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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. 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?

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24 May 2017, 22:35

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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?

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27 Oct 2016, 16:26

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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:

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25 May 2017, 04:14

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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 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.

Concentration: General Management, Entrepreneurship

GPA: 3.8

WE: Engineering (Energy and Utilities)

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23 Oct 2017, 22:22

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

*Kudos for all correct solutions.

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

Bill and Ted each competed in a 240-mile bike race.
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03 Nov 2017, 03:44

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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.

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

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19 Dec 2018, 17:56

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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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20 Dec 2018, 04:58

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.

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!

Answer (E)
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20 Dec 2018, 07:21

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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

Re: Bill and Ted each competed in a 240-mile bike race.
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20 Jan 2019, 16:36

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:

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