40. A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same
distance if its sp

2 thoughts on “40. A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same<br />distance if its sp”

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

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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 :-

   [tex] \longmapsto \sf\boxed{\bold{\pink{Time =\: \dfrac{Distance}{Time}}}}[/tex]

   Solution :-

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

   Given :

   • Distance = 360 km

   Then,

   Time taken by the train initially is [tex]\sf \dfrac{360}{x}[/tex]

   And, the speed was increased by 5 km/hr

   Then, the time taken by the train is [tex]\sf \dfrac{360}{x + 5}[/tex]

   According to the question,

   ↦ [tex]\sf \dfrac{360}{x} – \dfrac{360}{x + 5} =\: \dfrac{48}{60}[/tex]

   ↦ [tex]\sf \dfrac{1}{x} – \dfrac{1}{x + 5} =\: \dfrac{48}{60} \times \dfrac{1}{360}[/tex]

   ↦ [tex]\sf \dfrac{5}{{x}^{2} + 5x} =\: \dfrac{1}{450}[/tex]

   ↦ [tex]\sf {x}^{2} + 5x =\: 5 \times 450[/tex]

   ↦ [tex]\sf {x}^{2} + 5x =\: 2250[/tex]

   ↦ [tex]\sf {x}^{2} + 5x – 2250 =\: 0[/tex]

   ↦ [tex]\sf {x}^{2} – 45x + 50x – 2250 =\: 0[/tex]

   ↦ [tex]\sf x(x – 45) + 50(x – 45) =\: 0[/tex]

   ↦ [tex]\sf (x – 45) (x + 50) =\: 0[/tex]

   ↦ [tex]\sf x – 45 =\: 0[/tex]

   ➠ [tex]\sf\bold{\red{x =\: 45}}[/tex]

   Either,

   ↦ [tex]\sf x + 50 =\: 0[/tex]

   ➠ [tex]\sf\bold{\green{x = – 50}}[/tex]

   As we can’t take speed as negative (– ve).

   So , x = 45

   [tex]\therefore[/tex] The original speed of the train is 45 km/hr .

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