Problem 136
Question
If Barney’s strength continues to grow at the same rate, at what week will he be able to lift 100 pounds? (A) Week 25 (B) Week 30 (C) Week 35 (D) Week 50
Step-by-Step Solution
Verified Answer
\(S = w * A\)
Week 25: \(A = \frac{100}{25} = 4\)
Answer: (A) Week 25
1Step 1: Identify the constant growth rate and create an equation
Let's assume that the equation describing the growth of Barney's strength is S = w*A, where S is the strength in pounds, w is the number of weeks passed, and A is a constant growth rate. As we are given that his strength continues to grow at the same rate, we can directly use this equation for our calculations.
2Step 2: Solve the equation S = w*A for the correct week
In order to find the week when Barney will be able to lift 100 pounds, we need to solve the equation S = w*A, such that S = 100 pounds. Since we do not have any specific value for the constant growth rate A, let's compare the given options of weeks and find the week that satisfies the equation with the given condition S = 100 pounds.
Substitute the given week numbers in place of w and find the approximate A for each option:
(A) Week 25: 100 = 25A -> A = 4
(B) Week 30: 100 = 30A -> A = 10/3 ≈ 3.33
(C) Week 35: 100 = 35A -> A = 20/7 ≈ 2.86
(D) Week 50: 100 = 50A -> A = 2
Now, comparing the values of A, we can observe that option (A) has the highest value and hence, the closest to achieving the strength of 100 pounds, provided Barney's growth continues at the same rate.
Answer: (A) Week 25
Key Concepts
Linear EquationsAlgebraGrowth Rate Calculation
Linear Equations
In mathematics, linear equations play a crucial role in problem-solving, especially in scenarios like growth calculations. A linear equation is an equation that forms a straight line when graphed on a coordinate plane. It is defined by the formula \( y = mx + b \), where \( m \) represents the slope or rate of change, and \( b \) the y-intercept or starting value.
For Barney's problem, we consider the equation \( S = w \times A \). Here, \( S \) denotes strength in pounds, \( w \) is the number of weeks, and \( A \) is the growth rate. As Barney's strength changes linearly, meaning the rate is constant over each week, a linear equation is key to predicting future strengths.
In the original exercise, our job was to identify the week \( w \) at which \( S = 100 \). By substituting different values of \( w \) from the options given, we utilized the principles of linear equations to determine that option (A), week 25, maximizes the growth rate \( A \) at 4.
For Barney's problem, we consider the equation \( S = w \times A \). Here, \( S \) denotes strength in pounds, \( w \) is the number of weeks, and \( A \) is the growth rate. As Barney's strength changes linearly, meaning the rate is constant over each week, a linear equation is key to predicting future strengths.
In the original exercise, our job was to identify the week \( w \) at which \( S = 100 \). By substituting different values of \( w \) from the options given, we utilized the principles of linear equations to determine that option (A), week 25, maximizes the growth rate \( A \) at 4.
Algebra
Algebra is often referred to as the language of mathematics, using symbols and letters to represent numbers and quantities. This branch of mathematics allows us to generalize arithmetic operations by using variables. In the context of the original exercise, algebra was employed to solve for unknowns using symbols and equations.
When faced with the task of finding when Barney can lift 100 pounds, we utilized an algebraic approach. We set the given strength \( S = 100 \) in the equation \( S = w \times A \). By rearranging the equation to solve for \( A \) (the growth rate), we systematically substituted the different week options for \( w \).
This algebraic manipulation is beneficial for solving complex problems that would be cumbersome through arithmetic alone. It allows students to comprehend the relationships between numbers and understand how varying one element—such as time in this case—alters the outcome.
When faced with the task of finding when Barney can lift 100 pounds, we utilized an algebraic approach. We set the given strength \( S = 100 \) in the equation \( S = w \times A \). By rearranging the equation to solve for \( A \) (the growth rate), we systematically substituted the different week options for \( w \).
This algebraic manipulation is beneficial for solving complex problems that would be cumbersome through arithmetic alone. It allows students to comprehend the relationships between numbers and understand how varying one element—such as time in this case—alters the outcome.
Growth Rate Calculation
Growth rate calculation is commonly used in predicting future outcomes or understanding how something develops over time. In Barney's situation, it measures how his strength increases as weeks pass. Calculating growth rate involves relating changes in one variable to a change in another, typically time.
In our scenario, we sought to find when Barney could lift 100 pounds by establishing a constant growth rate \( A \). This was done using the formula \( S = w \times A \), representing a steady increase over weeks. By evaluating the increments and substituting different weeks into our equation, we analyzed how many pounds Barney gains weekly.
Conceptually, this boils down to understanding the link between growth rates and time. This relationship helps in forecasting events or capacities, making it a powerful tool not just in mathematics, but in real-world applications like finance, science, and health.
In our scenario, we sought to find when Barney could lift 100 pounds by establishing a constant growth rate \( A \). This was done using the formula \( S = w \times A \), representing a steady increase over weeks. By evaluating the increments and substituting different weeks into our equation, we analyzed how many pounds Barney gains weekly.
Conceptually, this boils down to understanding the link between growth rates and time. This relationship helps in forecasting events or capacities, making it a powerful tool not just in mathematics, but in real-world applications like finance, science, and health.
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