A bead shop sells wooden beads for 0.20 each and crystal...

A bead shop sells wooden beads for $\$ 0.20$ each and crystal beads for $\$ 0.50$ each. If a jewelry artist buys 127 beads total and pays $\$ 41$ for them, how much more did she spend on crystal beads than wooden beads? $$\$ 11$$ $$\$ \mathbb{25}$$ $$\begin{array}{l} \$ 33 \\ \$ 2 6 \end{array}$$

A bead shop sells wooden beads for $\$ 0.20$ each and crystal beads for $\$ 0.50$
each. If a jewelry artist buys 127 beads total and pays $\$ 41$ for them,
how much more did she spend on crystal beads than wooden beads?
$$\$ 11$$
$$\$ \mathbb{25}$$
$$\begin{array}{l}
\$ 33 \\
\$ 2 6
\end{array}$$

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Key Concepts

Algebraic Modeling

Algebraic Modeling involves translating a word problem into one or more equations using variables to represent unknown quantities. This process is essential for formulating problems in a mathematical language that can then be analyzed and solved.

Systems of Linear Equations

Systems of Linear Equations consist of multiple equations that simultaneously describe relationships between variables. Solving such systems allows one to find the exact values of the variables that satisfy all given conditions, a fundamental technique in algebra.

Solving Techniques (Substitution/Elimination)

Techniques such as substitution and elimination are used to solve systems of linear equations by reducing the number of variables in the equations. These methods streamline the process of finding solutions by isolating variables or combining equations to eliminate unknowns.

Cost and Quantity Analysis in Word Problems

Cost and Quantity Analysis in Word Problems involves setting up equations based on given totals and unit costs. This concept teaches how to balance different constraints, such as total cost and total items, to determine individual values, a common scenario in practical algebra applications.

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