Math
posted by Sanjana .
A student sarts for her school at the speed of 12 km / hr & reaches school 4 mins earlier than the asembily.next day she cycles at 10 km / hr & is late by 6 mins & missed assembly . Find distance b / w herschool & home

d/12 + 1/15 = d/10  1/10 ... 60 is LCD
5d + 4 = 6d  6 
12 km/hr * 1000 m/km * 1 hr/60 min = 200 meters/min
10 km/hr = 166.7 meters/min
distance = d
d = 200 (t4) = 200 t  800
d = 166.7(t+6) = 166.7 t + 1000
33.3 t = 1800
t = 54.1 minutes
d = 200(54.1)  800 = 10,010 meters
= 10 km
check
200(544) = 10,000
167(54+6) = 10,000 
her time difference is 10 minutes, or 1/6 hour
since distance = speed * time, and is the same both days, if she took t hours the first day, we have
12t = 10(t + 1/6)
t = 5/6 hr, or 50 minutes
so, the distance is
12km/hr * 5/6 hr = 10km
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