Question

Phillip received 75 points on a project for school. He can make changes and receive two-tenths of the missing points back. He can make corrections as many times as he wants. Create the formula for the sum of this geometric series, and explain your steps in solving for the maximum grade Phillip can receive. Identify this as converging or diverging.

Answer 1

@satellite73

Answer 2

@mathmale

Answer 3

120% of 75, or 75×1.2
which is 90.

Answer 4

that is what i got..is it correct?

Answer 5

I'm thinking out loud here:
The first term of this geometric series is 75.
What is the common ratio, r?
Is |r|<1, as it must be for convergence of this geometric series?
We both need to think about how we'd find that common ratio, r.

Answer 6

How would you express "two tenths of the missing points?"
If the student's score is 75, how many points are missing, and how did you calculate this result?

Answer 7

25 points are missing and .2?

Answer 8

What would two tenths of the number of points missing / lost come to?

Answer 9

5
which would give him 80...

Answer 10

Yes, that does make sense: (2/10)(25)=50/10=5.

Answer 11

Next: adding back those points, what's the current score?

Answer 12

75+5=80?

Answer 13

this is like the fifteenth time i have seen this exact same stupid problem
where does it come from?

Answer 14

Let's not label it "stupid," OK?

Answer 15

i over think these kind of word problems

Answer 16

you are going to add \[25\times \frac{1}{5}+25\times \left(\frac{1}{5}\right)^2+25\times \left(\frac{1}{5}\right)^3+...\]

Answer 17

what is wrong with "stupid" ? looks stupid to me

Answer 18

use \[\frac{a}{1-r}\] with \[a=5,r=\frac{1}{5}\]

Answer 19

you can do it in your head
what is \(1-\frac{1}{5}\) ?

Answer 20

We started with 75 and added 5 points, which was 2/10 of the missing points, which was 2/10 of 25, or 5.

@topstryker: See whether you can translate what we have done so far into satellite73's expression, which features 1/5 as the common ratio, r. It's important that you understand this (understand where that 1/5 came from).

@satellite73: As helpers, I think we ought to shed as much POSITIVE thought on any given problem as is possible; it's our job to help (not hinder) the problem solver.

Answer 21

@mathmale you are right,
it is positively stupid
still wondering where it comes from though

Answer 22

Bad mood today, Sat?

@topstryker: Satellite wants to know the CONTEXT in which you found this problem.

Answer 23

would it be converging or diverging?..

Answer 24

@topstryker : In your shoes my first question would be "where did that 1/5 come from?" After you're satisfied that you can answer that question, the next question would be "given the first term of this geometric series and the common ratio, r, how would I find the sum of this geometric series?
There is an important criterion for convergence or divergence, and that's that the magnitude of the common ratio, r, must be less than 1.

Answer 25

What is your "r" here? Is the magnitude of "r" less than 1, equal to 1 or greater than 1?

Answer 26

1/5 have to do with 5?

Answer 27

Where did we get 5 originally? Hint: 2/10 of the number of points missing.
Mind going through that calculation again: number of points earned: 75. Number of points missing: ???
2/10 of the numb er of points missing = ??

Answer 28

This is nothing new. We've already done it:

We started with 75 and added 5 points, which was 2/10 of the missing points, which was 2/10 of 25, or 5.

Answer 29

oh ok i see

Answer 30

so you simplified the 2/10

Answer 31

So we're going to add those 5 points to the original 75 to get ... how many points total, as the revised score?

That looks logical...2/10 does result in 1/5.

Answer 32

80 as the revised score

Answer 33

Right. what's missing (how many points short of a perfect score are we)?

Answer 34

20 now

Answer 35

Right. And what's 2/10 of that?

Answer 36

4

Answer 37

Exactly. Add that to 80.

Answer 38

84

Answer 39

Right. So, 100-84, or 16, points are missing. What's 2/10 of that?

Answer 40

Alternatively, what's 1/5 of 16?

Answer 41

3.2

Answer 42

right. If we're on the right track, we keep summing up all these increments in the score.
If you were to use 75 as the first term, a, and 1/5 as the common ratio, r, and if the sum of the series is

a
______ , then what is the sum of this series? This sum represents the max score
1-r

that the student can earn if his original grade was 75.

Answer 43

a=the first term of the series.
r= the common ratio.

Answer 44

93.75?

Answer 45

r=.2?

Answer 46

...which is the same as 1/5, right? Is the magnitude of 1/5 (the magnitude of 0.2) less than 1?

Answer 47

yes it is less than 1

Answer 48

If so, the series converges, and the sum is ....
which is the highest grade the student can earn after a first grade of 75.

Answer 49

So: think about what you have done. The hardest step, I thought, was identifying the value of r. Satellite73 got that right right away: r=1/5.
a is the initial value, obviously 75.

Answer 50

yes, only thing im confused is that formula you just gave me, answered 93.75..but i thought it was 80

Answer 51

that 80 was the student's score after he/she had made corrections for the FIRST time.
After the student had made corr. for the 2nd time, his/her score became 84, right?

Answer 52

We go on and on until the poor student has made corrections a million times. (Just kidding).

Answer 53

oh yes, because he can make as many changes?

Answer 54

Right...he can try fixing up his test paper as many times as he wants.

Answer 55

so the hightest score he can recieve is a 93.75?

Answer 56

Yes, I believe that's the case.

One last time: make certain you know how to identify the initial term of a geom. series and that you know how to figure / obtain the common ratio.

Answer 57

If the common ratio has a magnitude of less than 1, the series converges, and the sum of this infinite series is

a
------
1 - r

Answer 58

thank you so much! i really do appreciate it

Answer 59

My great pleasure. Thanks for sticking with this as long as you did!