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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 #1052 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

1036

10260

1026

1016

Correct answer:

1026

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, 27 is the multiplier and 38 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 8

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

Then, we multiply 7 and 3 and add the 5 that was carried

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

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.

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

Next, we multiply 2 and 8

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

Then, we multiply 2 and 3and add the 1 that was carried

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

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

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

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Example Question #1053 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

1628

1638

1648

16380

Correct answer:

1638

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 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 3 and 6

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

Then, we multiply 3 and 2 and add the 1 that was carried

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

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

Next, we multiply 6 and 6

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

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

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

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

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

Report an Error

Example Question #1051 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

3886

3876

3866

38760

Correct answer:

3876

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 68 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 8

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

Then, we multiply 7 and 6 and add the 5 that was carried

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

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

Next, we multiply 5 and 8

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

Then, we multiply 5 and 6and add the 4 that was carried

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

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

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

Report an Error

Example Question #1052 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

693

683

703

6930

Correct answer:

693

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 21 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 1

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

Then, we multiply 3 and 2

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

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

Next, we multiply 3 and 1

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

Then, we multiply 3 and 2

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

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

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

Report an Error

Example Question #91 : How To Multiply

Solve:

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

Possible Answers:

340

330

320

3300

Correct answer:

330

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 10 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

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

Then, we multiply 3 and 1

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

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

Next, we multiply 3 and 0

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

Then, we multiply 3 and 1

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

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

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

Report an Error

Example Question #1054 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

2356

2366

23560

2346

Correct answer:

2356

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, 62 is the multiplier and 38 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 8

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

Then, we multiply 2 and 3 and add the 1 that was carried

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

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

Next, we multiply 6 and 8

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

Then, we multiply 6 and 3and add the 4 that was carried

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

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

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

Report an Error

Example Question #1055 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

3898

3878

38880

3888

Correct answer:

3888

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, 81 is the multiplier and 48 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 8

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

Then, we multiply 1 and 4

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

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

Next, we multiply 8 and 8

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

Then, we multiply 8 and 4and add the 6 that was carried

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

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

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

Report an Error

Example Question #92 : How To Multiply

Solve:

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

Possible Answers:

352

342

3520

362

Correct answer:

352

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, 32 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 2 and 1

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

Then, we multiply 2 and 1

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

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

Next, we multiply 3 and 1

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

Then, we multiply 3 and 1

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

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

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

Report an Error

Example Question #92 : How To Multiply

Solve:

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

Possible Answers:

4400

450

430

440

Correct answer:

440

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 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 1 and 0

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

Then, we multiply 1 and 4

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

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

Next, we multiply 1 and 0

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

Then, we multiply 1 and 4

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

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

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

Report an Error

Example Question #93 : How To Multiply

Solve:

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

Possible Answers:

66000

6590

6600

6610

Correct answer:

6600

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 88 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 8

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

Then, we multiply 5 and 8 and add the 4 that was carried

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

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

Next, we multiply 7 and 8

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

Then, we multiply 7 and 8and add the 5 that was carried

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

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

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

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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 : 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 #1052 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1053 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1051 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1052 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #91 : How To Multiply

Example Question #1054 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1055 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #92 : How To Multiply

Example Question #92 : How To Multiply

Example Question #93 : How To Multiply

All Common Core: 5th Grade Math Resources

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