Add and Subtract Within 100: CCSS.Math.Content.2.NBT.B.5 - MAP 2nd Grade Math

Card 0 of 192

Question

Solve:

$\frac{\begin{array}[b]{r}47\ +\ 15\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:

Next, we add the numbers in the tens position, including the that was carried over:

Let's add the digits in the tens place:

Your final answer should be

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Question

Solve:

$\frac{\begin{array}[b]{r}31\ +\ 95\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Next, we add the numbers in the tens position.

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}59\ +\ 39\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:

Next, we add the numbers in the tens position, including the that was carried over:

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}42\ +\ 57\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Next, we add the numbers in the tens position.

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}77\ +\ 15\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:

Next, we add the numbers in the tens position, including the that was carried over:

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}81\ +\ 58\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Next, we add the numbers in the tens position.

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}58\ +\ 19\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:

Next, we add the numbers in the tens position, including the that was carried over:

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}38\ +\ 43\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:

Next, we add the numbers in the tens position, including the that was carried over:

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}48\ +\ 31\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Next, we add the numbers in the tens position.

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}28\ +\ 63\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:

Next, we add the numbers in the tens position, including the that was carried over:

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}41\ +\ 43\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Next, we add the numbers in the tens position.

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}39\ +\ 17\end{array}}{\space }$

Answer

When we add with multi-digit numbers, we start with the numbers in the ones place, and then move to the left.

Let's add the digits in the ones place:

Because the sum is greater than , we have to carry the to the tens place. So far, your work should look like the following:

Next, we add the numbers in the tens position, including the that was carried over:

Let's add the digits in the tens place:

Your final answer should be

Compare your answer with the correct one above

Question

Solve the following:

$\frac{\begin{array}[b]{r}32\ -\ 19\end{array}}{\space }$

Answer

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.

Let's look at the numbers in the ones place:

$\frac{\begin{array}[b]{r}2\ -\ 9\end{array}}{\space }$

When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:

$\frac{\begin{array}[b]{r}^{2} ot32\ -\ 19\end{array}}{\space }$

Remember, we've borrowed from the tens place, and one ten is equal to . We now add that to the digit in the ones place, . Your work should look something like this:

$\frac{\begin{array}[b]{r}^{2} ot3\ ^{12} ot2\ -\ 1 \ \ \ \ \ 9\end{array}}{\space }$

Now, we can subtract the numbers in the once place:

$\frac{\begin{array}[b]{r}12\ -\ 9\end{array}}{\ \ \ 3}$

Next, we can subtract the numbers in the tens place:

$\frac{\begin{array}[b]{r}2\ -\ 1\end{array}}{\ \ \ 1}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{2} ot3\ ^{12} ot2\ -\ 1 \ \ \ \ \ 9\end{array}}{\ \ \ 1\ \ \ \ \ 3}$

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}51\ -\ 36\end{array}}{\space }$

Answer

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.

Let's look at the numbers in the ones place:

$\frac{\begin{array}[b]{r}1\ -\ 6\end{array}}{\space }$

When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:

$\frac{\begin{array}[b]{r}^{4} ot51\ -\ 36\end{array}}{\space }$

Remember, we've borrowed from the tens place, and one ten is equal to . We now add that to the digit in the ones place, . Your work should look something like this:

$\frac{\begin{array}[b]{r}^{4} ot5\ ^{11} ot1\ -\ 3 \ \ \ \ \ 6\end{array}}{\space }$

Now, we can subtract the numbers in the once place:

$\frac{\begin{array}[b]{r}11\ -\ 6\end{array}}{\ \ \ 5}$

Next, we can subtract the numbers in the tens place:

$\frac{\begin{array}[b]{r}4\ -\ 3\end{array}}{\ \ \ 1}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{4} ot5\ ^{11} ot1\ -\ 3 \ \ \ \ \ 6\end{array}}{\ \ \ 1\ \ \ \ \ \ 5}$

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}83\ -\ 18\end{array}}{\space }$

Answer

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.

Let's look at the numbers in the ones place:

$\frac{\begin{array}[b]{r}3\ -\ 8\end{array}}{\space }$

When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:

$\frac{\begin{array}[b]{r}^{7} ot83\ -\ 18\end{array}}{\space }$

Remember, we've borrowed from the tens place, and one ten is equal to . We now add that to the digit in the ones place, . Your work should look something like this:

$\frac{\begin{array}[b]{r}^{7} ot8\ ^{13} ot3\ -\ 1 \ \ \ \ \ 8\end{array}}{\space }$

Now, we can subtract the numbers in the once place:

$\frac{\begin{array}[b]{r}13\ -\ 8\end{array}}{\ \ \ 5}$

Next, we can subtract the numbers in the tens place:

$\frac{\begin{array}[b]{r}7\ -\ 1\end{array}}{\ \ \ 6}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{7} ot8\ ^{13} ot3\ -\ 1 \ \ \ \ \ 8\end{array}}{\ \ \ \ 6\ \ \ \ \ 5 }$

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}90\ -\ 12\end{array}}{\space }$

Answer

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.

