# Formula Manipulation

• October 30th 2006, 01:11 AM
Mathematician
Formula Manipulation
Hello!!!!!!!!!!!

Can anyone help me understand a question:-
After months of constant practice, Tom manages to improve his time for the 1500m race by 10%. What is the pecentage change in his speed.

Thank You,
Mathematician.
• October 30th 2006, 01:27 AM
Glaysher
Quote:

Originally Posted by Mathematician
Hello!!!!!!!!!!!

Can anyone help me understand a question:-
After months of constant practice, Tom manages to improve his time for the 1500m race by 10%. What is the pecentage change in his speed.

Thank You,
Mathematician.

$Speed = \frac{Distance}{Time}$

$s = \frac{d}{t}$

Time is improved by 10% means time reduced by 10% from 100% to 90% which is represented as a decimal by 0.9. So new speed s':

$s' = \frac{d}{0.9t}= \frac{1}{0.9} \frac{d}{t} = 1 \frac{1}{9} s$

So speed has improved from 100% to 111.11...%

Speed increased by $11 \frac{1}{9}$%
• October 30th 2006, 04:03 AM
Soroban
Hello, Mathematician!

Glaysher is absolutely correct!
If you're desperate, insert some numbers and baby-talk through it.

Quote:

Tom manages to improve his time for the 1500m race by 10%.
What is the pecentage change in his speed?

Suppose Tom took 300 seconds to run 1500 meters.

. . His speed was: $\frac{1500}{300} \:=\:5$ m/sec.

I assume that a "10% improvement" means he takes only 90% of the time now.

. . So he takes only: $90\% \times 300 \:=\:270$ seconds.

His new speed is: $\frac{1500}{270} \;=\;\frac{50}{9} \;=\;5\frac{5}{9}$ m/sec.

His speed has increased by: $5\frac{5}{9} - 5 \:=\;\frac{5}{9}$ m/sec.

How does this compare with his original speed?

We have: $\frac{\frac{5}{9}}{5} \;= \;\frac{1}{9} \;= \;11\frac{1}{9}\%$ . . . see?