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Question

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 Days

TrainAverage Speed (in km/hr) of Trains  
MondayTuesdayWednesdayThursdayFridaySaturday
A7280-6454-
B88-80847290
C547072-6460
D-12095-90110
E72808475--

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.

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

$66 \frac{2}{3}$

 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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Important Questions from Tabulation

  1. An object is placed beyond the centre of curvature (C) of a concave mirror. The image formed will be now:
  2. Directions: 

    Using the data from the graph answer the question below: 

    Find the total number of males from sector P and females from sector R together. 

    Ratio of number of male and female employees

     MaleFemale
    P67
    Q59
    R98
    S78
    T35
  3. Directions: 

    Using the data from the graph answer the question below: 

    Number of Employees working in 5 different sector: 

    Find the number of males from sector Q.

     Ratio of number of male and female employees:

     MaleFemale
    P67
    Q59
    R98
    S78
    T35

     

  4. Directions: 

    Using the data from the graph answer the question below: 

    Find the ratio between number of males from the sector T to number of females from the sector S.

  5. Directions: 

    Using the data from the graph answer the question below: 

    Find the total number of males from sector Q and females from sector P together.

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