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Intern  Joined: 02 Jul 2009
Posts: 16
At a Certain school, the ratio of the number of second graders to the  [#permalink]

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Question Stats: 84% (02:18) correct 16% (02:33) wrong based on 602 sessions

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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 thenumber of third graders.

(A) 16 to 15
(B) 9 to 5
(C) 5 to 16
(D) 5 to 4
(E) 4 to 5

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/at-a-certain ... 28242.html

Originally posted by nickesha on 16 Jul 2009, 10:59.
Last edited by Bunuel on 14 Dec 2017, 05:14, edited 3 times in total.
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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5
3
I am not sure if I am allowed to post youtube videos on here, but here's a really quick and easy way to solve this problem.

My concern is if this approach has any drawbacks or limitations.

Before you use an approach, ensure you know why it works. This approach is exactly what has been done in posts above.

Say,
A:B = 3:4 = 6:8 (to make Bs equal in A:B and B:C)
B:C = 8:5
You get A:B:C = 6:8:5

Instead, if you have A:B = 3:4 and B:C = 5:6, how do you make Bs equal?
A:B = 3:4 = 15:20
B:C = 5:6 = 20:24
You multiply the B's to get the LCM.

That's what is done in the video too. He just multiplies the Bs to get a common value (even though it is not the least common value, it doesn't matter to us since the ratio is unchanged) and represents the whole thing in a table format. The method is no different from what is done above.
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Karishma
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Manager  Joined: 12 Oct 2009
Posts: 90
Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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nickesha wrote:
hi, could someone help me work this question?

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 thenumber of third graders.

N1,N2,N3,N4 be the number of students in 1st,2nd,3rd and 4th grade.

we have N2/N4 = 8/5, N1/N2 = 3/4, N3/N4 = 3/2. We need to find N1/N3
N1/N3 = N1/N2 * N2/N4 * N4/N3 = 3/4 * 8/5 * 2/3 = 4/5
##### General Discussion
Manager  Joined: 04 Jun 2008
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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nickesha wrote:
hi, could someone help me work this question?

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 thenumber of third graders.

Here's the method in general:

If A:B = 2:3
& B:C = 3:4
then A:B = 2:4 .....easy?

Now say B:D = 5:6

Then A:D = ??

The common one between them is B, so take such a value for B that the value is same for both A and D and gives integers.

ie, B is 3x wrt to A, and 5y wrt to D, so take lcm of 3 and 5 so that you get 15x for A and 15y for D.

make A:B = 2:3 or 10:15
make B:D = 5:6 or 15:18

so A:D = 10:18.... or 5:9

Ans to your question will be 12:15 or 4:5
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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nickesha wrote:
hi, could someone help me work this question?

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 thenumber of third graders.

is it 4:5?

I tried finding equal number of 4th graders to compare 2nd and 3rd with and got 32:20,30:20 and i guess 2nd to 3rd is 32:30 - reduced to 16:15. To get 16 for 2nd graders i multiply 3:4 ratio by 4 to get 12:16 and for 1st and 2nd and then switch to get 12:15 = 4:5

It is probably wrong... need to review on ratios...
Senior Manager  Joined: 18 Jun 2009
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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6
3
Based on the same principle as what rashminet84 has explained

Its simple

Step 1 ) write down the question as it states

Now to find the ratio between 4th grade and 1st grade, the right hand side of the above two ratios should have the same value for the 2nd grade position as of now it is 8 and 4

= 5:8::8:6

So we have 8 on both sides so canceling them we have

Combining them we have

two get the common value for 4th grade multiple the first ratio by 2 and second ratio by 5

Canceling the 10's out we have

Hope this helps, I tried to explain in real detail.

And yeah it looks long but I had to do it to give clear explanation. But it doesn't take more than 30-60 seconds if you know the concept
Manager  Joined: 27 Jun 2008
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3
An easier solution would be to choose a number

Info in q

Solution

use the values from above in next equ ( 1st grade:2nd grade = 3:4)
1st grader = 160*3/4 = 120

120: 150
Retired Moderator Joined: 02 Sep 2010
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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ChenggongMAS 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

I am having the most difficult time wrapping my mind around this problem. Can someone please explain a different approach than the OG uses? Thanks!

