A man and his dog are racing on a circular track at constant speed. John can run once around the track in six minutes, and Ben (his Dog) in four minutes. In how many minutes will Ben overtake John?

Ok, I think it can be solved this way: (hope that there is circular velocity here)
Assuming that both of them start at the same point and at the same moment, the dog overtake him, in time 0….01 cause he passed him, the dog velocity is greater than the man velocity, but I asumme that this mean when the dog meet again with his master.. so:
In an easy way:
John covers 1/3 of the lap distance in 2 minutes
Ben covers 1/2 of the lap distance in 2 minutes
1/2-1/3= 1/6 (distance that Ben gains over John every 2 min)
six time this time… Ben would have gained exactly 1 lap
that happen 6X2= 12 min… Ben meets John in 12 min so:
Ben would overtake John in 12 min and 1 nanosecond… jajaja, cool

One question: If John is running around the track, and the puzzle title says that it’s a bike race, does that mean that his dog is the one on the bicycle??

*Assume that the speed of the man is (Sm = 1/6 RPM)and speed of Dog is (Sd=1/4 RPM)
*Assume that they will meet again after (X) mins.
* Then the No. of rounds covered by Man = 1/6 X, and by the Dog = 1/4 X.
* As they will meet again, Then: (1/4 X = 1/6 X + 1).
* Then: (1/4-1/6) X = 1 ==> (1/12) X = 1 ==> X = 12 Mins.
THANKS FOR ALL

we must find at witch time man will have done x laps and dog x+1 laps
H 1 lap – 6 minutes/D 1 lap – 4 minutes
H 2 lap – 12 minutes/D 2 lap – 8 minutes
H 3 lap – 18 minutes/D 3 lap – 12 minutes

As we see at the time of 12 minutes the Human (H) has done 2 laps(x = 2) and the Dog(D) has made 3 (x + 1 = 2 + 1 = 3).
Answer is 12

12 minutes?? That answer doesn’t make any sense.
If the dog runs a lap in 4 minutes it is going to overtake the man before the man can finish his lap (in 6 min). The puzzle doesn’t say when they start, so we assume they start together. If so,the dog will immediately pull ahead and be ahead the entire race.

Oh, I see now… I thought you were all daft
Turns out the puzzle is just worded strangely. What you call “overtake” I would call “lap”.

When would the dog lap the man? Assuming they start together, the dog immediately overtakes the man and is in the lead. At 6 minutes when the man is finishing his first lap, the dog is already 1/2 around his second lap. At 12 the man has completed 2 laps and the dog is finishing his 3rd. The dog laps the man when he overtakes him for the second time.

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o.najaee| Profile August 19th, 2010 - 8:29 am2.4 min

funny

Shawn| PUZZLE GRANDMASTER | Profile August 19th, 2010 - 10:17 am12 minutes

suineg| PUZZLE MASTER | Profile August 19th, 2010 - 12:15 pmOk, I think it can be solved this way: (hope that there is circular velocity here)

Assuming that both of them start at the same point and at the same moment, the dog overtake him, in time 0….01 cause he passed him, the dog velocity is greater than the man velocity, but I asumme that this mean when the dog meet again with his master.. so:

In an easy way:

John covers 1/3 of the lap distance in 2 minutes

Ben covers 1/2 of the lap distance in 2 minutes

1/2-1/3= 1/6 (distance that Ben gains over John every 2 min)

six time this time… Ben would have gained exactly 1 lap

that happen 6X2= 12 min… Ben meets John in 12 min so:

Ben would overtake John in 12 min and 1 nanosecond… jajaja, cool

Bobo The Bear| PUZZLE MASTER | Profile August 19th, 2010 - 12:20 pm(Assuming they both start at the same point on the track …

After 12 minutes, John will have done 2 laps, and his dog will have done 3 laps and will then overtake him.

Bobo The Bear| PUZZLE MASTER | Profile August 19th, 2010 - 12:26 pmOne question: If John is running around the track, and the puzzle title says that it’s a bike race, does that mean that his dog is the one on the bicycle??

