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# The present ratio of students to teachers at a certain

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Manager
Joined: 14 Dec 2011
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GMAT 1: 630 Q48 V29
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The present ratio of students to teachers at a certain  [#permalink]

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Updated on: 07 Dec 2012, 05:59
9
72
00:00

Difficulty:

15% (low)

Question Stats:

83% (01:49) correct 17% (02:28) wrong based on 2350 sessions

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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?

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

Originally posted by Impenetrable on 26 Mar 2012, 02:48.
Last edited by Bunuel on 07 Dec 2012, 05:59, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 02 Sep 2009
Posts: 62290
Re: Slightly embarrasing  [#permalink]

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26 Mar 2012, 03:02
27
1
20
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 ?

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 $$x$$ --> $$x=15$$ --> $$t=x=15$$.

Answer: E.

Hope it helps.
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Status: May The Force Be With Me (D-DAY 15 May 2012)
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Re: The present ratio of students to teachers at a certain  [#permalink]

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26 Mar 2012, 03:16
14
8
+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
##### General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 62290
The present ratio of students to teachers at a certain  [#permalink]

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02 Aug 2014, 09:44
3
Pasqualo wrote:
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.

Hope it's clear.
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Re: The present ratio of students to teachers at a certain  [#permalink]

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06 Aug 2014, 21:19
3
$$\frac{s}{t} = \frac{30}{1}$$

$$\frac{s+50}{t+5} = \frac{25}{1}$$

Placing s=30t in the above equation

30t+50 = 25t+125

5t = 75

t=15

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

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09 Jun 2016, 13:10
2
2
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?

(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  [#permalink]

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10 Aug 2015, 12:09
1
ggurface wrote:
Hi, is there a quicker way to do this like using Alligation or something that would not involve setting up algebraic equations?

You can try plugging in the values from the options.

Lets say T = 10 ---> S = 300 ---> 350/15 = a bit less than 25 (A and B will only make this ratio lesser, eliminate A and B)

Go for T = 12 --> S = 30*12 = 360 ---> 360+50 / 12+5 = less than 25. Either you can now go further with T = 15 or mark E as the answer .

Check: T = 15 ---> S = 30*15=450 ---> 450+50/20 = 25.
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Re: The present ratio of students to teachers at a certain  [#permalink]

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22 Jun 2018, 06:42
1
Lets set an equation,
30x+50/x+5=25x/1x OR
x(30x+50)=25x(x+5)
30x^2=50x=25x^2+125x
5x^2=75x
x^2=15x
x(x-15)=0
x=15 (E)
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Re: The present ratio of students to teachers at a certain  [#permalink]

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26 Mar 2012, 03:14
How stupid of me. I should slap myself.

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

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25 Oct 2012, 01:28
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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The present ratio of students to teachers at a certain  [#permalink]

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02 Aug 2014, 03:23
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.

Thanks,

P.
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Posts: 62290
Re: The present ratio of students to teachers at a certain  [#permalink]

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02 Aug 2014, 09:05
Pasqualo wrote:
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.

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.

Does this make sense?
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Re: The present ratio of students to teachers at a certain  [#permalink]

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02 Aug 2014, 09:32
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.

Cheers
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Posts: 3
Re: The present ratio of students to teachers at a certain  [#permalink]

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02 Aug 2014, 09:56
Great, now it's crystal clear. Thanks again
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Re: The present ratio of students to teachers at a certain  [#permalink]

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10 Aug 2015, 12:01
Hi, is there a quicker way to do this like using Alligation or something that would not involve setting up algebraic equations?
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Re: The present ratio of students to teachers at a certain  [#permalink]

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11 Aug 2015, 02:26
s:t = 30:1
s= 30x
t = x

(30x + 50)/(x+5) = 25

So, we just solve for x

(30x + 50) = 25x + 125
5x = 75
x=15
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Re: The present ratio of students to teachers at a certain  [#permalink]

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11 Aug 2015, 03:48
$$\frac{s}{t} = \frac{30}{1}$$ => $$s = 30t$$ => $$s-30t = 0$$ ---> (i)

$$\frac{s+50}{t+5} = \frac{25}{1}$$

$$s + 50 = 25t + 125$$

$$s - 25t = 75$$ ---> (ii)

from (i) and (ii), $$t = 15$$.
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Re: The present ratio of students to teachers at a certain  [#permalink]

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25 Mar 2017, 05:18
ratio problems are always tricky for me

i tried to approach it this way:

(30x+50)/x+5=25/1
x=15
E is the answer
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Re: The present ratio of students to teachers at a certain  [#permalink]

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17 Dec 2017, 09:51
The wording is quite confusing, i calculated 15 and then read [ Present number of teachers ]. Question starts with [ Present number] i adjusted for change of 5 teachers :/.
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Re: The present ratio of students to teachers at a certain  [#permalink]

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27 Oct 2018, 10:32
Here the eqation

st=301st=301

s+50t+5=251s+50t+5=251

Placing s=30t in the above equation

5t = 75

t=15

Easy
Re: The present ratio of students to teachers at a certain   [#permalink] 27 Oct 2018, 10:32

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