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

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

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Updated on: 25 May 2017, 04:16
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Difficulty:

55% (hard)

Question Stats:

74% (02:58) correct 26% (03:17) wrong based on 183 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

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Originally posted by GMATPrepNow on 26 Oct 2016, 11:43.
Last edited by GMATPrepNow on 25 May 2017, 04:16, 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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26 Oct 2016, 17:04
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1
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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27 Oct 2016, 16:26
3
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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24 May 2017, 22:35
4
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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25 May 2017, 04:14
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Top Contributor
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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23 Oct 2017, 22:22
1
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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03 Nov 2017, 03:44
1
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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04 Nov 2017, 14:19
1
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
Bill and Ted each competed in a 240-mile bike race. &nbs [#permalink] 04 Nov 2017, 14:19
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