Randy and Wade are two friends living 60 km apart from each other. One spring morning they decided to meet each other for the afternoon

Question

they started riding their bikes at noon heading toward each other. If Randy was riding 4 km/h faster than Wade, how fast was each riding if they met each other at 1:30 pm? How many kilometers did each one ride?

Answer 1

We know distance=rate(time), or d=rt. Let person 1 travel distance 1 and bike at rate b: d₁ = b(1.5 hours), or d₁=1.5b.

Let person 2, who's faster, bike at rate (b+4). For this person, d₂ = (b+4)(1.5 hrs), or d₂=1.5b+6 after distributing 1.5.

We know that d₁ + d₂ = 60 km. Let's add the two equations and set them equal to 60.

d₁=1.5b

+

d₂=1.5b+6

---------------

d₁+d₂ = 3b+6. Since d₁+d₂=60, substitute that:

60 = 3b + 6

Subtract 6 from each side to get 54 = 3b, and divide both sides by 3 to get b=18.

Remember b is the rate, so Wade, who was riding more slowly, rode at 18 km/hr. Randy rides 4 km/hr faster, so that's 22 km/hr.

Answer 2

Hi Callie,

so first you need to know the r(t) of randy plus the r(t) of Wade=60

Let r= wade’s speed

let r+4 = randy speed

the time it took was noon to 1:30 so 1.5 hrs

so t=1.5

(r+4)1.5 + r(1.5) =60

1.5r+6 +1.5r=60

3r+6=60

3r=54

r=18

so wade’s speed is 18 kmh

that means Randy’s is 22 kmh or (18+4)

d=rt

Randy = 22(1.5) =33 km

Wade= 18(1.5) =27 km

33+27=60 ✅

I hope this helps!

Mrs. S

Answer 3

for this type of problem, first draw a quick sketch like this:

60 km

• ---> (x+4) (x)<------• And if they left at noon and met at 1:30, they were riding for 1.5 hours, so...

hours(speed) = distance

1.5(x+4) + 1.5(x) = 60 the distance each traveled to meet each other is equal to the full distance.

1.5x + 6 + 1.5x = 60

3x = 54

x = 18 kph

Randy is riding at 22 kph while Wade rides at 18 kph

Randy traveled 33 km, while Wade traveled 27

I hope this sequence of steps helps you!