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Manager  B
Joined: 17 Jul 2016
Posts: 54
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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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
VP  P
Joined: 07 Dec 2014
Posts: 1217
At a certain school, the ratio of the number of second grade  [#permalink]

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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

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

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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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Joined: 25 Feb 2013
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Re: At a certain school, the ratio of the number of second grade  [#permalink]

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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.
Intern  B
Joined: 31 May 2015
Posts: 1
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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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
Intern  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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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
Manager  P
Joined: 31 Jul 2017
Posts: 197
Location: Tajikistan
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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4
Lets set a ratio for this problem. We are told that ratio of sophomore to senior is 8:5, then we are told that ratio of freshman to sophomore is 3:4. Lastly, we are told that ratio of junior to senior is 3:2.
Let's conclude now the above given information.
freshmen : sophomore : junior : senior
__:8:__:5 (later we will multiply by 2)
3: 4:___:___ (later we will multiply by 4)
___:____:3:2 (later we will multiply by 5/2)
Lets combine it all
12:16:15:10.
Now, we are asked to find ratio of the number of first graders to the number of third graders, which is 12:15 or 4:5 (E)
Intern  B
Joined: 08 Jan 2013
Posts: 1
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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Please can someone help on how A/C=3/4=12/16. where did the 4 come from? Thanks
Non-Human User Joined: 09 Sep 2013
Posts: 13744
Re: At a certain school, the ratio of the number of second grade  [#permalink]

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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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