Candy drives to class in 30 minutes. By averaging 10 mph more, her roommate Fran drives to the same class in 20 minutes. What is Candy's average speed?
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Let the distance = x.
Candy drives there in 30 minutes = 0.5 hours.
Distance = Velocity multiplied by time.
d = vt.
d = x miles
t = 0.5 hours
Want to find v.
v = d/t
v = x/0.5 = 2x.
So Candy had an average speed of 2x mph.
Fran:
d = x (she still travels the same distance as they started at the same place - they are roommates).
t = 20 minutes = 1/3 hours.
v = d/t
v = x/(1/3) = 3x.
Now it said that her speed was 10mph more than Candy's which was 2x mph.
So it's:
3x = 2x + 10
x = 10.
Candy's average speed was 2x, which equals 2(10) = 20 mph.
Candy drives there in 30 minutes = 0.5 hours.
Distance = Velocity multiplied by time.
d = vt.
d = x miles
t = 0.5 hours
Want to find v.
v = d/t
v = x/0.5 = 2x.
So Candy had an average speed of 2x mph.
Fran:
d = x (she still travels the same distance as they started at the same place - they are roommates).
t = 20 minutes = 1/3 hours.
v = d/t
v = x/(1/3) = 3x.
Now it said that her speed was 10mph more than Candy's which was 2x mph.
So it's:
3x = 2x + 10
x = 10.
Candy's average speed was 2x, which equals 2(10) = 20 mph.
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Let S = Candy's average speed, in mph
S + 10 = Fan's average speed, in mph
Distance = Speed x time = (S + 10) x 20 / 60 = (S + 10)/3 (eqn 1)
Distance = S x 30 / 60 = S/2 (eqn 2)
Equate eqn 1 and eqn 2
S/2 = (S+10)/3
3S = 2S + 20
S = 20 mph
S + 10 = Fan's average speed, in mph
Distance = Speed x time = (S + 10) x 20 / 60 = (S + 10)/3 (eqn 1)
Distance = S x 30 / 60 = S/2 (eqn 2)
Equate eqn 1 and eqn 2
S/2 = (S+10)/3
3S = 2S + 20
S = 20 mph
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Speed is an illegal drug and should not be taken under any circumstances. Please tell Candy not to take any speed especially not a higher than average amount. Please tell Fran not to give in to peer pressure.