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

Answer:

690 km

Step-by-step explanation:

given,

speed= 40kmph, 50kmph

time= 11mins, 5 mins

d1= speed×time

40×11=440km

d2=speed ×time

50×5=250km

adding distance,

d1+d2

440+250= 690km

so total distance covered by train is 690km

Answer 2

${\huge{\underline{\underline{\rm{QuestiOn:}}}}}$

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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.

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${\huge{\underline{\underline{\rm{SolutiOn:}}}}}$

${\huge{\bf{\boxed{\red{Given\:that:-}}}}}$

✮Time delay when speed is 40 kmph = 11 minutes.

✮Time delay when speed is 50 kmph = 5 minutes.

✮The difference in time for different speeds

= 11 – 5

= 6 min.

=6/60 = 1/10 hrs.

⇛Let us suppose the distance covered by the train be x km.

We know that,

${} { \bf{ \boxed{Speed = \frac{Distance}{Time} }}}$

${ { \bf{ \boxed{∴Time = \frac{Distance}{Speed} }}}}$

${\bf{\boxed{\green{Acc.\:to\:the\: question;}}}}$

$\frac{x}{40} - \frac{x}{50} = \frac{1}{10} \\ \frac{5x - 4x}{200} = \frac{1}{10}$

______________(LCM of 40 & 50 is 200)

$\frac{x}{200} = \frac{1}{10} \\ \frac{x}{200} \times 200 = \frac{1}{10} \times 200$

_____________(Multiplying by 200 both sides)

${\huge{\bf{\boxed{\blue{x = 20}}}}}$

${\bf{\green{∴ The\: distance\: to\: be\: covered \:by \:the\: train\: is\: 20\: km.}}}$

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