ISEE Middle Level Quantitative : How to multiply

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

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Example Questions

Example Question #1 : Use Place Value Understanding And Properties Of Operations To Perform Multi Digit Arithmetic

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}10\\ \times\ 2\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 23\)

\(\displaystyle 20\)

\(\displaystyle 22\)

\(\displaystyle 21\)

Correct answer:

\(\displaystyle 20\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 10\times 2\); therefore, the rectangular array should have \(\displaystyle 10\) squares across the top row and \(\displaystyle 2\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 10.32.59 am

We can count the squares to find our answer. The correct answer is is \(\displaystyle 20\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 2\) is the multiplier and \(\displaystyle 10\) is the multiplicand. 

First, we multiply \(\displaystyle 2\) and \(\displaystyle 0\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 0}\\ \times\ {\color{Red} 2}\end{array}}{\ \ \ \ 0}\)

Next, we multiply\(\displaystyle 2\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}0\\ \times\ {\color{Red} 2}\end{array}}{\ \ \ 20}\)

Example Question #965 : Isee Middle Level (Grades 7 8) Quantitative Reasoning

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}10\\ \times\ 3\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 33\)

\(\displaystyle 31\)

\(\displaystyle 30\)

\(\displaystyle 32\)

Correct answer:

\(\displaystyle 30\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 10\times 3\); therefore, the rectangular array should have \(\displaystyle 10\) squares across the top row and \(\displaystyle 3\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 11.07.48 am

We can count the squares to find our answer. The correct answer is is \(\displaystyle 30\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 3\) is the multiplier and \(\displaystyle 10\) is the multiplicand. 

First, we multiply \(\displaystyle 3\) and \(\displaystyle 0\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 0}\\ \times\ {\color{Red} 3}\end{array}}{\ \ \ \ 0}\)

Next, we multiply \(\displaystyle 3\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}0\\ \times\ {\color{Red} 3}\end{array}}{\ \ \ 30}\)

Example Question #1 : How To Multiply

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}10\\ \times\ 4\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 43\)

\(\displaystyle 42\)

\(\displaystyle 41\)

\(\displaystyle 40\)

Correct answer:

\(\displaystyle 40\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 10\times 4\); therefore, the rectangular array should have \(\displaystyle 10\) squares across the top row and \(\displaystyle 4\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 12.00.41 pm

We can count the squares to find our answer. The correct answer is is \(\displaystyle 40\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 4\) is the multiplier and \(\displaystyle 10\) is the multiplicand. 

First, we multiply \(\displaystyle 4\) and \(\displaystyle 0\)

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

Next, we multiply \(\displaystyle 4\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}0\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ 40}\)

Example Question #3 : Use Place Value Understanding And Properties Of Operations To Perform Multi Digit Arithmetic

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}11\\ \times\ 5\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 53\)

\(\displaystyle 55\)

\(\displaystyle 54\)

\(\displaystyle 52\)

Correct answer:

\(\displaystyle 55\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 11\times 5\); therefore, the rectangular array should have \(\displaystyle 11\) squares across the top row and \(\displaystyle 5\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 12.14.20 pm

We can count the squares to find our answer. The correct answer is is \(\displaystyle 55\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 5\) is the multiplier and \(\displaystyle 11\) is the multiplicand. 

First, we multiply \(\displaystyle 5\) and \(\displaystyle 1\)

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

Next, we multiply \(\displaystyle 5\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}1\\ \times\ {\color{Red} 5}\end{array}}{\ \ \ 55}\)

Example Question #1 : Whole Numbers

Solve the following:

\(\displaystyle \frac{\begin{array}[b]{r}11\\ \times\ 6\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 63\)

\(\displaystyle 65\)

\(\displaystyle 66\)

\(\displaystyle 64\)

Correct answer:

\(\displaystyle 66\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 11\times 6\); therefore, the rectangular array should have \(\displaystyle 11\) squares across the top row and \(\displaystyle 6\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 12.20.54 pm

We can count the squares to find our answer. The correct answer is is \(\displaystyle 66\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 6\) is the multiplier and \(\displaystyle 11\) is the multiplicand. 

First, we multiply \(\displaystyle 6\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 1}\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ \ 6}\)

Next, we multiply \(\displaystyle 6\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}1\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ 66}\)

Example Question #4 : Use Place Value Understanding And Properties Of Operations To Perform Multi Digit Arithmetic

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}11\\ \times\ 4\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 44\)

\(\displaystyle 43\)

\(\displaystyle 46\)

\(\displaystyle 45\)

Correct answer:

\(\displaystyle 44\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 11\times 4\); therefore, the rectangular array should have \(\displaystyle 11\) squares across the top row and \(\displaystyle 4\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 12.31.20 pm

We can count the squares to find our answer. The correct answer is is \(\displaystyle 44\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 4\) is the multiplier and \(\displaystyle 11\) is the multiplicand. 

