Transcript for the Math Jam "MOEMS Teachers Math Jam" on Nov 1.

Math Jam hosted by JimMath-MOEMS (Jim Matthews ).

JimMath-MOEMS (19:30:47)
Hello Everyone - This is Jim Matthews in Albany NY. I'm tonight's moderator.

JimMath-MOEMS (19:31:45)
Please let me know if you are a teacher (grade level), a student (grade level) or other. Also let me know where you are geographically. No one else will see these responses.

JimMath-MOEMS (19:32:37)
Here's our first problem for the session.

JimMath-MOEMS (19:33:25)
A train is moving at the rate of 1 mile in 1 minute and 20 seconds. If the train continues at this rate, how far will it travel in 1 hour?

patrick Cao_2 (19:34:33)
8o miles

JimMath-MOEMS (19:35:09)
how did you get 80 miles?

Rocket95 (19:35:08)
80?!

patrick Cao_2 (19:36:19)
I changed 1 min. and 20 secs to 1 1/3 mins. then i multiplied by 60 mins. which is one hour to get 80 miles

crazyfrogz (19:37:06)
50?

mermaidtetty (19:35:15)
45 miles

JimMath-MOEMS (19:38:27)
how did you get 5o? how did you get 45?

thememan (19:38:26)
48 miles

JimMath-MOEMS (19:39:06)
and thememan, how did you get 48?

crazyfrogz (19:38:55)
well 1min and 20 sec is also 80sec

Rocket95 (19:39:20)
the mph is 60 min divided by 1 1/3 min which equals 45

thememan (19:39:34)
e=mc2

JimMath-MOEMS (19:40:17)
thememan,

JimMath-MOEMS (19:40:43)
we're not traveling at the speed of light. do you see the light?

VENI DAMODHARAN (19:40:18)
HI OUR ANSWER IS 45 MILES

mermaidtetty (19:39:24)
there are 3600 seconds in an hour, and 80 seconds to go one mile. 3600 divided by 80 is 45

JimMath-MOEMS (19:42:11)
it looks like you are coming to agreement that the answer is 45 miles which is correct. how far will the train travel in 4 minutes?

mermaidtetty (19:42:34)
3 miles

JimMath-MOEMS (19:42:55)
good

JimMath-MOEMS (19:43:24)
so 3 miles in 4 minutes in ? miles in 60 minutes

JimMath-MOEMS (19:44:30)
good work. i liked the conversion to seconds. let's see what you can do with the next one.

JimMath-MOEMS (19:45:11)
on May 6, 1954 Roger Bannister was the first human to be recorded as running a mile in less than 4 minutes

JimMath-MOEMS (19:45:29)
he did it in 3 minutes 59.4 seconds

JimMath-MOEMS (19:45:55)
to run a mile in 4 minutes what rate of speed must you average during the race?

boohenry (19:47:09)
2/8 miles per minute

JimMath-MOEMS (19:48:27)
boohenry, that is true. can you convert to mph?

eugenewon (19:47:32)
22 ft per second

JimMath-MOEMS (19:48:47)
eugenewon, can you convert to mph?

Rocket95 (19:48:28)
1 MILE= 5280 FEET
5280/4= HOW MANY FEET YOU MUST RUN IN 1 MINUTE
5280 FEET/4 MINUTES= 1320 FEET PER MINUTE

eugenewon (19:49:40)
15 mph

patrick Cao_2 (19:49:46)
15mph

crazyfrogz (19:50:00)
i didn't know that 1 mile was 5280 feet

mermaidtetty (19:50:04)
1/4 X 60 = 15

VENI DAMODHARAN (19:50:56)
.25mile/min*60

JimMath-MOEMS (19:51:49)
veni, that is correct so multiply .25 x 60 for your mph

JimMath-MOEMS (19:52:31)
so you all seem to be getting 15 mph which is correct------any questions on this one or are you ready for problem 3?

JimMath-MOEMS (19:53:24)
if a racehorse runs 1 and 1/4 mile in 2 minutes how far will it run at this rate in 1 hour?