~Chenggong

Let x1 be no of first graders
Let x2 be no of second graders
Let x3 be no of third graders
Let x4 be no of fourth graders

the ratio of the number of second graders to the number of fourth graders is 8 to 5 : (x2/x4) = (8/5)
the ratio of the number of first graders to the number of second graders is 3 to 4 : (x1/x2) = (3/4)
the ratio of the number of third graders to the number of fourth graders is 3 to 2 : (x3/x4) = (3/2)

what is the ratio of the number of first graders to the number of third graders? : We need to find out (x1/x3)

$$\frac{x3}{x2} = \frac{\frac{x3}{x4}}{\frac{x2}{x4}} = \frac{(3/2)}{(8/5)} = \frac{15}{16}$$
$$\frac{x1}{x3} = \frac{\frac{x1}{x2}}{\frac{x3}{x2}} = \frac{(3/4)}{(15/16)} = \frac{48}{60} = \frac{4}{5}$$

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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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(2nd gr/4th gr)= 8/5 -----(1)
(1st gr/2nd gr)=3/4-------(2)
3rd gr/4th gr = 3/2--------(3)

Multiply 1 nd 2: (2nd gr/4th gr)*(1st gr/2nd gr)=(1st gr/4th)=6/5---(4)
Inversing eq 3:(4th gr/3rd gr)=2/3----(5)

Multiply 4 nd 5:(1st gr/4th)(4th gr/3rd gr)=(1st/3rd)=(6/5)*(2/3)=(4/5)-------(Ans.E)
Intern  Joined: 21 Sep 2010
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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Manager  Joined: 03 Nov 2009
Posts: 54
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Ans E ,

Agree with denver, more simple approach.

8:5 (*) both with 2, and so on for every ratio.

1st /4th grader = 16/10 - A

1st/2 Nd = 12/16 - B

and 3 rd / 4 Th - You already have taken 4th as 10 from eqn. A , = 15/10 - C

Now just compare Eqn B and C , You get 12 / 15 = 4 : 5

I Believe the important thing to note here is all figures are given in ratios , makes it difficult to reach a concrete conclusion. So What we could do is just multiply the figures by either 2 or any number so we can get a different set of ratios.

Best
Manager  Joined: 08 Jun 2011
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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I am not sure if I am allowed to post youtube videos on here, but here's a really quick and easy way to solve this problem.

My concern is if this approach has any drawbacks or limitations.
Intern  Joined: 13 Jan 2012
Posts: 33
Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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1
If a,b,c,d represent first, second, third, fourth grades, it follows from the question, that:

1. $$b/d = 8/5$$
2. $$a/b = 3/4$$
3. $$c/d = 3/2$$ <-- this implies that $$d/c$$would be $$2/3$$...let's hang on to that, we will need this later on.

We need to find, a/c.

On such problems, the first thing I do is to try and build what the question's asking for (in this case "$$a/c$$") by adding/subtracting/multiplying/dividing the given ratios.

So, $$a/c = a/b x b/d x d/c = 3/4 x 8/5 x 2/3 = 4/5$$

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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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Here is my method: The key here is to cross cancel.

1) B/D = 8/5 or 16/10
2) A/B = 3/4
3) C/D = 3/2 or 15/10
A/C = ?

1) & 2) can be multiplied to get A/D
B/D X A/B = A/D
8/5 *3/4 = 6/5

2) Now we know A/D & C/D. Fine the LCM so that you have:

A/D = 6/5 or 12/10
C/D = 15/10
so A/C = 12/15 or 4/5
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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Can someone tell me if my simplistic approach is accurate?

We need to move from one ratio to the other until we have a "row" for all ratios, moving statement by statement:

a : b : c : d
1) a : 8 : c : 5

2) 3 : 4 : c : d
We notice that b is half of what it is on 1, or b is twice as much as in 2.
Hence we can rewrite 1 as
1) 6 : 8 : c : 5 (6 comes from 3x2 from statement 2)

moving on...

3) a : b : 3 : 2
Now, we need to bring d in statements 1 and 3 to be the same, by LCM: 5 x 2 = 10. So by multiplying statement 3 by x5 we get
3) a : b : 15 : 10
Now, d in statement 3 is twice as much as in statement 1, so we can say statement 3 is twice bigger than statement 1. So if we multiply statement 1 by x2...
1) 12 : 16 : c : 10, but we already know from statement 3 that c : 10 = 15 : 10, hence

a : b : c : d
12 : 16 : 15 : 10

And a/c = 12/15, divide both by 3, we get 4/5.
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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Skag55 wrote:
Can someone tell me if my simplistic approach is accurate?

