3rd Grade Math Puzzles Worksheets - Free Printable

Problem Explanation:

The "Total Product Puzzle" involves filling in missing numbers based on the relationship between two operations: addition (TOTAL) and multiplication (PRODUCT). The puzzle provides a formula:

\[
\text{TOTAL} = \text{Number 1} + \text{Number 2}
\]
\[
\text{PRODUCT} = \text{Number 1} \times \text{Number 2}
\]

We need to use these formulas to solve for the missing numbers in each puzzle.

---

Step-by-Step Solution:

#### Puzzle 1:
\[
\begin{aligned}
&\text{Number 1} = 4, \quad \text{Number 2} = 8 \\
&\text{TOTAL} = ?, \quad \text{PRODUCT} = ?
\end{aligned}
\]

1. Calculate the TOTAL:
\[
\text{TOTAL} = 4 + 8 = 12
\]

2. Calculate the PRODUCT:
\[
\text{PRODUCT} = 4 \times 8 = 32
\]

Answer:
\[
\boxed{12, 32}
\]

---

#### Puzzle 2:
\[
\begin{aligned}
&\text{Number 1} = 10, \quad \text{Number 2} = 8 \\
&\text{TOTAL} = ?, \quad \text{PRODUCT} = ?
\end{aligned}
\]

1. Calculate the TOTAL:
\[
\text{TOTAL} = 10 + 8 = 18
\]

2. Calculate the PRODUCT:
\[
\text{PRODUCT} = 10 \times 8 = 80
\]

Answer:
\[
\boxed{18, 80}
\]

---

#### Puzzle 3:
\[
\begin{aligned}
&\text{Number 1} = 5, \quad \text{Number 2} = 9 \\
&\text{TOTAL} = ?, \quad \text{PRODUCT} = ?
\end{aligned}
\]

1. Calculate the TOTAL:
\[
\text{TOTAL} = 5 + 9 = 14
\]

2. Calculate the PRODUCT:
\[
\text{PRODUCT} = 5 \times 9 = 45
\]

Answer:
\[
\boxed{14, 45}
\]

---

#### Puzzle 4:
\[
\begin{aligned}
&\text{Number 1} = 4, \quad \text{TOTAL} = 11 \\
&\text{Number 2} = ?, \quad \text{PRODUCT} = ?
\end{aligned}
\]

1. Solve for Number 2 using the TOTAL:
\[
\text{TOTAL} = \text{Number 1} + \text{Number 2} \implies 11 = 4 + \text{Number 2} \implies \text{Number 2} = 11 - 4 = 7
\]

2. Calculate the PRODUCT:
\[
\text{PRODUCT} = \text{Number 1} \times \text{Number 2} = 4 \times 7 = 28
\]

Answer:
\[
\boxed{7, 28}
\]

---

#### Puzzle 5:
\[
\begin{aligned}
&\text{Number 2} = 3, \quad \text{PRODUCT} = 18 \\
&\text{Number 1} = ?, \quad \text{TOTAL} = ?
\end{aligned}
\]

1. Solve for Number 1 using the PRODUCT:
\[
\text{PRODUCT} = \text{Number 1} \times \text{Number 2} \implies 18 = \text{Number 1} \times 3 \implies \text{Number 1} = \frac{18}{3} = 6
\]

2. Calculate the TOTAL:
\[
\text{TOTAL} = \text{Number 1} + \text{Number 2} = 6 + 3 = 9
\]

Answer:
\[
\boxed{6, 9}
\]

---

#### Puzzle 6:
\[
\begin{aligned}
&\text{Number 1} = 5, \quad \text{PRODUCT} = 35 \\
&\text{Number 2} = ?, \quad \text{TOTAL} = ?
\end{aligned}
\]

1. Solve for Number 2 using the PRODUCT:
\[
\text{PRODUCT} = \text{Number 1} \times \text{Number 2} \implies 35 = 5 \times \text{Number 2} \implies \text{Number 2} = \frac{35}{5} = 7
\]

2. Calculate the TOTAL:
\[
\text{TOTAL} = \text{Number 1} + \text{Number 2} = 5 + 7 = 12
\]

Answer:
\[
\boxed{12, 7}
\]

