Question

a man travels 300 km partly by train and partly by car . he takes 4 hours if he travels 60 km by train and rest by car . if he travels 100 km by train and remaining by car , he takes 10 minutes longer . find speed of train and car respectively​

Answer 1

Given:-

• Total distance travelled by a man = 300km
• Total time taken by a man = 4 hours

To Find :-

• The speed of car and train =?

Solution:-

We will solve this question by setting up equation as per the given clue in the question:-

Case - (i) :-

• Distance covered by train = 60km
• Let, Speed of train = x km/h

==> Time taken by train = Distance / speed

==> Time taken by train = 60/x

• Distance covered by car = 300-60 = 240km
• Let, speed of car = y km/h

==> Time taken by car = 240/y

==> Time by train + Time by car = Total time

==> 60/x + 240/y = 4

==> 60y + 240x/xy = 4

==> 240x + 60y = 4xy --------(i)

Case - (ii) :-

• Distance covered by train = 100km
• Let, Speed of train = x km/h

==> Time taken by train = Distance / speed

==> Time taken by train = 100/x

• Distance covered by car = 300-100 = 200km
• Let, speed of car = y km/h

==> Time taken by car = 200/y

==> Time by train + Time by car = Total time

==> 100/x + 200/y = 4 hours + 10 minutes

==> 100y + 200x / xy = 4 + 10/60

==> 100y + 200x / xy = 25/6

==> 1200x + 600y = 25xy -------(ii)

• Now eq (i) and (ii) solving here :-

==> 288000x + 72000y = 4800xy

==> 288000x + 144000y = 6000xy

• By subtracting we get here :-

==> - 72000y = - 1200xy

==> 72000 = 1200x

==> x = 60 km/h

• Now putting the value of x = 60 in eq (i)

==> 240x + 60y = 4xy

==> 240 × 60 + 60y = 4 × 60 × y

==> 14400 +60y = 240y

==> 14400 = 240y - 60y

==> 14400 = 180y

==> y = 80 km/h

Hence,

• The speed of train = 60 km/h
• The speed of car = 80 km/h

Answer 2

❍ Let's Consider the speed of train be x km/hr & speed of car be y km/hr .

⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀▪︎⠀⠀Total Distance travelled by men is 300 km

⠀⠀⠀⠀⠀CASE I : The man travells 60 km by train and rest by car .

$\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\\qquad\maltese\:\:\bf Formula\:for \:Time \::\:$

$\qquad \dag\:\:\bigg\lgroup \pmb{\frak{ Time \:=\:\dfrac{Distance}{Speed } \:\:} }\bigg\rgroup$

⠀⠀⠀⠀⠀Time taken by the man to cover 60 km by train = 60 / x km/hr .

⠀⠀☆ Time taken the man to to cover i.e. 300 -60 = 240 km by car = 240 / y km/hr .

Now ,

⠀⠀⠀⠀⠀Time taken by the man to cover 300 km = $\sf \dfrac{60}{x} + \dfrac{240}{y}$

Given that ,

⠀⠀▪︎⠀⠀Man takes 4 hrs to travel 300 km

$\qquad \therefore \: \sf 4 \:=\: \dfrac{60}{x} + \dfrac{240}{y} \:\\\\\qquad :\implies \: \sf 4 \:=\: \dfrac{60}{x} + \dfrac{240}{y} \:\\\\\qquad :\implies \: \sf 4 \:=\: \dfrac{240x + 60 y }{xy} \:\\\\\qquad :\implies \: \sf 4xy \:=\: 240x + 60 y \:\\\\\qquad :\implies \: \bf 4xy \:=\: 240x + 60 y \:\qquad \bigg\lgroup \sf{ Eq^n\:1 }\bigg\rgroup$

⠀⠀⠀⠀⠀CASE II : The man covers by train is 100 km and rest by car .

$\dag\:\:\pmb{ As,\:We\:know\:that\::}\\\\\qquad\maltese\:\:\bf Formula\:for \:Time \::\:$

$\qquad \dag\:\:\bigg\lgroup \pmb{\frak{ Time \:=\:\dfrac{Distance}{Speed } \:\:} }\bigg\rgroup$

⠀⠀⠀⠀⠀Time taken by the man to cover 100 km by train = 100 / x km/hr .

⠀⠀☆ Time taken the man to to cover i.e. 300 -100 = 200 km by car = 200 / y km/hr .

Now ,

⠀⠀⠀⠀⠀Time taken by the man to cover 300 km = $\sf \dfrac{100}{x} + \dfrac{200}{y}$

Given that ,

⠀⠀▪︎⠀⠀Man takes 4 hrs to travel 300 km and in case ii he takes 10 minutes longer .

