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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 #126 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

2300

23100

2320

2310

Correct answer:

2310

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, 66 is the multiplier and 35 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 5

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

Then, we multiply 6 and 3 and add the 3 that was carried

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

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.

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

Next, we multiply 6 and 5

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

Then, we multiply 6 and 3and add the 3 that was carried

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

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

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

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Example Question #127 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

684

6840

674

694

Correct answer:

684

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, 57 is the multiplier and 12 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 2

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

Then, we multiply 7 and 1 and add the 1 that was carried

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

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

Next, we multiply 5 and 2

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

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

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

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

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

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Example Question #128 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}6\\ \times\phantom{0}3\space{\,}4 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

29240

2914

2934

2924

Correct answer:

2924

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, 34 is the multiplier and 86 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 6

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

Then, we multiply 4 and 8 and add the 2 that was carried

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

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}6\\ \times \phantom{0} 3\space{\,}4 \\ \hline \phantom{\,}3\,4\,4 \\ \, 0 \end{array}$

Next, we multiply 3 and 6

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

Then, we multiply 3 and 8and add the 1 that was carried

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

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

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

Report an Error

Example Question #121 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

2221

22110

2211

2201

Correct answer:

2211

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, 33 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 3 and 7

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

Then, we multiply 3 and 6 and add the 2 that was carried

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

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

Next, we multiply 3 and 7

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

Then, we multiply 3 and 6and add the 2 that was carried

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

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

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

Report an Error

Example Question #122 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

4320

43200

4330

4310

Correct answer:

4320

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, 72 is the multiplier and 60 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

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

Then, we multiply 2 and 6

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

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

Next, we multiply 7 and 0

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

Then, we multiply 7 and 6

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

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

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

Report an Error

Example Question #131 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

4077

40870

4087

4097

Correct answer:

4087

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, 67 is the multiplier and 61 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 1

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

Then, we multiply 7 and 6

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

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

Next, we multiply 6 and 1

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

Then, we multiply 6 and 6

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

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

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

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Example Question #132 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

27470

2737

2747

2757

Correct answer:

2747

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, 41 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 1 and 7

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

Then, we multiply 1 and 6

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

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

Next, we multiply 4 and 7

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

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

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

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

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

Report an Error

Example Question #133 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\begin{array}{r}\overset{~ \space{\ }}{\ 8} \space{\,}4\\ \times\phantom{0}2\space{\,}3 \\ \hline \phantom{\,}\end{array}$

Possible Answers:

19320

1942

1922

1932

Correct answer:

1932

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, 23 is the multiplier and 84 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 4

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

Then, we multiply 3 and 8 and add the 1 that was carried

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

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

Next, we multiply 2 and 4

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

Then, we multiply 2 and 8

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

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

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

Report an Error

Example Question #1 : Fluently Divide Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.6

$3828\div 22=$

Possible Answers:

132

174

503

356

Correct answer:

174

Explanation:

Report an Error

Example Question #2 : Fluently Divide Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.6

Solve:

$13{\overline{\smash{)}143}}$

Possible Answers:

Correct answer:

Explanation:

Before we begin, let's review the pieces of a division problem:

${\begin{array}[b]{r} \ \textup{ quotient}\\ \textup{ divisor}{\overline{\smash{)}\ \textup{dividend}}}\\\end{array}}{ \ \ \ \space}$

The quotient is the answer to the division problem. The dividend is the number that gets divided by the divisor.

One way to solve a division problem is to create an area model. The divisor equals the number of squares that make up the base of the area model. The dividend equals the number of total squares used. Fill the squares up from the base until you've used the correct number of squares. The height of the area model will be the quotient:

Another way to solve a division problem is to think of it as a multiplication problem. What number times the divisor equals the dividend?

$13\times$ __________ =143

$\frac{\begin{array}[b]{r}13\\ \times \ \ \ 11\end{array}}{\frac{\begin{array}[b]{r}{\color{black} 13}\\ \ +\ {\color{black} 130} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 143}$

This means that $13\times {\color{Red} 11}=143$ ; thus, ${\begin{array}[b]{r} \ \textup{ {\color{Red} 11}}\\ \textup{ 13}{\overline{\smash{)}\ \textup{143}}}\\\end{array}}{ \ \ \ \space}$

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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

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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 #126 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #127 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #128 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #121 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #122 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #131 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #132 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #133 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #1 : Fluently Divide Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.6

Example Question #2 : Fluently Divide Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.6

All Common Core: 5th Grade Math Resources

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