Answer this for me?
A student at a speed of 3 km/hr reaches school 15 minutes late.While at a speed of 4km/hr he reaches 15 minutes bfore the right time.Find the distance of the school.
2006-11-09 02:37:34 · 8 answers · asked by Dark angel 1 in Science & Mathematics ➔ Mathematics
Can u show how u got these answers?
2006-11-09 02:51:15 · update #1
Speed = distance/time
let x be the time it takes to get to school on time
3=d/(x+15/60) 4=d/(x-15/60)
d=3x+3/4 d=4x-1
therefore 3x+3/4=4x-1
x=1 3/4 = 7/4
Hence distance=d=3(7/4) + 3/4 = 6
Therefore the school is 6 km away
2006-11-09 02:57:40 · answer #1 · answered by anonymous 2 · 0⤊ 0⤋
5
2006-11-09 10:45:16 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Let x = distance in kilometers and t = time (to arrive on time) in hours
x/3 = t +15/60
x/4 = t -15/60
solve both equations for x
x = 3t + 45/60 = 3t + 3/4
x = 4t - 60/60 = 4t -1
set the equations equal to each other
3t + 3/4 = 4t - 1
solve for t
t = 7/4
substitute t back into either of the original equations to solve for x.
x = 3t + 3/4
x = 3*(7/4) + 3/4
x = 21/4 + 3/4
x = 24/4
x = 6 km
2006-11-09 10:57:18 · answer #3 · answered by T 5 · 0⤊ 0⤋
distance = rate x time
Let D be the distance to the school
Let T be the time to get there at on time
15 minutes is 0.25 hours
D = (3 km/hr) x (T+.25)
D = (4 km/hr) x (T -.25)
3T + .75 = 4T -1
T = 1.75
D = 6 km
2006-11-09 10:53:52 · answer #4 · answered by Anonymous · 1⤊ 0⤋
x = distance in kilometers
t = time to arrive on time in hours
x/3 = t +15/60
x/4 = t -15/60
x (1/3-1/4) = 0.5
x (1/12) = 0.5 * 12 = 6 km
2006-11-09 11:06:04 · answer #5 · answered by Dr. J. 6 · 0⤊ 0⤋
Let t = time to get to school
Then 3(t+.25) = 4(t-.25) [.25 = 1/4 hour = 15 minutes]
3t+.75 = 4t -1
t = 1.75 hours
So distance = 3(1.75+.25) = 4(1.75-.25)=6 km
2006-11-09 10:53:07 · answer #6 · answered by ironduke8159 7 · 1⤊ 0⤋
9.6 km
I think :)
2006-11-09 10:50:00 · answer #7 · answered by WizardofID 3 · 0⤊ 0⤋
3.5?
2006-11-09 10:49:52 · answer #8 · answered by loodoogal 2 · 0⤊ 0⤋