| Transcript
for the Math
Jam "MOEMS Teachers Math Jam"
on Feb 6. |
| Math Jam hosted by JimMath-MOEMS
(Jim Matthews ). |
rrusczyk (19:29:50)
Greetings and welcome to a MOEMS Teachers Math Jam!
rrusczyk (19:29:55)
My name is Richard Rusczyk, and I work here at Art of Problem Solving.
rrusczyk (19:29:58)
MOEMS is an outstanding way for students in grades 4-8 to get started with problem solving mathematics. Therefore, we have invited the people from MOEMS to host Math Jams to discuss how to use MOEMS to inspire students to tackle challenging problems.
rrusczyk (19:30:02)
For those of you who may have showed up to see what AoPS classes are like, please view transcripts of other Math Jams from the past, by using the link below:
rrusczyk (19:30:06)
http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php
rrusczyk (19:30:10)
Today's Math Jam will be considerably different than AoPS classes, as it is both for a program that focuses on an earlier age than most of our classes, and because it is aimed at teachers involved with MOEMS, instead of being aimed more at students.
rrusczyk (19:30:14)
One note to our newcomers: This classroom is moderated, meaning that when you make a comment, it will come to the instructors. It will only appear in the classroom if the instructor decides to share it with the class. Therefore, some of your comments may appear after a considerable delay, or possibly not at all.
rrusczyk (19:30:20)
Now, on with the show!
rrusczyk (19:30:24)
I'll now turn things over to your instructor for today, Jim Matthews.
JimMath-MOEMS (19:30:48)
Hello everyone. This is Jim Matthews
in Albany, NY. I'm tonight's moderator.
Please let me know if you are a teacher,
a student or other. Teachers and students
can also indicate a grade level (thanks).
Also, let me know where you are geographically.
JimMath-MOEMS (19:32:35)
We have a nice mix tonight. One other thing. If you are a teacher could you please let me know if you are a PICO.
JimMath-MOEMS (19:33:01)
Warm-up Problem:
On May 6, 1954 Roger Bannister was the first
human to be recorded running a mile in less than
4 minutes. (He did it in 3 minutes 59.4 seconds.)
To run a mile in 4 minutes, what rate of speed must
you average during the race?
KingRoy (19:34:40)
5280/4=1320 feet/minute
goldendomer (19:34:43)
1320 feet per a minute
teatherboard (19:35:01)
1600 meters / 4 minutes --> 400 meters/minute
suzifrizzle (19:35:23)
5280/4 = 1320/60 22ft/sec
JimMath-MOEMS (19:35:54)
How many miles per hour do your answers translate to?
math maniak (19:34:03)
15 mph
hongyu6868 (19:36:14)
15 mph?
7h3.D3m0n.117 (19:36:17)
1320/5280 miles per hour =)
goldendomer (19:36:37)
15 mph
teatherboard (19:36:37)
1 mile / 4 minutes ---> 15 miles / 60 minutes
KingRoy (19:36:45)
15mph
JimMath-MOEMS (19:37:22)
whoops - 7h3 - can you explain your ratio?
7h3.D3m0n.117 (19:37:14)
ah other way
JimMath-MOEMS (19:37:30)
good
math maniak (19:37:41)
1mi/4min x 60min/1hr =15 mph
JimMath-MOEMS (19:38:14)
so it looks like you have the warm-up although bannister was more than a little warm at the end of his run
JimMath-MOEMS (19:38:25)
any questions on this one?
JimMath-MOEMS (19:38:54)
First Problem:
A car is moving at the rate of 2 miles in
3 minutes and 20 seconds. If it continues
at this rate, how far will it travel in
1 hour?
JimMath-MOEMS (19:40:15)
some people have an answer to this one but try to show how you got to your answer ---
teatherboard (19:40:16)
200 seconds : 3600 seconds = 2 miles: miles... 36 miles
KingRoy (19:40:20)
60/(10/3)=18 18*2=36
suzifrizzle (19:40:24)
36 miles 3600 seconds in an hour, 2 miles every 200 seconds
7h3.D3m0n.117 (19:40:36)
60 minutes/3.333333...=18
hongyu6868 (19:40:50)
2/200 = x/3600, 200x = 3600, x = 36 miles
JimMath-MOEMS (19:41:39)
you guys are good. i thought of it the same way as rich kal
RichKal-MOEMS (19:39:53)
6 miles in 10 minutes = 36 miles in 60 minutes.
math maniak (19:42:08)
2mi/200sec x 60 sec/1min x 60 min/1hr = 36 mph
JimMath-MOEMS (19:42:41)
so 36 miles.
