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2 workers,One Young And One Old, Live Together and Work at the same office . It takes 20 mins for the young man to walk to office . the old man takes 30 mins for the same distance. when will the young man catch up with the old man , if the old man starts at 10.00 am and the young man starts at 10.05 am ?

Thanks in advance :)

Earthdreams
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  • Hello, welcome to Math.SE. Please read http://meta.math.stackexchange.com/questions/9959/how-to-ask-a-good-question/9960#9960 and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. – Shobhit Oct 06 '13 at 08:46

4 Answers4

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From the condition we know that the old man passes $\frac {1}{30}$ of the way in 1 minute. So in 5 minutes, he'll pass $\frac{1}{6}$.

So we could think of the problem like this. They both start at the same time, but the old man need to pass $\frac{5}{6}$ of the way, while the young man should walk whole way. We aleady know that the old man pass $\frac 1{30}$ of the way in a minute, while the young man pass $\frac 1{20}$ of the way in a minute. So we need to find solution to the following equation:

$$1-\frac 1{20} t = \frac 56 - \frac1{30} t$$ $$1- \frac 56 = \left(\frac 1{20} - \frac 1{30}\right)t$$ $$\frac 16 = \frac 1{60} t \implies t = 10$$

So they'll meet 10 minutes after the young man starts his journey, i.e in $10:15 am$.

Stefan4024
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distance b/w office and room be x speed of young man x/20 speed of old man x/30 relative speed x/20-x/30=x/60 distance traveled by old man in 5 min =(x/30)*5 =x/6 so young man have to manage only the x/6 distance =(x/6)/x/60 =10 min so they will meet in 10:15 time

Neeraj Kr
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Let $r_y =$ rate of the young man

$r_o =$ rate of the old man

$d = $distance to the office

Then $r_y = d/20 minutes$ and $r_o = d/30 minutes$

Now, let $t_y =$ time that the young man travels after 10:05

$t_o = $time that the old man travels after 10:00

Since the distance that the young man travels is the same as the distance that the old man travels (as they meet up), then

$r_yt_y = r_ot_o$

But $t_0 = t_y + 5 minutes$

So, $r_o(t_y+5) = r_y*t_y$

And $r_o/r_y = (d/30)/(d/20) = 2/3 = t_y/(t_y+5)$

Then you just solve for $t_y$ and add it to 10:05 to get the time they meet up.

John Yetter
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Let the distance be $d$ KM

So, the speed of the younger worker is $\frac d{20}$ KM/minute and that of the older person is $\frac d{30}$ KM/minute

So, in $t$ minutes the young man will cover $\frac d{20}t$ KM

But, by that time, the old man will get $(t+5)$ minutes as he has left $5$ minutes earlier, when he will cover $\frac d{30}(t+5)$ KM

The young man catch up with the old man , if the distance covered by them are same

lab bhattacharjee
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