Run!

Let the length of the race be $x$ km. Then when she reached the halfway point, she traveled $\frac{x}{2}$ km which took $\frac{\frac{x}{2}}{8}=\frac{x}{16}$ hours. After the halfway point, she traveled $1-\frac{x}{2}=\frac{x}{2}$ km which took $\frac{\frac{x}{2}}{10}=\frac{x}{20}$ hours. We are given that the sum of these is $1$ hour therefore

$\frac{x}{16}+\frac{x}{20}=1 \Rightarrow \frac{(20+16)x}{16\times20}=1 \Rightarrow x=\frac{320}{36}=\frac{80}{9}=\frac{a}{b}$

$a+b=\boxed{89}$ .

Let the total distance be x km . We use here the formula, Time = $\frac{Distance}{Speed}$

Given, speed for 1st half path= 8 m/s and speed for 2nd half path= 10 m/s

Now, Time for 1st half path = $\frac{\frac{x}{2}}{8} = \frac{x}{16}$

Then, Time for 2nd half path = $\frac{\frac{x}{2}}{10} = \frac{x}{20}$

According to the question, we have,

$\frac{x}{16}+\frac{x}{20} = 1 \implies \frac{5x+4x}{80}=1 \implies 9x=80 \implies x=\frac{80}{9}$

Here x is in $\frac{a}{b}$ form with a and b as coprime integers, a=80 and b=9, so we have $a+b = \boxed{89}$

Let d = distance of the race

Write a time equation; time = dist/speed

8km time +10km time = 1 hour

(½d)/8 + (½d)/10 = 1

multiply by 40 to clear the denominators, resulting in:

5(½d) + 4(½d) = 40

2.5d + 2d = 40

4.5d = 40

d = 40/4.5

d = 80/9 is the distance of the race

d = a/b and a = 80 and b = 9

a + b = 89

T1+T2 = Ttotal

T=S/V

S1/V1+S2/V2=Ttotal=1 HOUR

0.5S/8+0.5S/10=1

9S/80=1

S=80/9=A/B

A+B=89

Let the distance be x.
NOTE: She realized that she had to increase her speed halfway.

Now, Recall or remember:distance/time = Speed... This can be also written as: distance/speed= time! We know total time = 1 hour

Now~!!!!! distance covered by 8km/hr speed = x/2 therefor, time = x/16 hr

distance covered by 10km/hr speed=x/2

therefore, time = x/20 km hr

Now, x/16 + x/20= 1

==> x = 80/9

let a = 80 and b = 9

therefore, a + b = 89

Use the Harmonic Mean:

$\frac{2}{\frac{1}{8}+\frac{1}{10}} = \frac{80}{9}$

Let the dist. be divided into x km and x km.

Time taken to run x km in first half= x/8 hr

Time taken to run x km in second half= x/10 hr

As she reached just on time x/8 +x/10 = 1hr on solving, we get x = 40/9 So 2x = 80/9 Hence the answer is 80+9=89....

total time = 1 h

let half distance = x

so 1 = (x/8) + (x/10)

=> x= 40/9

-> 2x = 80/9

hence a=80 b=9

a+b = 89

Let d = distance of the race

Write a time equation; time = dist/speed

8km time +10km time = 1 hour

(½d)/8 + (½d)/10 = 1

multiply by 40 to clear the denominators, resulting in:

5(½d) + 4(½d) = 40

2.5d + 2d = 40

4.5d = 40

d = 40/4.5

d = 80/9 is the distance of the race

d = a/b and a = 80 and b = 9

a + b = 89

Since the distance covered is the same as demonstrated in the question's language (halfway point) we can use the idea of average speed which in the present case is the harmonic mean of the values. The harmonic mean of the given values is 80/9 and the answer is 89

time1st=s/16, time2nd=s/20, time1st+time2nd=1. s: length of the race. A bit about me: i am a high school student, I need a lot of practice and advise. I will always appreciate all yours comment about my solution, my english and whatever about me. I just know the idea and I need the real solution cause i don't use english in my country. Thanks

LET THE DISTANCE TO BE TRAVELED BE Y. SHE COVERS Y/2 DISTANCE WITH SPEED OF 8KM/HR. SO TIME REQUIRED IS Y/2 DIVIDED BY 8 = Y/16. SHE COVERS ANOTHER Y/2 DISTANCE WITH SPEED OF 10KM/HR. SO TIME REQUIRED IS Y/2 DIVIDED BY 10= Y/20. SHE COVERED THE RACE IN TIME THAT MEANS SHE COVERED THE RACE IN 1 HOUR. SO TOTAL TIME REQUIRED BY HER TO COVER THE RACE IS 1 HOUR. THEREFORE Y/16 +Y/20 = 1 9Y/80=1 Y=80/9. Y IS THE LENGTH OF RACE. THEREFORE LENGTH OF RACE IS 80/9 KM. LENGTH OF RACE CAN BE REPRESENTED IN THE FORM OF A/B WHERE A=80 AND B=9 AND A,B ARE CO PRIME INTEGERS. THEREFORE A+B=80+9=89KM.