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Bill and Ted each competed in a 240-mile bike race. Bill’s a

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Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink] New post 06 Jun 2017, 13:00
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Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?

[Reveal] Spoiler:
15

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Re: Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink] New post 07 Jun 2017, 15:11
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GreenlightTestPrep wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?




Bill’s average speed was 5 miles per hour slower than Ted’s average speed.
Let B = Bill's travel speed
So, B + 5 = Ted's average speed

Ted completed the race 4 hours sooner than Bill did
Let's start with a word equation:
(Bill's travel time) = (Ted's travel time) + 4

time = distance/speed

So, we get: 240/B = 240/(B + 5) + 4
Rewrite 4 as 4(B + 5)/(B + 5) to get: 240/B = 240/(B + 5) + 4(B + 5)/(B + 5)
Simplify: 240/B = 240/(B + 5) + (4B + 20)/(B + 5)
Combine terms: 240/B = (4B + 260)/(B + 5)
Cross multiply: 240(B + 5) = (B)(4B + 260)
Expand and simplify: 240B + 1200 = 4B² + 260B
Set equal to zero: 4B² + 20B - 1200 = 0
Divide both sides by 4 to get: B² + 5B - 300 = 0
Factor: (B + 20)(B - 15) = 0
So, EITHER B = -20 OR B = 15
Since B (Bill's speed) cannot be a negative value, we can conclude that B = 15.

Answer:
[Reveal] Spoiler:
15


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Re: Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink] New post 05 Dec 2017, 18:19
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GreenlightTestPrep wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?

[Reveal] Spoiler:
15


We are given that Bill and Ted each competed in a 240-mile bike race. We are also given that Bill’s average speed was 5 miles per hour slower than Ted’s average speed.

Both Bill and Ted had distance of 240 miles. If we let Ted’s speed = r, we can let Bill’s speed = r - 5. We can use the above information to determine the time of Bill and Ted in terms of variable r.

Since time = distance/rate, Ted’s time = 240/r and Bill’s time = 240/(r - 5).

Since Ted completed the race 4 hours sooner than Bill did, we can create the following equation:

240/r + 4 = 240/(r - 5)

To eliminate the denominators of the fractions we can multiply the entire equation by r(r-5) and we have:

240(r - 5) + 4[r(r - 5)] = 240r

240r - 1,200 + 4r^2 - 20r = 240r

4r^2 - 20r - 1,200 = 0

r^2 - 5r - 300 = 0

(r - 20)(r + 15) = 0

r = 20 or r = -15

Since r must be positive, r = 20.

Thus, Bill’s rate = 20 - 5 = 15 mph.
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Re: Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink] New post 19 Dec 2018, 15:17
Expert's post
GreenlightTestPrep wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?

[Reveal] Spoiler:
15


Another option is to start with the following word equation:
(Bill's average speed) + 5 = (Ted's average speed) [since Bill’s average speed was 5 miles per hour slower than Ted’s average speed]

Let t = Bill's travel TIME (in hours)
So, t - 4 = Ted's travel TIME (in hours) [since Ted completed the race 4 hours sooner than Bill did]

speed = distance/time

So, our word equation becomes: 240/t + 5 = 240/(t - 4)
Multiply both sides by t to get: 240 + 5t = 240t/(t - 4)
Multiply both sides by (t - 4) to get: 240(t - 4) + 5t(t - 4) = 240t
Expand: 240t - 960 + 5t² - 20t = 240t
Subtract 240 t from both sides: -960 + 5t² - 20t = 0
Rearrange: 5t² - 20t - 960 = 0
Divide both sides by 5 to get: t² - 4t - 192 = 0
Factor to get: (t - 16)(t + 12) = 0

So, EITHER t = 16 OR t = -12

Since the time cannot be NEGATIVE, we can conclude that t = 16
In other words, Bill's travel time was 16 hours

What was Bill’s average speed in miles per hour?
speed = distance/time
So, Bill's speed = 240/16 = 15 miles per hour

Answer: 15

Cheers,
Brent
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Re: Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink] New post 20 Dec 2018, 07:20
Expert's post
GreenlightTestPrep wrote:
Bill and Ted each competed in a 240-mile bike race. Bill’s average speed was 5 miles per hour slower than Ted’s average speed. If Ted completed the race 4 hours sooner than Bill did, what was Bill’s average speed in miles per hour?

[Reveal] Spoiler:
15


As I mention in the video below, these kinds of questions can be solved in a variety of ways.

For our 3rd approach, let's start with a word equation involving distances traveled.
Since Bill and Ted both traveled 240 miles, we can write:
Bill's travel distance = Ted's travel distance

Let x = Bill's speed
So, x + 5 = Ted's speed

Let t = Ted's travel time (in hours)
So, t + 4 = = Bill's travel time (in hours)

Distance = (speed)(time)
Plug values into the word equation to get: (x)(t + 4) = (x + 5)(t)
Expand to get: xt + 4x = xt + 5t
Subtract xt from both sides to get: 4x = 5t
Divide both sides by 5 to get: 4x/5 = t

Where do we go from here?

Well, we know that Bill traveled 240 miles
So, (Bill's speed)(Bill's travel time) = 240
In other words: (x)(t + 4) = 240
Replace t with 4x/5 to get: (x)(4x/5 + 4) = 240
Expand: 4x²/5 + 4x = 240
Multiply both sides by 5 to get: 4x² + 20x = 1200
Divide both sides by 4 to get: x² + 5x = 300
Rewrite as: x² + 5x - 300 = 0
Factor: (x - 15)(x + 20) = 0
So, EITHER x = 15 OR x = -20
Since the speed cannot be negative, it must be the case that x = 15

Answer: 15

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Re: Bill and Ted each competed in a 240-mile bike race. Bill’s a   [#permalink] 20 Dec 2018, 07:20
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