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At a certain school, the ratio of the number of second grade
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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
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Re: At a certain school, the ratio of the number of second grade
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27 Feb 2012, 13:17
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
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11 Sep 2014, 04:07
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 26105 times ]
based on table, 1st: 3rd = 12:15 = 4:5
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table.jpg [ 18.75 KiB  Viewed 26098 times ]




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Re: At a certain school, the ratio of the number of second grade
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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
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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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Re: At a certain school, the ratio of the number of second grade
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17 Jan 2013, 02:40
12 is 3:4 (a) 24 is 8:5 (b) which means (aX2) 14 is 6:5 (c) 43 is 2:3 (d) which means (cX2 & dX5) 13 is 12:15 or 4:5
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Re: At a certain school, the ratio of the number of second grade
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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: \(\frac{A}{C}=\frac{12}{15}=\frac{4}{5}\).
Thanks in advance.



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Re: At a certain school, the ratio of the number of second grade
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11 Jun 2013, 13:04
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
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12 Jun 2013, 01:36
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
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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
(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.



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Re: At a certain school, the ratio of the number of second grade
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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
Asked: F/T
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
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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 i.e 4:5 answer



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At a Certain school, the ratio of the number of second graders t
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27 Dec 2014, 11:02
Hi All, While this question is wordy (and requires that we stay organized), it can be solved by TESTing VALUES. The answer choices themselves also offer something of a "hint" as to what VALUES we should be TESTing.... First, we have to get all of the information down on the pad: 2nd:4th 8:5 1st: 2nd 3:4 3rd: 4th 3:2 1st: 3rd ?:? From the answer choices, the number 16 stands out (maybe it's part of the correct answer, maybe it's not); I notice that it's a MULTIPLE of 8... Since ratio questions are all about MULTIPLES, it gets me thinking that I can TEST 16 as the number of 2nd graders... So... 2nd graders = 16 2nd:4th 8:5 So... 4th graders = 10 3rd:4th: 3:2 So.... 3rd graders = 15 1st:2nd 3:4 So... 1st gradres = 12 Thus... 1st:3rd 12:15 12:15 reduces to 4:5 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: At a certain school, the ratio of the number of second grade
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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. Thanks in advance!
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Re: At a certain school, the ratio of the number of second grade
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01 Apr 2015, 04:04
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. Thanks in advance! 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
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02 May 2016, 08:24
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 We are given the following: 1) The ratio of the number of second graders to the number of fourth graders is 8 to 5. 2) The ratio of the number of first graders to the number of second graders is 3 to 4. 3) The ratio of the number of third graders to the number of fourth graders is 3 to 2. We can use this information to create three different ratio expressions with variable multipliers. We have: 1) 2nd : 4th = 8x : 5x 2) 1st : 2nd = 3x : 4x 3) 3rd : 4th = 3x : 2x We must determine the ratio of 1st to 3rd. To determine this, we must manipulate our ratios. Let’s start with our first two ratios. We have: 1) 2nd : 4th = 8x : 5x 2) 1st : 2nd = 3x : 4x Notice “2nd” is common to both ratios. So if we make 2nd the same in both ratios we can create a 3 part ratio comparing 1st to 2nd to 4th. To make the 2nd the same in both ratios, we can multiply 1st : 2nd by 2. That is: 1st : 2nd = 3x : 4x 1st : 2nd = 2(3x : 4x) 1st : 2nd = 6x : 8x Because we know that 2nd : 4th = 8x : 5x, we can say that: 1st : 2nd : 4th = 6x : 8x : 5x Eliminate the “2nd”, and we have: 1st : 4th = 6x : 5x Now, recall that we were originally given: 3rd : 4th = 3x : 2x We should notice that the common term in our two ratios is “4th”. Thus if we can make those values the same, we can compare 1st to 4th to 3rd, and this will be enough for us to calculate the ratio of 1st to 3rd. The easiest way to do this is to turn 5x and 2x into 10x. Let’s first adjust 1st : 4th 1st : 4th = 6x : 5x 1st : 4th = 2(6x : 5x) 1st : 4th = 12x : 10x Next we can adjust 3rd : 4th. 3rd : 4th = 3x : 2x 3rd : 4th = 5(3x : 2x) 3rd : 4th = 15x : 10x Since the “4th” is now 10x in both ratios, we can say: 1st : 4th : 3rd = 12x : 10x : 15x By eliminating the “4th”, we can see that 1st : 3rd = 12x : 15x = 4x : 5x = 4 to 5 Answer E.
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Re: At a certain school, the ratio of the number of second grade
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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



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Re: At a certain school, the ratio of the number of second grade
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19 May 2017, 11:26
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 Attachment:
Capture.PNG [ 3.29 KiB  Viewed 16593 times ]
Thus, the ratio of the number of first graders to the number of third graders = \(12 : 15\) => \(4 : 5\) Hence, the correct answer must be (E) \(4 : 5\)
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At a certain school, the ratio of the number of second grade
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15 Sep 2017, 21:33
Let the first graders be represented by F, the second graders be represented by S, third graders by T, and fourth graders by Fo. Given data :\(\frac{S}{Fo} = \frac{8x}{5x}\)  \(\frac{F}{S} = \frac{3y}{4y}\)  \(\frac{T}{Fo} = \frac{3z}{2z}\) To find a common ratio, we need a common ratio \(\frac{S}{F} = \frac{16}{10}\)When \(x=2)\) \(\frac{F}{S}= \frac{12}{16}\)(when \(y=4)\) \(\frac{T}{Fo} = \frac{15}{10}\)(when \(z=5)\) The common ratio will be F : S : T : Fo = \(12 : 16 : 15 : 10\)Therefore, the ratio of first graders to third graders is 12:15 or 4:5(Option E)
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Re: At a certain school, the ratio of the number of second grade
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17 Dec 2017, 16:13
Theory
If you have two equations a:b 2:5 & b:c 3:7 Questions asks what is ratio of a:c?
[a/b]=2/5 & [b/c] = 3/7 [ Notice b is common in both ratios] We multiply value of b in other ratios
[a/b]= (2/5)*(3/3) & [b/c]= (3/7) *(5/5)
Ratios now are [a/b]= 6/15 & [b/c]= 15/35
is now common between ratios, hence the ratio a:b:c is 6:15:35
[b]Approach 2 if you have a/b & b/c , questions asks what is a/c?
[a/c]= [a/b] * [b/c] [ b cancels out and (a/c) remains ]
Solutions are pretty much already posted in thread, may be this would help some clarify what is happening/




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