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Old 10-01-2004, 03:02 PM   #1 (permalink)
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Math Question: (my term depends on this)

First off, this is a huge part of my grade. Half of it is the problems, the other half is explaining how we did them. I think I did them all right, but I am not very good at math, so please help a fellow TFP'er out.

There are four problems:

1) Find the average of the first:
A) 999 counting numbers
B) 999 even natural numbers
C) 999 odd whole numbers
D) 999 numbers in the set (4, 8, 12, 16...)

My answers:
A) 500
B) 1000
C) 1000
D) 2000
This problem is the one I was really most unsure about. It seems to simple to be right.

Problem 2:

Jane has 16 coins consisting of nickles, dimes, and quarters. Fourteen of her coins are not nickles and all but seven of her coins are quarters. How much money does she have?

My answer:
2 Nickles
9 quarters
5 dimes.

Seemed a little to easy for that to be the answer...


Problem 3:
A long document is printed with each page numbered. Suppose the total number of digits used in numbered is 2901. (For example, a ten page book uses 11 digits. 1-9 each use 1 digit, and ten used 2). Find the number of pages.

My answer:
914

Problem 4:

Suppose the top of a rectangular box measures six inches by eight inches. Suppose that that box is twelve inches tall. Find the distance from one of the bottom corners of the box to the point of intersection of the diagonals of the top of the box. (Hint: try finding the distance from one corner of the top to intersection of the diagonals of the top first.)

My answer:

13 inches.

I am not really that great at math, and I very well may be wrong on all these problems. I'm an English major, and my math skills really are lacking.(No math for over 3 years will do that to you). So, I really appreciate the help that you guys can offer, this is a huge part of my grade!

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Old 10-01-2004, 03:35 PM   #2 (permalink)
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1, 2 and 4 are right
3. is 1002
1002+(1002-9)+(1002-90-9)+(1002-900-90-9)
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Old 10-01-2004, 04:18 PM   #3 (permalink)
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Problem #3:
for pages 1-9 there are 9 digits
for 10 through 19 there are 20 digits. Proof:
10 (2 digits)
11 (2 digits)
12 (2 digits)
13 (2 digits)
14 (2 digits)
15 (2 digits)
16 (2 digits)
17 (2 digits)
18 (2 digits)
19 (2 digits)
2+2+2+2+2+2+2+2+2+2=20
Following the same logic, from 10-99 there are 180 digits, because 90x2=180.
So far you have 9 digits plus 180 digits. 189 digits.
From 100-199 you have 300 digits.
That's 300 digits plus 189 digits, 489 digits.
200-999 is 2400 digits, because 300x8=2400
2400 digits plus 489 digits gives you 2889 digits.
Your goal is to hit 2901. So from 1-999, you have 2889 digits so far.
You need another 12 digits to hit 2901, because 2901-2889=12.
1000=4 digits
1001=4 digits
1002=4 digits
4x3=12 digits
Your final answer should be: 1002 pages.
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Old 10-01-2004, 04:23 PM   #4 (permalink)
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:bounces: I love TFP.
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Old 10-01-2004, 05:24 PM   #5 (permalink)
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shouldn't 1.D be 1998 for the average? The set is increasing by 4 thus we have this:

4*999=3996/2=1998

It's not in the set but you are looking for the average off an odd # based on an even set therefore you should get an odd # (2 in this case being odd).

4*1000=4000/2=2000

B and C should be 999

2*999=1998/2=999

(1+2)+(2*998)/2=999 (because you are still using 2*999, however you need are starting at 1 and going up with odd #'s. The way you have 998 and not 999*2 is because the 1+2 counts as the 999th #.
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Old 10-01-2004, 08:38 PM   #6 (permalink)
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Quote:
Originally Posted by pan6467
shouldn't 1.D be 1998 for the average? The set is increasing by 4 thus we have this:

4*999=3996/2=1998

It's not in the set but you are looking for the average off an odd # based on an even set therefore you should get an odd # (2 in this case being odd).

4*1000=4000/2=2000

B and C should be 999

2*999=1998/2=999

(1+2)+(2*998)/2=999 (because you are still using 2*999, however you need are starting at 1 and going up with odd #'s. The way you have 998 and not 999*2 is because the 1+2 counts as the 999th #.
Can anyone verify?
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Old 10-01-2004, 09:00 PM   #7 (permalink)
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hope you don't wreck the curve. or have to take an ethics class next semester.
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Old 10-01-2004, 09:46 PM   #8 (permalink)
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We are always allowed to work in groups in class anyways, and on this as well. TFP is just my extended group
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Old 10-01-2004, 10:22 PM   #9 (permalink)
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Stare I've tried to find a site that would help, but couldn't. My wife is a math genius when we see each other tomorrow, I'll have her take a look at it.

What college are you at? I'm over at UofA taking Math (only Math I have for my major, but I realize I like it...lol I'm a geek). We just finished matrices and are getting ready for Logic (whoopee).

All I can say is I pray to God I never have to even see (let alone ever do) a Gauss-Jordan equation again.
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I just love people who use the excuse "I use/do this because I LOVE the feeling/joy/happiness it brings me" and expect you to be ok with that as you watch them destroy their life blindly following. My response is, "I like to put forks in an eletrical socket, just LOVE that feeling, can't ever get enough of it, so will you let me put this copper fork in that electric socket?"

