ISEE Middle Level Quantitative Reasoning - How to multiply

It takes Hannah minutes to get dressed in the morning. It takes her older sister, Sara, times as long. How long does it take Sara to get dressed?

$16\ minutes$

$18\ minutes$

$15\ minutes$

$19\ minutes$

$17\ minutes$

Explanation

The phrase " times as long" tells you that we are going to multiply by .

Let's have represent how long it takes Sara to get dressed.

$4\times4=t$

Think: minutes 4 times would be how many minutes?

Screen shot 2015 10 06 at 9.26.54 am

$4\times4=16$

Solve:

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

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, 13 is the multiplier and 85 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

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

Then, we multiply 3 and 8 and add the 1 that was carried

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

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.

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

Next, we multiply 1 and 5

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

Then, we multiply 1 and 8

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

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

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

Solve:

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

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 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 5 and 5

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

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

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

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

Next, we multiply 8 and 5

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

Then, we multiply 8 and 3and add the 4 that was carried

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

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

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

Solve:

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

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

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

Solve:

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

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, 12 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 2 and 0

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

Then, we multiply 2 and 1

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

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

Next, we multiply 1 and 0

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

Then, we multiply 1 and 1

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

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

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

Solve:

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

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, 40 is the multiplier and 71 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

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

Then, we multiply 0 and 7

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

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

Next, we multiply 4 and 1

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

Then, we multiply 4 and 7

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

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

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

Solve:

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

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

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

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.

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

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

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

Solve:

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

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, 40 is the multiplier and 78 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

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

Then, we multiply 0 and 7

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

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

Next, we multiply 4 and 8

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

Then, we multiply 4 and 7and add the 3 that was carried

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

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

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

Solve:

$\frac{\begin{array}[b]{r}13\ \times\ 2\end{array}}{\space }$

Explanation

We can make a rectangular array using squares to help us solve this problem. Our multiplication problem is $13\times 2$ ; therefore, the rectangular array should have squares across the top row and squares down the first column. The rectangular array should look like the image below:

Screen shot 2016 05 18 at 2.15.20 pm

We can count the squares to find our answer. The correct answer is is

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.

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

For this problem, is the multiplier and is the multiplicand.

First, we multiply and

$\frac{\begin{array}[b]{r}1{\color{Red} 3}\ \times\ {\color{Red} 2}\end{array}}{\ \ \ \ 6}$

Next, we multiply and

$\frac{\begin{array}[b]{r}{\color{Red} 1}3\ \times\ {\color{Red} 2}\end{array}}{\ \ \ 26}$

Solve:

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

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

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

How to multiply

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