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ISEE Middle Level Quantitative : ISEE Middle Level (grades 7-8) Quantitative Reasoning

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

Example Question #1001 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

1785

17850

1775

1795

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 #1002 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

3672

3662

36720

3682

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

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

Report an Error

Example Question #1003 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

4472

44820

4482

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

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

Report an Error

Example Question #1004 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

4464

4454

44640

4474

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

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

Report an Error

Example Question #1005 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

3082

3092

30820

3072

Correct answer:

3082

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, 67 is the multiplier and 46 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

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

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

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

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

Next, we multiply 6 and 6

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

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

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

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

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

Report an Error

Example Question #1006 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

32130

3203

3213

3223

Correct answer:

3213

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, 51 is the multiplier and 63 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 3

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

Then, we multiply 1 and 6

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

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

Next, we multiply 5 and 3

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

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

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

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

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

Report an Error

Example Question #1007 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

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

Possible Answers:

32760

3276

3286

3266

Correct answer:

3276

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, 78 is the multiplier and 42 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 2

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

Then, we multiply 8 and 4 and add the 1 that was carried

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

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

Next, we multiply 7 and 2

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

Then, we multiply 7 and 4and add the 1 that was carried

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 7\space{\,}8 \\ \hline \phantom{\,}3\,3\,6 \\ \, 2 \, 9 \, 4 \, 0\end{array}$

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

$\begin{array}{r}\overset{1 \space{\ }}{\ 4} \space{\,}2\\ \times \phantom{0} 7\space{\,}8 \\ \hline \phantom{\,}3\,3\,6 \\+\, 2 \, 9 \, 4 \, 0\\ \hline 3\, 2 \, 7\, 6\end{array}$

Report an Error

Example Question #21 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

10640

1074

1054

1064

Correct answer:

1064

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, 14 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 4 and 6

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

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

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

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

Next, we multiply 1 and 6

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

Then, we multiply 1 and 7

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

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

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

Report an Error

Example Question #22 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1914

19140

1924

1904

Correct answer:

1914

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 58 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 8

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

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

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

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

Next, we multiply 3 and 8

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

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

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

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

$\begin{array}{r}\overset{2 \space{\ }}{\ 5} \space{\,}8\\ \times \phantom{0} 3\space{\,}3 \\ \hline \phantom{\,}1\,7\,4 \\+\, 1 \, 7 \, 4 \, 0\\ \hline 1\, 9 \, 1\, 4\end{array}$

Report an Error

Example Question #23 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

5465

54750

5475

5485

Correct answer:

5475

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, 73 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 3 and 5

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

Then, we multiply 3 and 7 and add the 1 that was carried

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

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

Next, we multiply 7 and 5

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

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

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

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

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

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All ISEE Middle Level Quantitative Resources

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ISEE Middle Level Quantitative : ISEE Middle Level (grades 7-8) Quantitative Reasoning

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

All ISEE Middle Level Quantitative Resources

Example Questions

Example Question #1001 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1002 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1003 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1004 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1005 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1006 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #1007 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Example Question #21 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #22 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #23 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

All ISEE Middle Level Quantitative Resources

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