MAP 2nd Grade Math : Operations and Algebraic Thinking

Study concepts, example questions & explanations for MAP 2nd Grade Math

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Example Questions

Example Question #1 : Operations And Algebraic Thinking

Solve:

 

 

 

\(\displaystyle \frac{\begin{array}[b]{r}75\\ +\ 10\end{array}}{ \ \ \space}\)

Possible Answers:

\(\displaystyle 80\)

\(\displaystyle 75\)

\(\displaystyle 85\)

 

\(\displaystyle 70\)

Correct answer:

\(\displaystyle 85\)

 

Explanation:

When we add \(\displaystyle 10\) to a two digit number, the only number that changes in our answer is the tens position, and it will always go up by \(\displaystyle 1\). Mentally, we can add \(\displaystyle 1\) to the number in the tens place to find our answer. 

\(\displaystyle 7+1=8\)

\(\displaystyle \frac{\begin{array}[b]{r}75\\ +\ 10\end{array}}{ \ \ \ \space85}\)

Example Question #2 : Operations And Algebraic Thinking

David is \(\displaystyle 23\textup{ inches}\) taller than Alison. Alison is \(\displaystyle 53\textup{ inches}\) tall. How tall is David? 

Possible Answers:

\(\displaystyle 80\textup{ inches}\)

\(\displaystyle 76\textup{ inches}\)

\(\displaystyle 82\textup{ inches}\)

\(\displaystyle 84\textup{ inches}\)

Correct answer:

\(\displaystyle 76\textup{ inches}\)

Explanation:

This is an addition problem because we have the difference in height from the question. Alison is \(\displaystyle 53\) inches tall and David is \(\displaystyle 23\) inches taller than her, \(\displaystyle 23\) is our difference. We can add our difference to Alison's height to find out how tall David is. 

\(\displaystyle \frac{\begin{array}[b]{r}53\\ +\ 23\end{array}}{ \ \ \ \space 76}\)

Example Question #3 : Operations And Algebraic Thinking

Add:

\(\displaystyle \frac{\begin{array}[b]{r}51\\ +\ 12\end{array}}{ \ \ \space}\)

Possible Answers:

\(\displaystyle 64\)

\(\displaystyle 66\)

\(\displaystyle 65\)

\(\displaystyle 63\)

Correct answer:

\(\displaystyle 63\)

Explanation:

When we add two digit numbers, we start by adding the numbers in the ones place.

\(\displaystyle \frac{\begin{array}[b]{r}5{\color{Red} 1}\\ +\ 1{\color{Red} 2}\end{array}}{ \ \ \ \ \ \space3}\)

Next, we need to add the numbers in the tens place. 

\(\displaystyle \frac{\begin{array}[b]{r}{\color{Red} 5}1\\ +\ {\color{Red} 1}2\end{array}}{ \ \ \ \ \space63}\)

The final answer is \(\displaystyle 63\)

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