Keiko walks once around a track at exactly the same constant speed eve

Question

Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has width 6 meters, and it takes her 36 seconds longer to walk around the outside edge of the track than around the inside edge. What is Keiko's speed in meters per second?

A. $\frac{\pi}{3}$

B. $\frac{2\pi}{3}$

C. $\pi$

D. $\frac{4\pi}{3}$

E. $\frac{5\pi}{3}$

A.  \frac{\pi}{3}

B.  \frac{2\pi}{3}

C.  \pi

D.  \frac{4\pi}{3}

E.  \frac{5\pi}{3}

firas92 · Answer

R=Constant rate
t=Time to walk around inside edge
l=Length of straight paths
r=Radius of circular paths

Rate x Time = Distance
? ? 2? + 2??
? ?+36 2? + 2?(?+6)

Subtracting the times

(2? + 2??)/? - (2? + 2?(?+6))/? = 36

On simplification we get R=?/3

ScottTargetTestPrep · Answer

The straight sides of the track have equal lengths on the outside edge and on the inside edge; therefore, the only difference in length when Keiko is walking the outside edge rather than the inside edge will be at the two semicircle ends.

Since there are two semicircle ends, together they make up a full circle. If we let r be the radius of the circle making up the inner edge, the radius of the circle making up the outer edge will be (r + 6) meters since the width of the track is 6 meters.

The difference of the circumferences of the two circles is 2π(r + 6) - 2πr = 2πr + 12π - 2πr = 12π. So, the extra length that Keiko walks when she walks from the outer edge is 12π meters, which takes her 36 seconds to walk. To find her speed, let’s set up a proportion. If Keiko walks 12π meters in 36 seconds, then in 1 second, she will walk x meters:

12π/36 = x/1

x = π/3

Answer: A

rrdeo22 · Answer

May be it is silly question but I could not understand concept of inner and outer edge in this scenario and then how width played role as differentiator for length of both the tracker. Anyone can help with some sort of diagram ?

Fdambro294 · Answer

Not a silly question at all.

It took me several minutes to understand what the question was trying to say.

Think of a track and field type track, which has an “inside” lane and an “outside” lane. The outside lane is a longer distance (that is why they stagger the starts if you ever watch track and field).

Anyone please feel free to correct me if I’m wrong.

I agree with you. The question could be written a tad more clearly. I was under the assumption that the “inside” track referred to the straight line portions of the track.

If you just focus on the difference it takes her to run around the outside track vs. the inside track at the 2 semi-circles:

Each semi-circle will come together with its corresponding semi-circle to create 1 circle equal to the “curves” at the ends of the track.

Let the inside, smaller circle have Radius = R

Its circumference = 2(pi)R

Since the track has a width of 6 meters ——> this will be the length from the “inside track” to the “outside track”. Both the straight portions will have the same distance, only the curved semi-circles at the ends will have different dimensions.

the outside circle will have Radius = R + 6

Its circumference = 2(pi) (R + 6) = 2(pi)R + 12(pi)

(Time to cover outside circle) - (Time to cover inside circle) = 36 seconds

(Circumference outside circle/S) - (Circumference of inside circle/S) = 36 seconds

Where her constant speed = S

(2(pi)R + 12(pi)) / S

-

(2(pi)R) / S

= 36 seconds

After the 2(pi)R subtracts itself out in the numerator you have

12(pi) / S = 36

S = 12(pi) / 36

S = (pi) / 3

A

Posted from my mobile device

devdattakhoche · Answer

Ti = inner track time 
To = Outer track time
ri = radius of inner semicircle
ro = radius of outer semicircle
and difference in radius is $ r_o - r_i = 6 $ meters.

Given: 
- The time taken on the inner track is 36 seconds **less** than on the outer track (Ti = To + 36)
- The straight sections take the same time, so the extra time is only due to the semicircular paths. 
- Let the speed of the athlete be **s**.

Time on the inner track = $ \frac{2\pi r_i}{s} $ 
Time on the outer track = $ \frac{2\pi r_o}{s} $

According to the question: 
\

\

✅ **Final Answer: The speed of the athlete is $ \frac{\pi}{3} $ meters per second.**

Keiko walks once around a track at exactly the same constant speed eve

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Keiko walks once around a track at exactly the same constant speed eve

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