Question 278767: Jack and Jim start at the same time to ride from place A to place B, 60 miles away. Jack travels 4 miles an hour slower than Jim. Jim reaches B and at once turns back meeting Jack 12 miles from B.The rate of Jack was:
(A) 4 mph (B) 8 mph (C) 12 mph (D) 16 mph (E) 20 mph
Answer by Grinnell(63)   (Show Source):
You can put this solution on YOUR website! Good Morning!
Immediately, you have to recognize the fact that Jim travels 72 miles in the same time that Jack travels 48
miles. Don't get 'twisted' in the turn-around--they want to confuse 'ya. Sooo,
let x be the speed (or mph) that Jim travels
let x-4 be the speed Jack travels.
x TIMES (time) =72 miles
(x-4) TIMES (time) =48 miles
Note, the time is the Konstant, or 'the same' Hey why don't we just divide both
equations by the TIME or the Konstant?
let t represent (time)
x=72/t
x-4=48/t
Let's substitute, off to the races...
72/t-4=48/t
72-4t=48 (multiplied thru by t)
-4t=-24
t=6--not finished yet! NOW we go back and plug in our 't'.
x(6)=72
x is 12, go all the way back to the beginning ...this is Jim's speed
12-4, or 8 is Jack's speed.
Have a wonderful day!
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