Let t be total distance
distance at 40mph = tx/100
time at 40mph = tx/(100*40)
distance at 60mph = t(100-x)/100
time at 60mph = t(100-x)/(100*60)
Total distance/Total time = t/[tx/4000 + t(100-x)/6000]
=12000/(x+200)
gives E.
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. (180-x)/2
B. (x+60)/4
C. (300-x)/5
D. 600/(115-x)
E. 12000/(x+200)
Official GuideSPOILER: E
Let t be total distance
distance at 40mph = tx/100
time at 40mph = tx/(100*40)
distance at 60mph = t(100-x)/100
time at 60mph = t(100-x)/(100*60)
Total distance/Total time = t/[tx/4000 + t(100-x)/6000]
=12000/(x+200)
gives E.
Last edited by e.cartman; 11-25-2008 at 02:22 PM. Reason: more steps
Let p be the total distance,
Let t1 be the time he takes to travel x percent of p then t1=p(x/100))/40=p*x/4000
Let t2 be the time he takes to travel the rest of p then t2=p (1-x/100)/60=p*(100-x)/6000
Average= total distance/ total time= p/(t1+t2)= p/[px/4000+p(100-x)/6000] =12000/3x+2(100-x)= 12000/(200+x)
Distance Traveled (D) = Average Speed (S) * Time (T)
D = S * T
S = D / T
Time for which she traveled at 40 miles per hour
T1 = ((x/100)* D)/ 40 = xD/4000
Time for which she traveled at 60 miles per hour
T2 = (100-x)D/6000
Now Average speed of the journey = Total Distance/ (T1 + T2)
Total Distance is D
Substituting the values you will get the answer . Hope that helps
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