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by gmatkkvinu » Tue Sep 17, 2013 1:55 pm
Sam 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. Sam, unaware of this, started from home as usual and met his son on the way and returns home with him 15 minutes early. If the speed of Sam is 30 km\hr, find the speed of his son.
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by theCodeToGMAT » Tue Sep 17, 2013 8:03 pm
What is the answer? is it 30/7??
gmatkkvinu wrote:Sam 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. Sam, unaware of this, started from home as usual and met his son on the way and returns home with him 15 minutes early. If the speed of Sam is 30 km\hr, find the speed of his son.
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by vipulgoyal » Tue Sep 17, 2013 8:27 pm
yes it seems 30/7

distance = 30 from point A to B
let the point of meet x
15 minuts early (time) = 2-1/4= 7/4 for two way and 7/8 for one way
distance in 7/8 time = 30*7/8= 105/4
distance fron point B = 30 - 105/4 = 15/4
time is same for father from point A to x and from B for son to x
let speed of son = s
15/4 = 7/8*s
s=30/7
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by ganeshrkamath » Wed Sep 18, 2013 12:32 am
gmatkkvinu wrote:Sam 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. Sam, unaware of this, started from home as usual and met his son on the way and returns home with him 15 minutes early. If the speed of Sam is 30 km\hr, find the speed of his son.
Sam takes 1 hour to go to school.
Sam's speed = 30 km/hr
So the distance between the school and home = 30 km

Now, because his son covers a distance of x, Sam's total distance = 2(30-x)
Time = distance/speed
1.75 = 2(30-x)/30
1.75*30 = 2*30 - 2x
0.25*30 = 2x
x = 3.75 km
d-x = 26.25 km
Time taken to cover this distance = 26.25/30 hr

Speed of son = 3.75/(26.25/30) = 3.75*30/26.25 = 15*30/105
[spoiler]= 30/7 km/hr[/spoiler]

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by sanju09 » Wed Sep 18, 2013 2:44 am
gmatkkvinu wrote:Sam 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. Sam, unaware of this, started from home as usual and met his son on the way and returns home with him 15 minutes early. If the speed of Sam is 30 km\hr, find the speed of his son.
It means Sam by and large reaches the school at 5 PM, and this is the same time his son is there waiting, today his son was freed at 4 PM and started walking.

The time saved by the son walking was 15 minutes, which is the time that Sam required to reach the school and come back from the point of meeting, it means that the time required from the point he reached to the school is 7½ minutes, if Sam usually reaches the school on 5 PM sharp, and now they met 7½ minutes before reaching the school, this means they met on 04:52:30 PM.

The son started walking on 4 PM, hence he walked for 52 minutes and 30 seconds (i.e. 7/8 hours) from 4 PM to 04:52:30 PM. In this time, the son covered a distance that Sam could have covered in 7½ minutes at 30 km\hr (calculations give this distance equal to 15/4 km). Hence, the speed of his son is

= (15/4) Km ÷ (7/8) hours

= [spoiler]30/7 kmph
[/spoiler]
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by GMATGuruNY » Wed Sep 18, 2013 12:36 pm
gmatkkvinu wrote:Sam 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. Sam, unaware of this, started from home as usual and met his son on the way and returns home with him 15 minutes early. If the speed of Sam is 30 km\hr, find the speed of his son.
Since the father arrives home 15 minutes early -- traveling for 7/8 of his normal time -- he travels 7/8 of his normal distance in each direction.
Thus, when the father and the son meet, the father has traveled 7/8 of the distance between home and the school, while the son has traveled 1/8 of the distance between home and the school.
Father's distance : son's distance = (7/8) : (1/8) = 7:1.
Since the son travels 1 kilometer for every 7 kilometers traveled by the father, the son's rate is 1/7 of the father's rate:
(1/7) * 30 = 30/7.
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