RK| Founder | Profile August 19th, 2010 - 12:58 pmOops! title changed Bobo

Hex| PUZZLE MASTER | Profile August 19th, 2010 - 7:42 pmIn 12 minutes, John would have made 2 laps and Ben 3.

munna| Profile August 20th, 2010 - 3:40 amAssuming they John and Ben start from same point, Ben will overtake John in 12 minutes.

Falwan| Profile August 20th, 2010 - 10:15 pmx = time in hrs.

D = diameter

(xD/4) – (xD/6) = D

(x/4) – (x/6) = 1

3x – 2x = 12

x = 12

Mr. Ben will overtake Mr. John after 12 hrs.

Falwan| Profile August 20th, 2010 - 10:19 pmI considered the time in hours.

So 12 min. in that case.

RK| Founder | Profile August 23rd, 2010 - 10:22 pmVery good, 12 min is correct

Hendy| Profile September 14th, 2010 - 11:31 amJ = 1/6 rpm (1 revolution/ 6 minutes)

B = 1/4 rpm (1 revolution/ 4 minutes)

rJ = t * 1/6 rpm

rB = t * 1/4 rpm

rJ = x min * 1/6 r/min

rB = x min * 1/4 r/min

rJ = x * 1/6 r = x * 1/6 r

rB = x* 1/4 r = x * 1/4 r

rB = rJ + 1

x * 1/4 r = x * 1/6 r + 1 r

1/4-1/6 = 1/12

1 r = x * 1/12 r

1 = x * 1/12

x = 12

12 * 1/6 = 2

12 * 1/4 = 3

Overtaken in 12 laps (revolutions).

Hendy| Profile September 14th, 2010 - 11:55 amOops! That should be time in minutes, not revolutions. Not paying attention!

Hedaiet El-Sabbahy| Profile September 28th, 2010 - 2:12 am*Assume that the speed of the man is (Sm = 1/6 RPM)and speed of Dog is (Sd=1/4 RPM)

*Assume that they will meet again after (X) mins.

* Then the No. of rounds covered by Man = 1/6 X, and by the Dog = 1/4 X.

* As they will meet again, Then: (1/4 X = 1/6 X + 1).

* Then: (1/4-1/6) X = 1 ==> (1/12) X = 1 ==> X = 12 Mins.

THANKS FOR ALL

Tsopi| Profile October 18th, 2010 - 2:48 pmwe must find at witch time man will have done x laps and dog x+1 laps

H 1 lap – 6 minutes/D 1 lap – 4 minutes

H 2 lap – 12 minutes/D 2 lap – 8 minutes

H 3 lap – 18 minutes/D 3 lap – 12 minutes

As we see at the time of 12 minutes the Human (H) has done 2 laps(x = 2) and the Dog(D) has made 3 (x + 1 = 2 + 1 = 3).

Answer is 12

venkatg70| Profile December 16th, 2010 - 6:02 pmLet a,b be the time taken by the person and dog respectively.

Let a>b

Let x be the time when they meet

Then,

x/a = fraction of the circular track covered by person

x/b = fraction of the circular track covered by dog

As they both run, gap created between them. When the gap (fraction of the track length)

approaches “n” where n = 1,2,3…, they meet.

For initial meet n =1, so

So x/b – x/a = 1

Solving, we get x = ab/(a-b)

For eg: when a=6 min, b=4 min, x = 12 min

swejhammer| Profile January 20th, 2015 - 12:59 am12 minutes?? That answer doesn’t make any sense.

If the dog runs a lap in 4 minutes it is going to overtake the man before the man can finish his lap (in 6 min). The puzzle doesn’t say when they start, so we assume they start together. If so,the dog will immediately pull ahead and be ahead the entire race.

swejhammer| Profile January 20th, 2015 - 1:12 amOh, I see now… I thought you were all daft

Turns out the puzzle is just worded strangely. What you call “overtake” I would call “lap”.

When would the dog lap the man? Assuming they start together, the dog immediately overtakes the man and is in the lead. At 6 minutes when the man is finishing his first lap, the dog is already 1/2 around his second lap. At 12 the man has completed 2 laps and the dog is finishing his 3rd. The dog laps the man when he overtakes him for the second time.