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

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Difficulty:   55% (hard)

Question Stats: 71% (03:00) correct 29% (03:14) wrong based on 300 sessions

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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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 15

*Kudos for all correct solutions.

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Originally posted by GMATPrepNow on 26 Oct 2016, 12:43.
Last edited by GMATPrepNow on 25 May 2017, 05:16, edited 1 time in total.
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Re: Bill and Ted each competed in a 240-mile bike race.  [#permalink]

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 15

*Kudos for all correct solutions.

You can find our video solution here: https://www.gmatprepnow.com/module/gmat ... /video/947

Another way to build up the quadratic...

Let Bill's speed = x
and Ted's speech = x+5

When Ted covers the full distance, Bill has yet to travel for 4 more hours to complete the distance
$$\frac{240}{(240 - 4x)} = \frac{(x + 5)}{x}$$
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Re: Bill and Ted each competed in a 240-mile bike race.  [#permalink]

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Vb and Vt be the average speeds of Bill and Ted.
Vb = Vt - 5

Tb and Tt be the time take,
Tt = Tb - 4

We have Distance = Speed * Time
$$Vb* Tb = Vt * Tt = 240$$
$$(Vt - 5) * ( Tt + 4) = Vt * Tt$$
4Vt - 5 Tt = 20
Replacing value of Tt as 240/Vt

$$4 Vt - 5*(\frac{240}{Vt}) = 20$$
$$Vt^{2} - 5 Vt = 300$$
$$(Vt -20)* (Vt +15) = 0$$
Vt = 20

Average speed of Bill Vb = Vt-5 = 20- 5 = 15.
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Re: Bill and Ted each competed in a 240-mile bike race.  [#permalink]

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 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.  [#permalink]

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 15

*Kudos for all correct solutions.

Another approach....

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
(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.

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

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 15

*Kudos for all correct solutions.

Let Ted's average speed be x miles/hr.
Bill's average speed be x-5 miles/hr.

Distance = 240 miles

240/(x-5) - 240/x = 4
240*5/(x(x-5)) = 4
x(x-5) = 300
x = 20
x-5 = 15

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

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Easier method
Bills time be= t and speed=x
Ted's time=t-4 and speed=x+5
distance =240
Bill's time(t) = distance/speed=240/x
Teds time(t-4)=240/(x+5)

Plugin answer choices. you should start from c or answers which are whole numbers

Let bills speed be =15
we now have
Bill(t)=240/15=16
Now Plugin for ted
t-4=240/(15+5)=240/20=12 or t=12+4=16.
with above approach we can avoid quadratic equation.
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Bill and Ted each competed in a 240-mile bike race.  [#permalink]

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 15

*Kudos for all correct solutions.

let r=Bill's average speed
Bill: d=rt
Ted: d=(r+5)(t-4)
combining,
rt=(r+5)(t-4)➡
4r=5t-20
substituting,
4r=5(240/r)-20➡
r^2+5r-300=0
(r+20)(r-15)=0
r=15 mph
E
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Posts: 4015
Re: Bill and Ted each competed in a 240-mile bike race.  [#permalink]

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 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

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

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 15

*Kudos for all correct solutions.

Usually, the way I do these questions is this - Go to the options directly.

If Bill's speed is 12, time taken is 240/12 = 20 hrs
Then Ted's speed is 17 which is not divisible by 240 so is not 16 hrs.

If Bill's speed is 15, time taken is 240/15 = 16 hrs
Then Ted's speed is 20 and time taken is 240/20 = 12 hrs
Works!

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GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4015
Re: Bill and Ted each competed in a 240-mile bike race.  [#permalink]

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 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

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

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GMATPrepNow 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?

A) 7.5
B) 10
C) 12
D) 12.5
E) 15

*Kudos for all correct solutions.

My big issue with 700 level math questions isn't a matter of getting an answer, but more of a timing issue. Does anyone have any tips on how I can improve here? It takes me 2-3minutes to solve a 700 level question. My process is 1) read the question, 2) setup equations, 3) solve. Thats as simplified as I can make the process, but it still takes (more often than not) closer to 3 minutes to solve any question. On practice tests, most of my questions are in the 700 range, and as a result I never finish the test on time and always end up missing the last 5 or 6 questions.... Quant scores usually hover around 48-50.

Here's my thought process if it helps anyone:

distance traveled for both racers are the same hence we can set the two rates and times equal to each other:

Let t represent Bill's time, and s-5 represent Bill's speed.
Let t-4 represent Ted's time, and s represent Ted's speed.

$$t*(s-5) = (t-4)*s($$

$$s=\frac{5t}{4}$$

Let's plug this back into the equation above and solve for the time bill spent on the race:

$$\frac{5t}{4}*t - 5t = 240$$
$$t=16$$
$$Average speed = distance/time$$

$$240/16 = 15$$
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Bill and Ted each competed in a 240-mile bike race.  [#permalink]

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kchen1994 wrote:
My big issue with 700 level math questions isn't a matter of getting an answer, but more of a timing issue. Does anyone have any tips on how I can improve here? It takes me 2-3minutes to solve a 700 level question. My process is 1) read the question, 2) setup equations, 3) solve. Thats as simplified as I can make the process, but it still takes (more often than not) closer to 3 minutes to solve any question. On practice tests, most of my questions are in the 700 range, and as a result I never finish the test on time and always end up missing the last 5 or 6 questions.... Quant scores usually hover around 48-50.

Your solution is perfect. So, we need not discuss that.
I want to mention that guessing on (or, even worse, NOT answering) 5 or 6 questions in a row will kill your score. Your goal should be to spread out your guesses.

Please review my time-management strategy below:

If you consistently don't have enough time to answer all of the quant questions, it's better to allow yourself to guess on a few questions. IMO, this is better than hurrying and answering all of the questions haphazardly (which can result in a lot of careless mistakes).

Cheers,
Brent
_________________ Bill and Ted each competed in a 240-mile bike race.   [#permalink] 21 Jan 2019, 12:38
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