- Define the original expression as \(E = 2 \times (30 + 70)\).
- Set up the goal: find an expression \(E' = 3 \times E\).
- Substitute \(E\) into the goal: \(E' = 3 \times [2 \times (30 + 70)]\).
- Simplify using the associative property: \(E' = 6 \times (30 + 70)\).
- Identify the correct option as \(\boxed{B}\).
Explanation
- Analysis of the Problem
The given numerical expression is \(E = 2 \times (30 + 70)\). We are looking for an expression that has a value 3 times greater than \(E\). Let the new expression be \(E'\), so we require \(E' = 3 \times E\). Substituting the given value of \(E\) into this equation, we get \(E' = 3 \times [2 \times (30 + 70)]\).
- Simplifying the Expression
Using the associative property of multiplication, which states that \((a \times b) \times c = a \times (b \times c)\), we can simplify the expression for \(E'\). The equation becomes \(E' = (3 \times 2) \times (30 + 70)\). By multiplying the coefficients \(3\) and \(2\), we obtain \(E' = 6 \times (30 + 70)\).
- Selecting the Correct Option
We compare our result, \(E' = 6 \times (30 + 70)\), with the provided options:
A\(3 \times (30 + 70)\)
B\(6 \times (30 + 70)\)
C\(9 \times (30 + 70)\)
Clearly, our result matches option B. Therefore, the expression that has a value 3 times greater than \(2 \times (30 + 70)\) is \(6 \times (30 + 70)\). The final answer is \(\boxed{B}\).
Examples
Understanding how multiplication scales quantities is essential in many everyday situations. For example, if a recipe calls for 2 cups of ingredients, and you want to triple your production for a large party, you are multiplying your base requirement by 3, resulting in \(3 \times 2 = 6\) cups. This algebraic concept of scaling by a factor allows for efficient planning in cooking, manufacturing, or financial budgeting.