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Common Core: 2nd Grade Math : Add within 1000

Study concepts, example questions & explanations for Common Core: 2nd Grade Math

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Example Questions

Example Question #1 : Add Within 1000

Solve:

$\frac{\begin{array}[b]{r}235\\ +\ 204\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}235\\ +\ 204\end{array}}{\space }$

Possible Answers:

438 $\displaystyle 438$

439 $\displaystyle 439$

436 $\displaystyle 436$

437 $\displaystyle 437$

Correct answer:

439 $\displaystyle 439$

Explanation:

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

Add the numbers in the ones place:

5+4=9 $\displaystyle 5+4=9$

Add the numbers in the tens place:

3+0=3 $\displaystyle 3+0=3$

Add the numbers in the hundreds place:

2+2=4 $\displaystyle 2+2=4$

Your final answer should be 439: $\displaystyle 439:$

$\frac{\begin{array}[b]{r}235\\ +\ 204\end{array}}{\ \ \ \ 439 }$ $\displaystyle \frac{\begin{array}[b]{r}235\\ +\ 204\end{array}}{\ \ \ \ 439 }$

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Example Question #2 : Add Within 1000

Solve:

$\frac{\begin{array}[b]{r}330\\ +\ 126\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}330\\ +\ 126\end{array}}{\space }$

Possible Answers:

458 $\displaystyle 458$

459 $\displaystyle 459$

456 $\displaystyle 456$

457 $\displaystyle 457$

Correct answer:

456 $\displaystyle 456$

Explanation:

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

Add the numbers in the ones place:

0+6=6 $\displaystyle 0+6=6$

Add the numbers in the tens place:

3+2=5 $\displaystyle 3+2=5$

Add the numbers in the hundreds place:

3+1=4 $\displaystyle 3+1=4$

Your final answer should be 456: $\displaystyle 456:$

$\frac{\begin{array}[b]{r}330\\ +\ 126\end{array}}{\ \ \ 456 }$ $\displaystyle \frac{\begin{array}[b]{r}330\\ +\ 126\end{array}}{\ \ \ 456 }$

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Example Question #3 : Add Within 1000

Solve:

$\frac{\begin{array}[b]{r}387\\ +\ 410\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}387\\ +\ 410\end{array}}{\space }$

Possible Answers:

798 $\displaystyle 798$

797 $\displaystyle 797$

795 $\displaystyle 795$

796 $\displaystyle 796$

Correct answer:

797 $\displaystyle 797$

Explanation:

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

Add the numbers in the ones place:

7+0=7 $\displaystyle 7+0=7$

Add the numbers in the tens place:

8+1=9 $\displaystyle 8+1=9$

Add the numbers in the hundreds place:

3+4=7 $\displaystyle 3+4=7$

Your final answer should be 797: $\displaystyle 797:$

$\frac{\begin{array}[b]{r}387\\ +\ 410\end{array}}{\ \ \ 797}$ $\displaystyle \frac{\begin{array}[b]{r}387\\ +\ 410\end{array}}{\ \ \ 797}$

Report an Error

Example Question #4 : Add Within 1000

Solve:

$\frac{\begin{array}[b]{r}228\\ +\ 628\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}228\\ +\ 628\end{array}}{\space }$

Possible Answers:

855 $\displaystyle 855$

858 $\displaystyle 858$

856 $\displaystyle 856$

857 $\displaystyle 857$

Correct answer:

856 $\displaystyle 856$

Explanation:

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

Add the numbers in the ones place:

8+8=16 $\displaystyle 8+8=16$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 6$ from the ones place and carry the $\displaystyle 1$ from he tens place over to the left. Your work should look something like this:

$\frac{\begin{array}[b]{r}2^{{\color{Red} 1}}2\ 8\\ +\ 6\ 2\ 8\end{array}}{\ \ \ \ \ \ \ \ 6 }$ $\displaystyle \frac{\begin{array}[b]{r}2^{{\color{Red} 1}}2\ 8\\ +\ 6\ 2\ 8\end{array}}{\ \ \ \ \ \ \ \ 6 }$

Add the numbers in the tens place, including the $\displaystyle 1$ that was carried over:

1+2+2=5 $\displaystyle 1+2+2=5$

Add the numbers in the hundreds place:

2+6=8 $\displaystyle 2+6=8$

Your final answer should be 856: $\displaystyle 856:$

$\frac{\begin{array}[b]{r}228\\ +\ 628\end{array}}{\ \ \ 856 }$ $\displaystyle \frac{\begin{array}[b]{r}228\\ +\ 628\end{array}}{\ \ \ 856 }$

Report an Error

Example Question #2 : Add Within 1000

Solve:

$\frac{\begin{array}[b]{r}195\\ +\ 519\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}195\\ +\ 519\end{array}}{\space }$

Possible Answers:

711 $\displaystyle 711$

712 $\displaystyle 712$

713 $\displaystyle 713$

714 $\displaystyle 714$

Correct answer:

