Akash, Baichung and Charanjeet can walk at the speeds of 3, 4 and 5 km

1.	Akash, Baichung and Charanjeet can walk at the speeds of 3, 4 and 5 km per hour respectively. They starts form Pune at 1, 2, 3 o’clock respectively. When Baichung catches Akash, Baichung sends him back with a message to Charanjeet. When will Charanjeet get the message?1). 4:15 o’clock2). 5:15 o’clock3). 6:25 o’clock4). Can’t be determined
Answer» In order to find the NUMBER of hours it will take for Akash and Baichung to meet together, LET us take the LCM of their SPEEDS 3 kmph and 4 kmph. LCM (3, 4) = 12 KMS ∴ Akash will cover the distance of 12 kms in time = $(\frac{{12}}{3} = {\rm{\;}}4{\rm{\;hours}})$ ∴ Baichung will cover the distance of 12 kms in time = $(\frac{{12}}{3} = {\rm{\;}}3{\rm{\;hours}})$ ?Akash starts at 1 o’ clock and Baichung starts at 2 o’ clock, they will meet together at 5 o’ clock Or we can say that Baichung catches Akash at 5 o’ clock At 5 o’ clock, Charanjeet will have covered the distance of = 5kmph× 2 hrs = 10 kms ∴ Distance between Akash and Charanjeet is 2 kms now (12 kms - 10 kms), the time taken by Akash and Charanjeet to meet will be = $(\frac{2}{{\left( {5\; + \;3} \right)}}\;{\rm{hours\;}} = {\rm{\;}}\frac{1}{4}{\rm{\;hours\;}} = {\rm{\;}}15{\rm{\;mins}})$ ∴ Charanjeet will get the MESSAGE at 5:15 o’ clock

Akash, Baichung and Charanjeet can walk at the speeds of 3, 4 and 5 km per hour respectively. They starts form Pune at 1, 2, 3 o’clock respectively. When Baichung catches Akash, Baichung sends him back with a message to Charanjeet. When will Charanjeet get the message?1). 4:15 o’clock2). 5:15 o’clock3). 6:25 o’clock4). Can’t be determined

Answer»

In order to find the NUMBER of hours it will take for Akash and Baichung to meet together, LET us take the LCM of their SPEEDS 3 kmph and 4 kmph.

LCM (3, 4) = 12 KMS

∴ Akash will cover the distance of 12 kms in time = $(\frac{{12}}{3} = {\rm{\;}}4{\rm{\;hours}})$

∴ Baichung will cover the distance of 12 kms in time = $(\frac{{12}}{3} = {\rm{\;}}3{\rm{\;hours}})$

?Akash starts at 1 o’ clock and Baichung starts at 2 o’ clock, they will meet together at 5 o’ clock

Or we can say that Baichung catches Akash at 5 o’ clock

At 5 o’ clock, Charanjeet will have covered the distance of = 5kmph× 2 hrs = 10 kms

∴ Distance between Akash and Charanjeet is 2 kms now (12 kms - 10 kms), the time taken by Akash and Charanjeet to meet will be = $(\frac{2}{{\left( {5\; + \;3} \right)}}\;{\rm{hours\;}} = {\rm{\;}}\frac{1}{4}{\rm{\;hours\;}} = {\rm{\;}}15{\rm{\;mins}})$

∴ Charanjeet will get the MESSAGE at 5:15 o’ clock

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