A new train goes 20% further in 20% less time than an old train. By what % is the average speed of the new train GREATER THAN that of the old train?
Let Do = 1 mi = Distance of old train.
Dn = 1.2*Do = 1.2 * 1 = 1.2 mi = Dist.
of new train.
Let To = 1min = Time of old train.
Tn = 0.8*To = 0.8 * 1 = 0.8 min. = Time
of new train
Vo=Do/To = 1mi/1min=1mi/min Velocity of old train.
Vn = Dn/Tn = 1.2mi/0.8min = 1.5 mi/min = Velocity of new train.
Vn/Vo = 1.5/1 = 1.5 = 150%,
% Greater = 150% - 100% = 50%.
the answer before mine is wrong becaude it is actually 65%
It's 50%.
To find the percentage by which the average speed of the new train is greater than that of the old train, we first need to determine the average speeds of both trains.
Let's assume that the old train travels a distance of "d" units in "t" units of time. Therefore, the average speed of the old train would be given by:
Average speed of old train = d/t
According to the statement, the new train goes 20% further than the old train, which means the new train travels a distance of 1.2d units. Additionally, the new train takes 20% less time than the old train, which is equivalent to 0.8t units of time. Therefore, the average speed of the new train would be:
Average speed of new train = (1.2d) / (0.8t)
To compare the average speeds of the new and old trains, we can calculate the percentage difference using the formula:
Percentage Difference = ((New Speed - Old Speed) / Old Speed) × 100
Let's substitute the values into the formula:
Percentage Difference = (((1.2d) / (0.8t)) - (d / t)) / (d / t) × 100
Simplifying the expression:
Percentage Difference = ((1.5d - d) / d) × 100
Percentage Difference = (0.5d / d) × 100
Percentage Difference = 50%
Therefore, the average speed of the new train is 50% greater than that of the old train.