
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

g1:g2=3:4=12:16
g2:g4=8:5=16:10
g4:g3=2:3=10:15
g1:g3=12:15=4 to 5
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Re: At a certain school, the ratio of the number of second grade  [#permalink]

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12 Jan 2018, 23:05
Bunuel niks18 amanvermagmat

Based on fact that for equating two ratio I need common denominator, can I approach the problem
taking LCM of all no from 8 to 2?
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Re: At a certain school, the ratio of the number of second grade  [#permalink]

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12 Jan 2018, 23:29
Bunuel niks18 amanvermagmat

Based on fact that for equating two ratio I need common denominator, can I approach the problem
taking LCM of all no from 8 to 2?

can you explain your process in detail? I am not sure what are you going to do with LCM and how you are going to arrive at the final answer using the approach.
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Joined: 31 May 2015
Posts: 1
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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23 Jan 2018, 13:36
Let,

Given info:

B/D = 8/5
A/B = 3/4
C/D = 3/2
A/C = ?

Find two equations from above which have "A" and "C"

A/B/C/D = 3/4/3/2 => A*D/C*B = 3*2/4*3 => 1/2

Now substitute B/D information (given) in the above equation

A*5/C*8 = 1/2 => A/C = 4/5
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Joined: 13 Jan 2018
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At a certain school, the ratio of the number of second grade  [#permalink]

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04 Mar 2018, 05:05
Bunuel wrote:
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: $$\frac{A}{B}=\frac{3}{4}=\frac{12}{16}$$;

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

B/D=8/5, C/D=3/2, A/B=3/4
B/C = B/D *D/C = 8/5 * 2/3 = 16/15
A/C = A/B * B/C = 3/4 * 16/15 = 4/5
Intern
Joined: 08 Jan 2013
Posts: 1
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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18 Aug 2018, 04:04
Please can someone help on how A/C=3/4=12/16. where did the 4 come from? Thanks
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22 Aug 2019, 03:12
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Re: At a certain school, the ratio of the number of second grade   [#permalink] 22 Aug 2019, 03:12

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