Question

A man leaves from P at 6am and reaches Q at 2pm on the same day. another man leaves Q at 8am and reaches P at 3pm on the same day . at what time do they meet?​

Answer 1

Answer:

here's ur answer!!!!

Step-by-step explanation:

Required time = 10:48 a.m.

hope it helps u ☺️

Answer 2

Answer:

Let the distance between point P to point Q be x km.

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⇒ Time taken by the 1st man to reach point Q from point P = 6 am to 2 pm = 8 hrs

⇒ Speed of the 1st man = x/8 km/hr

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⇒ Time taken by the 2nd man to reach point P from point Q = 8 am to 3 pm = 7 hrs

⇒ Speed of the 2nd train = x/7 km/hr

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Let both the men meet after t hrs from the starting time of 1st man i.e. 6 am

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\textsf{Distance traveled by 1st man in t hrs + Distance}\\\textsf{traveled by 2nd man in (t - 2)hrs = Total Distance}\\\\\\:\implies\sf \bigg\lgroup\dfrac{x}{8} \times t\bigg\rgroup + \bigg\lgroup\dfrac{x}{7} \times (t - 2)\bigg\rgroup = x\\\\\\:\implies\sf \dfrac{tx}{8} + \dfrac{tx - 2x}{7} = x\\\\\\:\implies\sf \dfrac{7tx + 8tx - 16x}{56} = x\\\\\\:\implies\sf 15tx - 16x = 56x\\\\\\:\implies\sf 15tx = 56x+16x\\\\\\:\implies\sf 15tx = 72x\\\\\\:\implies\sf t = \dfrac{72x}{15x}\\\\\\:\implies\sf t = \dfrac{24}{5}\\\\\\:\implies\sf t =4\dfrac{4}{5}\:hrs\\\\\\:\implies\sf t =4 \:hrs \:and \:\dfrac{4}{5} \times 60\:minutes\\\\\\:\implies\sf t = 4\:hrs \:and \:(4 \times 12) \:minutes\\\\\\:\implies\sf t = 4\:hrs\:48\:minutes$

$\rule{200px}{.3ex}$

$\underline{\bigstar\:\textsf{Both men will meet at :}}$

$\dashrightarrow\sf\:\:Time = 6\:AM+ t\\\\\\\dashrightarrow\sf\:\: Time = 6 \:AM + 4\:hrs \:48\:minutes\\\\\\\dashrightarrow\sf\:\: Time = 10: 48\:AM$

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$\therefore\:\underline{\textsf{Hence, both men will meet at \textbf{10:48 AM}}}.$