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## Comment on

Population Increase## Hi Brent,

Please help me with the solution for one:

In a certain pond, 50 fish were caught, tagged, and returned to the pond. A few days later, 50 fish were caught again, of which 2 were found to have been tagged. If the percent of tagged fish in the second catch approximates the percent of the tagged fish in the pond, what is the approximate number of fish in the pond ?

(A) 400

(B) 625

(C) 1,250

(D) 2,500

(E) 10,000

## I answered that question here

I answered that question here: http://www.beatthegmat.com/og10-228-tagged-fish-t286938.html

## i did this using this logic

65008/19984

100%= 19984

65008= 200+

so 65008(1+200/100)

65008*3

19984

since real increase was greater than 200% so option d.

is this approach correct?

## I think that's correct,

I think that's correct, although it's not entirely clear to me when you write the following:

65008(1+200/100)

65008*3

19984

since real increase was greater than 200% so option d.

Can you elaborate?

Cheers,

Brent

## How do I solve this question?

A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least populated district could have?

10,700

10,800

10,900

11,000

11,100

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/a-certain-city-with-population-of-132-000-is-...

## Another way to calculate the

lets say 5, 10, what is next?

10(10/5)=20

## Yes, if the quantities

Yes, if the quantities increase by the same factor, then we can use that technique for find subsequent increases.

Nice work!

## I ended up getting 145,000

From 2007 to 2008 the percent increase was 225%. If I multiply 2.25 by 20 I'll get 65 and also if plug in both the original and new numbers in the percent change formula I end up with 225....(1+x/100)20=65

So to solve for 2008 to 2009 I multipled 65 by 2.25...which was wrong....

If you already know the percent change (225) can't you just multiply it by the original number?

exmple percent change of 200...with the original number of 20....is just simply 40? no? I wouldn't plug that into the formul becuse then I'd get 60.

A percent increase of 200 from 20 is 40, but a percent change of 200 from 20 equates to a percent increse of 300?

## Be careful. You are confusing

Be careful. You are confusing "PERCENT OF" with "PERCENT GREATER THAN"

For example, 40 is 200% OF 20, but 40 is not 200% GREATER THAN 20 (40 is 100% GREATER THAN 20).

Likewise, 20 is 100% OF 20, but 20 is not 100% GREATER THAN 20 (20 is 0% GREATER THAN 20)

One more example: If Joe weighs 80 kg, and Bob's weight is 10% GREATER THAN Joe's weight, then we can write:

Bob's weight = (Joe's weight) + (10% of Joe's weight)

= (80 kg) + (10% of 80 kg)

= 80 kg + 8 kg

= 88 kg

Now on to the question.....

We know that the 2009 population is 225% GREATER THAN the 2008 population.

We can write: 2009 population ≈ (2008 population) + (225% of the 2008 population)

≈ 65,000 + (225% of the 65,000 population)

≈ 65,000 + 146,000

≈ 211,000

Answer: D

Does that help?

## The answer has to be B

The confusion I guess is in the first part of the question a 225% increase from 20 is the same as 225% of 20. So why wouldn't 2008-2009 be 225% of 65?

## This is a situation of

This is a situation of "PERCENT OF" versus "PERCENT GREATER THAN"

We can say that 65,000 is 225 PERCENT GREATER THAN 20,000

Or we can say that 65,000 is 325 PERCENT OF 20,000

Please see my comments above for more information.

Cheers,

Brent

## 3.25 * 65000,

actually, I can use 3 * 6000 (both decreased) = 18000, but the original value must be greater than it, so D

## Yes! That's the GMAT-friendly

Yes! That's the GMAT-friendly solution. Great work!!