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Common Core: 5th Grade Math : Number & Operations in Base Ten

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

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All Common Core: 5th Grade Math Resources

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Example Question #651 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

2201

2211

2221

22110

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

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

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.

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

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

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

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Example Question #652 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

4330

43200

4310

4320

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

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

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Example Question #653 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

40870

4097

4077

4087

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 #651 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

2747

2757

2737

27470

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 #655 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

1942

19320

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

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

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

$3828\div 22=$

Possible Answers:

503

356

174

132

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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Example Question #3 : Fluently Divide Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.6

Solve:

$21{\overline{\smash{)}357}}$

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.

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

$21\times$ __________ =357

$\frac{\begin{array}[b]{r}21\\ \times \ \ \ 17\end{array}}{\frac{\begin{array}[b]{r}{\color{black} 147}\\ \ +\ {\color{black} 210} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 357}$

This means that $21\times {\color{Red} 17}=357$ ; thus, ${\begin{array}[b]{r} \ \textup{ {\color{Red} 17}}\\ \textup{ 21}{\overline{\smash{)}\ \textup{357}}}\\\end{array}}{ \ \ \ \space}$

Report an Error

Example Question #4 : Fluently Divide Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.6

Solve:

$12{\overline{\smash{)}432}}$

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.

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

$12\times$ __________ =432

$\frac{\begin{array}[b]{r}36\\ \times \ \ \ 12\end{array}}{\frac{\begin{array}[b]{r}{\color{black} 72}\\ \ +\ {\color{black} 360} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 432}$

This means that $12\times {\color{Red} 36}=432$ ; thus, ${\begin{array}[b]{r} \ \textup{ {\color{Red} 36}}\\ \textup{ 12}{\overline{\smash{)}\ \textup{432}}}\\\end{array}}{ \ \ \ \space}$

Report an Error

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

Solve:

$33{\overline{\smash{)}759}}$

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.

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

$33\times$ __________ =759

$\frac{\begin{array}[b]{r}33\\ \times \ \ \ 23\end{array}}{\frac{\begin{array}[b]{r}{\color{black} 99}\\ \ +\ {\color{black} 660} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 759}$

This means that $33\times {\color{Red} 23}=759$ ; thus, ${\begin{array}[b]{r} \ \textup{ {\color{Red} 23}}\\ \textup{ 33}{\overline{\smash{)}\ \textup{759}}}\\\end{array}}{ \ \ \ \space}$

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Common Core: 5th Grade Math : Number & Operations in Base Ten

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

All Common Core: 5th Grade Math Resources

Example Questions

Example Question #651 : Number & Operations In Base Ten

Example Question #652 : Number & Operations In Base Ten

Example Question #653 : Number & Operations In Base Ten

Example Question #651 : Number & Operations In Base Ten

Example Question #655 : Number & Operations In Base Ten

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

Example Question #3 : Fluently Divide Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.6

Example Question #4 : Fluently Divide Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.6

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

All Common Core: 5th Grade Math Resources

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