Charles ran the first leg of a 12 km road race at a speed of 9 km/hr and the remainder of the race at 10 km/hr. If his time for the whole race was 1.25 hours, how long was the first leg of the course?
work too please. thank you very much. appreciate it. :]
2006-11-12 16:53:44 · 7 answers · asked by bluenecklace 2 in Science & Mathematics ➔ Mathematics
4.5Km. The equation is simple assuming X Km first leg. we have 12-X as the second leg. So time taken in each leg X/9 and (X-12)/10 resp. The sum of which is the total time of 1.25 Hrs. So
(x/9)+((12-X)/10)=1.25. Solve the simpe linear equation to get X=4.5Km
2006-11-12 17:10:10 · answer #1 · answered by balstoall 2 · 1⤊ 0⤋
Mercy girl, you have too many questions! :^)
Let x = distance of first leg. Then since the total distance was 12, the length of the second leg must = 12 - x, right?
Distance = speed * time, so time = distance/speed. He ran the first leg at 9km/hr for a distance of x km, therefore the time he spent running the first leg was x/9
By the same token, the time spent on leg 2 was (12-x)/10.
The total time was 1.25 hr, so
x/9 + (12-x)/10 = 1.25
Can you take it from there?
2006-11-13 01:20:49 · answer #2 · answered by Gary H 6 · 0⤊ 0⤋
There are two separate times for this problem
T1 = Time for first leg
T2 = Time for second leg
9 * T1 + 10 * T2 = 12
T1 + T2 = 1.25
Now solve two equations and two unknowns
Multiply the second equation by 10 and subtract the first equation from it.
T1 = 0.5 hours
T2 = 0.75 hours
The first leg of the race was 9 * 0.5 = 4.5 km
The second leg of the race was 10 * 0.75 = 7.5 km
2006-11-13 01:12:00 · answer #3 · answered by z_o_r_r_o 6 · 1⤊ 0⤋
let the first leg be x km long.
The second leg is then 12-x.
using the speed, time and distance formula (t=d/s)
the time for first leg is x/9
time for second leg is (12-x)/10
sum of both is 1.25 hrs (given)
so x/9 + (12-x)/10 = 1.25
solve for x and you get x = 4.25
Answer 4.25 km
2006-11-13 01:23:11 · answer #4 · answered by SS 1 · 0⤊ 0⤋
Let the first leg be x km
=>2nd leg is (12-x)km
Let the time required to cover the first half and the second half be t1 and t2 respectively.
distance=speed*time
=>x=9*t1
=>t1=x/9....................(1)
=>12-x=10*t2
=>t2=(12-x)/10..........(2)
adding (1) and (2)
=>t1+t2=x/9 + (12-x)/10
=>1.25=[10x + (12-x)*9]/90
=>112.5=10x + 108 - 9x
=>x=4.5 km
2006-11-13 01:34:50 · answer #5 · answered by sushant 3 · 0⤊ 0⤋
I not sure...but i try...
(X*9+10*1.25)/2=12
12.5+ X9 =24
X9= 11.5
X= 11.5/9
2006-11-13 01:08:26 · answer #6 · answered by China Boy 1 · 0⤊ 0⤋
Its doesnt make sense
2006-11-13 00:58:05 · answer #7 · answered by Calvin 2 · 0⤊ 0⤋