# Help with distance catch up problem?

A train is going at 1/3 of its usual speed and it takes an extra 30 minutes to reach its destination. Find its usual time to cover the same distance.

Could you please give a few different examples of how to solve this? Best answer awarded to most helpful. Thank you.

### 6 Answers

- PuzzlingLv 72 months agoFavorite Answer
You know the distance formula is d = rt.

Let t be the usual time it takes (in minutes)

Let r be its usual speed

d = rt

Then when it is going slower:

The time it takes is t + 30.

The rate is r/3.

d = (t + 30)r/3

Equate these two:

rt = (t + 30)r/3

rt = rt/3 + 30r/3

rt = rt/3 + 10r

rt - rt/3 = 10r

rt(1 - 1/3) = 10r

rt(2/3) = 10r

Divide both sides by r (assuming it is non-zero):

(2/3)t = 10

t = 10 * 3/2

t = 30/2

t = 15

That's a longer, more algebraic way to solve this.

But here's an intuitive way to solve it.

The speed and the time are inversely proportional. In other words, if the speed is 1/3 the usual speed, the time will be 3 times as much as usual.

Let t be the usual time.

Let 3t be the new time.

3t = t + 30

2t = 30

t = 30/2

t = 15 minutes

Summary:

Normally it takes 15 minutes, but when it goes 1/3 as fast it will take 3 times as long (45 minutes) which is 30 minutes longer.

Answer:

15 minutes

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- la consoleLv 72 months ago
Recall: s = d/t → where s is the speed, d is the distance, t is the time

First case: usual speed

s = d/t → where d is the distance to reach the destination

s = d/t → where t is the required time to reach the destination at normal speed

Second case: usual speed divided by 3, time is 30 minutes more, i.e. (1/2) hours

s = d/t → adapt this formula to the present case

s/3 = d/[t + (1/2)] → recall: s = d/t

(d/t)/3 = d/[t + (1/2)]

d/(3t) = d/[t + (1/2)] → you can simplify by d both sides

1/(3t) = 1/[t + (1/2)]

3t = t + (1/2)

2t = 1/2

t = 1/4 ← in hours

t = 15 minutes

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- jacob sLv 72 months ago
usual time to cover the same distance is 15 minutes.

Step-by-step explanation:

Required Formula:

Time = Distance / Speed

Let's assume

The usual speed of the train be "s" km/hr

The distance travelled by the train be "D" km.

The usual time taken by the train to cover the same distance be "t" hr = D/s

Given that,

The reduced speed of the train = km/hr

Therefore, according to the question, we can write the eq.

- D/(s/3) -D/s=30min/60min

3D/s-D/s=1/2

D/s(3-1)=1/2

D/s(2)=1/2

D/s=1/4

Thus,

The usual time "t" taken by the to cover the same distance is,

= D/s

= 1/4*60min

= 15 minutes

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- stanschimLv 72 months ago
Let s = Usual speed in mph

Let s/3 = Speed now in mph

Let t = Usual time in hours

Let t + 1/2 = Time now in hours

Let d = Distance in miles.

Then.

t = d / s. This is the "usual" situation.

t + 1/2 = d / (s / 3). This is the current situation.

We can now solve these equations for t.

t = d / s

2t + 1 = 2d / (s / 3) = 6d / s = 6t

4t = 1

t = 1/4 hour or 15 minutes.

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- micatkieLv 62 months ago
Let t min be the usual time to cover the same distance.

When velocity = 3v units/min, time taken = t min:

Hence, distance = 3v • t units ... [1]

When velocity = (1/3)(3v) = v units/min, time taken (t + 30 min):

Hence, distance = v • (t + 30) units ... [2]

[1] = [2]:

3v • t = v • (t + 30)

3t = t + 30

2t = 30

t = 15

Usual time to cover the same distance = 15 min

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- Anonymous2 months ago
distance = speed * time. Let V be the speed and T be the time

distance = V * T

now with the new information remembering that the distance is the same no matter how fast or slow you travel at

distance = V / 3 * (T + 30)

setting the two equations equal to each other because distance is the same gives you

V * T = V/3 * (T + 30)

VT = VT/3 + 10V

2/3VT = 10V

T = 15 minutes or 0.25 hour

Try some numbers and see if it works. Say the speed = 60 miles/hr. Distance = 15 miles. If the speed is dropped to 20 mile/hr then it takes 0.75 hours, 0.5 hr or 30 minutes longer.

Try 10 miles/hr. Distance = 10 mile/hr * 0.25 hr = 2.5 miles. Drop the speed to 1/3 and it now takes 0.75 hours. Again, 30 minutes faster.

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