The following table shows the average speed (in km per hour) of five different trains A-E during six days of a week from Monday through Saturday. Some data is missing in the table ( indicated as '-') that you are expected to calculate, if required. Based on the data in the table, answer the question that follows: Train - wise speed (in km/hour) on different DaysTrain Average Speed (in km/hr) of Trains Monday Tuesday Wednesday Thursday Friday Saturday A 72 80 - 64 54 - B 88 - 80 84 72 90 C 54 70 72 - 64 60 D - 120 95 - 90 110 E 72 80 84 75 - -
For train E, if the ratio of average speed on Saturday and Friday is 5:3, then the average speed on Saturday is ___% more than that on Friday.
$66 \frac{2}{3}$
To solve this problem, let's analyze the given question about Train E's average speed on Friday and Saturday. The ratio between the average speeds on these two days is provided as 5:3.
We'll denote the average speed on Friday as \(S_F\) and on Saturday as \(S_S\). From the ratio given, we have:
\(\frac{S_S}{S_F} = \frac{5}{3}\)
This implies that:
\(S_S = \frac{5}{3} \times S_F\)
We need to find how much percentage more the average speed on Saturday is compared to Friday. The formula to calculate the percentage increase is:
\(\text{Percentage Increase} = \left(\frac{S_S - S_F}{S_F}\right) \times 100\%\)
Substituting \(S_S = \frac{5}{3} \times S_F\) into the formula, we get:
\(\text{Percentage Increase} = \left(\frac{\frac{5}{3} S_F - S_F}{S_F}\right) \times 100\%\)
Simplifying this expression:
\(\text{Percentage Increase} = \left(\frac{\frac{5}{3} S_F - \frac{3}{3} S_F}{S_F}\right) \times 100\% = \left(\frac{\frac{2}{3} S_F}{S_F}\right) \times 100\%\)
\(\text{Percentage Increase} = \left(\frac{2}{3}\right) \times 100\% = 66\frac{2}{3}\%\)
Therefore, the average speed on Saturday is 66\(\frac{2}{3}\)% more than that on Friday, making the correct answer the first option.
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