Problem 33
Question
The average weight of students in a class of 35 students is \(40 \mathrm{~kg}\). If the weight of the teacher be included, the average rises by \(\frac{1}{2} \mathrm{~kg}\); the weight of the teacher is (A) \(40.5 \mathrm{~kg}\) (B) \(50 \mathrm{~kg}\) (C) \(41 \mathrm{~kg}\) (D) \(58 \mathrm{~kg}\)
Step-by-Step Solution
Verified Answer
The teacher's weight is 58 kg.
1Step 1: Determine Total Weight of Students
Given that the average weight of 35 students is \(40 \mathrm{~kg}\), the total weight of all students can be calculated as follows: \[\text{Total weight of students} = \text{Average weight} \times \text{Number of students} = 40 \times 35 = 1400 \mathrm{~kg}.\]
2Step 2: Calculate New Average with Teacher
With the teacher included, the number of people becomes 36 and the new average weight is \(40 + \frac{1}{2} = 40.5 \mathrm{~kg}\). So the total weight with the teacher is: \[\text{Total weight with teacher} = 40.5 \times 36 = 1458 \mathrm{~kg}.\]
3Step 3: Find the Teacher's Weight
To find the teacher's weight, subtract the total weight of the students from the total weight including the teacher: \[\text{Teacher's weight} = \text{Total with teacher} - \text{Total of students} = 1458 - 1400 = 58 \mathrm{~kg}.\]
Key Concepts
Mean CalculationSolving Algebraic EquationsProblem-Solving in Mathematics
Mean Calculation
To calculate the average weight, or mean, of a group of students, you simply add up all the individual weights and divide by the number of students. In this exercise, we have 35 students, each with an average weight of \(40 \mathrm{~kg}\). Thus, the total weight is the product of the average weight and the number of students, which equals 1400 kg. This is a straightforward application of mean calculation:
Understanding how mean works is essential because it helps to determine a central value of data. This can be useful in various fields like statistics, finance, and daily life decision-making.
- Mean = Total Weight of Group \(\div\) Number of Individuals
Understanding how mean works is essential because it helps to determine a central value of data. This can be useful in various fields like statistics, finance, and daily life decision-making.
Solving Algebraic Equations
In this problem, calculating the teacher's weight involves solving an equation. After adding the teacher, the average changes, hinting at the presence of an unknown variable. Let's assume the teacher's weight as \(x\). When the class includes the teacher, the total number becomes 36, and the new average becomes 40.5 kg:
Algebraic equations like this use variables to represent unknown values and are key in problem-solving for many applications.
- New Total Weight = (Number of People) \(\times\) New Average = \(36 \times 40.5\)
- Total Weight with Teacher = Previous Total Weight + Teacher's Weight
- From this, solve for \(x\)
Algebraic equations like this use variables to represent unknown values and are key in problem-solving for many applications.
Problem-Solving in Mathematics
Problem-solving in mathematics brings together different concepts, like understanding averages and solving equations, to find solutions. It involves logical thinking and recognizing patterns. In this exercise, we need to:
By approaching problems step by step, you can make complex problems much simpler. Practice is essential to get better at mathematical problem-solving, which is a valuable skill not only in academics but also in real-life situations.
- Understand the role of averages in calculating weight increases.
- Use logical steps to determine what's missing, i.e., the teacher's weight.
- Apply mathematical operations by calculating new totals and subtracting known values.
By approaching problems step by step, you can make complex problems much simpler. Practice is essential to get better at mathematical problem-solving, which is a valuable skill not only in academics but also in real-life situations.
Other exercises in this chapter
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