Different Speed Distance and Time

lingping7

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Jan 6, 2013
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OK Here's the problem.
A boy rides a bicycle from town P to town Q with a constant speed. If he increases his speed by 3 m/sec, he will arrive at town Q 3 times faster.How many times faster will the boy arrive if he increases his speed by 6 m/sec?

Looking at the question I thought it was 6 times faster, but the answer says 5 times faster.
I have no clue how to solve this problem and would appreciate help. ;)
 
Hello, lingping7!

An interesting problem . . . certainly different!


A boy rides a bicycle from town P to town Q with a constant speed.
If he increases his speed by 3 m/sec, he will arrive at town Q three times faster.
. . This means "in one-third of the time".
How many times faster will the boy arrive if he increases his speed by 6 m/sec?

We will use the formula: .\(\displaystyle \text{Time} \:=\:\dfrac{\text{Distance}}{\text{Speed}}\)

Let \(\displaystyle x\) = boy's original speed (m/sec).
Let \(\displaystyle d\) = distance from P to Q.

He travels \(\displaystyle d\) meters at \(\displaystyle x\) m/sec.
This takes: .\(\displaystyle \dfrac{d}{x}\) seconds. .[1]

If he travels \(\displaystyle d\) meters at \(\displaystyle x+3\) m/sec,
. . it will take: .\(\displaystyle \dfrac{d}{x+3}\) seconds. .[2]

Since [2] is one-third of [1]: .\(\displaystyle \dfrac{d}{x+3} \:=\:\dfrac{1}{3}\cdot\dfrac{d}{x}\)

. . \(\displaystyle 3dx \:=\:dx + 3d \quad\Rightarrow\quad 2dx \:=\:3d \quad\Rightarrow\quad x \:=\:\dfrac{3}{2}\)

His original speed is \(\displaystyle 1\!\frac{1}{2}\) meters per second.
. . His original time is: .\(\displaystyle \dfrac{d}{\frac{3}{2}} \,=\,\dfrac{2d}{3}\) seconds.


If he travels \(\displaystyle d\) meters at \(\displaystyle \frac{3}{2} + 6\) m/sec,
. . his time would be: .\(\displaystyle \dfrac{d}{\frac{3}{2} + 6} \,=\,\dfrac{d}{\frac{15}{2}} \,=\,\dfrac{2d}{15}\) seconds.

How does this compare to his original time?
. . \(\displaystyle \dfrac{2d}{15} \;=\;k\cdot\dfrac{2d}{3} \quad\Rightarrow\quad 6d \:=\:30dk \quad\Rightarrow\quad k \:=\:\dfrac{1}{5}\)

His time is one-fifth of the original time.
. . He arrives five times faster.
 
Thanks a lot everyone. Love these forums. :p
Hello soroban,
I would like to know how you entered the fractions
 
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