The present ratio of students to teachers at a certain

Question

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

(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

Bunuel · Accepted Answer

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.

boomtangboy · Answer

+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

BrentGMATPrepNow · Answer

I like to use two variables for these "changing ratios" questions.

The PRESENT ratio of students to teachers at a certain school is 30 to 1  
Let S = # of students PRESENTLY
Let T = # of teachers PRESENTLY
We can write: S/T = 30/1
Cross multiply to get: (1)(S) = (30)(T)
Simplify:  S = 30T

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  
In this scenario...
S + 50 = # of students 
T + 5 = # of teachers 
So, we get: (S + 50)/(T + 5) = 25/1
Cross multiply to get: (S + 50)(1) = (25)(T + 5)
Expand: S + 50 = 25T + 125
Simplify to get:  S = 25T + 75

We now have 2 equations with 2 variables: 
 S = 30T 
 S = 25T + 75

To solve, replace  S  in the bottom equation with  30T 
We get:  30T  =  25T + 75 
So: 5T = 75
Solve:  T = 15 
So, there are PRESENTLY  15  teachers

Answer: E

Cheers,
Brent

Impenetrable · Answer

How stupid of me. I should slap myself. Thanks a lot!

Pasqualo · Answer

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 (25 x /1 x ). This is what I did wrong when I tried to solve the problem.

Thanks,

P.

Bunuel · Answer

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

Does this make sense?

Pasqualo · Answer

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,: 30 x  + 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

Bunuel · Answer

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.

Pasqualo · Answer

Great, now it's crystal clear. Thanks again

PareshGmat · Answer

\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

ggurface · Answer

Hi, is there a quicker way to do this like using Alligation or something that would not involve setting up algebraic equations?

ENGRTOMBA2018 · Answer

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.

JeffTargetTestPrep · Answer

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.

jabhatta2 · Answer

Hi  VeritasKarishma

Why can't we use an approach similar to like this used on another ratio related question ?

https://gmatclub.com/forum/the-number-of-stamps-that-kaye-and-alberto-had-were-in-the-98116.html#p1640044

KarishmaB · Answer

Saumya2403 :

You can use weighted average here but the figures are not conducive to that method.

A1 = 1/31 - Concentration of teachers in the group
A2 = 5/55 = 1/11 - Concentration of teachers in the added group
Aavg = 1/26 - Concentration of teachers in the combined group

w1/w2 = (1/11 - 1/26) / (1/26 - 1/31) = 15*31/5*11 = 93/11

Ratio of total group earlier : group added = 93 : 11 = x : 55
So there were 465 people earlier out of which 1/31 were teachers i.e. (1/31)*465 = 15 teachers

Answer (E)

KarishmaB · Answer

We can't use the LCM method here because the number in the group is different in the two cases. Teachers and students were added. So it is like 2 mixtures were added to give an avg.

jkolachi · Answer

https://www.youtube.com/watch?v=AsnYfIR3bWU

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=TiJIw9uDKG4

vmonferrino · Answer

The difficulty level of this question here on GMAT Club is  low  (15%). However, on the Official Guide (through the APP efficient learning) is categorized as a  hard  question.

I don't understand - how is such a huge discrepancy possible?

Thank you a lot in advance!
Best,
Vittorio

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