A mathematical problem ?

Question

Father and son go to work every day to the same factory. Father arrives to the factory from home in 30 minutes, while his son arrives in 20 minutes. If his son leaves home 5 minutes after his father, after how much time will he catch up with his father?

Possible solutions are a) 5 mins b) 10 c)15 d)20

Now you're confusing me completely, I'm wondering between 10 and 15 minutes, and both groups have reasonable arguments and formulas.

Answer 1

in 10 minutes he'll catch his father

Split there travel time into sections of 5 min. The father takes 6 five minute segments and the son takes 4. So since the dad left first....

d=1/6 for dad and d2=0 for son since he left later

d=2/6 d2=1/4 5 min after the son leaves
d=3/6 d2=2/4 10 min after the son leaves

d=d2 so he's caught his dad

Answer 2

After 15 min.

We assume they have to walk a unit-length of 1. If we write the progress of father and son as two functions, we will have:

F(x) = 2x
S(x) = 3x -1/4 , we subtract 1/4 because the son would have reached
1/4 of the unit-length after 5 min.

where x is measured in hours. We set F(x) = S(x) to find when they are at the same progress.

2x = 3x - 1/4 -> x = 1/4

The answer is 0.25 hours or 15 minutes.

[edit]: Some have answered that father and son met after 10 minutes. This is also correct, but the two methods of calculation assumes a different offset. Us who reached 15 minutes started counting after the father began walking, and those who reached 10 minutes after the son started walking.

Answer 3

After 15 minutes

Answer 4

30/20 = The son travels 1.5 times faster than the father or 50 percent faster.
Fifty percent = 0.5
5/(0.5)=10

10 minutes

The father has traveled 15 minutes and the son has traveled 10 minutes.

30/20 = 15/10 = 1.5

Answer 5

it's basically a trick question, since you don't have enough information. What you need to know, most importantly, is "are they travelling in the same direction?" Obviously if they're not, the son will never catch up. The second thing you need to know, is "are they travelling in comparable methods?" Obviously if the father is travelling by train and the son by car, they will never catch up in the human sense (being able to give one a message, etc.).

ASSUMING they are both travelling the same direction and method, here is the answer: in 12 1/2 minutes

Answer 6

15 min.
let office is at a dist of x from home, then,
speed of son = x/20
speed of father = x/30
let son catches father a distance of y from home.
let t be the time taken by father to travel y dist
this implies t= y/[x/30] and since son travels the same y dist in 5 min less than his father so
t-5 = y/[x/20]
substitue y/x value from one eq into other and get the value of t as 15min

Answer 7

12 minutes after the father starts or 8 minutes after the son starts is when they will meet.

Answer 8

given the father walks at a rate of x/30 per minute and the son moves at x/20 per minute

5x/30 + yx/30 = yx/20

where x is the distance from the factory to their home and y is the time when they meet

(5x+xy)/30 = yx/20

(10x+2xy)60 = 3yx/60

10x - xy = 0

x(10-y)=0

10-y=0

y=10

they meet after ten minutes given any amount of distance

Answer 9

15 min after the father starts
let the distance to factory be "d"
the speed of fatherV(f)=d/30 (units per minute)
and tha of the sonV(s)=d/20
Now assuming that they start from the same position, they meet when they have travelled equal distances let that distance be "X" and the time taken by father be "t" then
X=d/30*t=d/20(t-5)
i.e. t=15 minutes
after the father starts.

Answer 10

this is a simple linear equations problem

let d = distance to work.
let v1 = speed of father
let v2 = speed of son

v1 = d/30
v2 = d/20

The position at any time is given as

x=v*T

where t is the elapsed TRAVELLING time.

the question is, when x1=x2, what is t?

well, if the father leaves 5 minutes earlier at T=t-5,

x1= v1*T = v1*(t-5)

and the son leaves 5 minutes later , T=t

x2 = v2*T = v2*t

and so equating x1=x2:

v1*(t-5)=v2*t

and solving for t:

v1*t-5*v1=v2*t
v1*t-v2*t=5*v1
t*(v1-v2)=5*v1
t=5*v1/(v1-v2)
t=5*(d/30)/(d/30-d/20) Note that the d's cancel out here.
t=5*(1/30)/(20/600-30/600)
t=5*(1/30)/(10/600)
t=5*(1/30)*(600/10)
t=5*600/300
t=5*2
t=10

so the son will catch up to the father in 10 minutes