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View AllLet, the speed of the train during returning journey be x kmph
Speed during onward journey= kmph
Distance covered in onward journey =
Time taken by the train on onward journey= hours [As, time= Distance/Speed]
And time by the train in returning journey = hours
According to the question,
Speed of the train in the onward journey= kmph
āĻĒā§āϰāĻļā§āύ⧠āĻŦāϞāĻž āĻšāĻā§āĻā§, 800 āĻŽāĻŋāĻāĻžāϰ āĻāĻāĻāĻŋ āĻĻā§ā§ āĻĒā§āϰāϤāĻŋāϝā§āĻāĻŋāϤāĻžā§ āĻāĻ āĻŦā§āϝāĻā§āϤāĻŋ āĻ āĻāĻļ āĻĒāĻĨ 130 āĻŽāĻŋāĻāĻžāϰ/ āĻŽāĻŋāύāĻŋāĻ āĻāĻŦāĻ āĻĒāϰāĻŦāϰā§āϤ⧠āĻĒāĻĨ 145 āĻŽāĻŋāĻāĻžāϰ/ āĻŽāĻŋāύāĻŋāĻ āĻŦā§āĻā§ āĻā§āϞ⧠āĻŽā§āĻ āĻĻā§ā§ā§ āϤāĻžāϰ āĻā§ āĻāϤāĻŋāĻŦā§āĻ āĻāϤ āĻāĻŋāϞ?
Here, first part of race = = 400 meter
â´ Second part of race = 800-400 = 400 meter
Time taken to cross first half of race
meter/minute
Again, Time taken to cross second part of race = meter/minute
â´ Average speed
[nearest whole of 5.83 is 6]
= 133.333 meter per minute
= 133 meter per minute (Approximately)
āĻŽā§āĻ āϏāĻŽā§ āύā§ā§ ā§Š āĻāĻŖā§āĻāĻž
āϝāĻžāϤā§āϰāĻž āĻŦāĻŋāϰāϤāĻŋ ā§§ āĻāĻŖā§āĻāĻž
āύāĻŋāĻ āϏāĻŽā§ āύā§ā§ = ā§ - ā§§ = ā§Ŧ āĻāĻŖā§āĻāĻž
āĻŽā§āĻ āĻĻā§āϰāϤā§āĻŦ = ā§Šā§Šā§Ļ āĻāĻŋ.āĻŽāĻŋ
â´ āĻā§ āĻāϤāĻŋāĻŦā§āĻ āĻāĻŋ.āĻŽāĻŋ/ āĻāĻŖā§āĻāĻž = ā§Ģā§Ģ āĻāĻŋ.āĻŽāĻŋ / āĻāĻŖā§āĻāĻž
ā§§ āĻā§āϞāĻŋāĻā§ āĻĒā§āϰāĻļā§āύ, āĻļā§āĻ, āϏāĻžāĻā§āĻļāύ āĻ
āĻ
āύāϞāĻžāĻāύ āĻĒāϰā§āĻā§āώāĻž āϤā§āϰāĻŋāϰ āϏāĻĢāĻāĻāϝāĻŧā§āϝāĻžāϰ!
āĻļā§āϧ⧠āĻĒā§āϰāĻļā§āύ āϏāĻŋāϞā§āĻā§āĻ āĻāϰā§āύ â āĻĒā§āϰāĻļā§āύāĻĒāϤā§āϰ āĻ āĻā§āĻŽā§āĻāĻŋāĻ āϤā§āϰāĻŋ!