JimMath-MOEMS (19:43:11)
ready for the next one or do you want to talk about how to get 36miles
JimMath-MOEMS (19:43:42)
whooops i forgot a part of 7h3's answer - sorry
7h3.D3m0n.117 (19:40:41)
18 times 2 equals 36
JimMath-MOEMS (19:44:26)
Second Problem Part A.
Suppose you want to take pictures of
4 people every way that you can line
them up in a straight line. If it takes
an average of 15 seconds to take a picture,
how long will it take?
JimMath-MOEMS (19:45:25)
remember to simplify your answer. for example, 100 seconds simplifies to 1 minute and 40 seconds (and not 1.66666 minutes)
RichKal-MOEMS (19:45:23)
I assume all 4 are in every lineup?
JimMath-MOEMS (19:46:09)
yep, every way you can line up all 4 of them
goldendomer (19:44:57)
4*3*2=24
goldendomer (19:45:11)
24*15=360
goldendomer (19:46:08)
4*3*2=24 24*15=360 secs/60= 6 mins
7h3.D3m0n.117 (19:45:14)
4! times 15 = 360 seconds or 6 minutes?
KingRoy (19:46:11)
4*3*2*1=24 24*15=360 seconds=6 minutes
nitish (19:46:44)
6 minutes
hongyu6868 (19:46:56)
4! = 24, 24 * 15 = 360 mins = 6 mins
teatherboard (19:46:58)
ABCD, ABDC, ACBD, ACDB, ADBC, ADCB ... repeat for 4 different starting people ... So, 6 ways X 4 people = 24 ways ... And each 4 ways will take one whole minute .... so 6 minutes
JimMath-MOEMS (19:47:20)
you guys are fast.
JimMath-MOEMS (19:48:14)
Second Problem Part B.
Suppose you want to take pictures of
8 people every way that you can line
them up in a straight line. If it takes
an average of 15 seconds to take a picture,
how long will it take?
JimMath-MOEMS (19:49:48)
for people who entered the session after we started, please let me know if you are a teacher, student or other. also let me know where you are geographically. if a teacher let me know your grade level. also let me know if you are a pico. if you are a student let me know your grade level. thanks.
JimMath-MOEMS (19:50:23)
aha - this one has slowed you down a little.
RichKal-MOEMS (19:48:29)
Too long.
nitish (19:48:57)
8!/4
hongyu6868 (19:49:11)
8! = 40320, 40320 / 4 = 10080 mins
KingRoy (19:50:07)
8!*15=604800 seconds=10080 minutes[img id=em-10]
nitish (19:50:08)
40320/4 = 10080 minutes
7h3.D3m0n.117 (19:50:21)
8! times 15 = 604800 secs = 10080 hours? wow either the person is going to spend a loong time doing that or i did something wrong ;p
suzifrizzle (19:50:42)
8! /60 = 10080 minutes/60 = 168 hours
7h3.D3m0n.117 (19:51:31)
nope its minutes nvm so its 10080 minutes =)
JimMath-MOEMS (19:52:22)
but you haven't simplified minutes yet!!!!
JimMath-MOEMS (19:52:35)
100 minutes simplifies to 1 hour and 40 minutes
hongyu6868 (19:53:05)
In that case, 168 hours
JimMath-MOEMS (19:53:32)
hongyu - you can still simplify more!!
KingRoy (19:53:34)
Wow, people sure waste time taking photos[img id=em-7]
suzifrizzle (19:52:40)
7 days
suzifrizzle (19:52:46)
1 week
KingRoy (19:53:16)
10080/60=168 hours
7h3.D3m0n.117 (19:52:58)
10080/60=168 hours or 7 days or 1 week
hongyu6868 (19:53:49)
7 days
nitish (19:54:00)
one week
Silbaugh (19:54:21)
168 hours. I first tried to divide by 15, then I realized it was 4. 7 days.
KingRoy (19:54:25)
168/24=7 days=1 week
JimMath-MOEMS (19:55:16)
yep, 7 days or 1 week
JimMath-MOEMS (19:55:26)
pretty amazing
teatherboard (19:51:30)
5 people = 5 * 6 minutes = 30 minutes .... 6 people = 6 * 30 minutes = 180 min = 3hrs ... 7 people = 7 * 3 hrs = 21 hrs ... 8 people = 8 * 21 hrs = 168 hrs ... 168/24 = 7 days = 1 week (exactly)
JimMath-MOEMS (19:56:05)
teatherboard has a nice answer that shows the growth in the amount of time very nicely
JimMath-MOEMS (19:56:32)
ok so now try this last part
JimMath-MOEMS (19:56:45)
Second Problem Part C.
Suppose you want to take pictures of
12 people every way that you can line
them up in a straight line. If it takes
an average of 15 seconds to take a picture,
how long will it take?
