lnm87
Karen travelled x miles from her home to office, at a speed of y miles per hour and on her way back, she takes the same route, and travels for 10 miles at a speed of y miles per hour before stopping for an hour. Now, by what percentage should she increase her speed so that the overall time taken to reach back home from the office is the same as that taken to reach the office from home?
a. \(\frac{(x−10)×100y}{x−y−10 }\)
b. \(\frac{100y}{x−y−10}\)
c. \(\frac{100x}{y−x−10}\)
d. \(\frac{(x−10)×100y}{y−x−10}\)
e. \(\frac{100y}{y−x−10}\)
Given: Karen travelled x miles from her home to office, at a speed of y miles per hour and on her way back, she takes the same route, and travels for 10 miles at a speed of y miles per hour before stopping for an hour.
Asked: Now, by what percentage should she increase her speed so that the overall time taken to reach back home from the office is the same as that taken to reach the office from home?
From home to office :
Distance = x miles
Speed = y miles/hr
Time taken = x/y hrs
From office to home:
Distance 1= 10 miles
Speed 1= y miles/hr
Time taken 1 = 10/y hrs
Idle time = 1 hr
Distance 2 = (x -10) miles
Time taken 2 = x/y - 10/y - 1= (x-y-10)/y hrs
Speed 2 = (x-10)y/(x-y-10) miles/hr
Percentage increase in speed = {(x-10)y/(x-y-10) / y - 1}*100% = {(x-10)/(x-y-10) -1}*100 = 100(x-10-x+y+10)/(x-y-10)% = 100y/(x-y-10) %
IMO B