Question

What is the Speed of Train ( in Kmph) :
1) The Car takes 2 hours more than the Train to cover a distance of 264 Km.
2) The train moves 22 km/hr faster than the Car.

– SBI Bank PO​

Answer 1

AnswEr :

• Fixed Distance ( D ) = 264 Km
• Speed of Car ( s ) = C Km/hr
• Speed of Train ( S ) = Speed of Car + 22 = (C + 22) Km/hr
• Time Taken by Train ( T ) = T hr
• Time Taken by Car ( t ) = Time Taken by Train + 2 = (T + 2) hr
• Find the Speed of Train.

Relation Between Distance, Speed & Time

⋆ Distance ( d ) = Speed ( s ) × Time ( t )

• Let's Head to the Question Now :

⇒ Distance = Speed of Train × Time Taken

⇒ 264 = (C + 22) × T

⇒ 264 = CT + 22T ⠀⠀⠀⠀⠀—eq. ( I )

⇒ Distance = Speed of Car × Time Taken

⇒ 264 = C × (T + 2)

⇒ 264 = CT + 2C⠀⠀⠀⠀⠀⠀—eq. ( II )

• From eq.( I ) and eq.( II ) :

⇒ CT + 22T = CT + 2C

• LHS are Equal, therefore RHS will be Equal too

⇒ CT - CT + 22T = 2C

⇒ 22T = 2C

⇒ 11T = C⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ —eq. ( III )

_________________________________

• Putting the Value of C in eq.( I ) :

⇒ 264 = CT + 22T

⇒ 264 = (11T × T) + 22T

⇒ 264 = 11T² + 22T

⇒ 264 = 11(T² + 2T)

• Dividing Both term by 11

⇒ 24 = (T² + 2T)

⇒ T² + 2T - 24 = 0

⇒ T² + 6T - 4T - 24 = 0

⇒ T(T + 6) - 4(T + 6) = 0

⇒ (T - 4)(T + 6) = 0

⇒ T = 4 hr ⠀,or ⠀T = - 6 hr

⋆ We will Ignore the Negative Time ;

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• Finally we will calculate Speed of Train :

⇒ Speed of Train = (C + 22)

⇒ Speed of Train = (11T + 22)

• From eq.( III )

⇒ Speed of Train = (11 × 4) + 22

⇒ Speed of Train = 44 + 22

⇒ Speed of Train = 66 Km/hr

∴ Speed of the Train will be 66 Km/hr.

Answer 2

Answer:

$\large\bold\red{66\:Km\:{hr}^{-1}}$

Step-by-step explanation:

Let's assume that,

• Speed of car = ‘x’ km/hr
• Time take by train = ‘t’ hrs.

Now,

It's given that,

The train moves 22 km/hr faster than the car.

And,

The car takes 2 hrs more than the train.

Therefore,

We get,

• Speed of train = ‘(x+22)’ km/hr
• Time taken by car = ‘(t+2)’ hrs

Also,

• Distance = 264 Km

Now,

We know that,

$\large \boxed{ \bold \purple{distance = speed \times time}}$

Therefore,

We get,

$= > (x + 22)t = 264 \: \: \: \: \: \: \: .......(1)$

And,

$= > x(t + 2 ) = 264 \: \: \: \: \: \: \: \: \: .........(2)$

Now,

Subtracting eqn (2) from (1) ,

We get,

$= > ( x + 22)t - x(t + 2) = 264 - 264 \\ \\ = > xt + 22t - xt - 2x = 0 \\ \\ = > 22t - 2x = 0 \\ \\ = > 11t - x = 0 \\ \\ = > x = 11t \: \: \: \: \: \: \: ........(3)$

Now,

Putting this value of x from eqn (3) in (2),

We get,

$= > 11t(t + 2) = 264 \\ \\ = > {t}^{2} + 2t = \frac{264}{11} \\ \\ = > {t}^{2} + 2t = 24 \\ \\ = > {t}^{2} + 2t - 24 = 0 \\ \\ = > {t}^{2} + 6t - 4t - 24 = 0 \\ \\ = > t(t + 6) - 4(t + 6) = 0 \\ \\ = > (t + 6)(t - 4) = 0$

Therefore,

We get,

$= > t + 6 = 0 \\ \\ = > t = - 6$

And,

$= > t - 4 = 0 \\ \\ = > t = 4$

But,

We know that,

• Time can't ever be negative.

Therefore,

We get,

• t = 4 hrs.

Thus,

We get,

• Speed of car, x = 44 km/hr

Hence,

Speed of train, (x+22) = 66 km/hr