Let's look at the numbers in the ones place:

$\frac{\begin{array}[b]{r}0\ -\ 2\end{array}}{\space }$

When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with an to replace the . So far, your work should look something like this:

$\frac{\begin{array}[b]{r}^{8} ot90\ -\ 12\end{array}}{\space }$

Remember, we've borrowed from the tens place, and one ten is equal to . We now add that to the digit in the ones place, . Your work should look something like this:

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{10} ot0\ -\ 1 \ \ \ \ \ 2\end{array}}{\space }$

Now, we can subtract the numbers in the once place:

$\frac{\begin{array}[b]{r}10\ -\ 2\end{array}}{\ \ \ 8}$

Next, we can subtract the numbers in the tens place:

$\frac{\begin{array}[b]{r}8\ -\ 1\end{array}}{\ \ \ 7}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{8} ot9\ ^{10} ot0\ -\ 1 \ \ \ \ \ 2\end{array}}{\ \ \ \ 7\ \ \ \ \ 8 }$

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}26\ -\ 17\end{array}}{\space }$

Answer

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.

Let's look at the numbers in the ones place:

$\frac{\begin{array}[b]{r}6\ -\ 7\end{array}}{\space }$

When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:

$\frac{\begin{array}[b]{r}^{1} ot26\ -\ 17\end{array}}{\space }$

Remember, we've borrowed from the tens place, and one ten is equal to . We now add that to the digit in the ones place, . Your work should look something like this:

$\frac{\begin{array}[b]{r}^{1} ot2\ ^{16} ot6\ -\ 1 \ \ \ \ \ 7\end{array}}{\space }$

Now, we can subtract the numbers in the once place:

$\frac{\begin{array}[b]{r}16\ -\ 7\end{array}}{\ \ \ 9}$

Next, we can subtract the numbers in the tens place:

$\frac{\begin{array}[b]{r}1\ -\ 1\end{array}}{\ \ \ 0}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{1} ot2\ ^{16} ot6\ -\ 1 \ \ \ \ \ 7\end{array}}{\ \ \ \ \ \ \ \ \ \ \ 9}$

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}31\ -\ 18\end{array}}{\space }$

Answer

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.

Let's look at the numbers in the ones place:

$\frac{\begin{array}[b]{r}1\ -\ 8\end{array}}{\space }$

When the top number is smaller than the bottom number, we have to borrow from the number to the left because we can't take away from since is the smaller number. In this case, we are going to look to the . We only ever need to take away from the number to the left. For this problem, that will leave us with a to replace the . So far, your work should look something like this:

$\frac{\begin{array}[b]{r}^{2} ot31\ -\ 18\end{array}}{\space }$

Remember, we've borrowed from the tens place, and one ten is equal to . We now add that to the digit in the ones place, . Your work should look something like this:

$\frac{\begin{array}[b]{r}^{2} ot3\ ^{11} ot1\ -\ 1 \ \ \ \ \ 8\end{array}}{\space }$

Now, we can subtract the numbers in the once place:

$\frac{\begin{array}[b]{r}11\ -\ 8\end{array}}{\ \ \ 3}$

Next, we can subtract the numbers in the tens place:

$\frac{\begin{array}[b]{r}2\ -\ 1\end{array}}{\ \ \ 1}$

Your final answer should be

$\frac{\begin{array}[b]{r}^{2} ot3\ ^{11} ot1\ -\ 1 \ \ \ \ \ 8\end{array}}{\ \ \ 1\ \ \ \ \ 3}$

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}13\ -\ 10\end{array}}{\space }$

Answer

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.

Let's look at the numbers in the ones place:

$\frac{\begin{array}[b]{r}1{\color{Red} 3}\ -\ 1{\color{Red} 0}\end{array}}{\ \ \ \ \ {\color{Red} 3} }$

Next, we can subtract the numbers in the tens place:

$\frac{\begin{array}[b]{r}{\color{Red} 1}3\ -\ {\color{Red} 1}0\end{array}}{\ \ \ \ {\color{Red} 0}3}$

Your final answer should be

Compare your answer with the correct one above

Question

Solve:

$\frac{\begin{array}[b]{r}32\ -\ 11\end{array}}{\space }$

Answer

When we subtract multi-digit numbers, we start with the digits in the ones place and move to the left.

Let's look at the numbers in the ones place:

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

Next, we can subtract the numbers in the tens place:

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

Your final answer should be

Compare your answer with the correct one above

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