First, we multiply \(\displaystyle 4\) and \(\displaystyle 1\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 1}\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ \ 4}\)

Next, we multiply \(\displaystyle 4\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}1\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ 44}\)

Example Question #5 : Use Place Value Understanding And Properties Of Operations To Perform Multi Digit Arithmetic

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}12\\ \times\ 4\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 50\)

\(\displaystyle 48\)

\(\displaystyle 49\)

\(\displaystyle 47\)

Correct answer:

\(\displaystyle 48\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 12\times 4\); therefore, the rectangular array should have \(\displaystyle 12\) squares across the top row and \(\displaystyle 4\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 12.36.48 pm

We can count the squares to find our answer. The correct answer is is \(\displaystyle 48\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 4\) is the multiplier and \(\displaystyle 12\) is the multiplicand. 

First, we multiply \(\displaystyle 4\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}1{\color{Red} 2}\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ \ 8}\)

Next, we multiply \(\displaystyle 4\) and \(\displaystyle 1\)

 \(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 1}2\\ \times\ {\color{Red} 4}\end{array}}{\ \ \ 48}\)

Example Question #7 : Multiply Multi Digit Numbers: Ccss.Math.Content.4.Nbt.B.5

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}12\\ \times\ 5\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 61\)

\(\displaystyle 60\)

\(\displaystyle 63\)

\(\displaystyle 62\)

Correct answer:

\(\displaystyle 60\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 12\times 5\); therefore, the rectangular array should have \(\displaystyle 12\) squares across the top row and \(\displaystyle 5\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 12.45.25 pm

We can count the squares to find our answer. The correct answer is is \(\displaystyle 60\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 5\) is the multiplier and \(\displaystyle 12\) is the multiplicand. 

First, we multiply \(\displaystyle 5\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}^11{\color{Red} 2}\\ \times\ {\color{Red} 5}\end{array}}{\ \ \ \ 0}\)

Because the product is greater than \(\displaystyle 9\), we carry the \(\displaystyle 1\) and add it to the next product. 

Next, we multiply \(\displaystyle 5\) and \(\displaystyle 1\) and then add the \(\displaystyle 1\) we carried. 

 \(\displaystyle \frac{\begin{array}[b]{r}^1{\color{Red} 1}2\\ \times\ {\color{Red} 5}\end{array}}{\ \ \ 60}\)

Example Question #6 : Use Place Value Understanding And Properties Of Operations To Perform Multi Digit Arithmetic

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}12\\ \times\ 6\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 72\)

\(\displaystyle 71\)

\(\displaystyle 73\)

\(\displaystyle 74\)

Correct answer:

\(\displaystyle 72\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 12\times 6\); therefore, the rectangular array should have \(\displaystyle 12\) squares across the top row and \(\displaystyle 6\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 12.53.41 pm

We can count the squares to find our answer. The correct answer is is \(\displaystyle 72\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 6\) is the multiplier and \(\displaystyle 12\) is the multiplicand. 

First, we multiply \(\displaystyle 6\) and \(\displaystyle 2\)

\(\displaystyle \frac{\begin{array}[b]{r}^11{\color{Red} 2}\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ \ 2}\)

Because the product is greater than \(\displaystyle 9\), we carry the \(\displaystyle 1\) and add it to the next product. 

Next, we multiply \(\displaystyle 6\) and \(\displaystyle 1\) and then add the \(\displaystyle 1\) we carried. 

 \(\displaystyle \frac{\begin{array}[b]{r}^1{\color{Red} 1}2\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ 72}\)

Example Question #7 : Use Place Value Understanding And Properties Of Operations To Perform Multi Digit Arithmetic

Solve:

\(\displaystyle \frac{\begin{array}[b]{r}13\\ \times\ 6\end{array}}{\space }\)

Possible Answers:

\(\displaystyle 79\)

\(\displaystyle 77\)

\(\displaystyle 76\)

\(\displaystyle 78\)

Correct answer:

\(\displaystyle 78\)

Explanation:

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is \(\displaystyle 13\times 6\); therefore, the rectangular array should have \(\displaystyle 13\) squares across the top row and \(\displaystyle 6\) squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 1.11.14 pm

We can count the squares to find our answer. The correct answer is is \(\displaystyle 78\)

In order to solve this problem using multiplication, we must multiply the multiplier by each digit of the multiplicand to get the answer, which is called the product.

\(\displaystyle \frac{\begin{array}[b]{r}\textup{multiplicand}\\ \times\ \ \ \ \textup{multiplier}\end{array}}{\ \ \ \ \textup{product}}\)

For this problem, \(\displaystyle 6\) is the multiplier and \(\displaystyle 13\) is the multiplicand. 

First, we multiply \(\displaystyle 6\) and \(\displaystyle 3\)

\(\displaystyle \frac{\begin{array}[b]{r}^11{\color{Red} 3}\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ \ 8}\)

Because the product is greater than \(\displaystyle 9\), we carry the \(\displaystyle 1\) and add it to the next product. 

Next, we multiply \(\displaystyle 6\) and \(\displaystyle 1\) and then add the \(\displaystyle 1\) we carried. 

 \(\displaystyle \frac{\begin{array}[b]{r}^1{\color{Red} 1}3\\ \times\ {\color{Red} 6}\end{array}}{\ \ \ 78}\)

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