A bus travels 1/3 distance at 10 kmph,next 1/3 at 20 mph. what is the average speed
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This is a D-I-R-T (Distance-Is-Rate-times-Time) problem. The problem asks for overall rate, so you need (total distance)/(total time).
You let x=total distance.
OK so far.
D=RT and you want to find R (that is, the overall average speed)
R = D/T
R = x/T [total distance is x]
Now, for each segment, T = D/R for that portion.
R = x / ( (x/3)/10 + (x/3)/20 + (x/3)/60) ) [insert the distances and rates]
R = 60x / ( 6(x/3) + 3(x/3) + (x/3) ) [multiply by 1 = 60/60 to get rid of fractions]
R = 60x / (10x/3)
[collect terms]
R = 60 * 3/10 [cancel 'x'; to divide by fraction, invert and multiply]
R = 18 km/hr
You let x=total distance.
OK so far.
D=RT and you want to find R (that is, the overall average speed)
R = D/T
R = x/T [total distance is x]
Now, for each segment, T = D/R for that portion.
R = x / ( (x/3)/10 + (x/3)/20 + (x/3)/60) ) [insert the distances and rates]
R = 60x / ( 6(x/3) + 3(x/3) + (x/3) ) [multiply by 1 = 60/60 to get rid of fractions]
R = 60x / (10x/3)
[collect terms]
R = 60 * 3/10 [cancel 'x'; to divide by fraction, invert and multiply]
R = 18 km/hr
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