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ISEE Middle Level Quantitative : Whole Numbers

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

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All ISEE Middle Level Quantitative Resources

6 Diagnostic Tests 110 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #8 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}40\\ \times\ 10\end{array}}{\ \ \ \ }$

Possible Answers:

400

440

420

444

410

Correct answer:

400

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, is the multiplier and 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 and

$\frac{\begin{array}[b]{r}4{\color{Red} 0}\\ \times\ {1\color{Red} 0}\end{array}}{\ \ \ \ \ 0}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red}4}0\\ \times\ {1\color{Red} 0}\end{array}}{\ \ \ 00}$

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 as a place holder in the ones position.

$\frac{\begin{array}[b]{r}40\\ \times\ 10\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}4 {\color{Red} 0}\\ \times\ {\color{Red} 1}0\end{array}}{\ \ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ \ 00}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 4} {\color{Black} 0}\\ \times\ {\color{Red} 1}0\end{array}}{\ \ \ \ 00 }\\ {\ \ \ \ \ \ \ \ 400}$

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

$\frac{\begin{array}[b]{r}40\\ \times \ \ \ 10\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 00}\\ \ +\ {\color{Red} 400} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 400}$

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Example Question #9 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}52\\ \times\ 43\end{array}}{\ \ \ \ }$

Possible Answers:

$1\textup,328$

$2\textup,136$

$1\textup,824$

$2\textup,236$

$2\textup,080$

Correct answer:

$2\textup,236$

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, is the multiplier and 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 and

$\frac{\begin{array}[b]{r}5{\color{Red} 2}\\ \times\ {4\color{Red} 3}\end{array}}{\ \ \ \ \ 6}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 5}2\\ \times\ {4\color{Red} 3}\end{array}}{\ \ 156}$

$\frac{\begin{array}[b]{r}52\\ \times\ 43\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}5 {\color{Red} 2}\\ \times\ {\color{Red} 4}3\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ \ \ \ 80}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 5} {\color{Black} 2}\\ \times\ {\color{Red} 4}3\end{array}}{\ \ 156 }\\ {\ \ \ \ \ \ \ 2080}$

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

$\frac{\begin{array}[b]{r}52\\ \times \ \ \ 43\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 156}\\ \ +\ {\color{Red} 2080} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 2236}$

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Example Question #10 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

$\frac{\begin{array}[b]{r}56\\ \times\ 17\end{array}}{\ \ \ \ }$

Possible Answers:

808

781

752

952

874

Correct answer:

952

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, is the multiplier and 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 and

$\frac{\begin{array}[b]{r}^45{\color{Red} 6}\\ \times\ {1\color{Red} 7}\end{array}}{\ \ \ \ \ 2}$

Then, we multiply and and add the that was carried

$\frac{\begin{array}[b]{r}^{\color{Red} 4}{\color{Red} 5}6\\ \times\ {1\color{Red} 7}\end{array}}{\ \ 392}$

$\frac{\begin{array}[b]{r}5 6\\ \times\ 17\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}5 {\color{Red} 6}\\ \times\ {\color{Red} 1}7\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ \ \ 60}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 5} {\color{Black} 6}\\ \times\ {\color{Red} 1}7\end{array}}{\ \ 392 }\\ {\ \ \ \ \ \ \ \ 560}$

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

$\frac{\begin{array}[b]{r}56\\ \times \ \ \ 17\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 392}\\ \ +\ {\color{Red} 560} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 952}$

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Example Question #31 : Whole Numbers

Solve:

$\frac{\begin{array}[b]{r} 47\\ \times\ 12\end{array}}{\ \ \ \ }$

Possible Answers:

582

564

704

673

622

Correct answer:

564

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, is the multiplier and 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 and

$\frac{\begin{array}[b]{r}^14{\color{Red} 7}\\ \times\ {1\color{Red} 2}\end{array}}{\ \ \ \ \ 4}$

Then, we multiply and and add the that was carried

$\frac{\begin{array}[b]{r}^{\color{Red} 1}{\color{Red} 4}7\\ \times\ {1\color{Red} 2}\end{array}}{\ \ \ \ 94}$

$\frac{\begin{array}[b]{r}47\\ \times\ 12\end{array}}{\ \ \ 94 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}4 {\color{Red} 7}\\ \times\ {\color{Red} 1}2\end{array}}{\ \ \ 94 }\\ {\ \ \ \ \ \ \ \ \ 70}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 4} {\color{Black} 7}\\ \times\ {\color{Red} 1}2\end{array}}{\ \ \ 94 }\\ {\ \ \ \ \ \ \ \ 470}$