Rocket95 (19:53:54)
37.5 MILES

mermaidtetty (19:53:59)
37.5 miles

patrick Cao_2 (19:54:16)
37 1/2 miles

crazyfrogz (19:54:34)
37.5 miles

Rocket95 (19:55:20)
THERE ARE 30 PAIRS OF 2 MINUTES. 1 1/4=1.25. 1.25*30=37.5

JimMath-MOEMS (19:56:27)
no correct answers yet.

thememan (19:56:54)
42 miles

JimMath-MOEMS (19:57:13)
nope , not 42 miles either

eugenewon (19:57:10)
24 miles

JimMath-MOEMS (19:57:28)
nope, not 24

mermaidtetty (19:57:34)
what did you get then?

JimMath-MOEMS (19:57:59)
come on. really think about this one.

SASWATS (19:58:01)
5/4 * 30 = 37.5 miles

thememan (19:58:10)
39 miles

JimMath-MOEMS (19:58:38)
no you still don't get it!!!

boohenry (19:58:38)
how can ahorse run that long

JimMath-MOEMS (19:59:02)
YESSSSSSSS boohenry got it

mermaidtetty (19:58:41)
the horse cant run that long then

JimMath-MOEMS (19:59:33)
Now you have it! A horse can't run at that rate for that long without dropping dead.

boohenry (19:58:46)
it will die

boohenry (19:59:03)
faint

thememan (19:59:12)
hahahahahaha

Rocket95 (19:59:55)
VERY FUNNY

Rocket95 (20:00:03)
THAT IS NOT MATH!

JimMath-MOEMS (20:00:38)
I just wanted to see if you guys were really thinking. Math people have a sense of humor too.

boohenry (20:00:07)
that was a trick question

Rocket95 (20:00:15)
IT IS A TRICK QUESTION!

JimMath-MOEMS (20:01:25)
yes it was. if the horse was a mechanical horse your answers of 37.5 miles were correct. good.

JimMath-MOEMS (20:01:55)
calm down. next problem coming.

JimMath-MOEMS (20:02:48)
A kangaroo chases arabbit which starts 150 feet ahead of the kangaroo. For every 12 leap of the kangaroo, the rabbit makes a 7 ft leap

JimMath-MOEMS (20:03:08)
How many leaps will the kangaroo have to make to catch up to the rabbit?

eugenewon (20:03:49)
do you mean 12 leap or 12 ft?

JimMath-MOEMS (20:04:31)
thanks. yes it should have bee 12 foot leap

boohenry (20:05:02)
192 feet

boohenry (20:05:18)
16 leaps

JimMath-MOEMS (20:05:35)
which answer and why?

Rocket95 (20:05:35)
DO THE LEAPS OCCUR AT THE SAME TIME?

JimMath-MOEMS (20:05:51)
Yes they do

RichKal-MOEMS (20:06:38)
Stop that.

Rocket95 (20:06:08)
30 LEAPS
12-7=5
150/5=30

patrick Cao_2 (20:06:32)
30 leaps becasue for each leap the kangaroo gains 5 feet on the rabbit. it would take 30 leaps to gain 150 ft

JimMath-MOEMS (20:07:39)
we have one participant who needed to stop. the rest of you can ignore the red remark

eugenewon (20:07:14)
30 leaps

mermaidtetty (20:07:50)
30 leaps

crazyfrogz (20:07:14)
I got 30 leaps because the difference between 12 and 7 is 5 and 5 goes into 150 30 times

JimMath-MOEMS (20:08:33)
nice explanation crazyfrog

Rocket95 (20:08:53)
is this a trick question again?

JimMath-MOEMS (20:09:29)
no. this one is legit.

JimMath-MOEMS (20:10:02)
any questions on crazyfrog's answer?

JimMath-MOEMS (20:10:40)
ready for the next question?

JimMath-MOEMS (20:11:24)
At the equator there is 12 hours of
daylight and 12 hours of darkness each
day.
The sun shines from 6:00am to 6pm
and there is darkness from
6:00pm to 6:00am.
At the equator on November 1`at 6:00am
a snail at the bottom of a 30 foot deep
well starts to climb up.

JimMath-MOEMS (20:11:53)
The snail climbs up 3 feet during the
12 hours of sunlight.
He slowly slides back down 2 feet
during the 12 hours of darkness.
WHEN DOES THE SNAIL GET OUT
OF THE WELL?

patrick Cao_2 (20:12:35)
november 31

mermaidtetty (20:12:38)
November 30

JimMath-MOEMS (20:13:17)
don't forget to put the date and time

mermaidtetty (20:13:14)
november 29 actually

boohenry (20:13:36)
november 28* at 6:00 p.m.

