Question

If a train runs at 40 kmph it reaches its destination late by 11 minutes. But if it runs at
50 kmph it is late by 5 minutes only. Find the distance to be covered by the train.​

Answer 1

In first condition,

Speed of train = 40km/hr

Delay timing = 11 minutes

Let distance covered by train = x km

Time taken by train = y

$\sf \implies \: Time = Distance/Speed \\ \\ \sf\sf \implies \frac{x}{40} = y+ \frac{11}{60} \\ \\ \sf \implies \: x = 40y + 40 [\frac{22}{3} ] \\ \\ \: \sf \implies x - 40y = \frac{22}{3} \\ \\ \: \sf \implies\: 5x-200y = \frac{110}{3}...eq.1 \: \{multiplied \: by \: 5 \}$

In the second condition,

Speed of train = 50km/hr

Delay timing = 5 minutes

$\sf \implies \frac{x}{50} = y + \frac{5}{60} \\ \\ \: \sf \implies x-50y= \frac{25}{6} \\ \\ \sf \implies \: 4x-200y = \frac{100}{6} \: \{multiply \: by \: 4 \} \\ \\ \: \sf \implies 4x-200y= \frac{50}{3} ...eq.2$

Now, Subtract equation 2 from equation 1,

$\sf \implies \: x= \frac{(110-50)}{3} \\ \\ \sf\implies 3x = 110-50 \\ \\ \implies \boxed{\boxed{ \sf{ \purple{ x=20km}}}}$

Hence, distance covered by train = x = 20km/hr

Answer 2

Answer:

• Let the Time taken be t hours.

◙ When train runs at 40 km/hr the time (t) will be :]

• $\sf Time\:(t) = \bigg \lgroup t + \dfrac{11}{60} \bigg \rgroup$

Therefore,

$\mapsto \sf s = 40 + \bigg \lgroup t + \dfrac{11}{60} \bigg \rgroup$

$\mapsto \sf s = 40t + \dfrac{22}{3} \qquad \Bigg \lgroup \bf{ Equation \: (i)} \Bigg \rgroup$

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◙ When train runs at 50 km/hr the time (t) will be :]

• $\sf Time\:(t) = \bigg \lgroup t + \dfrac{5}{60} \bigg \rgroup$

$\leadsto \sf s = 50 + \bigg \lgroup t + \dfrac{5}{60} \bigg \rgroup$

$\leadsto \sf s = 50t + \dfrac{25}{6} \qquad \Bigg \lgroup \bf{ Equation \: (ii)} \Bigg \rgroup$

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$\qquad \: \: \: \: \tiny \dag \: \underline{\tt From \: Equation \: (i) \: and \: (ii) \: we get :}$

$\dashrightarrow\:\:\sf 40t + \dfrac{22}{3} = 50t + \dfrac{25}{6}$

$\dashrightarrow\:\:\sf \dfrac{22}{3} - \dfrac{25}{6} = 50t - 40t$

$\dashrightarrow\:\:\sf \dfrac{44 - 25}{6}= 10t$

$\dashrightarrow\:\:\sf t = \dfrac{19}{60} \: hours$

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$\qquad \:\tiny \dag \: \underline{\tt Putting \: Time \: (t) = 19/60 \: hrs \: in \: equation \: (i) \: we \: get :}$

$:\implies \sf Distance = 40 \times \dfrac{19}{60} + \dfrac{22}{3}$

$:\implies \sf Distance =\dfrac{38}{3} + \dfrac{22}{3}$

$:\implies \sf Distance =\dfrac{60}{3}$

$:\implies \underline{ \boxed{ \textsf{ \textbf{Distance = 20 km}}}}$