12 Days of Christmas GMAT Competition 2025 - Day 8: At 9:00 AM, a man

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Bunuel

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24 Dec 2025, 10:00

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Bunuel

GMAT Club Official Explanation:

When the man and the dog meet at 10:00 AM, the man has been walking for 60 minutes, while the dog has been running from 9:50 AM to 10:00 AM for 10 minutes. The distance from home to the park is the total distance covered by the man in 60 minutes and the dog in 10 minutes.

We know the dog covered the entire distance to the park in 30 minutes, so in 10 minutes, the dog would have covered one-third of this distance. This means that in 60 minutes, the man covered two-thirds of the distance to the park.

If the man covers two-thirds of the distance in 60 minutes, it would take him 90 minutes to cover the entire distance. Therefore, the man will reach the park at 9:00 AM plus 90 minutes, which means he will arrive at 10:30 AM.

Answer: D.

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24 Dec 2025, 09:30

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Let speed of the man and the dog be Sm & Sd respectively and distance from home to park be D

From 9:00 AM to 9:30 AM: -
D/Sd = .5
D = .5Sd = Sd/2

Dog starts at 9:50 and reaches man at 10:00 AM
Distance travelled by dog = Sd/6
Distance travelled by man from 9:00 AM to 10:00 AM = D - Sd/6 = Sd/2 - Sd/6 = 2Sd/6 = Sd/3
Speed of the man = Sd/3 = Sm

Distance travelled by both = Sd/6
Time taken = Sd/6 / Sm = (Sd/6)/(Sd/3) = 1/2 hours

Man will reach the park at 10:30 AM

IMO D

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24 Dec 2025, 09:30

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	Speed	Time	Distance
Dog to the park	d	30	30d
Dog at the park	0	20	0
Dog to the man	d	t=10	dt=10d
Man	m	60	60m

Distance covered by man = 60m = 30d-10d
60m= 20d
m=d/3
Dog took 30mins to reach the park
Man's 3 times slower and will take 3x time so man will take 30*3 = 90mins
9AM +90mins = 10:30PM

Bunuel

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24 Dec 2025, 09:43

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At 9:00 AM, a man and his dog leave their home and walk along the same path toward a park. The man walks at a constant speed, while the dog runs at a faster constant speed. The dog reaches the park at 9:30 AM, spends 20 minutes playing with other dogs, and then runs back toward the man at its usual constant rate. They meet at 10:00 AM, and from there, both walk together at the man’s speed to the park. When will the man reach the park?

A. 10:10 AM
B. 10:15 AM
C. 10:20 AM
D. 10:30 AM
E. 10:40 AM

time man & dog start is 9 am
speed of man be a and speed of dog be b
distance home to park be d
dog reaches park in 30 mins and plays 20 mins , runs back , reaches man at 10 am

d= 0.5*b
ran back 10 mins then distance is b*(10/60) ; b*1/6

in 1 hours man has covered distance
a * 1 and dog distance from home is D-b*(1/6)
now that both man & dog are at same positions
a = d-b*(1/6)
d= 0.5*b
a= 0.5b-b/6
a = b/2-b/6
a=3b-b/6
a=2b/6
a=b/3

speed of man is 1/3 of dog
d= 0.5*b
a= b/3
.5b/b/3 = 1.5 hours
time it takes for man to reach park is 9 + 1.5 hours ; 10:30 am

OPTION D is correct; 10:30 am

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24 Dec 2025, 10:01

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Let total distance be d, dog speed be s1 and man speed be s2 where s1>s2
Dog run from 9 am to 9;30 am taking a total of 30 minutes to cover distance
so d=30s1
By the time the dog runs back to the man at 9;50 am the man has been walking for 60 minutes covering 60s2 distance
In return journey the dog covered 10s1 distance
The remaining distance upon the meetup is d-10s1
So 60s2=d-10s1 since d=30s1
60s2=30s1-10s1
60s2=20s1
s2=s1/3
Time required by man equals total distance divide by his speed 30s1/(s1/3)= 90 minutes (1 hour 30 minutes)
Departure time 9 am plus 90 minutes = 10:30 am
Ans D

