# Dev cover a distance of 15 km, partly by foot at 6km/h & partly by cycle at 18km/h. If he takes 1h 10min to finish, find how far he walks?

Start off by picking some variables.  I walk for "x" km.  I cycle for "y" km.  So, x+y=15.  It's a start.

Now, I start playing with the speeds and time.

x km at 6km/hour.  If I multiply that, I get km squared over hours. That doesn't help. But if I travel 6 km per hour, that also means in one hour I travel 6 km. I can switch the fraction.  That's important.  If I multiply x km times 1hr/6km, I get that I travel x/6 hours by walking.

So, I travel x/6 hours walking and y/18 hours cycling.  I do have a time reference.  I traveled a total of one hour and ten minutes.  But I want to have everything in only hours.  So, I traveled 1 and 10/60th of an hour.  So, I can use the improper fraction 1 1/6 or 7/6.

x/6 + y/18 = 7/6    solving that out....

3x/18 + y/18 = 21/18

3x + y = 21

y = 21 - 3x

This seems useless until we look at the first equation we wrote... x+y=15

Every time you see "y," put in what you know y is... 21 - 3x

x + (21-3x) = 15  Solving, you get x = 3

So, what the heck is x?  Right up front, we said that I walk for "x" km.  So, if x = 3, I walk for 3 km.

The hard part to remember is to flip those km/hr fractions so that the "km" measurements cancel each other out.

revolution | Student

Let's say Dev travelled xkm on foot and y km by cycling his bike.

So, x+y=15 km- eqn 1

You know that he walk by foot at 6km/h and cycle at 18km/h, so the time taken for each would be x/6h and y/18h

So, x/6+y/18= 7/6h

3x+y= 21- eqn 2

Multiply eqn 1 by x3

so, 3x+3y=45 - eqn 4

Minus eqn 2 from eqn 4 so:

3x-3x+3y-y=45-21

2y=24

y=12 km (by cycling)

So, his walking distance (x) is= 15-12

= 3km