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ISEE Middle Level Quantitative : Numbers and Operations

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 #131 : Whole Numbers

Solve:

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

Possible Answers:

1410

14000

1400

1390

Correct answer:

1400

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, 50 is the multiplier and 28 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

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

Then, we multiply 0 and 2

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

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.

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

Next, we multiply 5 and 8

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

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

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

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

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

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Example Question #821 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

972

9720

982

962

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, 54 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 4 and 8

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

Then, we multiply 4 and 1 and add the 3 that was carried

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

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

Next, we multiply 5 and 8

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

Then, we multiply 5 and 1and add the 4 that was carried

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

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

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

Report an Error

Example Question #822 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

10560

1066

1056

1046

Correct answer:

1056

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, 66 is the multiplier and 16 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 6

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

Then, we multiply 6 and 1 and add the 3 that was carried

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

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

Next, we multiply 6 and 6

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

Then, we multiply 6 and 1and add the 3 that was carried

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

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

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

Report an Error

Example Question #823 : Common Core Math: Grade 5

Solve:

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

Possible Answers:

9360

946

936

926

Correct answer:

936

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 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 6 and 6

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

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

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

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

Next, we multiply 3 and 6

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

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

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

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

$\begin{array}{r}\overset{1 \space{\ }}{\ 2} \space{\,}6\\ \times \phantom{0} 3\space{\,}6 \\ \hline \phantom{\,}1\,5\,6 \\+\, 7 \, 8 \, 0\\ \hline 9\, 3 \, 6\end{array}$

Report an Error

Example Question #641 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

4085

4105

4095

40950

Correct answer:

4095

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 65 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

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

Then, we multiply 3 and 6 and add the 1 that was carried

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

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

Next, we multiply 6 and 5

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

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

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

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

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

Report an Error

Example Question #642 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

410

420

4200

430

Correct answer:

420

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, 20 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 0 and 1

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

Then, we multiply 0 and 2

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

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

Next, we multiply 2 and 1

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

Then, we multiply 2 and 2

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

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

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

Report an Error

Example Question #121 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

3722

3702

3712

37120

Correct answer:

3712

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, 58 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 8 and 4

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

Then, we multiply 8 and 6 and add the 3 that was carried

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

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

Next, we multiply 5 and 4

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

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

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

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

$\begin{array}{r}\overset{2 \space{\ }}{\ 6} \space{\,}4\\ \times \phantom{0} 5\space{\,}8 \\ \hline \phantom{\,}5\,1\,2 \\+\, 3 \, 2 \, 0 \, 0\\ \hline 3\, 7 \, 1\, 2\end{array}$

Report an Error

Example Question #641 : Number & Operations In Base Ten

Solve:

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

Possible Answers:

1785

1775

1795

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, 51 is the multiplier and 35 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 5

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

Then, we multiply 1 and 3

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

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

Next, we multiply 5 and 5

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

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

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

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

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

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Example Question #123 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

1726

17160

1716

1706

Correct answer:

1716

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 52 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

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

Then, we multiply 3 and 5

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

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

Next, we multiply 3 and 2

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

Then, we multiply 3 and 5

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

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

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

Report an Error

Example Question #124 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Solve:

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

Possible Answers:

21000

2110

2100

2090

Correct answer:

2100

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, 25 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

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

Then, we multiply 5 and 8 and add the 2 that was carried

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

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

Next, we multiply 2 and 4

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

Then, we multiply 2 and 8

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

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

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

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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 #131 : Whole Numbers

Example Question #821 : Common Core Math: Grade 5

Example Question #822 : Common Core Math: Grade 5

Example Question #823 : Common Core Math: Grade 5

Example Question #641 : Number & Operations In Base Ten

Example Question #642 : Number & Operations In Base Ten

Example Question #121 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #641 : Number & Operations In Base Ten

Example Question #123 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

Example Question #124 : Fluently Multiply Multi Digit Whole Numbers: Ccss.Math.Content.5.Nbt.B.5

All ISEE Middle Level Quantitative Resources

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