Question

A train left from a station.After 1 hour of leaving the train delayed for 1 hour for a special reason and after that the train went on with the speed of 3/5 of first speed.Going with that speed the train reached its destination 3 hours late from certain time.If the special reason was held 50k.m. faraway from first place, the train reached its destination 1 hour 20 minutes ago from first time of reaching.How much way in total the train exceeded and what was the train's first speed?​

Answer 1

$\huge{\boxed{\boxed{\green{\bf{Answer:-}}}}}$

$\large{\bigstar{\underline{\underline{\bf{Given:-}}}}}$

• Original Speed = 100 km/Hr

• Distance of Journey = 1200 km

$\huge{\boxed{\boxed{\green{\bf{Explainationtion:-}}}}}$

• Let say original speed of train = S km/Hr

• Original time to reach station = T hr

• Total Distance = ST km

$\bigstar{\underline{\underline{\bf{Now:-}}}}$

• Distance covered in 3 Hours = 3S km

• Reduced Speed = (75/100)S = 3S/4 km/Hr

• Time run with reduced speed = ( T + 4 -3 - 1) = T hr

$\bigstar{\underline{\underline{\bf{Point\:to\: remember}}}}$

• Total time taken increased by 4 hour, 3 hrs already run and one hour stop time.

Distance covered with reduced speed = 3ST/4

• ST = 3S + 3ST/4

• 4ST = 12S + 3ST

• ST = 12S

• $\large{\boxed{\boxed{\green{\sf{T = 12}}}}}$

Original time to journey = 12 hr

• If accident was 150 km ahead of the point then time taken with Original Speed = 150/S

Time taken with Reduced Speed = 150/(3S/4) = 200/S

• 200/S - 150/S = 1/2 (4 - 3.5 = 0.5 = 1/2)

• 200 - 150 = S/2

• $\large{\boxed{\boxed{\green{\sf{S = 100}}}}}$

Original Speed = 100 km/Hr

• Distance of Journey = ST = 100 × 12 = 1200 km

Therefore, the correct answer is:-

$\huge{\boxed{\boxed{\green{\sf{ST = 1200km}}}}}$

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Answer 2

Answer : )

$\Large{\boxed{\rm{\red{ST\:=\:1200km}}}}$

Solution : )

$\Large{\underline{\rm{\green{Given\::-}}}}$

★ Original speed = 100 km/hr

★ Distance of journey = 1200 km.

$\Large{\underline{\rm{\green{Explaination\::-}}}}$

★ Let say original speed of train = S km/hr

★ Original time to reach station = T hr

★ Total distance = ST km

$\Large{\underline{\rm{\green{Now\::-}}}}$

★ Distance cover in three Hours = 3S km

★ Reduced speed = (75/100)S = 3S/4 km/hr

★ Time run with reduced speed = ( T + 4 - 3 - 1 ) = T hr

Distance covered with reduced speed = 3ST/4

• ST = 3S + 3ST/4
• 4ST = 12S + 3ST
• ST = 12S

T = 12

Time taken with reduced speed :-

★ (150/(3S/4) = 200/S

★ 200/S - 150/S = 1/2 ( 4 - 3.5 = 0.5 = 1/2 )

★ 200 - 150 = S/2

$\Large{\boxed{\rm{\purple{S=100}}}}$

Original speed = 100 km/hr

★ Distance of journey = ST = 100 × 12 = 1200 km/hr

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