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

$\frac{\begin{array}[b]{r}47\\ \times \ \ \ 12\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 94}\\ \ +\ {\color{Red} 470} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 564}$

Report an Error

Example Question #532 : Number & Operations In Base Ten

Solve:

$\frac{\begin{array}[b]{r}15\\ \times\ 11\end{array}}{\ \ \ \ }$

Possible Answers:

170

160

165

150

175

Correct answer:

165

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, is the multiplier and 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 and

$\frac{\begin{array}[b]{r}1{\color{Red} 5}\\ \times\ {1\color{Red} 1}\end{array}}{\ \ \ \ \ 5}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 1}5\\ \times\ {1\color{Red} 1}\end{array}}{\ \ \ 15}$

$\frac{\begin{array}[b]{r}15\\ \times\ 11\end{array}}{\ \ \ 15 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}1 {\color{Red} 5}\\ \times\ {\color{Red} 1}1\end{array}}{\ \ \ 15 }\\ {\ \ \ \ \ \ \ \ \ 50}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 1} {\color{Black} 5}\\ \times\ {\color{Red} 1}1\end{array}}{\ \ \ 15 }\\ {\ \ \ \ \ \ \ \ 150}$

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

$\frac{\begin{array}[b]{r}15\\ \times \ \ \ 11\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 15}\\ \ +\ {\color{Red} 150} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 165}$

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Example Question #533 : Number & Operations In Base Ten

Solve:

$\frac{\begin{array}[b]{r}30\\ \times\ 20\end{array}}{\ \ \ \ }$

Possible Answers:

400

500

300

600

200

Correct answer:

600

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, is the multiplier and 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 and

$\frac{\begin{array}[b]{r}3{\color{Red} 0}\\ \times\ {2\color{Red} 0}\end{array}}{\ \ \ \ \ 0}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 3}0\\ \times\ {2\color{Red} 0}\end{array}}{\ \ \ 00}$

$\frac{\begin{array}[b]{r}30\\ \times\ 20\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ \ \ 0}$

Next, we multiply and

$\frac{\begin{array}[b]{r}3 {\color{Red} 0}\\ \times\ {\color{Red} 2}0\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ \ 00}$

Then, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 3} {\color{Black} 0}\\ \times\ {\color{Red} 2}0\end{array}}{\ \ \ 00 }\\ {\ \ \ \ \ \ \ \ 600}$

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

$\frac{\begin{array}[b]{r}30\\ \times \ \ \ 20\end{array}}{\frac{\begin{array}[b]{r}{\color{Red} 00}\\ \ +\ {\color{Red} 600} \end{array}}{\ }} \\ {\ \ \ \ \ \ \ \ \ \ \ 600}$

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Example Question #534 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

1795

1785

1775

17850

Correct answer:

1785

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 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 5 and 1

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

Then, we multiply 5 and 5

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

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.

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

Next, we multiply 3 and 1

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

Then, we multiply 3 and 5

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

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

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

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Example Question #535 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

3662

36720

3682

3672

Correct answer:

3672

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, 68 is the multiplier and 54 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 4

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

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

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

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

Next, we multiply 6 and 4

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

Then, we multiply 6 and 5and add the 2 that was carried

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

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

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

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Example Question #536 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

4472

4482

44820

4492

Correct answer:

4482

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, 83 is the multiplier and 54 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

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

Then, we multiply 3 and 5 and add the 1 that was carried

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

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

Next, we multiply 8 and 4

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

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

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

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

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

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Example Question #537 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

4454

4464

4474

44640

Correct answer:

4464

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 72 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 2

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

Then, we multiply 2 and 7

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

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

Next, we multiply 6 and 2

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

Then, we multiply 6 and 7and add the 1 that was carried

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

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

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

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All ISEE Middle Level Quantitative Resources

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ISEE Middle Level Quantitative : Whole Numbers

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

All ISEE Middle Level Quantitative Resources

Example Questions

Example Question #8 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #9 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #10 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #31 : Whole Numbers

Example Question #532 : Number & Operations In Base Ten

Example Question #533 : Number & Operations In Base Ten

Example Question #534 : Number & Operations In Base Ten

Example Question #535 : Number & Operations In Base Ten

Example Question #536 : Number & Operations In Base Ten

Example Question #537 : Number & Operations In Base Ten

All ISEE Middle Level Quantitative Resources

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