714 $\displaystyle 714$

Explanation:

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

Add the numbers in the ones place:

5+9=14 $\displaystyle 5+9=14$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 4$ from the ones place and carry the $\displaystyle 1$ from he tens place over to the left. Your work should look something like this:

$\frac{\begin{array}[b]{r}1^{{\color{Red} 1}}9\ 5\\ +\ 5\ 1\ 9\end{array}}{\ \ \ \ \ \ \ \ 4 }$ $\displaystyle \frac{\begin{array}[b]{r}1^{{\color{Red} 1}}9\ 5\\ +\ 5\ 1\ 9\end{array}}{\ \ \ \ \ \ \ \ 4 }$

Add the numbers in the tens place, including the $\displaystyle 1$ that was carried over:

$\displaystyle 1+9+1=11$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 1$ from the ones place and carry the $\displaystyle 1$ from the tens place over to the left. Your work should look something like this:

$\frac{\begin{array}[b]{r}^{{\color{Red} 1}}1^{{\color{Red} 1}}9\ 5\\ +\ 5\ 1\ 9\end{array}}{\ \ \ \ \ \ 1\ 4 }$ $\displaystyle \frac{\begin{array}[b]{r}^{{\color{Red} 1}}1^{{\color{Red} 1}}9\ 5\\ +\ 5\ 1\ 9\end{array}}{\ \ \ \ \ \ 1\ 4 }$

Add the numbers in the hundreds place, including the $\displaystyle 1$ that was carried over:

1+1+5=7 $\displaystyle 1+1+5=7$

Your final answer should be 714: $\displaystyle 714:$

$\frac{\begin{array}[b]{r}195\\ +\ 519\end{array}}{\ \ \ 714}$ $\displaystyle \frac{\begin{array}[b]{r}195\\ +\ 519\end{array}}{\ \ \ 714}$

Report an Error

Example Question #6 : Add Within 1000

Solve:

$\frac{\begin{array}[b]{r}312\\ +\ 496\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}312\\ +\ 496\end{array}}{\space }$

Possible Answers:

810 $\displaystyle 810$

808 $\displaystyle 808$

809 $\displaystyle 809$

807 $\displaystyle 807$

Correct answer:

808 $\displaystyle 808$

Explanation:

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

Add the numbers in the ones place:

2+6=8 $\displaystyle 2+6=8$

Add the numbers in the tens place:

1+9=10 $\displaystyle 1+9=10$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 0$ from the ones place and carry the $\displaystyle 1$ from the tens place over to the left. Your work should look something like this:

$\frac{\begin{array}[b]{r}^{{\color{Red} 1}}312\\ +\ 496\end{array}}{\ \ \ \ \ 08 }$ $\displaystyle \frac{\begin{array}[b]{r}^{{\color{Red} 1}}312\\ +\ 496\end{array}}{\ \ \ \ \ 08 }$

Add the numbers in the hundreds place, including the $\displaystyle 1$ that was carried over:

1+3+4=8 $\displaystyle 1+3+4=8$

Your final answer should be 808: $\displaystyle 808:$

$\frac{\begin{array}[b]{r}^{1}312\\ +\ 496\end{array}}{\ \ \ \ 808 }$ $\displaystyle \frac{\begin{array}[b]{r}^{1}312\\ +\ 496\end{array}}{\ \ \ \ 808 }$

Report an Error

Example Question #3 : Add Within 1000

Solve:

$\frac{\begin{array}[b]{r}269\\ +\ 221\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}269\\ +\ 221\end{array}}{\space }$

Possible Answers:

491 $\displaystyle 491$

493 $\displaystyle 493$

490 $\displaystyle 490$

492 $\displaystyle 492$

Correct answer:

490 $\displaystyle 490$

Explanation:

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

Add the numbers in the ones place:

9+1=10 $\displaystyle 9+1=10$

$\frac{\begin{array}[b]{r}2^{{\color{Red} 1}}6\ 9\\ +\ 2\ 2\ 1\end{array}}{\ \ \ \ \ \ \ \ 0 }$ $\displaystyle \frac{\begin{array}[b]{r}2^{{\color{Red} 1}}6\ 9\\ +\ 2\ 2\ 1\end{array}}{\ \ \ \ \ \ \ \ 0 }$

Add the numbers in the tens place, including the $\displaystyle 1$ that was carried over:

1+6+2=9 $\displaystyle 1+6+2=9$

Add the numbers in the hundreds place:

2+2=4 $\displaystyle 2+2=4$

Your final answer should be 490: $\displaystyle 490:$

$\frac{\begin{array}[b]{r}269\\ +\ 221\end{array}}{\ \ \ 490 }$ $\displaystyle \frac{\begin{array}[b]{r}269\\ +\ 221\end{array}}{\ \ \ 490 }$

Report an Error

Example Question #2131 : Common Core Math: Grade 2

Solve:

