a man cycles to his office from his house at a speed 5 km/ hour and reaches 10 minutes late. if he cycles at a speed of 6km/hour, he reaches 10 minutes early. what is the distance between the office and his house?
Answers
Let +d+ = the distance in km between house and office
Let +t+ = The travel time in hrs which is on time for him
given:
From office to house at 5 km/hr
(1) +d+=+5%2A%28+t+%2B+10%2F60+%29+
From office to house at 6 km/hr
(2) +d+=+6%2A%28+t+-+10%2F60+%29+
-----------------------
(1) +d+=+5t+%2B+5%2F6+
(2) +d+=+6t+-+1+
----------------
Substitute (1) into (2)
(2) +5t+%2B+5%2F6+=+6t+-+1+
(2) +t+=+5%2F6+%2B+1+
(2) +t+=+11%2F6+
Plug this result back into (1) or (2)
(1) +d+=+5%2A%2811%2F6%29+%2B+5%2F6+
(1) +d+=+55%2F6+%2B+5%2F6+
(1) +d+=+10+
the distance between his house and office is 10 km
check:
(2) +d+=+6t+-+1+
(2) +10+=+6t+-+1+
(2) +6t+=+11+
(2) +t+=+11%2F6+
OK
(1) +d+=+5t+%2B+5%2F6+
(1) +10+=+5t+%2B+5%2F6+
(1) +5t+=+10+-+5%2F6+
(1) +5t+=+60%2F6+-+5%2F6+
(1) +5t+=+55%2F6+
(1) +t+=+11%2F6+