xcelxp1 wrote:Aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. How many miles from home can Aaron job so that he spends a total of t hours jogging and walking?

a) xt/y
b) x+t/xy
c) xyt/x+y
d) x+y+t/xy
e) y+t/x - t/y

OA C

It doesn't say Aaron jogs faster than he walks... So i set x and y both equal to 1, T = 2

I got the right answer...

you can also do this algebraically:

let's say the distance of the route is d miles each way.

-by jogging at x miles per hour along d miles, Aaron takes d/x hours for the route from home.
-by walking at y miles per hour along d miles, Aaron takes d/y hours for the route back home.

t = d/x + d/y
t = d(1/x + 1/y)
d = t/(1/x + 1/y)
d = t/((x+y)/xy)
d = xyt/(x + y)

choose C.

-BM-

xcelxp1 wrote:Aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. How many miles from home can Aaron job so that he spends a total of t hours jogging and walking?

a) xt/y
b) x+t/xy
c) xyt/x+y
d) x+y+t/xy
e) y+t/x - t/y

OA C

It doesn't say Aaron jogs faster than he walks... So i set x and y both equal to 1, T = 2

I got the right answer...

You cannot take the two speeds same unless specified. A right answer is not by chance every time.

Let’s say Aaron needs to jog m miles from home to meet the situation. In that case his jogging time will be m/x hours and his walking time will be m/y hours, such that:

(m/x) + (m/y) = t

or m = (x y t)/(x + y).

I’m happy with [spoiler]C[/spoiler].