Bunuel

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24 Dec 2025, 10:01

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answer is d 10:30
total distance be 30 km
assume that dog travels at 1km/min
at 9:30 AM dog has covered 30 km and played for 20 mins and then met the man at 10 am hence dog would have traveled 40 km in total (30 km - to reach park and 10 km backwards to meet man)
in the mean time by 10 AM man has walked 20 km to meet the dog and then both walked at man's speed. to cover 10 km man will take another 30 mins as he took 1 hr to cover 20 km.

hence he will reach the park by 10:30

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24 Dec 2025, 10:28

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the dog took 30 mins to reach park

let the distance be "d'' from home to park

after 20 mins of playing it started towards man and they met at 10 am, that means the dog after travelling 30mins+20mins playing it took 10 mins to reach man

in 30 mins it covered "d" distance & in 10 mins it covers "d/3"

also it given that from the point of meeting both goes to park at the man's speed.

if the distance the dog travelled is "d/3" to meet the man -> calculated from the 10 mins time it travelled to meet man after playing

it says that man travelled "2d/3" in 1 hour and remaining is "d/3" which he can do in 30 mins if walking at the same speed

there both of them reaches by 10:30

Bunuel

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24 Dec 2025, 10:36

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This one took me a very long time.

I plugged in using answer choices.

The dog spends a total of 50 minutes before it heads back to meet the man, implying that it takes him 10 minutes to reach the man, and that the man has been walking for 50 minutes.

Since the dog arrives at the park in 30 mins, the distance he travels back to the man is 1/3 of the entire distance of the park. We should analyze the answer choices to decide which one suggests the man has traveled 2/3s of the distance of the park.

If the man will reach the park at 10:10, this implies it takes him 70 minutes at his rate to reach the park from 9:00. So if he walked 50 minutes, he will be 5/7s of the way to the park.

If the man will reach the park at 10:15, this implies it takes him 75 minutes to reach the park. So if he walked for 50 minutes, he will have walked 2/3s of the way to the park which matches the 1/3 the dog must come to meet him. So choice B.

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24 Dec 2025, 10:51

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Let the total distance be x.

Speed of dog, v1 = x / 30. Also, after playing, the dog returned to the man from the park in 10 min, thus, his speed can also be computed as v1 = (x - y)/10 where y is the distance travelled by man from home and where he meets the dog.

From above we get, x/30 = (x-y)/10 => 3y = 2x or y = 2x/3

This means that the man covered 2/3rd distance in 60 min (ie, by 10am). Hence, he'll reach the park in another 30 min (ie, 10:30am).

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24 Dec 2025, 10:57

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let's say the distance park-home is 1km.
dog's speed -> d = 2km/h
man's speed -> m = ?

in 1h (30 min from home to park, 20 playing, 10 going back) the man has walked a fraction m of the distance.

You can consider the problem as a "meeting" problem, where the man has been walking for an hour and the dog for 10 minutes (1/6 hour), each at their respective speeds. When they meet, it means that they combine distance is equal to the total distance (1km).
m * 1 + d * 1/6 = 1
m + 1/3 = 1
m = 2/3 km/h

this means that in 1hour the man has walked at 2/3 km/h and that the remaining distance is 1/3 km.

Going at his speed, it will take 30 more minutes.

Answer D. 10.30 AM

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24 Dec 2025, 11:27

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LET, speed of man be s1 & of dog s2
60s1+10s2=30s2
60s1=20s2
s2/s1 = 3

man takes = 10s2 /s1 = 10*3 =30 min

so reaching time 10 hrs +30min = 10:30A.M

Ans D

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24 Dec 2025, 11:50

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let the man's velocity be m and dog's velocity be d.
acc. to question -> dist. b/w home to park = 30d (30 min * dog's velocity) - (i)
also by 10:00 -> man covered 50m + 10m and dog covered 10d so total distance from home to park = 60m + 10d - (ii)
equating (i) and (ii) -> 60m + 10d = 30d => d = 3m hence the distance the dog and man meet -> 10d = 30m -> 30 minutes of man's time to cover the distance hence post 10:00 + 30 more minutes = 10:30 AM

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24 Dec 2025, 12:33

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Bunuel

Let d=distance from home to park, a=distance from park to man after 60 min, dog's speed = y and man's speed =x.

