Comments on: ALGEBRA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!? http://www.mathstudenthelp.info/algebra/algebra-9 Let us help you add it up! Sat, 11 Feb 2012 19:06:36 +0000 http://wordpress.org/?v=2.7 hourly 1 By: morgan http://www.mathstudenthelp.info/algebra/algebra-9/comment-page-1#comment-7077 morgan Fri, 12 Feb 2010 23:11:59 +0000 http://www.mathstudenthelp.info/algebra/algebra-9#comment-7077 Let: r1 = rate of first runner t1 = time taken by first runner r2 = rate of second runner t2 = time taken by second runner Using: rate * time = distance Given: r1 = 300m/min r2 = 360m/min t1 + t2 = 16 Solve the time equation for t1: t1 = 16 - t2 The total distance is the sum of each runner's distance. The distance for each runner is simply their rate * time. r1*t1 + r2*t2 = 5160 Substitute in the given values for their rates: 300*t1 + 360*t2 = 5160 Substitute in the equation for t1: 300*(16 - t2) + 360*t2 = 5160 Solve for t2: 4800 - 300*t2 + 360*t2 = 5160 60*t2 = 360 t2 = 360/60 t2 = 6 Solve for t1: t1 = 16 - t2 t1 = 16 - 6 t1 = 10 Answer: Runner #1 ran for 10 minutes Runner #2 ran for 6 minutes .<br><b>References : </b><br> Let:
r1 = rate of first runner
t1 = time taken by first runner
r2 = rate of second runner
t2 = time taken by second runner

Using:
rate * time = distance

Given:
r1 = 300m/min
r2 = 360m/min
t1 + t2 = 16

Solve the time equation for t1:
t1 = 16 - t2

The total distance is the sum of each runner’s distance.
The distance for each runner is simply their rate * time.
r1*t1 + r2*t2 = 5160

Substitute in the given values for their rates:
300*t1 + 360*t2 = 5160

Substitute in the equation for t1:
300*(16 - t2) + 360*t2 = 5160

Solve for t2:
4800 - 300*t2 + 360*t2 = 5160
60*t2 = 360
t2 = 360/60
t2 = 6

Solve for t1:
t1 = 16 - t2
t1 = 16 - 6
t1 = 10

Answer:
Runner #1 ran for 10 minutes
Runner #2 ran for 6 minutes

.
References :

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