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ISEE Middle Level Quantitative : How to multiply

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All ISEE Middle Level Quantitative Resources

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Example Questions

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Example Question #743 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

7472

7492

74820

7482

Correct answer:

7482

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, 87 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 7 and 6

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

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

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

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.

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

Next, we multiply 8 and 6

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

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

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

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

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

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Example Question #744 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

20800

2070

2080

2090

Correct answer:

2080

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 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 2 and 0

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

Then, we multiply 2 and 4

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

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

Next, we multiply 5 and 0

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

Then, we multiply 5 and 4

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

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

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

Report an Error

Example Question #745 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

426

416

4160

406

Correct answer:

416

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, 13 is the multiplier and 32 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 2

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

Then, we multiply 3 and 3

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

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

Next, we multiply 1 and 2

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

Then, we multiply 1 and 3

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

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

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

Report an Error

Example Question #746 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

2284

2304

22940

2294

Correct answer:

2294

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 37 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 7

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

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

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

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

Next, we multiply 6 and 7

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

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

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

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

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

Report an Error

Example Question #747 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

24640

2464

2454

2474

Correct answer:

2464

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, 28 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 8 and 8

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

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

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

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

Next, we multiply 2 and 8

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

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

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

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

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

Report an Error

Example Question #748 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

2420

2430

24300

2440

Correct answer:

2430

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, 54 is the multiplier and 45 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 5

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

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

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

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

Next, we multiply 5 and 5

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

Then, we multiply 5 and 4and add the 2 that was carried

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

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

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

Report an Error

Example Question #749 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

972

9720

962

982

Correct answer:

972

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, 12 is the multiplier and 81 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

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

Then, we multiply 2 and 8

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

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

Next, we multiply 1 and 1

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

Then, we multiply 1 and 8

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

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

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

Report an Error

Example Question #750 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

19040

1894

1904

1914

Correct answer:

1904

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 34 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 4

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

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

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

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

Next, we multiply 5 and 4

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

Then, we multiply 5 and 3and add the 2 that was carried

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

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

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

Report an Error

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

Solve:

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

Possible Answers:

35200

3530

3510

3520

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

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

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

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

Solve:

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

Possible Answers:

4420

4430

4410

44200

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

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

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All ISEE Middle Level Quantitative Resources

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ISEE Middle Level Quantitative : How to multiply

Study concepts, example questions & explanations for ISEE Middle Level Quantitative

All ISEE Middle Level Quantitative Resources

Example Questions

Example Question #743 : Common Core Math: Grade 5

Example Question #744 : Common Core Math: Grade 5

Example Question #745 : Common Core Math: Grade 5

Example Question #746 : Common Core Math: Grade 5

Example Question #747 : Common Core Math: Grade 5

Example Question #748 : Common Core Math: Grade 5

Example Question #749 : Common Core Math: Grade 5

Example Question #750 : Common Core Math: Grade 5

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

All ISEE Middle Level Quantitative Resources

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