Rocket95 (20:13:53)
november 29 6 pm

patrick Cao_2 (20:14:02)
November 31 6:00 A.M.

mermaidtetty (20:14:56)
november 27 6pm

boohenry (20:15:02)
november 27

boohenry (20:15:17)
november 27 6pm

crazyfrogz (20:15:29)
wait november 27 6:00 A.M.

Rocket95 (20:15:39)
november 27 6 pm

crazyfrogz (20:15:54)
my bad 6:00 P.M.

SASWATS (20:16:17)
November 27 6am.

patrick Cao_2 (20:16:20)
november 30 6 pm

JimMath-MOEMS (20:17:07)
no one's last guess is correct. be careful with this one.

JimMath-MOEMS (20:18:03)
hint: suppose the well is 3 feet deep. when will the snail reach the top?

boohenry (20:18:23)
november 1 at 6pm

JimMath-MOEMS (20:19:32)
right - the snail is out in just 12 hours. so be careful with your time and date!!

JimMath-MOEMS (20:20:22)
remember, the snail starts on november 1 at 6am and he doesn't need food

mermaidtetty (20:20:16)
november 28 at 6 pm?

boohenry (20:20:16)
november 28 6pm

JimMath-MOEMS (20:21:27)
why do you think it's nov 28 at 6pm?

crazyfrogz (20:21:12)
november 28 6:00 P.M.

eugenewon (20:22:02)
November 28 at 6 PM because he starts at November 1st not October 31st

eugenewon (20:22:21)
So 1 day would be November 1st

boohenry (20:22:34)
on the 28 the snail only has 3fett left

Rocket95 (20:22:31)
must be 3 days before 30 days

eugenewon (20:22:35)
So 1 day will be November 2nd i mean

mermaidtetty (20:22:52)
well, the snail starts on november 1, and by november 28 6am, it should be 27 feet up the well, and will reach the top by the end of the day

JimMath-MOEMS (20:23:26)
very nice you guys.

JimMath-MOEMS (20:23:48)
ok - here is our last problem of the session

eugenewon (20:23:30)
i feel bad for the snail

crazyfrogz (20:23:42)
so 1 +30 -3 is november 28 but he just climed and reached the top so it's 6:00 P.M.

JimMath-MOEMS (20:25:19)
Suppose you drive to grandma's house at an average speed of 60mph. You return on the same route at an average speed of 40mph. What speed did you average for your round trip?

boohenry (20:24:18)
ohwell

boohenry (20:24:31)
there are millions of snails in the world anyways

boohenry (20:25:03)
but still poor snail :[

mermaidtetty (20:25:34)
i cant drive

JimMath-MOEMS (20:26:25)
ok - suppose you are in the car that is traveling at those average speeds

Rocket95 (20:25:37)
what is the distance?

Rocket95 (20:26:13)
waht is the distance to granny's house

eugenewon (20:26:18)
yeah, me neither

JimMath-MOEMS (20:27:39)
this is what makes this problem interesting. you don't need to know the distance?

patrick Cao_2 (20:25:54)
50 mph?

mermaidtetty (20:26:24)
50 mph

boohenry (20:26:24)
50 mph

boohenry (20:26:40)
50 mph is the mean or average

eugenewon (20:27:30)
50 mph

boohenry (20:27:46)
50 mph is the average for the round trip

crazyfrogz (20:28:17)
50mph? you add 60 and 40 and you get 100 divided by 2 because there are 2 numbers and you get 50

crazyfrogz (20:26:43)
wait i got 50mph

boohenry (20:26:52)
50mph is the average

eugenewon (20:28:10)
50 mph

SASWATS (20:28:26)
60 + 40 = 100, 100/2 = 50 mph

Rocket95 (20:28:30)
it is not 50 mph

JimMath-MOEMS (20:29:11)
ok - you guys are smart - 50 would be too easy for your last question!!!

patrick Cao_2 (20:28:49)
why not?

JimMath-MOEMS (20:30:22)
why not 50?

crazyfrogz (20:29:32)
wait what?!?!?!?!?!?!?!?!?!?!

mermaidtetty (20:29:40)
does that mean we're right or we're wrong?

boohenry (20:29:53)
is 50 the anser

JimMath-MOEMS (20:31:08)
50 mph is NOT correct. and you don't need to know the distance to grandma's house

Rocket95 (20:30:50)
you are wrong guys

SASWATS (20:31:03)
yes, i think so.

boohenry (20:31:31)
then wats the correct answer

JimMath-MOEMS (20:32:11)
boohenry, can't you get it? do you need a hint or can you do it without the hint?

