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Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink]
06 Jun 2017, 13:00

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Question Stats:

53% (03:02) correct
46% (02:34) wrong based on 13 sessions

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?

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

Re: Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink]
05 Dec 2017, 18:19

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?

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:

Re: Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink]
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?

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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Brent Hanneson – Creator of greenlighttestprep.com Sign up for our free GRE Question of the Dayemails

Re: Bill and Ted each competed in a 240-mile bike race. Bill’s a [#permalink]
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?

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

RELATED VIDEO FROM OUR COURSE

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Brent Hanneson – Creator of greenlighttestprep.com Sign up for our free GRE Question of the Dayemails

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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