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The present ratio of students to teachers at a certain [#permalink]

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26 Mar 2012, 03:48

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The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers?

The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers ?

5 8 10 12 15

This might be really embarrasing, since I normally score Q47 or higher, but this question really made me stumble.

What is wrong with my approach (I always use this approach on ratio questions and it normally works well):

Plug in E:

25:1 is the new ratio -> 1 = 1*15 = 15 teachers and 25 = 25*15 = 375 students

15 - 5 = 10 teachers before the increase 375 - 50 = 325 students before the increse

so the old ratio is 325 : 10 which is unequal to 30:1

Where is my mistake?

PS. No need to tell me others ways how to solve it. I am familiar with the 'equations ways'.

Cheers a lot and I am happy ti give away Kudos Lars

The very first step is not correct: the present ratio of students to teachers is 30 to 1, not 25 to 1. So if you want to plug t=15 you should use 30:1 ratio, not 25:1.

Complete solution:

Given: \(\frac{s}{t}=\frac{30x}{x}\) and \(\frac{30x+50}{x+5}=\frac{25}{1}\). Solve \(\frac{30x+50}{x+5}=\frac{25}{1}\) for \(x\) --> \(x=15\) --> \(t=x=15\).

Re: The present ratio of students to teachers at a certain [#permalink]

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26 Mar 2012, 04:16

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+1 E

Here's what I did s/t = 30/1 => s=30T

S+50/t+5 = 25/1

Substitute S=30T in the above equation

30t+50 = 25t+125

5t = 75 t=15

I hope this helps. By the way I was facing the same situation & it helps if you just take a break. I couldnt even recognise simple s v errors :-p
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Re: The present ratio of students to teachers at a certain [#permalink]

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25 Oct 2012, 02:28

I actually thought it would be easier to tackle this problem by plugging in the answers choices, but I was VERY wrong, it's just tedious and prone to errors for this exercise.

Straight algebra saves you at least 1 full minute and it's very simple in this case.

Thanks to Bunuel for the clarifications.
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Re: The present ratio of students to teachers at a certain [#permalink]

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25 Jun 2014, 22:30

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The present ratio of students to teachers at a certain [#permalink]

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02 Aug 2014, 04:23

Hello,

I am referring to Bunuel's solution. Can somebody explain why we are setting up the equation using a variable for the first ratio (30x/1x), but not for the second one (25x/1x). This is what I did wrong when I tried to solve the problem.

I am referring to Bunuel's solution. Can somebody explain why we are setting up the equation using a variable for the first ratio (30x/1x), but not for the second one (25x/1x). This is what I did wrong when I tried to solve the problem.

Thanks,

P.

If you write the first ratio as 30x/x, then you cannot write the second one as 25x/x, because x's there are not the same. You could write it as 25y/y though.

Re: The present ratio of students to teachers at a certain [#permalink]

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02 Aug 2014, 10:32

Thanks for the quick answer Bunuel. I understand that the two ratios are different and should be translated into two different variables (say x and y), but how do we solve for two variables if we do that?

e.g,: 30x + 50 = 25 y (x+5)

I hope I am making sense here. Basically, I don't understand why we use the first ratio with a variable and the second one as a normal numeric value. By the way, thanks for your amazing work on this forum.

Thanks for the quick answer Bunuel. I understand that the two ratios are different and should be translated into two different variables (say x and y), but how do we solve for two variables if we do that?

e.g,: 30x + 50 = 25 y (x+5)

I hope I am making sense here. Basically, I don't understand why we use the first ratio with a variable and the second one as a normal numeric value. By the way, thanks for your amazing work on this forum.

Cheers

If you use two variables, the second one will be simply reduced, that's why we don't need it.

\(\frac{30x+50}{x+5}=\frac{25y}{y}\) --> reduce by y: \(\frac{30x+50}{x+5}=\frac{25}{1}\) --> solve for x.

Re: The present ratio of students to teachers at a certain [#permalink]

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02 Aug 2014, 23:45

E is the answer. The best way to solve this problem is using equations I think. You can work a different way by using the answer choice to check.
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Re: The present ratio of students to teachers at a certain [#permalink]

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11 Sep 2014, 10:49

Impenetrable wrote:

The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers?

The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers?

(A) 5 (B) 8 (C) 10 (D) 12 (E) 15

We are given that the ratio of students to teacher is 30 to 1. We can rewrite this using variable multipliers.

students : teachers = 30x : x

We are next given that student enrollment increases by 50 and the number of teachers increases by 5. With this change the new ratio becomes 25 to 1. We can put all this into an equation:

Students/Teachers 25/1 = (30x + 50)/(x + 5)

If we cross multiply we have:

25(x + 5) = 30x + 50

25x + 125 = 30x + 50

75 = 5x

15 = x

Since x is the present number of teachers, currently there are 15 teachers.

Answer E.
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Re: The present ratio of students to teachers at a certain
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09 Jun 2016, 14:10

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