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ISEE Middle Level Quantitative : Numbers and Operations

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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

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

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

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.

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

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

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

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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

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

Report an Error

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

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

Report an Error

Example Question #111 : How To Multiply

Solve:

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

Possible Answers:

3516

3526

3536

35260

Correct answer:

3526

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, 43 is the multiplier and 82 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

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

Then, we multiply 3 and 8

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

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

Next, we multiply 4 and 2

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

Then, we multiply 4 and 8

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

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

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

Report an Error

Example Question #1077 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

2399

24090

2409

2419

Correct answer:

2409

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 73 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 3

$\begin{array}{r}\overset{~ \space{\ }}{\ 7} \space{\,}3\\ \times \phantom{0} 3\space{\,}3 \\ \hline \phantom{\,} \,9\end{array}$

Then, we multiply 3 and 7

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

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

Next, we multiply 3 and 3

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

Then, we multiply 3 and 7

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

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

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

Report an Error

Example Question #1078 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

661

681

6710

671

Correct answer:

671

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, 61 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 1 and 1

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

Then, we multiply 1 and 1

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

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

Next, we multiply 6 and 1

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

Then, we multiply 6 and 1

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

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

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

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Example Question #1079 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

3600

370

350

360

Correct answer:

360

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, 36 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 6 and 0

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

Then, we multiply 6 and 1

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

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

Next, we multiply 3 and 0

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

Then, we multiply 3 and 1

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

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

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

Report an Error

Example Question #1080 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

6385

6375

63750

6365

Correct answer:

6375

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, 85 is the multiplier and 75 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 5

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

Then, we multiply 5 and 7 and add the 2 that was carried

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

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

Next, we multiply 8 and 5

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

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

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

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

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

Report an Error

Example Question #112 : Whole Numbers

Solve:

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

Possible Answers:

470

490

480

4800

Correct answer:

480

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, 10 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 0 and 8

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

Then, we multiply 0 and 4

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

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

Next, we multiply 1 and 8

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

Then, we multiply 1 and 4

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

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

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

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All ISEE Middle Level Quantitative Resources

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

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ISEE Middle Level Quantitative : Numbers and Operations

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

All ISEE Middle Level Quantitative Resources

Example Questions

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

Example Question #111 : How To Multiply

Example Question #1077 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1078 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1079 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1080 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #112 : Whole Numbers

All ISEE Middle Level Quantitative Resources

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