Problems and Solutions

irspow · 2005-12-07 14:41:58

Using ESP, I posted the answer to this today in HELP ME!>>calculus help: maximum/minimum problem. Well for a circle with a radius of four anyway. My ESP isn't perfect. I thought that it was funny that you asked the same question as someone else.

Last edited by irspow (2005-12-07 14:44:05)

irspow · 2005-12-07 14:47:40

I have the answer to the square question too, but I can't figure out how to use the bleeping hide tag thing. Old guys like me have issues with computers you know?

Could someone please tell me step by step how to use the hide tag, the description in this thread didn't get through my thick skull. I tried what I thought it said to no avail.

Last edited by irspow (2005-12-07 15:09:49)

ryos · 2005-12-07 15:49:54

I fixed mine (I think).

irspow: Here's an example of the hide tag:

[ hide=Hide Tag Example ]Examples are nifty.[/hide]

Here's what that code does:

Just leave out the extra spaces I put inside the opening hide tag--I did that so the forum wouldn't convert it into a button.

Note that the text following "hide=" is placed in the button.

Last edited by ryos (2005-12-07 15:51:27)

justlookingforthemoment · 2005-12-07 15:52:57

Five easy steps to using the hide tag!

1. Type your text.

e.g. This is going to be hidden.

2. Type 'hide' in square brackets (shown below) just before the text you want to be hidden. This converts all text after it into hidden text.

type <hide> but replace the <> with []

3. Type '/hide' in square brackets (shown below) just after the text you want to be hidden. This stops converting the text to hidden text.

type </hide> but replace the <> with []

4. If you want to change the text on the actual button, to something like 'my answer', add '=my answer' in the tag made in step 2. So now the first tag should look something like this:

<hide=my answer> but once again, replacing the <> with []

The closing hide tag still should stay the same.

5. Click submit and have fun with your button!

Here's one I prepared earlier:

---

Okay, my post is now long and boring and pointless, because ryos got there first...

Last edited by justlookingforthemoment (2005-12-07 16:05:34)

Jai Ganesh · 2005-12-07 16:37:03

ryos wrote:

I fixed mine (I think).

The first part?

You have calculated the area of one triangle. Read the question again.

ryos · 2005-12-07 16:48:44

The only thing I can think of is that I didn't get the biggest possible triangle. My thinking is that the largest equilateral triangle will have all three vertices touch the circle, but I can't prove that to be the case.

I'll wait for irspow to post his solution and see where I went wrong.

Jai Ganesh · 2005-12-07 19:27:43

You needn't wait for long. You have found the area of one third of the bigger equilateral triangle. You had forgot to multiply the result by 3 to give the final answer

Jai Ganesh · 2005-12-07 21:40:04

Problem # k + 59

A sum of money invested for a certain number of years at 8% p.a. simple interest grows to $180. The same sum of money invested for the same number of years at 4% p.a. simple interest grows to $120 only. For how many years was the sum invested?

irspow · 2005-12-08 14:02:10

Thanks a lot ryos and justlookingforthemoment. I truly appreciate the time you spent explaining to an old man that new fangled computer stuff....really thanks!

irspow · 2005-12-08 15:20:44

irspow · 2005-12-08 16:03:50

Jai Ganesh · 2005-12-08 16:10:38

To irspow:
Sir,
In your solution for Problem # k + 58,
you have posted
A = sh/2 = [3a(3a^2)^(1/2)] / 4 = a( 3 + 3^(1/2)) / 4 ≈ 1.183a
It should be [3√(3)*a²]/4.
Solution to problem # k + 59 is not correct.

mathsyperson · 2005-12-09 05:24:24

ganesh wrote:

A sum of money is invested... at simple interest...

irspow · 2005-12-09 07:38:37

Sorry about that. I had [3a(3a^2)^(1/2)] / 4 before I blew in on the last step of simplification which equals the same thing that you posted. darn!

Jai Ganesh · 2005-12-09 20:24:46

Problem # k + 60

Three men walk into a motel and ask for a room. The desk clerk says a room is $30 so each man pays $10 towards the cost. Later, the clerk realizes he made a mistake, that the room should have been $25. He calls the bell boy over and asks him to refund the other $5 to the three men. The bellboy, not wanting to mess with a lot of change dividing the $5 three ways, decides to lie about the price, refunding each man $1 and keeping the other $2 for himself. Ultimately each man paid $9 towards the room and the bellboy got $2, totaling $29. But the original charge was $30, where did the extra $1 go?

