Mrehul and Mridank started walking from their home to go to their scho

1.	Mrehul and Mridank started walking from their home to go to their school both walked for 2 hrs after that Mridank injured his legs and walked at 60% of his previous speed, but Mrehul kept moving at his original speed and he reached his school 2 hrs before his brother Mridank. At what distance from his home does Mridank injured his legs if the distance between their home and school is 25 km? 1. 10 km2. 15 km3. 12.5 km4. 8 km5. None of the above
Answer» Correct Answer - Option 1 : 10 km Let the original speed of Mridank and Mrehul be x km/hr They walked for 2 hrs when Mridank ijured his legs ∴ Distance travelled by both of them = Speed × Time ⇒ Distance = x × 2 = 2x km ∴ Remaining distance = 125 – 2x km Now, speed of Mridank = x × (60/100) = 3x/5 ∴ According to question (125 – 2x)/(3x/5) – {(125 –2x)/x} = 2 ⇒ (125 – 10x – 75 + 6x)/3x = 2 ⇒ 50 – 4x = 6x ⇒ x = 5 ∴ Speed of Mridank and Mrehul = 5 km/hr ∴ Distance at which Mridank injured his legs = 5 × 2 = 10 km

Mrehul and Mridank started walking from their home to go to their school both walked for 2 hrs after that Mridank injured his legs and walked at 60% of his previous speed, but Mrehul kept moving at his original speed and he reached his school 2 hrs before his brother Mridank. At what distance from his home does Mridank injured his legs if the distance between their home and school is 25 km? 1. 10 km2. 15 km3. 12.5 km4. 8 km5. None of the above

Answer» Correct Answer - Option 1 : 10 km

Let the original speed of Mridank and Mrehul be x km/hr

They walked for 2 hrs when Mridank ijured his legs

∴ Distance travelled by both of them = Speed × Time

⇒ Distance = x × 2 = 2x km

∴ Remaining distance = 125 – 2x km

Now, speed of Mridank = x × (60/100) = 3x/5

∴ According to question

(125 – 2x)/(3x/5) – {(125 –2x)/x} = 2

⇒ (125 – 10x – 75 + 6x)/3x = 2

⇒ 50 – 4x = 6x

⇒ x = 5

∴ Speed of Mridank and Mrehul = 5 km/hr

∴ Distance at which Mridank injured his legs = 5 × 2 = 10 km