Last edited by pan6467; 10-01-2004 at 10:24 PM..
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Old 10-02-2004, 09:15 AM   #10 (permalink)
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All of your answers for the first problem are right, even D.
For problem 1D I crunched some #'s and found that if you plot the points of the averages, you get a linear graph, and the function is avg=2x+2, or y=2x+2. So you want the avg of the sum of 999th # in the series, just plug in 2(999)+2, 1998+2=<u>2000</u>. Try it. If you want the sum of the 5th number and all the numbers in the series before it... let's try it out. The numbers on the left represent x, they just indicates which number in the series we're at. The number in the middle column will be the actual numbers in the series, and the numbers on the right column are the sum so far:
<pre>
x|#in series|sum so far
1| 4 | 4
2| 8 | 12
3| 12 | 24
4| 16 | 40
5| 20 | 60
</pre>
For the 5th #, the sum is 60. So avg is 60/5=<font color=red>12</font>.
Of course, it would take forever to make a table that goes all the way up to 999,
but first let's prove that avg=2x+2 works.
x=5, so avg=2*5+2=10+2=<font color=red>12</font>. Voila! Try it with any number you want.
So again, x=999. avg=2*999+2=1998+2=2000. ^^ Hope this helps.

If you make a similar table for 1B and 1C, you can use avg=x+1.
Taking this a bit further, I think whenever you have a series like these, say you start with 3 and go up in increments of 3, like 3, 6, 9, 12... and you want the average of the sums, you just take the interval, in this case 3, and divide it by 2. That's 1.5. Then y=1.5x+1.5, just in case you need to do something like this again with a different interval.
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Last edited by Stiltzkin; 10-02-2004 at 09:32 AM..
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Old 10-02-2004, 02:42 PM   #11 (permalink)
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I must admit to having spent no time trying to work out your answers, but I was always taught - show ALL your working - even if you don't get the correct end result, at least you will get some marks for your thinking - if you just write answers down and they are wrong, you get nothing...
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Old 10-02-2004, 07:16 PM   #12 (permalink)
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Quote:
Originally Posted by Stiltzkin
All of your answers for the first problem are right, even D.
For problem 1D I crunched some #'s and found that if you plot the points of the averages, you get a linear graph, and the function is avg=2x+2, or y=2x+2. So you want the avg of the sum of 999th # in the series, just plug in 2(999)+2, 1998+2=<u>2000</u>. Try it. If you want the sum of the 5th number and all the numbers in the series before it... let's try it out. The numbers on the left represent x, they just indicates which number in the series we're at. The number in the middle column will be the actual numbers in the series, and the numbers on the right column are the sum so far:
<pre>
x|#in series|sum so far
1| 4 | 4
2| 8 | 12
3| 12 | 24
4| 16 | 40
5| 20 | 60
</pre>
For the 5th #, the sum is 60. So avg is 60/5=<font color=red>12</font>.
Of course, it would take forever to make a table that goes all the way up to 999,
but first let's prove that avg=2x+2 works.
x=5, so avg=2*5+2=10+2=<font color=red>12</font>. Voila! Try it with any number you want.
So again, x=999. avg=2*999+2=1998+2=2000. ^^ Hope this helps.

If you make a similar table for 1B and 1C, you can use avg=x+1.
Taking this a bit further, I think whenever you have a series like these, say you start with 3 and go up in increments of 3, like 3, 6, 9, 12... and you want the average of the sums, you just take the interval, in this case 3, and divide it by 2. That's 1.5. Then y=1.5x+1.5, just in case you need to do something like this again with a different interval.
I understand how you got what you did but why are you adding 2?
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I just love people who use the excuse "I use/do this because I LOVE the feeling/joy/happiness it brings me" and expect you to be ok with that as you watch them destroy their life blindly following. My response is, "I like to put forks in an eletrical socket, just LOVE that feeling, can't ever get enough of it, so will you let me put this copper fork in that electric socket?"
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Old 10-02-2004, 11:08 PM   #13 (permalink)
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Question 2 is ambiguous. It says that fourteen of his coins are not nickels, but that does not necessarily imply that the other two <I>are</I> nickels (If he has 15 coins that are not nickels, it is a valid statement that he has 14 coins that are not nickels). So his coins could be {9,2,5}, {9,1,6} or {9,0,7} (the third option may not be possible, depending on how you interpret the first statement in the question, but either of the first two definitely fit the description). Also, be sure to report your answer as a dollars-and-cents amount, not as a list of coins, since that's what the question is asking for.
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Old 10-06-2004, 12:56 PM   #14 (permalink)
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Location: London, UK
Hope this answer isn't too late...

For Question 1, seeing as you are taking an average, you can you the formula for summing an arithmetic progression.

This is given by:

n/2(2a + (n-1)d)

Where n is the number of terms, 999 in this case
a is the first term, d is the difference in the progression.

Thus the average is then:

1/2(2a + (n-1)d)

So, the answers are:

A 500
B 1000
C 999
D 2000

Note that C is different to B as a=1, n=999, d = 2 for C so the above expression is:

1/2(2 + 998*2) = 1 + 998 = 999

whereas a = 2, n = 999, d = 2 for B

Perhaps someone can double check this, but I reckon it is right
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