# Math problem **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [September 22, 2009, 12:53pm UTC](https://boards.straightdope.com/t/math-problem/511071 "2009-09-22T12:53:01Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Keweenaw](https://avatars.discourse-cdn.com/v4/letter/k/da6949/32.png) [@Keweenaw](https://boards.straightdope.com/u/Keweenaw)\ **Post date:** [September 22, 2009, 12:53pm UTC](https://boards.straightdope.com/t/math-problem/511071/1 "2009-09-22T12:53:01Z")
Got this in an email from a friend yesterday. I’ll post my answer after awhile. You have a 12.00 in diameter record. The dead zone in the center is 3.50 in dia. there is a 0.25 in dead zone around the outside. The groove pitch is 0.10 in. If the needle is started at the exact beginning of the groove, how far will it have traveled when it reaches the end? ---
**Author:** ![Aspidistra](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/aspidistra/32/3894_2.png) [@Aspidistra](https://boards.straightdope.com/u/Aspidistra)\ **Post date:** [September 22, 2009, 1:18pm UTC](https://boards.straightdope.com/t/math-problem/511071/2 "2009-09-22T13:18:10Z")
4 inches. The oldies are the best ones 🙂 ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [September 22, 2009, 1:58pm UTC](https://boards.straightdope.com/t/math-problem/511071/3 "2009-09-22T13:58:04Z")
To be really precise, a little over 4 inches, depending on how long the tonearm is ---
**Author:** ![Machine\_Elf](https://avatars.discourse-cdn.com/v4/letter/m/82dd89/32.png) [@Machine\_Elf](https://boards.straightdope.com/u/Machine_Elf)\ **Post date:** [September 22, 2009, 2:37pm UTC](https://boards.straightdope.com/t/math-problem/511071/4 "2009-09-22T14:37:03Z")
> [@Keweenaw](#): > > Got this in an email from a friend yesterday. I’ll post my answer after awhile. > > You have a 12.00 in diameter record. The dead zone in the center is 3.50 in dia. there is a 0.25 in dead zone around the outside. The groove pitch is 0.10 in. If the needle is started at the exact beginning of the groove, how far will it have traveled when it reaches the end? This is either an honest math problem, or an asshole word problem. Honest math problem? grooved area divided by groove width equals groove length. pi /4 \* (11.5^2 - 3.5^2 ) / 0.10 = 942.5 inches. (incidentally, 0.1 inches is a pretty damn fat record groove…) This strategy is also helpful when trying to figure out how far you’ve walked while mowing your lawn. Asshole word problem? Groove length may not be the same as “how far the needle traveled.” If the record spins beneath a needle on a tonearm - as in a traditional phonograph - then the needle hardly moves at all; it just moves from the outer edge of the record to the inner edge, in this case a distance of 4 inches. ---
**Author:** ![Keweenaw](https://avatars.discourse-cdn.com/v4/letter/k/da6949/32.png) [@Keweenaw](https://boards.straightdope.com/u/Keweenaw)\ **Post date:** [September 23, 2009, 3:56pm UTC](https://boards.straightdope.com/t/math-problem/511071/5 "2009-09-23T15:56:54Z")
> [@Joe Frickin Friday](#): > > This is either an honest math problem, or an asshole word problem. > > Honest math problem? grooved area divided by groove width equals groove length. > > pi /4 \* (11.5^2 - 3.5^2 ) / 0.10 = 942.5 inches. (incidentally, 0.1 inches is a pretty damn fat record groove…) > > This strategy is also helpful when trying to figure out how far you’ve walked while mowing your lawn. > > Asshole word problem? Groove length may not be the same as “how far the needle traveled.” If the record spins beneath a needle on a tonearm - as in a traditional phonograph - then the needle hardly moves at all; it just moves from the outer edge of the record to the inner edge, in this case a distance of 4 inches. My first answer was the 4" of travel with respect to the ground. I also used the area = length\*pitch to get the needle travel with respect to the record surface. The expected (trick) answer was indeed 4" ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [September 23, 2009, 4:18pm UTC](https://boards.straightdope.com/t/math-problem/511071/6 "2009-09-23T16:18:24Z")
But is the record player on a treadmill? ---
