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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

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

Example Questions

Example Question #102 : How To Multiply

Solve:

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

Possible Answers:

3472

34720

3462

3482

Correct answer:

3472

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 62 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 2

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

Then, we multiply 6 and 6 and add the 1 that was carried

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

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.

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

Next, we multiply 5 and 2

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

Then, we multiply 5 and 6and add the 1 that was carried

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

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

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

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Example Question #601 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

2110

2090

2100

21000

Correct answer:

2100

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, 35 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 5 and 0

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

Then, we multiply 5 and 6

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

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

Next, we multiply 3 and 0

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

Then, we multiply 3 and 6

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

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

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

Report an Error

Example Question #781 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

8250

835

825

815

Correct answer:

825

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, 75 is the multiplier and 11 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 1

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

Then, we multiply 5 and 1

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

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

Next, we multiply 7 and 1

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

Then, we multiply 7 and 1

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

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

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

Report an Error

Example Question #1069 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

6090

6070

6080

60800

Correct answer:

6080

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, 80 is the multiplier and 76 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 6

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

Then, we multiply 0 and 7

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

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

Next, we multiply 8 and 6

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

Then, we multiply 8 and 7and add the 4 that was carried

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

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

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

Report an Error

Example Question #1070 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

46110

4601

4611

4621

Correct answer:

4611

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, 53 is the multiplier and 87 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

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

Then, we multiply 3 and 8 and add the 2 that was carried

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

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

Next, we multiply 5 and 7

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

Then, we multiply 5 and 8and add the 3 that was carried

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

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

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

Report an Error

Example Question #1071 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

3366

3376

33660

3356

Correct answer:

3366

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 51 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 1

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

Then, we multiply 6 and 5

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

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

Next, we multiply 6 and 1

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

Then, we multiply 6 and 5

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

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

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

Report an Error

Example Question #1072 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

10080

1008

1018

998

Correct answer:

1008

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 18 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 8

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

Then, we multiply 6 and 1 and add the 4 that was carried

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

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

Next, we multiply 5 and 8

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

Then, we multiply 5 and 1and add the 4 that was carried

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

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

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

Report an Error

Example Question #1073 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

1232

1212

12220

1222

Correct answer:

1222

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, 47 is the multiplier and 26 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 6

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

Then, we multiply 7 and 2 and add the 4 that was carried

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

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

Next, we multiply 4 and 6

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

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

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

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

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

Report an Error

Example Question #81 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1270

12600

1260

1250

Correct answer:

1260

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, 15 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 5 and 4

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

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

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

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

Next, we multiply 1 and 4

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

Then, we multiply 1 and 8

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

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

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

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Example Question #1075 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

386

3960

396

406

Correct answer:

396

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, 11 is the multiplier and 36 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 6

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

Then, we multiply 1 and 3

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

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

Next, we multiply 1 and 6

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

Then, we multiply 1 and 3

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

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

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

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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 : 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 #102 : How To Multiply

Example Question #601 : Number & Operations In Base Ten

Example Question #781 : Common Core Math: Grade 5

Example Question #1069 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1070 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1071 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1072 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1073 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #81 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #1075 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

All Common Core: 5th Grade Math Resources

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