Anwar leaves home everyday at 4 p.m to pick his son from

thanks a lot it is really helpful i need one help of your i have a big tabboo for Time distance and speed problem can you please give me some tips to improve it

Theory on Distance/Rate Problems: distance-speed-time-word-problems-made-easy-87481.html

DS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=44
PS Distance/Rate Problems to practice: search.php?search_id=tag&tag_id=64

Hope it helps.

BTW, the question you've posted has quite ambiguous wording, so I believe it's not from reliable GMAT source. My advice would be not to study from such sources.

Anwar leaves home everyday at 4 p.m to pick his son from school and returns home at 6 p.m. One day, the school was over at 4 p.m and the son started walking home from school. Anwar, unaware of this, starts from home as usual and meets his son on the way and returns home with him 15 minutes early. If the speed of Anwar is 30 km\hr, find the speed of his son.

The distance from home to school is 30 km. The round trip distance = 60 km. Because he returns home 15 minutes early, I thought it might be easier to break his trip up into 8 identical 15 minute blocks, each of which he traveled at 7.5 km/hour.

Because we are dealing with a round trip here, Anwar would have stopped 7.5 minutes short of the school. This means he traveled Distance=rate* time .5Km/hour*52.5 minutes or 26.25 KM. His son therefore traveled 3.75km/hour by the time he met his dad at 4:52.5. It took the son 52.5 minutes to travel 3.75 km. Because we are trying to find the son's rate per hour (i.e. 60 minutes) and we have his speed for 52.5 minutes it's best to convert to km/minute then multiply by 60 to get km/hour

3.75km/52.5 = .07 km/minute
.07km/minute*60 minutes = 4.28 km/hour.

Answer: C. 4.28 km/hr

Since it normally takes him 2 hours round trip at 30 km/hr, we know that each leg is 30 km. Anwar's position at time t is 30t, while his son's is 30 - x*t.

\(30t = 30 - xt\)

\((30+x)t = 30\)

\(t = \frac{30}{30+x}\).

So they meet at time t. Since the total time is 2*t = 1 hour and 45 minutes, we know that:

\(\frac{60}{30+x} = \frac{7}{4}\)

\(240 = 210 + 7x\)

\(30 = 7x\)

\(x = \frac{30}{7} = 4.28\) km/hr.

Answer: C

Anwar normally picks up son after driving 1 hour for 30 km
on this day he saves 15 minutes/120 minutes=1/8 time and distance
he drives only 7/8*1 hour before meeting son
he drives only 7/8*30 km before meeting son
therefore, son, who has started at the same time, walks the equivalent of 1/8 *30 km=3.75 km in 7/8 hour
3.75/(7/8)=4.28 km/hr
C

1. Let Anwar travel distance y and his son travel distance x when they meet.
2. Normally Anwar travels 2x + 2y distance, now he travels only 2y distance. He saves 15/2= 7.5 min in not traveling x distance or in other words he takes 7.5 min to travel x but we know he takes 120/2= 60 min to travel x+y.
3. The ratio of x and y is therefore 1:7.
4. So the speed is also in that ratio. Anwar's son speed is therefore 30/7 km/hr = 4.28 km/hr

As for father
t=2 r=30 d=60
As for son
t-15min *R =60
(120-15)*R=60
R=34.28km/h

NOW relative speed traveling at same direction returning home
34.28-30=4.28km/h

Can someone please explain, how is time 7/4?

Normally it takes 120 mins for Anwar but this time it took 105 mins that is 7/4 hours.

However, as per my understanding the equation should be as below in the context of relative speed.

\(t=\frac{d}{s}\)

here t is the time at which Anwar meets his son, d is the distance between them and s is their relative speed, i.e, speed of Anwar plus his son's
Normally, it takes 2 hours for Anwar with the speed of 30 km/hr to do a round trip. From this it can be derived that round trip distance is 60 km and the distance between anwar and his son d =30 km
When it took 105 mins for Anwar to complete his trip, he met his son at 105/2= 52.5 mins = 7/8 hrs
Considering the speed of Anwar's son as x, the relative speed is 30 + x

The above equation becomes,
\(\frac{7}{8}=\frac{30}{(30+x)}\)
\(x=30/7=4.28 km/hr\)

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chetan2u VeritasKarishma Bunuel
The son starts at 4 PM , meets Anwar somewhere in the return journey and returns back 15 min earlier than 6PM...ie at 5.45 PM...So it means the son completed the return journey(30km) in 1hr 45 mins....then his speed should be 30/1.75 ...

Can someone please elaborate where am I wrong?

When you say "30/1.75", you are considering that the distance travelled by son in time 1.75 is 30 km. However, son travelled only the distance d till he met Anwar. After he met Anwar, they returned with the speed of 30 km/hr, which is the speed of Anwar and the time changed accordingly.

It should be speed of son = ds/ts

Ds = 30 - Da = 30 - 26.25 = 3.75 km
ts = time taken by Anwar/2 = 1.75/2 = 0.875 hrs,

speed of son = 3.75/0.875 = 4.28 km/hr

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where from are u getting Da = 26.25?

Da is the distance covered by Anwar that is normally 60 kms. But on this specific day, Anwar travelled for 105 mins(7/4 hrs) and at 30 km/hr. So distance covered by Anwar on this day is D = 30*7/4= 52.5 km.
But 52.5 is round trip distance(home to school to home), so one way(home to school) is 52.5/2 = 26.25 km. In simple words, the distance between Anwar and his son is 30 km but on this day Anwar travelled only for 26.25 km because his son already started and covered 3.75 km by walk.

Bunuel

If both Anwar and his some have different speeds then how come they reach together. If he is coming back with father then the average speed change either for Anwar or for the son or both.

Posted from my mobile device

I guess Anwar is driving and he picks up his son when they meet. The question asks the speed of the son when he is walking by himself.