The following table shows the number of students in each of the five classes (A-E) of a school and the percentage (%) of students in these classes who like to play cricket, tennis, handball and football. Based on the data in the table, answer the questions. (A student can play one or more games or no game at all) Class - wise students participation in sports.Class Number of students Percentage (%) of students who like Cricket Tennis Handball Football A 240 60% 70% 50% 60% B 280 50% 60% 60% 50% C 320 40% 65% 55% 45% D 360 65% 75% 65% 55% E 480 70% 80% 75% 45%
The number of students who like to play Tennis in Class A is _______ % less than those who like to play Handball in Class E.
53.33
Understanding Student Sports Preferences Data
The question asks us to analyze the provided table showing the number of students in five classes (A-E) and the percentage of students in each class who like to play certain sports: Cricket, Tennis, Handball, and Football.
We need to find the percentage difference between the number of students who like Tennis in Class A and the number of students who like Handball in Class E. Specifically, we need to find how much the number of Tennis players in Class A is less than the number of Handball players in Class E, expressed as a percentage of the Handball players in Class E.
| Class | Number of students | Percentage (%) of students who like | |||
|---|---|---|---|---|---|
| Cricket | Tennis | Handball | Football | ||
| A | 240 | 60% | 70% | 50% | 60% |
| B | 280 | 50% | 60% | 60% | 50% |
| C | 320 | 40% | 65% | 55% | 45% |
| D | 360 | 65% | 75% | 65% | 55% |
| E | 480 | 70% | 80% | 75% | 45% |
Calculating Students Liking Tennis in Class A
First, let's find the number of students in Class A who like to play Tennis.
- Total students in Class A = 240
- Percentage of students in Class A who like Tennis = 70%
The number of students who like Tennis in Class A is calculated as:
\[ \text{Number of students (Tennis, Class A)} = \text{Total students in Class A} \times \frac{\text{Percentage liking Tennis}}{100} \] \[ \text{Number of students (Tennis, Class A)} = 240 \times \frac{70}{100} = 240 \times 0.70 = 168 \]So, 168 students in Class A like to play Tennis.
Calculating Students Liking Handball in Class E
Next, let's find the number of students in Class E who like to play Handball.
- Total students in Class E = 480
- Percentage of students in Class E who like Handball = 75%
The number of students who like Handball in Class E is calculated as:
\[ \text{Number of students (Handball, Class E)} = \text{Total students in Class E} \times \frac{\text{Percentage liking Handball}}{100} \] \[ \text{Number of students (Handball, Class E)} = 480 \times \frac{75}{100} = 480 \times 0.75 = 360 \]So, 360 students in Class E like to play Handball.
Calculating the Percentage Less
We need to find by what percentage the number of students liking Tennis in Class A (168) is less than the number of students liking Handball in Class E (360). The percentage is calculated relative to the number of students liking Handball in Class E.
First, find the difference between the two numbers:
\[ \text{Difference} = \text{Number of students (Handball, Class E)} - \text{Number of students (Tennis, Class A)} \] \[ \text{Difference} = 360 - 168 = 192 \]Now, calculate the percentage difference relative to the number of students liking Handball in Class E:
\[ \text{Percentage Less} = \frac{\text{Difference}}{\text{Number of students (Handball, Class E)}} \times 100 \] \[ \text{Percentage Less} = \frac{192}{360} \times 100 \]Simplify the fraction:
\[ \frac{192}{360} = \frac{19.2}{36} = \frac{9.6}{18} = \frac{4.8}{9} = \frac{2.4}{4.5} = \frac{24}{45} = \frac{8}{15} \]Now, calculate the percentage:
\[ \text{Percentage Less} = \frac{8}{15} \times 100 = \frac{800}{15} = \frac{160}{3} \] \[ \text{Percentage Less} = 53.333... \]Rounding to two decimal places, the percentage less is approximately 53.33%.
Therefore, the number of students who like to play Tennis in Class A is 53.33% less than those who like to play Handball in Class E.
Revision Table: Key Calculations
| Calculation Item | Formula / Value | Result |
|---|---|---|
| Students Liking Tennis (Class A) | \(240 \times 70\%\) | 168 |
| Students Liking Handball (Class E) | \(480 \times 75\%\) | 360 |
| Difference | \(360 - 168\) | 192 |
| Percentage Less (vs. Handball E) | \(\frac{192}{360} \times 100\%\) | 53.33% (approx.) |
Additional Information: Percentage Calculations
Understanding percentage calculations is crucial for data analysis questions like this one. Here are a few key concepts:
- Percentage of a Quantity: To find X% of a quantity Y, you calculate \( Y \times \frac{X}{100} \). This is used to find the number of students playing a specific sport from the total number of students.
- Percentage Increase/Decrease: To find the percentage increase or decrease from a value A to a value B, the formula is \( \frac{|B - A|}{A} \times 100 \). Here, A is the original value or the base for comparison.
- "A is X% less than B": This means \( X\% = \frac{B - A}{B} \times 100 \). The value B is the base for the percentage comparison. In our problem, B is the number of students liking Handball in Class E.
Always pay attention to which value is used as the base (denominator) in percentage difference calculations, as it significantly affects the result.
Important Questions from Miscellaneous
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f1V (f1 < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f2V. Then:
A railway wagon (open at the top) of mass M1 is moving with speed v1 along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M2 and speed becomes v2. Taking the rain to be falling vertically and the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :
Consider the following statements:
1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
One block of 2⋅0 kg mass is placed on top of another block of 3⋅0 kg mass. The coefficient of static friction between the two blocks is 0⋅2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s2, the maximum value of the frictional force is :
