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Sajjad1994
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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.
­
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­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?­
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prajjwal_verma
­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?­
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!

 
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Sajjad1994

prajjwal_verma
­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?­
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!


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

Shubham.Sharma
GMATNinja, shouldn't the equation be 8800(1+r1/100)^7=15100(1+r2/100)^7 ?
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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 ?
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Correct me if I'm wrong but imo the wording in the question should be constant number instead of constant rate.
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can you explain how did you do it ???

Sajjad1994
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.
­
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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

Biswajeet1930
can you explain how did you do it ???

Sajjad1994
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.
­
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