Question
1. Simplify. \( \begin{array}{lll}\text { a) } 15+4 \times 2 & \text { b) } 0 \times 50+0 \div 18 & \text { c) }(125 \div 1)+32 \times 1\end{array} \)
Ask by Lewis Phillips. in South Africa
Jan 30,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
a) 23
b) 0
c) 157
Solution
Sure, let's simplify each of the given expressions step by step.
### a) \( 15 + 4 \times 2 \)
**Step 1:** Follow the order of operations (PEMDAS/BODMAS), which dictates that multiplication comes before addition.
\[
4 \times 2 = 8
\]
**Step 2:** Add the result to 15.
\[
15 + 8 = 23
\]
**Simplified Expression:** \( 23 \)
---
### b) \( 0 \times 50 + 0 \div 18 \)
**Step 1:** Perform the multiplication and division operations.
\[
0 \times 50 = 0
\]
\[
0 \div 18 = 0
\]
**Step 2:** Add the results.
\[
0 + 0 = 0
\]
**Simplified Expression:** \( 0 \)
---
### c) \( (125 \div 1) + 32 \times 1 \)
**Step 1:** Perform the division and multiplication.
\[
125 \div 1 = 125
\]
\[
32 \times 1 = 32
\]
**Step 2:** Add the results.
\[
125 + 32 = 157
\]
**Simplified Expression:** \( 157 \)
---
**Summary of Simplified Expressions:**
a) \( 15 + 4 \times 2 = 23 \)
b) \( 0 \times 50 + 0 \div 18 = 0 \)
c) \( (125 \div 1) + 32 \times 1 = 157 \)
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution The Deep Dive
First, remember the order of operations: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right), commonly referred to as PEMDAS. For (a) \( 15 + 4 \times 2 \), first do the multiplication: \( 4 \times 2 = 8 \), then add: \( 15 + 8 = 23 \). For (b) \( 0 \times 50 + 0 \div 18 \), both components yield zero, so the result is \( 0 + 0 = 0 \). Finally, (c) \( (125 \div 1) + 32 \times 1 \) simplifies to \( 125 + 32 = 157 \). So the simplified results are: a) 23 b) 0 c) 157
