Peter is a high school math teacher. He walks to work from his house everymorning. He figures out that if he walks 60meters/minute, he will be 8 minutes late; if he walks faster up to 120meters/minute, he will be 16 minutes late; ask:
How far from Peter's house to his school?
--- This is a real math for my 11-year-old cousin's test paper !
2007-03-02 19:04:57 · 6 answers · asked by Shanghai Amanda 2 in Science & Mathematics ➔ Mathematics
Dont get this one. How can he be later if he walks faster ?
does he passes his school and then walk back ? is that why although he goes faster he arrives later. it makes no sense in my head. A better question would be : What is the name of the teacher ?
2007-03-02 19:28:18 · answer #1 · answered by gjmb1960 7 · 0⤊ 0⤋
You must have typed the problem incorrectly.It should be if he walks 120m/minute,he will be 16 minutes earlier.
I am solving the problem treating my assumption as correct
If the distance between Peter's house and the school was 1meter
In the first case he would have taken 1/60 minutes to cover the distance
iWhile travelling 120 m/sec,he would have taken 1/120 minute to cover the distance
Difference of time=1/60 -1/120
=2/120 - 1/120
=1/120 minute
now,the difference of time reaching the school 6 minutes late and 12 minutes earlier is 6+12=18 minutes
Therefore,If the difference of time is 1/120 minutes,the distance is 1 meter
So,when the difference of time is 1 minute,the distance is120m
The distance in case of 18 minutes differenceis
120*18m
=2160m
=2.16km
You can solve the problem by algebrical equation also
let the distance be x meter
By the problem
x/60 -x/120=18
=>2x-x=2160 [multiplying the terms of both sides by 120]
=>x=2160 meter
kkkkkTherefore,the distanc is 2160 metr or 2.16 Km.
2007-03-03 03:34:26 · answer #2 · answered by alpha 7 · 0⤊ 0⤋
The problem as stated doesn't make sense - there must be a misprint. If the "16 minutes" is replaced by something less than 8 minutes, then the problem could be solved. For example, let's replace the 16 minutes with 6 minutes. Then the time required to walk to school is 2 minutes less at the faster pace. The formula for motion is
(rate) times (time) = (distance)
If we let D represent the distance to school and T represent the time required to walk to school at the slower pace, then we get
(60)(T) = D
(120)(T - 2) = D
So we can solve for T and D
60T = 120(T - 2)
60T = 120T - 240
60T + 240 = 120T
240 = 60T
4 = T
Substituting, we get
D = 60T = (60)(4) = 240
So it takes 4 minutes to walk to school at the slower pace, and the distance to school is 240 meters.
Even if the problem had been stated correctly, it seems pretty advanced for an 11-year-old.
2007-03-03 03:45:28 · answer #3 · answered by jim n 4 · 0⤊ 0⤋
x : distance
s : speed to arrive on time
x/60 -x/s = 8 minutes
x/s - x/120 = 16 minutes
solving these two equations yealds:
x/60 -x/120 = 24
x = 2880 meters
this is based on the assumption that in case of 120 minutes/minutes he arrives 16 miutes earlier, not later which is not logical.
2007-03-03 07:56:23 · answer #4 · answered by reza 2 · 0⤊ 0⤋
how is it that if he walks faster he will be later then if he walks slower? not logical.
peter should get a car or take the bus!
60m = 0.037 miles x 8 minutes = .30 miles
120m = 0.075 miles x 16 minutes = 1.2 miles
like i said. not logical if you use the information you gave us
2007-03-03 03:15:03 · answer #5 · answered by Kynnie 6 · 0⤊ 0⤋
A trick question...or a dumb teacher...:) For mr the question is...what time does he leave home and what time does school start?
2007-03-03 03:31:01 · answer #6 · answered by Roy W 2 · 0⤊ 0⤋