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

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27 Feb 2012, 09:02

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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

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 \(\frac{C}{D}=\frac{3}{2}\);

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.
_________________

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

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12 Jun 2013, 00:36

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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 [#permalink]

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11 Sep 2014, 00:42

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

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

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15 Apr 2016, 20:58

is it possible to equate A/B=3/4 and B/D=8/5 to A/B=6/8 and B/D=8/5 since you have 2 numbers for the 2nd grade or do both numbers have to be in the numerator/denominator?

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

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02 May 2016, 07:24

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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