John rode 70 miles on his bike in 4 1/2 hours. He started out at the rate of 20mph; however, after a while he got tired and reduced his speed to 12mph. How far did he travel before reducing his speed?
2007-01-09 19:23:12 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Suppose he rode x miles at 20 mph and (70-x) miles at 12 mph. The total time taken, in hours, will be (x / 20) + (70 - x) / 12. This must equal 4 1/2 = 9/2. So we have
x / 20 + (70 - x) / 12 = 9/2.
Multiply through by 60 (LCM of 20 and 12) to make it look nicer:
3x + 5(70 - x) = 270
=> 3x + 350 - 5x = 270
=> -2x = -80
=> x = 40.
So he rode 40 miles before reducing speed.
2007-01-09 19:30:41 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
Let
This problem utilizes the formula
distance = rate*time
t = time he travelled before reducing speed.
70 = 20t + 12(9/2 - t) = 20t + 54 - 12t
16 = 8t
2 = t
Distance travelled before reducing speed
d = 20t = 20*2 = 40 miles
2007-01-09 21:56:16 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
let x miles he travelled at 20mph and remaining (70-x ) mileshe tavelled at 12mph
x/20 + (70-x ) / 12 = 4.5
solving we get x = 40
that is he travelled 40 miles in2 hrs at 20 mph and remaining 30 miles he traveled in2.5 hrs @ 12mph
2007-01-09 19:37:39 · answer #3 · answered by aneesh 1 · 0⤊ 0⤋
time spent going 20mph = x
time spent going 12 mph = y
You get 2 equations:
x + y = 4.5
20x + 12y = 70
y = 4.5-x
20x + 12*4.5 + 4.5x = 70 ...
im too lazy but u should be able to finish
2007-01-09 19:32:11 · answer #4 · answered by Anonymous · 0⤊ 0⤋
my best guess is 20 for 2 hours and 12 for 2.5 hours.
2007-01-09 19:31:31 · answer #5 · answered by hermespgc 2 · 0⤊ 0⤋