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Common Core: 5th Grade Math : Common Core Math: Grade 5

Study concepts, example questions & explanations for Common Core: 5th Grade Math

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All Common Core: 5th Grade Math Resources

7 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #372 : Numbers And Operations

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 6} \space{\,}4\\ \times\phantom{0}5\space{\,}5 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

35200

3530

3520

3510

Correct answer:

3520

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 55 is the multiplier and 64 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 5 and 4

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}4\\ \times \phantom{0} 5\space{\,}5\\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 5 and 6 and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}4\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}3\,2\,0\end{array}$

Now, we multiply the digit in the tens place of the multiplier by each digit of the multiplicand. In order to do this, we need to start a new line in our math problem and put a 0 as a place holder in the ones position.

$\begin{array}{r}\overset{~ \space{\ }}{\ 6} \space{\,}4\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}3\,2\,0 \\ \, 0 \end{array}$

Next, we multiply 5 and 4

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}4\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}3\,2\,0 \\ \, 0 \, 0\end{array}$

Then, we multiply 5 and 6and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}4\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}3\,2\,0 \\ \, 3 \, 2 \, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}4\\ \times \phantom{0} 5\space{\,}5 \\ \hline \phantom{\,}3\,2\,0 \\+\, 3 \, 2 \, 0 \, 0\\ \hline 3\, 5 \, 2\, 0\end{array}$

Report an Error

Example Question #373 : Numbers And Operations

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}5\\ \times\phantom{0}5\space{\,}2 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

4420

44200

4430

4410

Correct answer:

4420

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 52 is the multiplier and 85 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 2 and 5

$\begin{array}{r}\overset{1 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 5\space{\,}2\\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 2 and 8 and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,7\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,7\,0 \\ \, 0 \end{array}$

Next, we multiply 5 and 5

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,7\,0 \\ \, 5 \, 0\end{array}$

Then, we multiply 5 and 8and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,7\,0 \\ \, 4 \, 2 \, 5 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{2 \space{\ }}{\ 8} \space{\,}5\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,7\,0 \\+\, 4 \, 2 \, 5 \, 0\\ \hline 4\, 4 \, 2\, 0\end{array}$

Report an Error

Example Question #562 : Number & Operations In Base Ten

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times\phantom{0}4\space{\,}0 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

1690

16800

1670

1680

Correct answer:

1680

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 40 is the multiplier and 42 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 0 and 2

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 0 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 0 \end{array}$

Next, we multiply 4 and 2

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 8 \, 0\end{array}$

Then, we multiply 4 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\ \, 1\, 6\, 8 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 4\space{\,}0 \\ \hline \phantom{\,} \,0\,0 \\+\, 1\, 6\, 8 \, 0\\ \hline 1\, 6 \, 8\, 0\end{array}$

Report an Error

Example Question #41 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 6} \space{\,}7\\ \times\phantom{0}4\space{\,}4 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

2938

2958

29480

2948

Correct answer:

2948

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 44 is the multiplier and 67 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 4 and 7

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}7\\ \times \phantom{0} 4\space{\,}4\\ \hline \phantom{\,} \,8\end{array}$

Then, we multiply 4 and 6 and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}7\\ \times \phantom{0} 4\space{\,}4 \\ \hline \phantom{\,}2\,6\,8\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 6} \space{\,}7\\ \times \phantom{0} 4\space{\,}4 \\ \hline \phantom{\,}2\,6\,8 \\ \, 0 \end{array}$

Next, we multiply 4 and 7

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}7\\ \times \phantom{0} 4\space{\,}4 \\ \hline \phantom{\,}2\,6\,8 \\ \, 8 \, 0\end{array}$

Then, we multiply 4 and 6and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}7\\ \times \phantom{0} 4\space{\,}4 \\ \hline \phantom{\,}2\,6\,8 \\ \, 2 \, 6 \, 8 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}7\\ \times \phantom{0} 4\space{\,}4 \\ \hline \phantom{\,}2\,6\,8 \\+\, 2 \, 6 \, 8 \, 0\\ \hline 2\, 9 \, 4\, 8\end{array}$

Report an Error

Example Question #42 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}0\\ \times\phantom{0}7\space{\,}7 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

5400

5380

5390

53900

Correct answer:

5390

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 77 is the multiplier and 70 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 7 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}0\\ \times \phantom{0} 7\space{\,}7 \\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 7 and 7

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}0\\ \times \phantom{0} 7\space{\,}7 \\ \hline \phantom{\,}4\,9\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}0\\ \times \phantom{0} 7\space{\,}7 \\ \hline \phantom{\,}4\,9\,0 \\ \, 0 \end{array}$

Next, we multiply 7 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}0\\ \times \phantom{0} 7\space{\,}7 \\ \hline \phantom{\,}4\,9\,0 \\ \, 0 \, 0\end{array}$

Then, we multiply 7 and 7

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}0\\ \times \phantom{0} 7\space{\,}7 \\ \hline \phantom{\,}4\,9\,0 \\ \, 4\, 9\, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}0\\ \times \phantom{0} 7\space{\,}7 \\ \hline \phantom{\,}4\,9\,0 \\+\, 4\, 9\, 0 \, 0\\ \hline 5\, 3 \, 9\, 0\end{array}$

Report an Error

Example Question #51 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}0\\ \times\phantom{0}6\space{\,}3 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

25200

2530

2510

2520

Correct answer:

2520

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 63 is the multiplier and 40 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 3 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}0\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 3 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}0\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,2\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}0\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,2\,0 \\ \, 0 \end{array}$

Next, we multiply 6 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}0\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,2\,0 \\ \, 0 \, 0\end{array}$

Then, we multiply 6 and 4

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}0\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,2\,0 \\ \, 2\, 4\, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 4} \space{\,}0\\ \times \phantom{0} 6\space{\,}3 \\ \hline \phantom{\,}1\,2\,0 \\+\, 2\, 4\, 0 \, 0\\ \hline 2\, 5 \, 2\, 0\end{array}$

Report an Error

Example Question #52 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}0\\ \times\phantom{0}5\space{\,}2 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

26000

2590

2600

2610

Correct answer:

2600

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 52 is the multiplier and 50 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 2 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}0\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 2 and 5

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}0\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,0\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}0\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,0\,0 \\ \, 0 \end{array}$

Next, we multiply 5 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}0\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,0\,0 \\ \, 0 \, 0\end{array}$

Then, we multiply 5 and 5

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}0\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,0\,0 \\ \, 2\, 5\, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}0\\ \times \phantom{0} 5\space{\,}2 \\ \hline \phantom{\,}1\,0\,0 \\+\, 2\, 5\, 0 \, 0\\ \hline 2\, 6 \, 0\, 0\end{array}$

Report an Error

Example Question #52 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}6\\ \times\phantom{0}7\space{\,}3 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

4098

4088

40880

4078

Correct answer:

4088

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 73 is the multiplier and 56 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 3 and 6

$\begin{array}{r}\overset{1 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 7\space{\,}3\\ \hline \phantom{\,} \,8\end{array}$

Then, we multiply 3 and 5 and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 7\space{\,}3 \\ \hline \phantom{\,}1\,6\,8\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 7\space{\,}3 \\ \hline \phantom{\,}1\,6\,8 \\ \, 0 \end{array}$

Next, we multiply 7 and 6

$\begin{array}{r}\overset{4 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 7\space{\,}3 \\ \hline \phantom{\,}1\,6\,8 \\ \, 2 \, 0\end{array}$

Then, we multiply 7 and 5and add the 4 that was carried

$\begin{array}{r}\overset{4 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 7\space{\,}3 \\ \hline \phantom{\,}1\,6\,8 \\ \, 3 \, 9 \, 2 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{4 \space{\ }}{\ 5} \space{\,}6\\ \times \phantom{0} 7\space{\,}3 \\ \hline \phantom{\,}1\,6\,8 \\+\, 3 \, 9 \, 2 \, 0\\ \hline 4\, 0 \, 8\, 8\end{array}$

Report an Error

Example Question #54 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}0\\ \times\phantom{0}5\space{\,}6 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

1690

16800

1670

1680

Correct answer:

1680

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 56 is the multiplier and 30 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 6 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}0\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,} \,0\end{array}$

Then, we multiply 6 and 3

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}0\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}1\,8\,0\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}0\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}1\,8\,0 \\ \, 0 \end{array}$

Next, we multiply 5 and 0

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}0\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}1\,8\,0 \\ \, 0 \, 0\end{array}$

Then, we multiply 5 and 3

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}0\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}1\,8\,0 \\ \, 1\, 5\, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{~ \space{\ }}{\ 3} \space{\,}0\\ \times \phantom{0} 5\space{\,}6 \\ \hline \phantom{\,}1\,8\,0 \\+\, 1\, 5\, 0 \, 0\\ \hline 1\, 6 \, 8\, 0\end{array}$

Report an Error

Example Question #381 : Numbers And Operations

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{\,}5\\ \times\phantom{0}4\space{\,}5 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

685

6750

665

675

Correct answer:

675

Explanation:

In order to solve this problem using multiplication, we must multiply each digit of the multiplier by each digit of the multiplicand to get the answer, which is called the product.

$\frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}$

For this problem, 45 is the multiplier and 15 is the multiplicand. We start with the multiplier, and multiply the digit in the ones place by each digit of the multiplicand.

First, we multiply 5 and 5

$\begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5\\ \hline \phantom{\,} \,5\end{array}$

Then, we multiply 5 and 1 and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5\end{array}$

$\begin{array}{r}\overset{~ \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5 \\ \, 0 \end{array}$

Next, we multiply 4 and 5

$\begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5 \\ \, 0 \, 0\end{array}$

Then, we multiply 4 and 1and add the 2 that was carried

$\begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5 \\ \, 6 \, 0 \, 0\end{array}$

Finally, we add the two products together to find our final answer

$\begin{array}{r}\overset{2 \space{\ }}{\ 1} \space{\,}5\\ \times \phantom{0} 4\space{\,}5 \\ \hline \phantom{\,} \,7\,5 \\+\, 6 \, 0 \, 0\\ \hline 6\, 7 \, 5\end{array}$

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Common Core: 5th Grade Math : Common Core Math: Grade 5

Study concepts, example questions & explanations for Common Core: 5th Grade Math

All Common Core: 5th Grade Math Resources

Example Questions

Example Question #372 : Numbers And Operations

Example Question #373 : Numbers And Operations

Example Question #562 : Number & Operations In Base Ten

Example Question #41 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #42 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #51 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #52 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #52 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #54 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #381 : Numbers And Operations

All Common Core: 5th Grade Math Resources

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