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During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine's average speed for the entire trip?
A. \(\frac{(180-x)}{2}\)
B. \(\frac{(x+60)}{4}\)
C. \(\frac{(300-x)}{5}\)
D. \(\frac{600}{(115-x)}\)
E. \(\frac{12,000}{(x+200)}\)
PS08570
A. \(\frac{(180-x)}{2}\)
B. \(\frac{(x+60)}{4}\)
C. \(\frac{(300-x)}{5}\)
D. \(\frac{600}{(115-x)}\)
E. \(\frac{12,000}{(x+200)}\)
PS08570
13 Jun 2010, 04:40
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During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine’s average speed for the entire trip?
A. (1800 - x) /2
B. (x + 60) /2
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)
Algebraic approach:
\(Average \ speed=\frac{distance}{total \ time}\), let's assume \(distance=40\) (distance \(d\) will cancel out from the equation, so we can assume distance to be some number.) so we should calculate total time.
Francine traveled \(x\) percent of the total distance at an average speed of 40 miles per hour --> time needed for this part of the trip: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*40}{40}=\frac{x}{100}\);
Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);
\(Total \ time=t_1+t_2=\frac{x}{100}+\frac{100-x}{150}=\frac{x+200}{300}\);
\(Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}\).
Answer: E.
A. (1800 - x) /2
B. (x + 60) /2
C. (300 - x ) / 5
D. 600 / (115 - x )
E. 12,000 / ( x + 200)
Algebraic approach:
\(Average \ speed=\frac{distance}{total \ time}\), let's assume \(distance=40\) (distance \(d\) will cancel out from the equation, so we can assume distance to be some number.) so we should calculate total time.
Francine traveled \(x\) percent of the total distance at an average speed of 40 miles per hour --> time needed for this part of the trip: \(t_1= \frac{distance_1}{speed_1}=\frac{\frac{x}{100}*40}{40}=\frac{x}{100}\);
Timed needed for the rest of the trip: \(t_2= \frac{distance_2}{speed_2}=\frac{(1-\frac{x}{100})*40}{60}=\frac{100-x}{150}\);
\(Total \ time=t_1+t_2=\frac{x}{100}+\frac{100-x}{150}=\frac{x+200}{300}\);
\(Average \ speed=\frac{distance}{total \ time}=\frac{40}{\frac{x+200}{300}}=\frac{12000}{x+200}\).
Answer: E.
17 Aug 2010, 05:26
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Easier approach:
Formula
:When an body covers m part of journey at speed p and next n part of the journey at speed q then the Average speed of the total journey is:(m+n)*pq / (np+mq).
Using above formula:
initial part of journey =x
remaining part 100-x (since x is in percent)
m+n=100
so we have => 100*40*60/x*60+(100-x)40 -> solves to 12000/x+200 -E is the Answer
