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At a Certain school, the ratio of the number of second [#permalink]

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16 Jul 2009, 10:59

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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

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.

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...
_________________

Based on the same principle as what rashminet84 has explained

Its simple

Step 1 ) write down the question as it states a) 2nd grade:4th grade = 8:5 b) 1st grade:2nd grade = 3:4 c) 3rd grade:4th grade = 3:2

We need to find out 1st grade:3rd grade

Step 2) The only way to tie 1st grade and 3rd grade is by getting a common grade between the ratios 1st grade:4th grade and 4th grade:3rd grade

So lets find out what is 1st grade:4th grade

Reverse ratio a) and b) as 4th grade:2nd grade = 5:8 and 2nd grade:1st grade = 4:3

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

So 4th grade:2nd grade::2nd grade:1st grade = 5:8::4*2:3*2 = 5:8::8:6

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

4th grade:1st grade = 5:6

Step 3) Similar to above step we need to get a common grade between the ratios 1st grade:4th grade and 4th grade and 3rd grade

So 4th grade:1st grade = 5:6 Reversing them we have 1st grade:4th grade = 6:5

and 3rd grade:4th grade = 3:2 Reversing them we have 4th grade:3rd grade = 2:3

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

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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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, \(a/c = a/b x b/d x d/c = 3/4 x 8/5 x 2/3 = 4/5\)

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

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30 Sep 2013, 01:46

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

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.
_________________

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

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22 Nov 2013, 05:04

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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.

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

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27 Dec 2014, 09:37

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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...

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

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03 Jan 2015, 04:37

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).

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

At a Certain school, the ratio of the number of second [#permalink]

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28 Mar 2015, 00:20

asterixmatrix wrote:

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

It's certainly the most time saving and efficient method - thank you Asterixmatrix!

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

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21 May 2015, 17:25

VeritasPrepKarishma wrote:

Lstadt 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!

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

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02 Aug 2016, 22:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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