$\qquad \therefore \: \sf 4 \:hrs + \: 10 \: min\:=\: \dfrac{100}{x} + \dfrac{200}{y} \:\\\\\qquad :\implies \: \sf 4 \:hrs\: + 10\:min \:=\: \dfrac{100}{x} + \dfrac{200}{y} \:\\\\\qquad :\implies \: \sf 4 \:\: + \dfrac{10}{60}\: \:=\: \dfrac{100}{x} + \dfrac{200}{y} \:\\\\\qquad :\implies \: \sf 4 \:\: + \cancel {\dfrac{10}{60}}\: \:=\: \dfrac{100}{x} + \dfrac{200}{y} \:\\\\\qquad :\implies \: \sf 4 \:\: + \dfrac{1}{6}\: \:=\: \dfrac{100}{x} + \dfrac{200}{y} \:\\\\\qquad :\implies \: \sf \dfrac{24 + 1}{6}\: \:=\: \dfrac{100}{x} + \dfrac{200}{y} \:\\\\\qquad :\implies \: \sf \dfrac{25}{6}\: \:=\: \dfrac{100}{x} + \dfrac{200}{y} \:\\\\\qquad :\implies \: \sf \dfrac{25}{6}\: \:=\: \dfrac{100y + 200x }{xy} \:\\\\\qquad :\implies \: \sf xy(25)\: \:=\: 6(100y + 200x) \:\\\\\qquad :\implies \: \sf 25xy\: \:=\: 600y + 1200x \:\\\\\qquad :\implies \: \bf 25xy \:=\: 600y + 1200x \:\qquad \bigg\lgroup \sf{ Eq^n\:2 }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Solving \: the \: Eq^n \: 1 \: and \:Eq^n 2 \::}}$

$\qquad \leadsto \: \bf 4xy \:=\: 240x + 60 y \:\qquad \bigg\lgroup \sf{ Eq^n\:1 }\bigg\rgroup$

$\qquad \dashrightarrow \sf 4xy \:=\: 240x + 60 y$

$\qquad \dashrightarrow \sf \bigg( 4xy \:=\: 240x + 60 y\bigg) \times 1200$

$\qquad \dashrightarrow \sf 4800xy \:=\: 288000x + 72000 y$

$\qquad \leadsto \: \bf 25xy \:=\: 600y + 1200x \:\qquad \bigg\lgroup \sf{ Eq^n\:2 }\bigg\rgroup$

$\qquad \dashrightarrow \sf 25xy \:=\: 600x + 1200 y$

$\qquad \dashrightarrow \sf \bigg( 25xy \:=\: 600x + 1200 y \bigg) \times 240$

$\qquad \dashrightarrow \sf 6000xy \:=\:288000 + 144000y$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Subtracting \: the \: Eq^n \: 1 \: and \:Eq^n 2 \::}}\\\\\qquad \dashrightarrow \sf -72000y \:=\:-1200xy \\\\\qquad \dashrightarrow \sf -72000\cancel {y} \:=\:-1200x\cancel {y} \\\\\qquad \dashrightarrow \sf -72000 \:=\:-1200x \\\\\qquad \dashrightarrow \sf x \:=\:60 \\\\\qquad \therefore \pmb{\underline{\purple{\frak{\:x = 60\:\:km/hr }}} }\:\:\bigstar$

⠀⠀▪︎⠀⠀Here x denotes speed of train which is 60 km/hr .

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \:value \:of \: x \:in Eq^n \: 1 \: and \: \::}}\\\qquad \leadsto \: \bf 4xy \:=\: 240x + 60 y \:\qquad \bigg\lgroup \sf{ Eq^n\:1 }\bigg\rgroup\\\\\qquad \dashrightarrow \sf 4xy \:=\: 240x + 60 y \\\\\qquad \dashrightarrow \sf 4(60 )y \:=\: 240(60) + 60 y \\\\\qquad \dashrightarrow \sf 240y\:=\: 240(60) + 60 y \\\\\qquad \dashrightarrow \sf 240y \:=\: 14400 + 60 y \\\\\qquad \dashrightarrow \sf 240y - 60y \:=\: 14400 \\\\\qquad \dashrightarrow \sf y \:=\: 80 \\\\\qquad \therefore \pmb{\underline{\purple{\frak{\:y= 80\:\:km/hr }}} }\:\:\bigstar$

⠀⠀▪︎⠀⠀Here y denotes speed of car which is 80 km/hr .

$\qquad \therefore \:\underline {\sf Hence , \: The \: Speed \:of \:car \:and \:train \: are \: \bf \: 60 \:km/hr \:\sf and \:\bf 80 \:km/hr \: \sf , respectively.} \:$