Question

On a journey, a cyclist travelled 1 kilometer in x minutes. On as second journey, the cyclist travelled for y hours at the same average speed as on the first journey.
Find an expression, in terms of x and y, for the number of kilometers he travelled on the second journey?

Answer 1

@FoolForMath @robtobey @SmoothMath @AccessDenied @ash2326

Answer 2

@radar @stormfire1 @KingGeorge

Answer 3

For the first journey, d=rt 1km=rx(minutes) so avg speed is 1(km)/x(min)
for the second journey, the distance in km expressed in terms of x and y would be:
d(km)=(60y)/x

Answer 4

thanks i got , that...:)

Answer 5

Since two different units are used one is minutes and the other hours, you would convert to similar units, I converted to minutes. Others may convert to hours.

Answer 6

yes i did thaht...thanks for your help :D

Answer 7

Maybe storfire1 has further info regarding this problem.

Answer 8

I got the same thing:
\[\frac{1km}{60x}=\frac{km}{3600y}\]\[\frac {(3600y)(1km)}{60x}=km\]\[d(km)=\frac{60y}{x}\]

Answer 9

I went to seconds...which was more work in retrospect :)

Answer 10

There it is in a step by step manner, good job stormfire1

Answer 11

i have a doubt, since they have given minutes we have to convert them to hours, so we have to divide it right
so it becomes 1/(1/60)

Answer 12

In the final equation it doesn't matter...the relationship is the same.

Answer 13

As you can see, radar and I did it converting things two different ways and came up with the same final equation.

Answer 14

speed on journey 1
s = d/t
s = 1/(1/60)= 60x
so substitute that in equation 2
avg speed = d/t
60x = d/y
d = 60xy?

where did i go wrong?

Answer 15

\[1min*\frac{60s}{1min}=60s\]

Answer 16

okay check if i am now
s = d/t
s = 1/(x/60) = 60/x
substitute that is equation 2
60/x = d/y
60y/x = d

Answer 17

wait\[s= d/t ---> \frac{1}{\frac{x}{60}}\]

1= kilometer
x/60 = time since we are dividing it by 60
so it become 60/x

Answer 18

Now you're getting ME confused!

If you're traveling 1km/minute...then you're traveling (1/60)km per second. Not 1km per second.

Answer 19

i hr = 60 minutes so i am converting minutes into hours not seconds

Answer 20

AHHH...lemme look again then

Answer 21

Well, if you're converting minutes to hours it should work out this way:

\[s=\frac{d}{t}=\frac{1km}{x~mins}=\frac{60km}{x~hr}\]\[\frac{60km}{x~hr}=\frac{?km}{y~hrs}\]You still end up with \[\frac{(60km)y~hrs}{x~hrs}=\frac{60y}{x}=d(km)\]

Answer 22

i also ended up with the same answer check and see?

Answer 23

then you're right :P

Answer 24

can u help me with this
Given that 198 = 2 * 3^2 * 11
find
the smallest integer, k , such that 198k is a perfect square?

Answer 25

probably not...I'm much better at physics :P

Answer 26

Post it as a separate question and I'm sure one of the math folks will answer it