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## Comment on

Growth Tables## An engineer designed a ball

A. 5

B. 6

C. 7

D. 8

E. 9

How do you design the equation for this problem using the doubling formula?

## An equation is a little

An equation is a little trickier, because we're calculating the SUM of a series of UPs and DOWNs, which will require the sum of two different expressions.

I'd stick with a version of the "growth" table.

Drop #1: 16 m.......TOTAL = 15

BOUNCE #1

Rise #1: 8 m.......TOTAL = 24

Drop #2: 8 m.......TOTAL = 32

BOUNCE #2

Rise #2: 4 m.......TOTAL = 36

Drop #3: 4 m.......TOTAL = 40

BOUNCE #3

Rise #3: 2 m.......TOTAL = 42

Drop #4: 2 m.......TOTAL = 44

BOUNCE #4

Rise #4: 1 m.......TOTAL = 45

Drop #5: 1 m.......TOTAL = 46

BOUNCE #5

Rise #5: 0.5 m.......TOTAL = 46.5

CATCH!

Answer: 5 bounces

## How many times will the digit

Will we use table method or some different method . If we use table method then how ?

Thanks Brent for the help and making such a great site with great content .

## Check out my solution here:

Check out my solution here: http://www.beatthegmat.com/how-many-times-will-the-digit-7-t293940.html

I'm not sure if the Table Method would be of much use in this question.

## For the question below, how

A farmer constructs a fence along the northern edge of his property, using materials such that he places a post every 7 meters. if he uses 100 posts, how many meters will the fence span?

## If we're going to build a

If we're going to build a fence, we must place the first post somewhere. With only 1 post in place, there's no fence.

When we place the 2nd post, we can create a fence that's 7 meters long. So, WHEREVER we place the first post can be considered the 0 meter mark.

BTW, my solution to this question is here: https://gmatclub.com/forum/a-farmer-constructs-a-fence-along-the-norther...

## Dear Brent, how if the

Please correct my approach :

1. First, I define the pattern which is 2^(n-1) X 11.

2. Second, we must calculate how much is n.

3. So, I use the same approach to count how many number in consecutive. I calculate = [(51-0)/3] + 1 = 18.

Thus, the number after 51 months is : 2^(17) X 11.

## That's a perfectly valid

That's a perfectly valid approach.

A slightly faster approach is to recognize that, after t months, the population = (11)[2^(t/3)]

So, after 51 months, the population = (11)[2^(51/3)] = (11)(2^17)

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