On a 1-mile test course, a vehicle is travelling at a constant rate of

Question

On a 1-mile test course, a vehicle is travelling at a constant rate of r miles per hour. If it were traveling at a rate of r+20 miles per hour, it would take 4 min less time to travel the test course. What is the value of r?

A. 5
B. 10
C. 20
D. 25
E. 30

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Attachment:

2024-01-24_19-31-47.png

MartyMurray · Accepted Answer

On a 1-mile test course, a vehicle is travelling at a constant rate of r miles per hour. If it were traveling at a rate of r+20 miles per hour, it would take 4 min less time to travel the test course. What is the value of r?

A. 5
B. 10
C. 20
D. 25
E. 30

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Attachment:

2024-01-24_19-31-47.png

A. 5
B. 10
C. 20
D. 25
E. 30

We can quickly solve this one by checking the answer choices.

The car is going around a 1-mile course.

At 20 miles per hour, the car would go around the course in 3 minutes since 60/20 is 3 minutes.

Thus, r can't be 20, 25, or 30 miles per hour since, in each of those case, the car would go around the track in less than 4 minutes at r. So, there's no way for it to take 4 minutes less to go around the track at r + 20.

So, 10 and 5 are the only possible options.

Check 10.

At 10 miles per hour, a car goes a mile every 60/10 = 6 minutes.

At 10 + 20 = 30 miles per hour, a car goes a mile every 60/30 = 2 minutes.

6 - 2 = 4

Correct answer:  B

gmatophobia · Answer

Given

\frac{1}{r} - \frac{1}{r+20} = \frac{4}{60}

Taking the denominator as  r(r+20)

r + 20 - \frac{r}{r(r+20)} = \frac{1}{15}

20 * 15 = r(r+20)

300 = r(r+20)

Answer choice elimination

A.  5 → 5(5+20) = 5 * 25 
eq 300  ⇒  Eliminate

B.  10 → 10(10+20) = 10 * 30 = 300   ⇒ Keep

Option B

Archit3110 · Answer

distance = speed * time
D is 1 mile 
rate is r miles per hour
1/r = 1/( r+20) + 4/60
solve for r
r = 10
option B

rey1009 · Answer

On a 1-mile test course, a vehicle is travelling at a constant rate of r miles per hour. If it were traveling at a rate of r+20 miles per hour, it would take 4 min less time to travel the test course. What is the value of r?

A. 5
B. 10
C. 20
D. 25
E. 30

Show Spoiler

Attachment:

2024-01-24_19-31-47.png

distance is equal
let t be hours at r miles/hour

(r+ 20)(t-4/60) =rt

rt=1 => t=1/r
20t-4/3-r/15=0

20/r -4/3 -r/15 =0
20-4r/3-r^2/15=0

r^2+ 20r-300=0

r =  -20 +- \sqrt{400+1200}  2
(-20 +-40)/2
r=10 (only positive value)

Bunuel · Answer

(time)(rate) = (distance)

On a 1-mile test course, a vehicle is traveling at a constant rate of r miles per hour. Hence, tr = 1;

If it were traveling at a rate of r + 20 miles per hour, it would take 4 minutes less time to travel the test course. Hence, (t - 1/15)(r + 20) = 1 (because 4 minutes is 1/15 of an hour).

Since t = 1/r, then (1/r - 1/15)(r + 20) = 1. Multiplying both sides by 15r gives (15 - r)(r + 20) = 15r

At this point it's better to plug in the options rather than to solve quadratics. Options B, 10 works: (15 - 10)(10 + 20) = 15*10 = 150.

Answer: B.

GmatKnightTutor · Answer

Instead of a 1-mile course, let's imagine it's a 60-mile course. That would now mean we're looking for two speeds that do this distance with a difference of 4 minutes times 60 = 240 minutes. Basically, look for two speeds that have a difference of 4 hours.

10 mph can do 60 miles in 6 hours.
30 mph (10 + 20) can do 60 miles in 2 hours.

That's a difference of 4 hours!

B is your answer.

Fish181 · Answer

Solved in about 1 minute.

Since you're looking for a r to r+20 relationship, the only answer choices that contain those pairs are 5 and 25 or 10 and 30.

Since we're looking for R we know it must be either 5 or 10. We can eliminate all other answer choices.

Then we can use the formula d = rt to find time. Rearrange to d/r = t. We subtract (1/r) - (1/r+20) = four minutes less or 1/15th of an hour.

1/5 - 1/15 DNE 1/15

Therefore B.

MBAToronto2024 · Answer

KarishmaB 
The way I solved it: 
(x+20)(y-1/15) = xy

Got equation: 300y - 20 = r -> 300y^2 - 20y = 1. 
Too long...Is there any better any logical means to solve it? What about weighted averages/ratios?

KarishmaB · Answer

When dealing with multiples of speeds (twice as fast, 20% reduced speed etc), then ratios and weighted average come in handy.

When dealing with number added to speed (s and s + 5 or s - 2 etc) then we must make equations.

\frac{1}{r} - \frac{1}{(r+20)} = \frac{4}{60}

\frac{1}{r} - \frac{1}{(r+20)} = \frac{1}{15}

But we can easily find the answer without solving it in the typical quadratic way.

I need 15 in the denominator on RHS so I must have a 3 and a 5 as factors of my denominators on LHS. (anyway not in the options)
If r were 3, I would get a 23 in the denominator of LHS which is not there in the denominator on RHS. Ignore.
If r were 5, I would get a 1/5 and 1/25 so I have a 5 but not a 3. Ignore.
If r were 10, I would get 1/10 and 1/30 - looks promising so check.

\frac{1}{10} - \frac{1}{30} = \frac{1}{15} 
 \frac{2}{30} = \frac{1}{15}  -> works

Answer (B)
 
Note that we must be careful while using this method in Data Sufficiency. It is possible that a quadratic may give us two viable values of x. In problem solving, only one answer is correct so we don't have to worry.

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On a 1-mile test course, a vehicle is travelling at a constant rate of

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