$\frac{\begin{array}[b]{r}400\\ +\ 224\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}400\\ +\ 224\end{array}}{\space }$

Possible Answers:

621 $\displaystyle 621$

623 $\displaystyle 623$

624 $\displaystyle 624$

622 $\displaystyle 622$

Correct answer:

624 $\displaystyle 624$

Explanation:

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

Add the numbers in the ones place:

0+4=4 $\displaystyle 0+4=4$

Add the numbers in the tens place:

0+2=2 $\displaystyle 0+2=2$

Add the numbers in the hundreds place:

4+2=6 $\displaystyle 4+2=6$

Your final answer should be 624: $\displaystyle 624:$

$\frac{\begin{array}[b]{r}400\\ +\ 224\end{array}}{\ \ \ 624 }$ $\displaystyle \frac{\begin{array}[b]{r}400\\ +\ 224\end{array}}{\ \ \ 624 }$

Report an Error

Example Question #2132 : Common Core Math: Grade 2

Solve:

$\frac{\begin{array}[b]{r}128\\ +\ 548\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}128\\ +\ 548\end{array}}{\space }$

Possible Answers:

678 $\displaystyle 678$

674 $\displaystyle 674$

675 $\displaystyle 675$

676 $\displaystyle 676$

Correct answer:

676 $\displaystyle 676$

Explanation:

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

Add the numbers in the ones place:

8+8=16 $\displaystyle 8+8=16$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 6$ from the ones place and carry the $\displaystyle 1$ from the tens place over to the left. Your work should look something like this:

$\frac{\begin{array}[b]{r}1^{{\color{Red} 1}}2\ 8\\ +\ 5\ 4\ 8\end{array}}{\ \ \ \ \ \ \ \ 6 }$ $\displaystyle \frac{\begin{array}[b]{r}1^{{\color{Red} 1}}2\ 8\\ +\ 5\ 4\ 8\end{array}}{\ \ \ \ \ \ \ \ 6 }$

Add the numbers in the tens place, including the $\displaystyle 1$ that was carried over:

1+2+4=7 $\displaystyle 1+2+4=7$

Add the numbers in the hundreds place:

1+5=6 $\displaystyle 1+5=6$

Your final answer should be 676: $\displaystyle 676:$

$\frac{\begin{array}[b]{r}128\\ +\ 548\end{array}}{\ \ \ 676}$ $\displaystyle \frac{\begin{array}[b]{r}128\\ +\ 548\end{array}}{\ \ \ 676}$

Report an Error

Example Question #10 : Add Within 1000

Solve:

$\frac{\begin{array}[b]{r}407\\ +\ 328\end{array}}{\space }$ $\displaystyle \frac{\begin{array}[b]{r}407\\ +\ 328\end{array}}{\space }$

Possible Answers:

733 $\displaystyle 733$

735 $\displaystyle 735$

734 $\displaystyle 734$

732 $\displaystyle 732$

Correct answer:

735 $\displaystyle 735$

Explanation:

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

Add the numbers in the ones place:

7+8=15 $\displaystyle 7+8=15$

Because this sum is greater than $\displaystyle 9$, we write the $\displaystyle 5$ from the ones place and carry the $\displaystyle 1$ from the tens place over to the left. Your work should look something like this:

$\frac{\begin{array}[b]{r}4^{{\color{Red} 1}}0\ 7\\ +\ 3\ 2\ 8\end{array}}{\ \ \ \ \ \ \ \ 5 }$ $\displaystyle \frac{\begin{array}[b]{r}4^{{\color{Red} 1}}0\ 7\\ +\ 3\ 2\ 8\end{array}}{\ \ \ \ \ \ \ \ 5 }$

Add the numbers in the tens place, including the $\displaystyle 1$ that was carried over:

1+0+2=3 $\displaystyle 1+0+2=3$

Add the numbers in the hundreds place:

4+3=7 $\displaystyle 4+3=7$

Your final answer should be 735: $\displaystyle 735:$

$\frac{\begin{array}[b]{r}407\\ +\ 328\end{array}}{\ \ \ 735}$ $\displaystyle \frac{\begin{array}[b]{r}407\\ +\ 328\end{array}}{\ \ \ 735}$

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All Common Core: 2nd Grade Math Resources

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Common Core: 2nd Grade Math : Add within 1000

Study concepts, example questions & explanations for Common Core: 2nd Grade Math

All Common Core: 2nd Grade Math Resources

Example Questions

Example Question #1 : Add Within 1000

Example Question #2 : Add Within 1000

Example Question #3 : Add Within 1000

Example Question #4 : Add Within 1000

Example Question #2 : Add Within 1000

Example Question #6 : Add Within 1000

Example Question #3 : Add Within 1000

Example Question #2131 : Common Core Math: Grade 2

Example Question #2132 : Common Core Math: Grade 2

Example Question #10 : Add Within 1000

All Common Core: 2nd Grade Math Resources

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