ibm - a man moves from his house to office with bike at a constant speed if he increases his speed by 10000 meters per hour he would reach office by 60 minutes earlier and if he increases his speed by 20000 meters per hourhe would reach by 74 hours earlier fin - skillgun

Assume the required distance be x km and the usual speed be y km/hr.
Then actual time x/y hours.
If Speed is increased by (10000 meters/hr)10 km/hr then the time taken = x /(y + 10) hrs
Given that x /y - x /(y + 10) = 1 hour(60 minutes)
x(y + 10) - xy = 1(y)(y + 10)
xy + 10x - xy = Y(y + 10)
10x = y(y + 10) ....(1)
Similarly, if the speed is increased (20000 meters/hr) 20 km/hr, he would reach 7/4 hours earlier.
i.e., x/ y - x /(y + 20) = 7/4
x(y + 20) - xy = 7/4 (y)(y + 20)
xy+ 20X - xy = 7/4 (y)(y + 20)
80x/7 = y(y + 20) ...(2)
Dividing (1) by (2)
y(y + 10) /y(y + 20) = 10x /(80x / 7)
7(y + 20) = 8(y + 10)
y = 60 So by using y value we get x = 420 km.so the distance between house and office is 420km.