Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

At a Certain school, the ratio of the number of second [#permalink]
16 Jul 2009, 10:59

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

84% (02:33) correct
16% (01:58) wrong based on 150 sessions

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

Re: ratio and proportion formula [#permalink]
16 Jul 2009, 11:13

1

This post received KUDOS

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.

Re: ratio and proportion formula [#permalink]
16 Jul 2009, 11:20

1

This post received KUDOS

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... _________________

Re: ratio and proportion formula [#permalink]
16 Jul 2009, 13:25

3

This post received KUDOS

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

Re: ratio and proportion formula [#permalink]
03 Nov 2009, 08:27

2

This post received KUDOS

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

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. _________________

Re: ratio and proportion formula [#permalink]
21 Feb 2012, 21:33

1

This post received KUDOS

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/cwould 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.

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

Re: At a Certain school, the ratio of the number of second [#permalink]
30 Sep 2013, 02:36

Expert's post

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. _________________

Re: At a Certain school, the ratio of the number of second [#permalink]
22 Nov 2013, 05:04

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! J

Kudos Rain!!

gmatclubot

Re: At a Certain school, the ratio of the number of second
[#permalink]
22 Nov 2013, 05:04