Question

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

Answer 1

Answer:

• The original speed of the train is 45 km/hr.

Step-by-step explanation:

Let us assume:

• Original speed = S
• Time taken = T

Formula:

• 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:

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.

∴ The original speed of the train = 45 km/hr

Answer 2

Answer:

Given :-

• A train travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if it's speed were 5 km/h more.

To Find :-

• What is the original speed of the train.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{Time =\: \dfrac{Distance}{Time}}}}$

Solution :-

Let, the original speed of the train be x km/hr.

Given :

• Distance = 360 km

Then,

Time taken by the train initially is $\sf \dfrac{360}{x}$

And, the speed was increased by 5 km/hr

Then, the time taken by the train is $\sf \dfrac{360}{x + 5}$

According to the question,

↦ $\sf \dfrac{360}{x} - \dfrac{360}{x + 5} =\: \dfrac{48}{60}$

↦ $\sf \dfrac{1}{x} - \dfrac{1}{x + 5} =\: \dfrac{48}{60} \times \dfrac{1}{360}$

↦ $\sf \dfrac{5}{{x}^{2} + 5x} =\: \dfrac{1}{450}$

↦ $\sf {x}^{2} + 5x =\: 5 \times 450$

↦ $\sf {x}^{2} + 5x =\: 2250$

↦ $\sf {x}^{2} + 5x - 2250 =\: 0$

↦ $\sf {x}^{2} - 45x + 50x - 2250 =\: 0$

↦ $\sf x(x - 45) + 50(x - 45) =\: 0$

↦ $\sf (x - 45) (x + 50) =\: 0$

↦ $\sf x - 45 =\: 0$

➠ $\sf\bold{\red{x =\: 45}}$

Either,

↦ $\sf x + 50 =\: 0$

➠ $\sf\bold{\green{x = - 50}}$

As we can't take speed as negative (- ve).

So , x = 45

$\therefore$ The original speed of the train is 45 km/hr .