Lesson: Growth Tables

Comment on Growth Tables

An engineer designed a ball so that when it was dropped, it rose with each bounce exactly one-half as high as it had fallen. The engineer dropped the ball from a 16-meter platform and caught it after it had traveled 46.5 meters. How many times did the ball bounce?

A. 5
B. 6
C. 7
D. 8
E. 9

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

gmat-admin's picture

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

Rise #1: 8 m.......TOTAL = 24
Drop #2: 8 m.......TOTAL = 32

Rise #2: 4 m.......TOTAL = 36
Drop #3: 4 m.......TOTAL = 40

Rise #3: 2 m.......TOTAL = 42
Drop #4: 2 m.......TOTAL = 44

Rise #4: 1 m.......TOTAL = 45
Drop #5: 1 m.......TOTAL = 46

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

Answer: 5 bounces

How many times will the digit 7 be written when listing the integers from 1 to 1000?
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 .
gmat-admin's picture

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 do you know that you should place a post at 0 meters (starting point) when the question says he places a post for every 7 meters?

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?
gmat-admin's picture

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 question asks the population after 51 months?

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.
gmat-admin's picture

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)

The population of a bacteria culture doubles every 2 minutes. Approximately how many minutes will it take for the population to grow from 1,000 to 500,000 bacteria?
gmat-admin's picture

Hi Brent,

The below question is from Gmat prep test need your help in this one:

Last month a phone company charged $0.07 per minute for the first 300 minutes of calling time and $0.05 per minute for each minute of calling time in excess of the first 300 minutes. For how many minutes of calling time was Bob charged by the phone company last month?

(1) Last month the phone company charged Bob $21.00 for his calling time that was charged at the rate of $0.07 per minute.
(2) Last month the phone company charged Bob $40.00 for his calling time that was charged at the rate of $0.05 per minute.

gmat-admin's picture

Thanks! I didn't knew the answer to this one would be that simple, couldn't decode the language of the first sentence.

Hi Brent,

I could not understand the question stem and the solution to this problem - https://gmatclub.com/forum/on-the-first-even-day-of-the-month-steve-roggers-saves-278807.html

1. What do even days stand for?
2. Why, in your solution, you move from day #2 to #4 and so on in intervals of 2?

Warm Regards,
gmat-admin's picture

Question link: https://gmatclub.com/forum/on-the-first-even-day-of-the-month-steve-rogg...

ODD integers: ...-5, -3, -1, 1, 3, 5, 7,...
EVEN integers: ... -6, -4, -2, 0, 2, 4, 6, 8,...

1. Even days will be days in which the date is an even number.
For example, April 2, May 24, June 30 all qualify since 2, 24 and 30 are even integers.

2. I am considering only the even dates (2, 4, 6, 8,...)

Does that help?


Hey Brent,

a quick question:


it says "the ratio of a frequency to the next higher frequency is a fixed constant". from that I would have never inferred that the constant is *k. Couldn´t it have been +k? +2k,+3k etc.?


gmat-admin's picture

Question link: https://gmatclub.com/forum/a-certain-musical-scale-has-has-13-notes-each...

Great question.

The key word here RATIO.

Let's say we have a sequence of 5 numbers: a, b, c, d, e

If we say the ratio of one number and the number before it is a fixed constant, then that means:
b/a = c/b = d/c = e/d = k

So, if term1 is a, then:
term2 = b = ak
term3 = c = ak^2
term4 = d = ak^3
term5 = e = ak^4

Does that help?


Thanks. I can kind of see the why behind it but I honestly don´t really understand it. It doesn´t make "click" in my head. Is there another way you could explain it?


gmat-admin's picture

Here are some sequences in which the ratio of one number and the number before it is a fixed constant:

Sequence #1) 3, 6, 12, 24, 48, 96,....

Notice that, if we take any term (other than term1) and divide it by the PREVIOUS term, we always get 2.
For example, 6/3 = 2, 12/6 = 2, 24/12 = 2, etc

Notice that we can take the sequence (3, 6, 12, 24, 48, 96,...) and rewrite the terms as follows:
3 = 3
6 = (3)(2)
12 = (3)(2)(2)
24 = (3)(2)(2)(2)
48 = (3)(2)(2)(2)(2)
96 = (3)(2)(2)(2)(2)(2)

Sequence #2) 8, 80, 800, 8000, 80000, ,....

Notice that, if we take any term (other than term1) and divide it by the PREVIOUS term, we always get 10.
For example, 80/8 = 10, 800/80 = 10, 8000/800 = 10 etc

Notice that we can take the sequence and rewrite the terms as follows:
8 = 8
80 = (8)(10)
800 = (8)(10)(10)
8000 = (8)(10)(10)(10)
80000 = (8)(10)(10)(10)(10)

Does that help?


Okayyy, now I understand better. Considering the Q I refered to:


It said that the last term is bigger than the first term. for that reason we know that there is a constant k multiplied each time, right? If it said the last term is smaller than the first time, would that mean that we needed to divide each term with a certain constant?


gmat-admin's picture

Link: https://gmatclub.com/forum/a-certain-musical-scale-has-has-13-notes-each...

Great question.
Division and multiplication are actually pretty similar.

Consider this sequence: 300, 30, 3, 0.3, 0.03

We COULD say that we get each term by DIVIDING the previous term by 10.
That is 300/10 = 30, 30/10 = 3, 3/10 = 0.3, etc

We COULD also say that we get each term by MULTIPLYING the previous term by 0.1
That is 300 x 0.1 = 30, 30 x 0.1 = 3, 3 x 0.1 = 0.3, etc

Notice that MULTIPLYING the previous terms by 0.1 is the same as MULTIPLYING the previous terms by 1/10, and MULTIPLYING the previous terms by 1/10 is the same as DIVIDING the previous terms by 10

So, if the sequence is decreasing, we may be DIVIDING the previous terms by some value greater than 1, OR we may be MULTIPLYING the previous terms by some value between 0 and 1

Does that help?


Yeah, I see now, thanks :D

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