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

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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Table approach would suit to the problem quick and accurate. factor all the ratios such that two ratio have common number to match up.



based on table, 1st: 3rd = 12:15 = 4:5

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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Bunuel, could you please explain this a little more wordy.
I don't get the step from this

to this


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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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

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.

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

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

Bunuel

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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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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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

Best approach here is to generate a compound ratio that shows the relation between each grade to each other.

Second:Fourth = 8:5
First:Second = 3:4
Third:Fourth = 3:2
First:Third = ?

Let's start with the first two...the common term is "Second":
First:Second = 3:4 ---> 6:8
First:Second:Fourth = 6:8:5 <-----Combine the two inequalities after the transformation

Now we look at the third ratio 3:2...the common term is "Fourth"
First:Second:Fourth = 6:8:5 ----> 12:16:10
Third:Fourth = 3:2 ----> 15:10

Combine both ratios
12:16:15:10
So the ratio of first to third is 12:15 = 4/5

Answer is E.
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I totally agree to the explanations given by Bunuel and others on how to combine these ratios. But, I would like to add an other shortcut method with which you can save some more time .

Step 1. Assume first,second,third,fourth graders as a,b,c and d respectively.

Step 2: Analyze the ratio's given in the question.

b:d = 8:5, a:b = 3:4 , c:d = 3:2

Since ratios are nothing but fractions, we have values of \(\frac{b}{d}\), \(\frac{a}{b}\) and \(\frac{c}{d }\)and we need to find \(\frac{a}{c}\) ?

If we closely analyze the fractions, we can identify that there are few common terms in numerators and denominators. So try to arrange the fractions, in such a way that common terms will get cancelled , giving the end result \(\frac{a}{c.}\)

\(\frac{b}{d} * \frac{a}{b} * \frac{d}{c}\).

In the above step, if you look closely, I have taken the reciprocal of \(\frac{c}{d}\) , because I figured out that d will get cancelled in this case when we simplify.

Note: \(\frac{c}{d} =\frac{3}{2}\) so \(\frac{d}{c} = \frac{2}{3}\)

On cancelling the common terms, we will get \(\frac{a}{c }\) as the result, which is asked in the question.

So instead of combing the ratios, just arranging and multiplying the fractions wisely will give you the same result in very less time.

\(\frac{b}{d }* \frac{a}{b} * \frac{d}{c}\) = \(\frac{a}{c}\) = \(\frac{8}{5} * \frac{3}{4} * \frac{2}{3 } = \frac{4}{5} \)

Option E is the answer.

Thanks,
Clifin J Francis,
GMAT QUANT SME
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On point as always. Great solution.
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Is there any other way?

??? Have you checked the discussion above?
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