Diagram looks a bit like this:

Home--------------------- Dog Returns to Man ---------- a ------ Park (20 min wait)
X------------------------------------------X-----------------------------X
Entire distance = d

Some formulas we can make using r=d/t formula, everything is in minutes.

$t=\frac{d}{x}$ <- This is what we want to find. Eq 1
$d=30y$ <- Dog takes 30 min to get to park. Eq 2
$a=10y$ <- Takes the dog 10 min to return to the man. Eq 3
$x=\frac{d-a}{60}$ <- How far the man has travelled when the dog returns to him. Eq 4

We can sub in Equation 2 & 3 into 4 and get:
$x=\frac{30-10y}{60}=\frac{1}{3}y$ <- Eq 5

We can sub in Eq 5 to Eq 1 to get d in terms of x: $d=90x$

We now have everything we need to put back into Eq 1 and solve for t.
$t=\frac{90x}{x}=90$

It takes the man 90 min from leaving from home to reach the park, so he would have arrived at 10:30AM. Answer is D.

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24 Dec 2025, 13:14

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Let the speed of the dog => D
and speed of the man => M
Let the total distance from the home to park => x meters.

Speed = Distance / time. Therefore, Distance = Speed * Time.

Let us consider all the units in meters and minutes for ease of calculation.

The dog reached the park at 9:30. So time taken for the dog to reach the park is 30mins.
Therefore the total distance, x = D*30 or 30D.

Now the dog plays for 20mins, i.e. it plays from 9:30 to 9:50.
Then meets the man at 10:00 am, i.e. it travels from the park to the man's position in 10mins.

Therefore, the distance yet to be traveled by the man = 10D.
And distance already travelled by the man = (Total Distance - Distance yet to be travelled) = 30D - 10D = 20D.
The man has covered 20D meter from 9:00 am to 10:00am, i.e. 60mins.
So we can calculate the speed of the man = 20D/60 = D/3 meter/min.

Therefore, time required to travel the remaining distance = (Distance yet to be travelled)/(Speed of the man) = (10D)/(D/3) = 30mins.
So the man will reach the park 30mins after he met with the dog.
i.e. 30mins after 10:00 am
i.e. at 10:30 am.

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24 Dec 2025, 15:00

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dog speed= y km/h
man speed= x km/h
y>x
distance = d km

dog took 30 min to travel whole distance. so d= y/2
they both meet at 10 am. so in 1 hr man travelled x km
dog travelled = y/6
total distance d = x + y/6
we know d= y/2
y = 3x
d= 3x/2

now left distance is d-x= (3x/2)- x= x/2

now they walk at mans speed.
so time = (x/2) / x= 1/2= 30 mim
so 10:30

Bunuel

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24 Dec 2025, 15:24

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Bunuel

let man speed = m
dog speed = d
let total lenth of path = A
a= dx30/60=d/2
what dog covers in last 10 mins =dx10/60= d/6
what distance left to cover backwards is what the man covers= d/2-d/6=d/3

what man covers in 1 hr
mx1=d/3
d/m=3/1-----d:m= 3:1
let m= k, d= 3k

distance they both cover together= distance dog covers backward= d/6=3k/6=k/2
time to cover this at man 's speed= (k/2)/k= 1/2 hrs= 30 min
so man reaches park at 10.00+30 min
Ans-A

distance they both cover together=

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24 Dec 2025, 16:21

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Answer Choice: D

Conceptual refresh: Distance = Rate x Time where the total distance to the park is fixed, so the man and the dog each cover portions of that same distance at their own constant rates.

The man: walks from 9:00 to 10:00, so he has been walking for 60 mins when he meets the dog. To determine what fraction of the total distance he has covered by then we use the dog's motion.

The dog: reaches the park in 30 mins, so its rate is D/30. After playing for 20 mins the time is 9:50 and as the dog runs back and meets the man at 10:00 it means it has been running back for 10 mins. In those 10 mins using the dog's rate it covers D/30 x 10 = D/3 of the total distance.

This means the meeting point is 1/3 of the distance from the park, so the man has already covered 2/3 of the distance from home in 60 mins. Therefore, each third of the distance takes the man 30 mins. With 1/3 of the distance remaining, the man needs 30 more mins after 10:00 to reach the park therefore the man arrives at the park at 10:30 AM.

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24 Dec 2025, 19:00

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