---

#### Puzzle 7:
\[
\begin{aligned}
&\text{TOTAL} = 14, \quad \text{PRODUCT} = 40 \\
&\text{Number 1} = ?, \quad \text{Number 2} = ?
\end{aligned}
\]

1. Let the two numbers be \( x \) and \( y \). We have:
\[
x + y = 14 \quad \text{(Equation 1)}
\]
\[
x \times y = 40 \quad \text{(Equation 2)}
\]

2. From Equation 1, express \( y \) in terms of \( x \):
\[
y = 14 - x
\]

3. Substitute \( y \) into Equation 2:
\[
x \times (14 - x) = 40 \implies 14x - x^2 = 40 \implies x^2 - 14x + 40 = 0
\]

4. Solve the quadratic equation:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{14 \pm \sqrt{14^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1} = \frac{14 \pm \sqrt{196 - 160}}{2} = \frac{14 \pm \sqrt{36}}{2} = \frac{14 \pm 6}{2}
\]

5. Calculate the two solutions:
\[
x = \frac{14 + 6}{2} = 10 \quad \text{or} \quad x = \frac{14 - 6}{2} = 4
\]

6. If \( x = 10 \), then \( y = 14 - 10 = 4 \). If \( x = 4 \), then \( y = 14 - 4 = 10 \).

Answer:
\[
\boxed{10, 4}
\]

---

#### Puzzle 8:
\[
\begin{aligned}
&\text{TOTAL} = 11, \quad \text{PRODUCT} = 18 \\
&\text{Number 1} = ?, \quad \text{Number 2} = ?
\end{aligned}
\]

1. Let the two numbers be \( x \) and \( y \). We have:
\[
x + y = 11 \quad \text{(Equation 1)}
\]
\[
x \times y = 18 \quad \text{(Equation 2)}
\]

2. From Equation 1, express \( y \) in terms of \( x \):
\[
y = 11 - x
\]

3. Substitute \( y \) into Equation 2:
\[
x \times (11 - x) = 18 \implies 11x - x^2 = 18 \implies x^2 - 11x + 18 = 0
\]

4. Solve the quadratic equation:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} = \frac{11 \pm \sqrt{121 - 72}}{2} = \frac{11 \pm \sqrt{49}}{2} = \frac{11 \pm 7}{2}
\]

5. Calculate the two solutions:
\[
x = \frac{11 + 7}{2} = 9 \quad \text{or} \quad x = \frac{11 - 7}{2} = 2
\]

6. If \( x = 9 \), then \( y = 11 - 9 = 2 \). If \( x = 2 \), then \( y = 11 - 2 = 9 \).

Answer:
\[
\boxed{9, 2}
\]

---

#### Puzzle 9:
\[
\begin{aligned}
&\text{TOTAL} = 13, \quad \text{PRODUCT} = 40 \\
&\text{Number 1} = ?, \quad \text{Number 2} = ?
\end{aligned}
\]

1. Let the two numbers be \( x \) and \( y \). We have:
\[
x + y = 13 \quad \text{(Equation 1)}
\]
\[
x \times y = 40 \quad \text{(Equation 2)}
\]

2. From Equation 1, express \( y \) in terms of \( x \):
\[
y = 13 - x
\]

3. Substitute \( y \) into Equation 2:
\[
x \times (13 - x) = 40 \implies 13x - x^2 = 40 \implies x^2 - 13x + 40 = 0
\]

4. Solve the quadratic equation:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{13 \pm \sqrt{13^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1} = \frac{13 \pm \sqrt{169 - 160}}{2} = \frac{13 \pm \sqrt{9}}{2} = \frac{13 \pm 3}{2}
\]

5. Calculate the two solutions:
\[
x = \frac{13 + 3}{2} = 8 \quad \text{or} \quad x = \frac{13 - 3}{2} = 5
\]

6. If \( x = 8 \), then \( y = 13 - 8 = 5 \). If \( x = 5 \), then \( y = 13 - 5 = 8 \).

Answer:
\[
\boxed{8, 5}
\]

---

Final Answers:

\[
\boxed{
\begin{array}{ccc}
12, 32 & 18, 80 & 14, 45 \\
7, 28 & 6, 9 & 12, 7 \\
10, 4 & 9, 2 & 8, 5
\end{array}
}
\]