Clark University currently enrolls 8,800 students per year. Talbot Col

Question

Clark University currently enrolls 8,800 students per year. Talbot College currently enrolls 15,100 students per year. The numbers of students enrolled by both schools are increasing each year at a constant rate. If each of these schools continues to enroll an increased number of students annually at its constant rate, in seven years both schools will enroll the same number of students for the first time. Each year after seven years, Clark University will enroll more students per year than Talbot College.

In the table below, identify the rates of increase, in annual students enrolled, for each school that together meet the enrollment forecasts described above. Select only one option in each column.

Apt0810 · Accepted Answer

8800+ 7C = 15100 + 7T
C-T=900

So per the options, Clark = 1800,Talbot=900

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prajjwal_verma · Answer

The number of students are increasing at a constant rate. Wont that imply that students are increasing by a constant numerical factor, which would make the series a GP instead of an AP?

Sajjad1994 · Answer

Official Explanation

If Clark University increases its enrollment by 1,800 students per year, in 7 years it will enroll 21,400 students per year. If Talbot College increases its enrollment by 900 students per year, in 7 years it will enroll 21,400 students per year as well. After the 7-year mark, Clark University will enroll more students per year than Talbot College.

The correct answer is 1,800 enrollments per year for Clark University and 900 enrollments per year for Talbot College.

prajjwal_verma · Answer

Ohh yes Im not arguing that the answer is wrong, in fact when I use the logic I stated above I dont get an answer at all.
My query was regarding the wording of the question, whenever we are told that something increases by a factor (or at a constant rate as stated here), aren't we supposed to assume that the series will be a Geometric Progression instead of an Arithmetic like it is here?
Or should I consider both when approaching these questions?

Sajjad1994 · Answer

If it is increasing by a "constant" factor, then you can think of geometric progression; otherwise, assume it is a simple increase in value.

PS: The above Official explanation was not an answer to your query rather for a generic referrence.

Thank you!

illegallyblonde · Answer

Sajjad1994  is there a general rule that'd help us identify whether constant rate, factor, amount etc. are to be assumed as additions or multiplications?

For this question, post reading 'constant rate' I created my algebraic expression for the 7th year as 8800y^7 instead of 8800 + 7y. Only after looking at the options did I realise that it had to do with addition. Is there anything to keep in mind for this?

pratiksha1998 · Answer

I got very confused with this question, I thought that the question mentions Rate, I automatically started thinking on percentages. It looks like here rates means number of students increasing per year. Can someone direct me to the part of the question which equates reates as number of students? I guess I am missing something in the question.

Shubham.Sharma · Answer

GMATNinja , shouldn't the equation be 8800(1+r1/100)^7=15100(1+r2/100)^7 ?

charlie1195 · Answer

If you think of it in terms of simple interest and compound interest, you are using the formula for compound interest, whereas the problem statement is about SI.

Increasing a constant rate = SI

napolean92728 · Answer

Even I formed the same equation and then got stuck with the large exponents , is there a way we can identify whether it refers to simple interest of compound interest, cause reading the language of the question even I got puzzled

Shubham.Sharma

GMATNinja, shouldn't the equation be 8800(1+r1/100)^7=15100(1+r2/100)^7 ?

napolean92728 · Answer

Correct me if I'm wrong but imo the wording in the question should be constant number instead of constant rate.

Biswajeet1930 · Answer

can you explain how did you do it ???

cheshire · Answer

Given: 
 
 Clark University currently enrolls  8,800  students per year. 
 Talbot College currently enrolls  15,100  students per year. 
 Both increase enrollment each year at a  constant rate . 
 In  7 years , they will  enroll the same number  of students. 
 After that,  Clark  will enroll  more  students than Talbot.  
 Step 1: Set up the equation 
Let:
 
 x = Clark’s annual increase 
 y = Talbot’s annual increase

In 7 years: 8800 + 7x = 15100 + 7y

Step 2: Solve for the difference in growth rates

7x - 7y = 15100 - 8800
7x - 7y = 6300
x - y = 900

So, Clark’s annual rate is  900 more  than Talbot’s.

Step 3: Use answer choices to find matching pair 
From the table, we try values where:

Clark’s rate - Talbot’s rate = 900

Try:
Clark = 1800
Talbot = 900

This works: 
1800 - 900 = 900

Final Answer: 
 
 Clark’s increase =  1800  
 Talbot’s increase =  900

Clark University currently enrolls 8,800 students per year. Talbot Col

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Clark University	Talbot College	Rate of increase (enrollments per year)
		100
		250
		400
		900
		1200
		1800

GMAT Two-part Analysis (TPA) Questions

Clark University currently enrolls 8,800 students per year. Talbot Col

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