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Dev cover a distance of 15 km, partly by foot at 6km/h & partly by cycle at 18km/h....
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High School Teacher
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.
Posted by jlcannad on January 31, 2009 at 12:05 PM (Answer #1)
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:
y=12 km (by cycling)
So, his walking distance (x) is= 15-12
Posted by revolution on August 8, 2010 at 10:04 PM (Answer #2)
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