JimMath-MOEMS (20:34:41)
you guys must be thinking

JimMath-MOEMS (20:35:24)
since we are past our class time i will give you a hint. make up a distance to grandma's house and then see what happens

JimMath-MOEMS (20:36:16)
by the way. the average speed has to be greater or equal to the slower speed and less than or equal to the faster speed!

JimMath-MOEMS (20:37:46)
OK last big hint. what if it is 120 or 240 miles to grandma's house?

eugenewon (20:35:40)
i assumed that i had 100 miles for one trip, so that makes 1 and 2/3 plus 2 and 1/2 hours which equal 4 and 1/6 hours for round trip and divided 200 miles by 4 and 1/6

eugenewon (20:32:45)
48 mph

Rocket95 (20:35:59)
you find the time to get there and back.
in this case, it is 2hours going 3hours back
you add the times and you get 5hours.

Rocket95 (20:37:16)
120 miles one way so it must be 240 round trip.
you divide 240 by 5 to get 48

JimMath-MOEMS (20:39:33)
will it always be 48mph or does it depend on the distance?

JimMath-MOEMS (20:40:31)
can you explain to someone in simple terms why the anwer is less than 40 + 60 divided by 2 which is 50?

Rocket95 (20:40:19)
it is going to be the same average no matter what

eugenewon (20:39:45)
it doesn't matter

mermaidtetty (20:39:44)
always 48

boohenry (20:39:49)
always 48

JimMath-MOEMS (20:41:39)
it is always 48. i will send an explanation of that in a minute or so. why is it less than 50?

eugenewon (20:41:46)
it takes longer to travel at 40 mph than 60 mph

JimMath-MOEMS (20:42:43)
yes, you are spending more time at 40mph then you are at 60mph

patrick Cao_2 (20:42:21)
because you traveled longer at 40 mph than you did 60 mph

eugenewon (20:40:18)
i found it interesting

Rocket95 (20:41:48)
it is

Rocket95 (20:42:58)
if it was 50 mph bothways, then the average would be 50mph

JimMath-MOEMS (20:43:34)
rocket95 - that is true.

mermaidtetty (20:43:35)
you spend more time at 40 mph than 60 mph

JimMath-MOEMS (20:44:22)
In general D = R*T

To find the average speed we can use
AVG_SPEED = DIST_TRAVELED / HOURS_TRAVELED


so for our speeds we have
D = (60 * T1) and D = (40 * T2)
or
T1 = D / 60 and D = T2 / 40

JimMath-MOEMS (20:44:51)
DOES THIS MAKE SENSE?

JimMath-MOEMS (20:45:31)
T1 is the time it takes to drive to grannies and T2 is the time it takes to return

JimMath-MOEMS (20:46:12)
also - the distance you travel D is equal to the rate you are traveling at multiplied by the time you are traveling at that rate

patrick Cao_2 (20:46:13)
can't you just take the total distance and divide it by the total amount of time?

JimMath-MOEMS (20:48:09)
In general D = R*T

To find the average speed we can use
AVG_SPEED = DIST_TRAVELED / HOURS_TRAVELED


so for our speeds we have
D = (60 * T1) and D = (40 * T2)
or
T1 = D / 60 and D = T2 / 40

JimMath-MOEMS (20:49:56)
OK this was cool. We're going to stop. I'll post the whole solution for you to look over. Nice job. Maybe I'll see you next week.

JimMath-MOEMS (20:50:20)
In general D = R*T

To find the average speed we can use
AVG_SPEED = DIST_TRAVELED / HOURS_TRAVELED


so for our speeds we have
D = (60 * T1) and D = (40 * T2)
or
T1 = D / 60 and D = T2 / 40

JimMath-MOEMS (20:52:16)
The AVG_SPEED we are looking for is

2*D / (T1 + T2)

= 2*D / (D/60 + D/40)

= 2*D / ((2*D + 3D) / 120)

= 2*D / (5*D /120)

JimMath-MOEMS (20:52:33)
and
= 2*D * (120 / 5*D)

= 240*D / 5*D

= 240 / 5

= 48

JimMath-MOEMS (20:53:01)
48 yes.

JimMath-MOEMS (20:53:11)
See you next week. Good job.