JimMath-MOEMS (19:58:21)
let me know if you have any questions about simplifying your answer
JimMath-MOEMS (20:00:35)
some people are answering but some are still calculating!!!!
JimMath-MOEMS (20:01:11)
remember if you get over 52 and 1/7 weeks you need to keep simplifying
JimMath-MOEMS (20:01:35)
you can assume that there are 365 days in a year
math maniak (19:58:57)
But won't the people all be dead by the time you finish?
JimMath-MOEMS (20:02:43)
i suppose you could take pictures of dead people (sorry) but you may need to train new photographers
i_like_pie (19:58:50)
15! x 12 = 7185024000 seconds = 119750400 minutes = 1995840 hours = 83160 days = 11880 weeks
KingRoy (20:00:42)
12!/4=119750400 minutes=1995840 hours=83160 days=11880 weeks
nitish (19:58:58)
11x 3 x 7 years = 231 years
hongyu6868 (19:59:06)
12!/240 = 1995840 hours = 83160 days = 11880 weeks = 231 years, WOW!
Silbaugh (20:01:50)
228.4 years?
math maniak (20:03:41)
I got 227 yrs 10 mo. 5 days
JimMath-MOEMS (20:04:12)
WOW is right!!!!!!
JimMath-MOEMS (20:04:28)
4 people take 6 minutes
JimMath-MOEMS (20:04:38)
8 people take a week
JimMath-MOEMS (20:04:49)
12 people take over 225 years
JimMath-MOEMS (20:05:40)
isn't that a cool one? much better than how many different lunches can you make with 4 kinds of sandwiches, 3 kinds of cookies and 2 kinds of fruit
teatherboard (20:05:39)
12*11*10*9 * 1 week = 99 * 120 weeks ..... (12000 - 120) weeks = 11880 weeks / 52 weeks ... 228 years 24 weeks (using 364 days/yr) ... so ... 227 years 305 days
JimMath-MOEMS (20:06:51)
some of you guys are so into it you answered the lunch problem!!! pretty funny.
JimMath-MOEMS (20:07:47)
ok ok 24 lunches
JimMath-MOEMS (20:08:04)
did you like the picture taking problem?
JimMath-MOEMS (20:09:28)
for the teachers in our session - there's nothing wrong with the lunch question. it's a basic counting principle. it's just not very inspiring, at least to me.
nitish (20:08:12)
yes
RichKal-MOEMS (20:08:32)
You were right. It's a wonderful problem.
goldendomer (20:08:43)
yeah
RichKal-MOEMS (20:08:44)
But not a snap...
JimMath-MOEMS (20:10:35)
Third Problem:
If a racehorse runs 1 and 1/4 miles in 2 minutes,
how far will it run at this rate in 1 hour?
nitish (20:11:15)
37.50 miles
JimMath-MOEMS (20:12:20)
nope
suzifrizzle (20:11:40)
1.25 * 30 = 37.5 miles
hongyu6868 (20:11:46)
1.25/2 = x/60, 2x = 75, x = 37 1/2
i_like_pie (20:11:49)
(5/4)/2 = x/60 ... x = 150/4 = 37.5 miles
KingRoy (20:12:01)
60/2=30 30*1.25=37.5 miles
goldendomer (20:12:55)
1.25/2=x/60 x=37.5
JimMath-MOEMS (20:13:34)
no it won't run 37.5 miles
suzifrizzle (20:14:08)
37 and 1/2 miles?
JimMath-MOEMS (20:14:34)
no not 37 and 1/2 miles
i_like_pie (20:14:56)
75/2 ?
JimMath-MOEMS (20:15:16)
nope
JimMath-MOEMS (20:15:51)
the poor horse would die before it got to 5 miles at that rate
suzifrizzle (20:16:05)
groan
teatherboard (20:16:56)
Is this an ACTUAL horse - or one of those VIRTUAL horses???
JimMath-MOEMS (20:17:28)
thanks suzifrizzle - i was hoping to get more groans on this one.
JimMath-MOEMS (20:17:50)
but now you have the answer to the fourth problem
JimMath-MOEMS (20:18:04)
Fourth Problem:
If a mechanical horse run 1 and 1/4 miles in 2 minutes,
how far will it run at this rate in 1 hour?
suzifrizzle (20:17:37)
hahaha
hongyu6868 (20:17:45)
Alright, my new answer is 4.99..... miles
JimMath-MOEMS (20:19:22)
Fifth Problem:
A cheetah chases an impala which starts 200 feet ahead
of the cheetah. For every 12 foot leap of the cheetah,
the impala makes a 5 foot leap. How many leaps will the
cheetah have to make to catch up to the impala?
