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Together, Noodles and Freckles eat a 50 -pound bag of dog food in 30 days. Noodles by himself eats a 50 -pound bag in 2 weeks less time than Freckles does by himself. How many days to the nearest whole day would a 50 -pound bag of dog food last Freckles?

Short Answer

Expert verified
Freckles takes about 68 days to eat the 50-pound bag alone.

Step by step solution

01

Define Variables

Let \( x \) be the number of days it takes for Freckles to eat a 50-pound bag of dog food alone. Thus, Noodles eats the same amount in \( x - 14 \) days since he eats it in two weeks (14 days) less time than Freckles.
02

Express Rates of Eating

Freckles' rate of eating is \( \frac{50}{x} \) pounds per day, and Noodles' rate is \( \frac{50}{x-14} \) pounds per day.
03

Combined Eating Rate

Together, they eat \( \frac{50}{30} = \frac{5}{3} \) pounds per day. Combine their eating rates: \( \frac{50}{x} + \frac{50}{x-14} = \frac{5}{3} \).
04

Solve the Equation

Multiply through by \(3x(x-14)\) to eliminate the fractions: \[ 3(x)(x-14) \left( \frac{50}{x} + \frac{50}{x-14} = \frac{5}{3} \right) \]Giving:\[ 150(x-14) + 150x = 5x(x-14) \]Expanding and simplifying:\[ 150x - 2100 + 150x = 5x^2 - 70x \]\[ 300x - 2100 = 5x^2 - 70x \]
05

Rearrange and Solve Quadratic

Rearrange the equation:\[ 5x^2 - 370x + 2100 = 0 \]Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b = -370 \), \( c = 2100 \).
06

Apply Quadratic Formula

Calculate discriminant: \( b^2 - 4ac = 370^2 - 4 \times 5 \times 2100 \);\( = 136900 - 42000 \);\( = 94900 \).Solve \( x = \frac{-(-370) \pm \sqrt{94900}}{10} \);\( x = \frac{370 \pm 308 \approx}{10} \).The solution closest to realistic time is \( \frac{370 + 308}{10} = 67.8 \).
07

Round to Nearest Whole Day

Thus, Freckles eats the 50-pound bag in approximately 68 days when rounded to the nearest whole day.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equations
Quadratic equations form the backbone of many algebraic problems. They are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Solving them often involves finding the values of \( x \) that make the equation true. In our dog food problem, the quadratic arises from calculating the combined eating rate of Noodles and Freckles.

The equation we used is \( 5x^2 - 370x + 2100 = 0 \). Here, solving the quadratic involved using the quadratic formula:
  • Calculate the discriminant \( b^2 - 4ac \)
  • Apply it in the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This method allowed us to find the time it takes for Freckles to eat the bag of dog food alone. Remember, the discriminant helps determine the nature of the roots. If it's positive, we have two real solutions, which is the case here. Quadratic equations are prevalent in solving similar real-world problems.
Rate Problems
Rate problems involve finding how quickly an action occurs. In the context of our problem, it relates to how fast Noodles and Freckles can eat the bag of dog food. Rates are typically expressed as a quantity over time, such as miles per hour or pounds per day. Here, the eating rates were expressed as \( \frac{50}{x} \) for Freckles and \( \frac{50}{x-14} \) for Noodles.

Using their combined rate \( \frac{5}{3} \) pounds per day, we set up an equation that represents the total amount of food they ate together over 30 days. These types of problems often require the understanding of proportions and how rates can be summed when people or machines are working together. It's crucial to clearly define each rate and understand the relationship between the tasks being performed.
Variables and Expressions
Variables and expressions are fundamental in algebra as they allow us to represent unknowns and construct equations. In this exercise, \( x \) was used to denote the days Freckles takes to eat a 50-pound bag on his own. Expressions like \( \frac{50}{x} \) and \( \frac{50}{x-14} \) represent their eating rates.

Understanding how to manipulate expressions is key to forming equations that model real-world situations. The power of algebra lies in its ability to generalize situations such that once a variable is defined, the rest of the problem can be expressed in terms of this single unknown. This approach simplifies complex scenarios and allows for the application of a variety of mathematical tools and techniques to find solutions.
Systems of Equations
Systems of equations occur when we have multiple equations that work together to describe a situation. Although this problem doesn't involve multiple equations directly interacting, the essence of understanding how different parts of an equation system relate is crucial.

You can think of Freckles' and Noodles' rates as solving a system where their combined effort equals a known result \( \frac{5}{3} \) pounds per day over 30 days. This provides a platform to solve for unknowns jointly. Tackling systems of equations often means isolating variables, substitutions, or using elimination techniques. Even single equations, like the one used here, encapsulate principles from systems when tackled in layers or break the overall problem into simpler, related parts.

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