**Author:** ![Enderw24](https://avatars.discourse-cdn.com/v4/letter/e/ba9def/32.png) [@Enderw24](https://boards.straightdope.com/u/Enderw24)\ **Post date:** [September 23, 2009, 4:23pm UTC](https://boards.straightdope.com/t/math-problem/511071/7 "2009-09-23T16:23:43Z")
Well, if we’re going with asshole word problem, then it should be pointed out that the needle moves every single time it hits something inside the groove. Indeed, that’s the entire point of the needle. So try calculating that! ---
**Author:** ![Rigamarole](https://avatars.discourse-cdn.com/v4/letter/r/77aa72/32.png) [@Rigamarole](https://boards.straightdope.com/u/Rigamarole)\ **Post date:** [September 23, 2009, 4:41pm UTC](https://boards.straightdope.com/t/math-problem/511071/8 "2009-09-23T16:41:54Z")
What the hell is a record? Is that like a Zune? ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [September 23, 2009, 4:53pm UTC](https://boards.straightdope.com/t/math-problem/511071/9 "2009-09-23T16:53:05Z")
> [@Rigamarole](#): > > What the hell is a record? Is that like a Zune? It’s like a CD, only bigger and black instead of silver. (You must remember CDs). ---
**Author:** ![Rigamarole](https://avatars.discourse-cdn.com/v4/letter/r/77aa72/32.png) [@Rigamarole](https://boards.straightdope.com/u/Rigamarole)\ **Post date:** [September 23, 2009, 5:41pm UTC](https://boards.straightdope.com/t/math-problem/511071/10 "2009-09-23T17:41:46Z")
I see my attempt at sarcasm failed. ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [September 23, 2009, 5:59pm UTC](https://boards.straightdope.com/t/math-problem/511071/11 "2009-09-23T17:59:27Z")
> [@Rigamarole](#): > > I see my attempt at sarcasm failed. Join the club. ---
**Author:** ![ivylass](https://avatars.discourse-cdn.com/v4/letter/i/9de053/32.png) [@ivylass](https://boards.straightdope.com/u/ivylass)\ **Post date:** [September 23, 2009, 7:24pm UTC](https://boards.straightdope.com/t/math-problem/511071/12 "2009-09-23T19:24:44Z")
Mark is playing poker at a casino. Mark starts playing with 140 chips, 20% of which are $100 chips and 80% of which are $20 chips. For his first bet, Mark places chips, 10% of which are $100 chips, in the center of the table. If 70% of Mark’s remaining chips are $20 chips, how much money did Mark bet? A. $1,960 B. $1,740 C. $1,540 D. $3,080 E. $2,640 I got that Mark has 28 $100 chips and 112 $20 chips. But if his bet consists of 10% $100 chips, then it consists of 90% $20 chips. I can’t figure out the math from there. (I know Part = Percent x Whole, but I don’t know if that applies here.) ---
**Author:** ![RTFirefly](https://avatars.discourse-cdn.com/v4/letter/r/c77e96/32.png) [@RTFirefly](https://boards.straightdope.com/u/RTFirefly)\ **Post date:** [September 23, 2009, 8:00pm UTC](https://boards.straightdope.com/t/math-problem/511071/13 "2009-09-23T20:00:52Z")
> [@Joe Frickin Friday](#): > > Honest math problem? grooved area divided by groove width equals groove length. > > pi /4 \* (11.5^2 - 3.5^2 ) / 0.10 = 942.5 inches. (incidentally, 0.1 inches is a pretty damn fat record groove…) That’s what I came up with, too, for this interpretation of the problem. > [@](#): > > This strategy is also helpful when trying to figure out how far you’ve walked while mowing your lawn. Oh, I know the answer to this one: too damned far, considering that I’m pushing a lawn mower around the yard. ---
**Author:** ![Taber](https://avatars.discourse-cdn.com/v4/letter/t/ac91a4/32.png) [@Taber](https://boards.straightdope.com/u/Taber)\ **Post date:** [September 23, 2009, 8:10pm UTC](https://boards.straightdope.com/t/math-problem/511071/14 "2009-09-23T20:10:39Z")
> [@ivylass](#): > > Mark is playing poker at a casino. Mark starts playing with > 140 chips, 20% of which are $100 chips and 80% of which are > $20 chips. For his first bet, Mark places chips, 10% of > which are $100 chips, in the center of the table. If 70% of > Mark’s remaining chips are $20 chips, how much money did > Mark bet? > > A. $1,960 > B. $1,740 > C. $1,540 > D. $3,080 > E. $2,640 > > I got that Mark has 28 $100 chips and 112 $20 chips. But if his bet consists of 10% $100 chips, then it consists of 90% $20 chips. I can’t figure out the math from there. (I know Part = Percent x Whole, but I don’t know if that applies here.) I got A, but it is very possible that I did something wrong. I first set up the equation 112 - .9x = 2.33(28-.1x), where x is the number of chips bet. Solve for x to get 70 chips bet, 10% of that is 7, so 7 $100 chips are bet, 63 $20 chips are bet. That comes out to 1,920 ---