JimMath-MOEMS (20:21:46)
remember - is there any such thing as 5/7 of a leap?
KingRoy (20:20:57)
200/(12-5)=28 and 4/7
JimMath-MOEMS (20:22:31)
or 4/7 of a leap
hongyu6868 (20:20:54)
12x = 200 + 5x, 7x = 200, so 29 leaps for cheetah to get ahead
suzifrizzle (20:21:07)
the cheetah gains 7 feet with each leap. to overtake the 200 ft lead, 200/7 or 29 leaps
i_like_pie (20:21:15)
5x + 200 = 12x ... 29 full leaps
nitish (20:22:32)
29 leaps
suzifrizzle (20:22:40)
the cheetah has to make that 29th leap
RichKal-MOEMS (20:22:48)
So 29 leaps, the last being shorter than 12 feet.
teatherboard (20:23:31)
12-5 =7 foot 'catch-up' / leap ... 200 / 7 = 28.428571... so, on the 29th leap, the cheetah should catch the impala (assuming the impala has not been trained in standard evasive maneuvers!)
JimMath-MOEMS (20:23:51)
you guys are goooood
JimMath-MOEMS (20:24:07)
try this one
JimMath-MOEMS (20:25:01)
Sixth Problem:
At the equator there is 12 hours of daylight and
12 hours of darkness each day.
The sun shines from 6:00am to 6:00pm and there is
darkness from 6:00pm to 6:00am.
At the equator on February 1 at 6:00am a
snail at the bottom of a 30 foot deep well
start to climb up.
JimMath-MOEMS (20:25:24)
The snail climbs up 3 feet during the 12 hours
of daylight. He slowly slides back down 2 feet
during the 12 hours of darkness.
WHEN DOES THE SNAIL GET OUT OF THE WELL?
suzifrizzle (20:26:49)
He makes a net gain of 1 foot each day, but on the last day he gets out... doesn't wait around to slide back. 29 days
hongyu6868 (20:27:02)
30/(3-2) = 30 days, BUT the snail does not have to slide back down on the 29th day, so 29 days
JimMath-MOEMS (20:27:42)
similar answers but you are a little off
nitish (20:25:45)
in 27 days
JimMath-MOEMS (20:27:59)
also a little off - be careful
math maniak (20:26:52)
28 days because if he's out he doesn't slide back
JimMath-MOEMS (20:28:14)
also a little off
math maniak (20:27:44)
Is it a Leap year?
JimMath-MOEMS (20:28:43)
it won't matter!!
KingRoy (20:26:45)
30-3=27 27/1=27 27 and 1/2 days
cat810930 (20:28:17)
for every 24 hours, the snail gains 1 foot, so 30/1=1 The snail gets out in 30 days
JimMath-MOEMS (20:29:27)
cat81 - nope. be careful
i_like_pie (20:27:54)
The snail advances 1 foot every day, so 27 days to get the first 27 feet. Then, on the 28th day, 30 feet. The snail gets out on March 1.
JimMath-MOEMS (20:30:55)
how many days are in february?
JimMath-MOEMS (20:31:30)
remember - When not how long?
KingRoy (20:31:23)
On Febuary 27th at 6:00 pm
JimMath-MOEMS (20:32:14)
nope
suzifrizzle (20:31:26)
He gets out on February 28th
JimMath-MOEMS (20:32:36)
be more precise
nitish (20:32:37)
February 28th 6:00 am
JimMath-MOEMS (20:32:52)
nope
i_like_pie (20:32:38)
February 28 6:00 pm
JimMath-MOEMS (20:33:29)
is this it?
nitish (20:33:35)
28th 6 p.m
teatherboard (20:29:55)
Whoops - 27 1/2 days from Feb 1st (6am) ... Feb 28th (6pm) ...
JimMath-MOEMS (20:34:45)
yes - after 27 days the snail is at the 27 foot mark
JimMath-MOEMS (20:34:53)
so what happens on the 28th day?
hongyu6868 (20:35:44)
The snail climbs 3 ft. and is out
JimMath-MOEMS (20:36:17)
in 12 hours he climbs 3 feet, he's out ---- so after 27 1/2 days of climbing the snail rests
JimMath-MOEMS (20:36:31)
outside of the well
JimMath-MOEMS (20:36:50)
So - if we stop here we can say that all's well that ends well
math maniak (20:37:02)
groan!!
suzifrizzle (20:37:05)
Is there some way we can all groan together?
teatherboard (20:37:07)
ha ha ha
JimMath-MOEMS (20:37:59)
hey you guys - thanks for a great session. you were all fantastic - smart kids and insightful teachers
