Pascal grade 9 math contest question(can you ratio speed and time)

[kevinshen18]

I was given this question from the Pascal Math contest 2001:

__________________________________________
Cindy leaves school at the same time every day. If she cycles at 20 km/h, she arrives home at 4:30 in the afternoon. If she cycles at 10 km/h, she arrives home at 5:15 in the afternoon. At what speed, in km/h, must she cycle to arrive home at 5:00 in the afternoon?
__________________________________________

This was how I attempted to solve it:

A difference in 10km/h (20km/h - 10km/h) results in her being 45 minutes late(5:15 - 4:30).
So I ratio it:
$$\frac{10km/h}{45min} = \frac{Xkm/h}{30min}$$

If a difference of 10km/h results in a 45 minutes delay then what's the speed difference(X) that results in a 30 minutes delay(5:00 - 4:30).

I cross multiplied and got x = 6.6666...km/h.
Subtracting that from 20km/h my answer was 13.3333...km/h

The correct answer was 12 km/h.

Now I have no idea how I got it wrong but I would assume that the mistake was the speed to time ratio. Is it correct to do that? Can you even ratio speed and time or is that incorrect and not allowed? If it's not possible then why?

 

[jz92wjaz]

You need to find the total time it took her to get home, not just the additional time after the 20km/h trip.

Edit: Here are some additional hints of how to set it up correctly.
Distance = velocity x time.
In this case, because Distance1 = Distance2 = Distance3 = Distance,
Distance = Velocity1*Time1 = Velocity2*Time2 = Velocity3*Time3
Time1: Unknown
Time2: Time1 + 45 minutes
Time3: Time1 + 30 minutes.

You'll have to evaluate Time1 before you can solve this.

 

[kevinshen18]

I see. Thank you!
So if total time was given that I could ratio?

 

[Ray Vickson]

I was given this question from the Pascal Math contest 2001:

__________________________________________
Cindy leaves school at the same time every day. If she cycles at 20 km/h, she arrives home at 4:30 in the afternoon. If she cycles at 10 km/h, she arrives home at 5:15 in the afternoon. At what speed, in km/h, must she cycle to arrive home at 5:00 in the afternoon?
__________________________________________

This was how I attempted to solve it:

A difference in 10km/h (20km/h - 10km/h) results in her being 45 minutes late(5:15 - 4:30).
So I ratio it:
$$\frac{10km/h}{45min} = \frac{Xkm/h}{30min}$$

If a difference of 10km/h results in a 45 minutes delay then what's the speed difference(X) that results in a 30 minutes delay(5:00 - 4:30).

I cross multiplied and got x = 6.6666...km/h.
Subtracting that from 20km/h my answer was 13.3333...km/h

The correct answer was 12 km/h.

Now I have no idea how I got it wrong but I would assume that the mistake was the speed to time ratio. Is it correct to do that? Can you even ratio speed and time or is that incorrect and not allowed? If it's not possible then why?

Set up two equations for T = starting time (hrs) and X = distance traveled (km). Remember,
time traveled = distance/speed, so
(1) 4.5 - T = X/20
(2) 5.25 - T = X/10
Solve these two equations to get T and X.

Now you need to find v that gives 5-T = X/v, where T and X are known at this point.

 

[jz92wjaz]

I see. Thank you!
So if total time was given that I could ratio?

Sort of. Your ratio would be Velocity1/Time2 = Velocity2/Time1, but it's easier to leave it as Velocity1*Time1 = Velocity2*Time2.

 

[kevinshen18]

Sort of. Your ratio would be Velocity1/Time2 = Velocity2/Time1, but it's easier to leave it as Velocity1*Time1 = Velocity2*Time2.

Yea I agree, I'll just solve it with your method then.