**Author:** ![ivylass](https://avatars.discourse-cdn.com/v4/letter/i/9de053/32.png) [@ivylass](https://boards.straightdope.com/u/ivylass)\ **Post date:** [September 23, 2009, 8:11pm UTC](https://boards.straightdope.com/t/math-problem/511071/15 "2009-09-23T20:11:53Z")
I’m sorry, where do you get 2.33? ---
**Author:** ![muldoonthief](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/muldoonthief/32/3188_2.png) [@muldoonthief](https://boards.straightdope.com/u/muldoonthief)\ **Post date:** [September 23, 2009, 8:17pm UTC](https://boards.straightdope.com/t/math-problem/511071/16 "2009-09-23T20:17:00Z")
I got the same answer as **Taber**. 2.33 = 70%/30%. The remaining chips after the bet is placed are 70% twenties and 30% hundreds. ---
**Author:** ![Taber](https://avatars.discourse-cdn.com/v4/letter/t/ac91a4/32.png) [@Taber](https://boards.straightdope.com/u/Taber)\ **Post date:** [September 23, 2009, 8:18pm UTC](https://boards.straightdope.com/t/math-problem/511071/17 "2009-09-23T20:18:00Z")
> [@ivylass](#): > > I’m sorry, where do you get 2.33? To explain how I got that equation. 112 - .9x is equal to how many $20 chips are left after the bet, and 28-.1x is the number of $100 dollar after the bet. Since we know that $20 chips make up 70% of the total after the bet, we know there are 2.3333 times as many $20 chips as $100 chips. 30% X 2.33 = 70% ---
**Author:** ![RTFirefly](https://avatars.discourse-cdn.com/v4/letter/r/c77e96/32.png) [@RTFirefly](https://boards.straightdope.com/u/RTFirefly)\ **Post date:** [September 23, 2009, 8:23pm UTC](https://boards.straightdope.com/t/math-problem/511071/18 "2009-09-23T20:23:59Z")
> [@ivylass](#): > > Mark is playing poker at a casino. Mark starts playing with > 140 chips, 20% of which are $100 chips and 80% of which are > $20 chips. For his first bet, Mark places chips, 10% of > which are $100 chips, in the center of the table. If 70% of > Mark’s remaining chips are $20 chips, how much money did > Mark bet? > > A. $1,960 > B. $1,740 > C. $1,540 > D. $3,080 > E. $2,640 > > I got that Mark has 28 $100 chips and 112 $20 chips. But if his bet consists of 10% $100 chips, then it consists of 90% $20 chips. I can’t figure out the math from there. (I know Part = Percent x Whole, but I don’t know if that applies here.) Full algebraic workup, in case anyone needs a quick cure for insomnia: Setup, part 1: let B = number of chips bet, and R = number of chips remaining. Info: We know that Mark starts off with 28 $100 chips and 112 $20 chips. So B+R=140. We know that 10% of the chips in Mark’s first bet (0.1_B) consist of the $100 chips, so 90% of the chips in the bet (0.9_B) are $20 chips. And we know that 70% of the chips he’s still got (0.7_R) after placing his bet are $20 chips, so 30% of his remaining chips (0.3_R) are $100 chips. Setup, cont: [spoiler] $100 chips: 28 - 0.1_B = 0.3_R. $20 chips: 112 - 0.9_B = 0.7_R. But B = 140 - R, so: 28 - 0.1(140-R) = 0.3\*R. Which reduces to: 28 - 14 + 0.1_R = 0.3_R. Or: 14 = 0.2\*R, or 70 = R. And since B = 140 - R, B = 70 as well. So 0.1\*B (the number of $100 chips in the bet) = 7, and there are 63 $20 chips in the bet. And the bet is $700 + $1260 = $1960. [/spoiler] ---
**Author:** ![Keweenaw](https://avatars.discourse-cdn.com/v4/letter/k/da6949/32.png) [@Keweenaw](https://boards.straightdope.com/u/Keweenaw)\ **Post date:** [September 24, 2009, 12:08pm UTC](https://boards.straightdope.com/t/math-problem/511071/19 "2009-09-24T12:08:58Z")
Thought about making this a new thread, but I’ll just add it here. This one took me a several lunchtimes to work out. It takes 70 cows 24 days to eat all the grass in a pasture. The same pasture can sustain 30 cows for 60 days. How many cows would take 96 days to eat all the grass? Assume that all the grass grows at the same rate and that all the cows eat grass at the same rate. Also assume that the initial height of the grass is the same in all cases. ---
**Author:** ![ivylass](https://avatars.discourse-cdn.com/v4/letter/i/9de053/32.png) [@ivylass](https://boards.straightdope.com/u/ivylass)\ **Post date:** [September 25, 2009, 11:19pm UTC](https://boards.straightdope.com/t/math-problem/511071/20 "2009-09-25T23:19:15Z")
Okay, I’ve got 70/30 = 24/60. I believe we’re trying to find out an inverse proportion, so we invert it to 70/30 = 60/24, right? Maybe not. Dammit, I’ve been studying this too. It’s like the answer’s there, but I can’t get the math to work. [Next page](https://boards.straightdope.com/t/math-problem/511071.md?page=2)