We need to move from one ratio to the other until we have a "row" for all ratios, moving statement by statement:

a : b : c : d
1) a : 8 : c : 5

2) 3 : 4 : c : d
We notice that b is half of what it is on 1, or b is twice as much as in 2.
Hence we can rewrite 1 as
1) 6 : 8 : c : 5 (6 comes from 3x2 from statement 2)

moving on...

3) a : b : 3 : 2
Now, we need to bring d in statements 1 and 3 to be the same, by LCM: 5 x 2 = 10. So by multiplying statement 3 by x5 we get
3) a : b : 15 : 10
Now, d in statement 3 is twice as much as in statement 1, so we can say statement 3 is twice bigger than statement 1. So if we multiply statement 1 by x2...
1) 12 : 16 : c : 10, but we already know from statement 3 that c : 10 = 15 : 10, hence

a : b : c : d
12 : 16 : 15 : 10

And a/c = 12/15, divide both by 3, we get 4/5.

Yes, this is correct. The steps you are taking are the same as in other methods - just the format is different in each method.
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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nickesha 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 thenumber of third graders.

(A) 16 to 15
(B) 9 to 5
(C) 5 to 16
(D) 5 to 4
(E) 4 to 5

Best approach is to express all ratios as fractions and try to get a fraction of the form A/C (Target) given A for First Grades and C for Third graders.
For example we would have B/D = 8/5, A/B = 3/4, C/D = 3/2 etc.. Find A/C =

So we need to find the division of two fractions that can then cancel and leave us with A/C for example I'm thinking of (A/B)/(B/C)

And B/C could also be (B/D)/(C/D). So all in all this method takes some practice but once you master it it gives the answer straight away in less than 30 secs.

Cheers!
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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I use the multiple ratios approcah the share in the manhattan prep-world translations-ratios chapter (pg. 58). In this approach, you calculate the ratios based on the LCM, in order to get same denominators.

1) Line up your ratio relationships (F - first, S - second, T - Third and Fo - Fourth).

F:__S:__T:__Fo
8__S___T___5
3__4___T___Fo
F__S___3___2

We will calculate the rations in groups, starting with the first 2 rows. We see that we have 2 values for F, but they do not have a common denominator. In order to correct for that, we find the LCM of 8 and 3, which is 24, and calculate each row so that we end with 24 for F, leaving the unknown vaariables as they are:
F:__S:__T:__Fo---------------->F:__S:__T:__Fo
8__S___T___5-------X3------>24:__S:__T:__15
3__4___T___Fo------X8------>24:_32:__T:__15 We keep this final row. 15 was just moved down, as it is already done.

Now, we will do the same, using the new row above and the remaining ratio, between T and Fo, using the ratio given.
So, we need to find the LCM of 15 and 2, which is 30. We multiply, so that T and Fo result in 30.
F:___S:__T:__Fo--------------->F:___S:__T:__Fo
24:_32:__T:__15------X2------>24:_32:__T:__30
F:__S:___3___2------X10------>24:_32:_30:__20

So, we now have all of the ratios and we need F:T
F:T = 24:30 = 12/15 = 4/5. E
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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VeritasPrepKarishma wrote:
I am not sure if I am allowed to post youtube videos on here, but here's a really quick and easy way to solve this problem.

My concern is if this approach has any drawbacks or limitations.

Before you use an approach, ensure you know why it works. This approach is exactly what has been done in posts above.

Say,
A:B = 3:4 = 6:8 (to make Bs equal in A:B and B:C)
B:C = 8:5
You get A:B:C = 6:8:5

Instead, if you have A:B = 3:4 and B:C = 5:6, how do you make Bs equal?
A:B = 3:4 = 15:20
B:C = 5:6 = 20:24
You multiply the B's to get the LCM.

That's what is done in the video too. He just multiplies the Bs to get a common value (even though it is not the least common value, it doesn't matter to us since the ratio is unchanged) and represents the whole thing in a table format. The method is no different from what is done above.

It's not any different, but it helps to visualize things to keep from being scrambled and making careless mistakes. The guy in the video went really slow to make it easy to understand, but during the test itself, I would immediately draw a table with a ratio question like this and get it done quickly!
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Re: At a Certain school, the ratio of the number of second graders to the  [#permalink]

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

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/at-a-certain ... 28242.html
_________________ Re: At a Certain school, the ratio of the number of second graders to the   [#permalink] 14 Dec 2017, 05:15

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