mathsyperson · 2005-12-10 00:29:19

krassi_holmz · 2005-12-10 08:41:18

For k+55 problem.
I can give you smaller set than yours. Let P(x) is the product of all prime numbers less than x. Then all the numbers of the set
P(1001)+2
P(1001)+3
...
P(1001)+1000
are non-prime.
Thus the first number of your set is 999!+2 =
4023872600770937735437024339230039857
1937486421071463254379991042993851239
8629020592044208486969404800479988610
1971960586316668729948085589013238296
6994459099742450408707375991882362772
7188732519779505950995276120874975462
4970436014182780946464962910563938874
3788648733711918104582578364784997701
2476632889835955735432513185323958463
0755574091142624174743493475534286465
7661166779739666882029120737914385371
9588249808126867838374559731746136085
3795345242215865932019280908782973084
3139284440328123155861103697680135730
4216168747609675871348312025478589320
7671691324484262361314125087802080002
6168315102734182797770478463586817016
4365024153691398281264810213092761244
8963599287051149649754199093422215668
3257208082133318611681155361583654698
4046708975602900950537616475847728421
8896796462449451607653534081989013854
4248798495995331910172335555660213945
0399736280750137837615307127761926849
0343526252000158885351473316117021039
6817592151090778801939317811419454525
7223865541461062892187960223838971476
0885062768629671466746975629112340824
3920816015378088989396451826324367161
6762179168909779911903754031274622289
9880051954444142820121873617459926429
5658174662830295557029902432415318161
7210465832036786906117260158783520751
5162842255402651704833042261439742869
3306169089796848259012545832716822645
8066526769958652682272807075781391858
1788896522081643483448259932660433676
6017699961283186078838615027946595513
1156552036093988180612138558600301435
6945272242063446317974605946825731037
9008402443243846565724501440282188525
2470935190620929023136493273497565513
9587205596542287497740114133469627154
2284586237738753823048386568897646192
7383814900140767310446640259899490222
2217659043399018860185665264850617997
0235619389701786004081188972991831102
1171229845901641921068884387121855646
1249607987229085192968193723886426148
3965738229112312502418664935314397013
7428531926649875337218940694281434118
5201580141233448280150513996942901534
8307764456909907315243327828826986460
2789864321139083506217095002597389863
5542771967428222487575867657523442202
0757363056949882508796892816275384886
3396909959826280956121450994871701244
5164612603790293091208890869420285106
4018215439945715680594187274899809425
4742173582401063677404595741785160829
2301353580818400969963725242305608559
0370062427124341690900415369010593398
3835777939410970027753472000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
0000000000000000000000000000000000000
000000000002.

And the first number in my list is:
1959034064499908343126250819820638104
6123972390589368223882605328968666316
3798706618519516487894823215962295591
1543601914918952972521526672829228299
0852649023362731392404017939142010958
2613936349594714837571967216722434100
6711851622766113313519248884898991489
2157188308679896875137439519338903968
0949055497503864071060338365866606835
3920101163591790003990449506520329974
9542985993134669814805318474080581207
891125912.

Last edited by krassi_holmz (2005-12-10 08:46:14)

krassi_holmz · 2005-12-10 09:30:47

k+60
In my opinion Mathsyperson is right.
There's a logic mistake. From refuding $1 to each man and keeping $2 for himself doesn't follow that each man paid $9 towards the room and the bellboy got $2 because he(the bellboy) has already taken from the first $30, if you understand me

Last edited by krassi_holmz (2005-12-10 09:31:20)

irspow · 2005-12-10 11:04:49

This is where the logic fails.

krassi_holmz · 2005-12-10 11:48:56

Yes, I wanted to say exactly the same and you said it in more mathematical way.

Jai Ganesh · 2005-12-13 00:53:17

Problem # k + 61

A pole is standing vertically in a lake in such a way that one-fifth of the pole is in the mud, two-thirds is in the water, one-eighth above the water, and a piece of the top, measuring one foot and three inches, is broken off. What is the depth of the lake?

mathsyperson · 2005-12-13 01:04:12

John E. Franklin · 2005-12-13 05:38:42

mathsyperson, how do you know the unaccounted for part is the broken off
piece? ganesh said one-eight is above the water, but he didn't say one-eighth was
currently above the water with respect to the original size?
I think it's a trick question.

mathsyperson · 2005-12-13 06:11:08

That's certainly possible, but ganesh's puzzles don't tend to be tricky like that. And with an answer as nice and round as the one I got, that's too good to be coincidence. Besides, if all the fractions were in relation to the new pole then the information about the broken piece would be useless and you wouldn't be able to calculate the depth because you wouldn't have any lengths to work with.

Jai Ganesh · 2005-12-13 16:13:37

mathsyperson is correct!

Problem # k + 62

In a 5 by 12 rectangle, one of the diagonals is drawn and circles are inscribed in both right triangles thus formed. Find the distance between the centers of the two circles.

Math Is Fun Forum

#251 2005-12-07 14:41:58

Re: Problems and Solutions

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