Question

A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed was 5 km/ h more. Find the original speed of train.

Answer 1

:Let the speed of train be x km/hrDistance to be travelled  = 360kmWe know that, Time take by the train intially = If the speed was increased by 5km/hrTime taken by train = Difference in the time taken is 48 minutes,x2+5x -2250 = 0(x+50)(x-45)Speed = 45 km/hr as speed cannot be negative

Answer 2

$\small{\textsf{\textbf{\underline{\underline{∴ The original speed of the train = 45 km/hr\::}}}}}$

Step-by-step explanation:

Let us assume:

• Original speed = S
• Time taken = T

Formulas used :

• Distance = Speed × Time

Given that:

• A train travelling at a uniform speed for 360 km.
• ⟶ S × T = 360
• ⟶ T = 360/S ______(i)

• If speed speed increased by 5 km/hr it takes 48 minutes less.
• ⟶ (S + 5) × (T - 48/60) = 360
• ⟶ (S + 5) × (T - 4/5) = 360 ______(ii)

To Find:

• The original speed of the train.

Finding the original speed of the train:

Finding the original speed of the train:In equation (ii).

⇒ (S + 5) × (T - 4/5) = 360

Substituting the value of T from eqⁿ(i).

⇒ (S + 5) × (360/S - 4/5) = 360

⇒ (360/S - 4/5) = 360/(S + 5)

Taking 5S as LCM in LHS.

⇒ (1800 - 4S)/5S = 360/(S + 5)

Cross multiplication.

⇒ (S + 5)(1800 - 4S) = 360 × 5S

⇒ S(1800 - 4S) + 5(1800 - 4S) = 1800S

⇒ 1800S - 4S² + 9000 - 20S = 1800S

Cancelling 1800S.

⇒ 4S² + 20S - 9000 = 0

Taking 4 common.

⇒ 4(S² + 5S - 2250) = 0

⇒ S² + 5S - 2250 = 0

⇒ S² + 50S - 45S - 2250 = 0

⇒ S(S + 50) - 45(S + 50) = 0

⇒ (S - 45) (S + 50) = 0

⇒ S = 45 or S = - 50

We know that:

• Speed is always taken positive.

$\small{\textsf{\textbf{\underline{\underline{∴ The original speed of the train = 45 km/hr\::}}}}}$