# Math Help - tracks

1. ## tracks

A jogging park has 2 identical circular tracks touching each other, and a rectangular track enclosing the 2 circles.The edges of the rectangles are tangential to the circles.Two friends , A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight.Approximately, how much faster than A does B have top run, so that they take the same time to return to their starting point.

options are:
a. 3.88%
b. 4.22%
c. 4.44%
d. 4.72%

2. ## I will give you a hint

Originally Posted by manisha
A jogging park has 2 identical circular tracks touching each other, and a rectangular track enclosing the 2 circles.The edges of the rectangles are tangential to the circles.Two friends , A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight.Approximately, how much faster than A does B have top run, so that they take the same time to return to their starting point.

options are:
a. 3.88%
b. 4.22%
c. 4.44%
d. 4.72%

find how much larger the permiter of one track is to the other

3. i did but it doesn't match with any of the options.Sir can u plz do it...

4. ## Ok here is what you do

Lets say the Perimeter of the one is $\frac{3}{4}$ greater than the other then using that information we use our formula $d=rt$...and we solve for time to get $t=\frac{d}{r}$...now we want there times to be equal...so lets say the distance around one is 450 and the other is 300 then we would have that the one girl's time is given by $t=\frac{300}{r}$ and the others is $t=\frac{450}{r_1}$...now set their times equal say the girl who has to travel fasters rate(r) is 1 find the other girls rate and see what percentage bigger than one it is

5. i still din't get it...sorry to bother u like this...

6. ## Ok

Originally Posted by manisha
i still din't get it...sorry to bother u like this...
THe one runner has to run around the rectangle which is two circles wide so its with is $2{d}$ where d is the diameter and it is one circle high so its height is ${d}$ so its permiter is [tex]P=2{d}+{d}+{d}+{d}=6{d}[tex] and the circles circumference is given by $C=\pi{d}$ and there are two circles so there total P is $P_{t}=2\pi{d}$...now using that equation eariler we have that the first runners time is $t=\frac{6d}{r}$ and the second runners is given by $t=\frac{2\pi{d}}{r_1}$...now by setting they two t's together we get $\frac{6d}{r}=\frac{2\pi{d}}{r_1}$...now we want to see how much faster the runner who runs around the rectangle (r not r_1) goes so we solve for r and we get $r=\frac{3r_1}{\pi}=.954$...so the one has to go 95.4% of the others speed...so the other one must travel approximately 4.5%...so your answer is C