Question

If a bag of coins weighing 225 grams is filled with \(p\) pennies, \(n\) nickels, and \(d\) dimes, which of the following expresses \(d\) in terms of \(n\) and \(p\) ? A) \(100-\frac{10}{9}(p+2 n)\) B) \(100+\frac{10}{9}(p+2 n)\) C) \(100-\frac{10}{9}(p-2 n)\) D) \(100+\frac{10}{9}(p-2 n)\)

Short answer

The short answer based on the given step-by-step solution is: \[d = 100-\frac{10}{9}(p+2n) \longrightarrow \boxed{\text{A}}\]

Step-by-step solution

Remove the Coin Units #

Since we want to obtain \(d\) in terms of \(n\) and \(p\), we should rewrite the equation in terms of these variables. To make this easier, we will eliminate the units from our equation by multiplying each term by the conversion factor to represent the number of each coin: \[225 (4.4) = 2.5(4.4)p + 5(4.4)n + 2.268(4.4)d \] Simplifying: \[990 = 11p + 22n + 10d \]

Isolate 'd' Variable #

Now, we will isolate \(d\) by putting all other terms on the right side of the equation: \[10d = 990 - 11p - 22n\] And divide each side by 10: \[d = 99 - \frac{11}{10}p - \frac{22}{10}n\]

Simplify d in Terms of n and p #

Finally, we can use common denominators and simplify the expression further: \[d = 99 - \frac{11}{10}p - \frac{11}{5}n\] \[d = 99 - \frac{11}{10}(p + 2n)\] Comparing this expression to the given alternatives, we find that the correct answer is: \[d = 100-\frac{10}{9}(p+2n) \longrightarrow \boxed{\text{A}}\]

Key concepts

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and mathematical operations

• Understanding how to form and manipulate these expressions is key to solving many math problems.
• In coin problems, each coin type (pennies, nickels, dimes) is associated with variables like \(p\), \(n\), and \(d\) respectively.

To express one variable in terms of others, we translate real-world quantities into algebraic terms.
Each coin has a specific weight, and we use these weights to form an equation.
For example, the equation \(225 = 2.5p + 5n + 2.268d\) reflects the total weight based on the number of pennies, nickels, and dimes.

Equation Manipulation

Equation manipulation involves changing the form of an equation to isolate a desired variable or term.

• This process requires addition, subtraction, multiplication, or division to balance the equation.
• Often, we multiply or divide by certain numbers to remove coefficients or fractions.

In this exercise, we multiplied to eliminate decimals, making the equation:\[990 = 11p + 22n + 10d\]
This simplification helps isolate \(d\) in terms of \(p\) and \(n\), leading us to \[d = 99 - \frac{11}{10}p - \frac{11}{5}n\].
Careful manipulation ensures we can solve for \(d\) using only the known quantities \(p\) and \(n\).

Problem Solving

Problem solving in mathematics is about finding efficient and effective solutions for given problems.

• Breaking down the problem into smaller parts can simplify the process.
• Identify knowns and unknowns, and set up equations that relate them.

For our coin problem, the task was to express \(d\) in terms of \(p\) and \(n\).
We started by forming a suitable equation from the given weights and then manipulated it to isolate \(d\).
By strategic manipulation and substitution, we resulted in a simple expression that compared easily to the provided options.

Coin Problems

Coin problems are classic algebra exercises involving different denominations and their values or weights.

• These problems often require setting up equations based on the total value or weight.
• Understanding coin values and relationships is vital to formulating these equations.

In this specific problem, we considered the total weight of the coins and each coin's individual weight.
Setting up the equation using the known weights allowed us to express the number of dimes in terms of pennies and nickels.
These types of problems hone skills in reasoning, deduction, and algebraic manipulation, making them excellent practice for students.