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3. solve 667 + 295 using two different strategies. a. b. c. explain whi…

Question

  1. solve 667 + 295 using two different strategies.

a.
b.
c. explain which strategy you prefer to use when solving and why.

Explanation:

Response
Part a: Strategy 1 - Direct Addition (Place Value)

Step1: Add ones place

Add the digits in the ones place: \(7 + 5 = 12\). Write down \(2\) and carry over \(1\) to the tens place.

Step2: Add tens place with carry

Add the digits in the tens place and the carried over \(1\): \(6 + 9 + 1 = 16\). Write down \(6\) and carry over \(1\) to the hundreds place.

Step3: Add hundreds place with carry

Add the digits in the hundreds place and the carried over \(1\): \(6 + 2 + 1 = 9\).

Calculation:

\[

$$\begin{align*} &\quad\ 6\ 6\ 7\\ &+\quad2\ 9\ 5\\ &\overline{\quad9\ 6\ 2} \end{align*}$$

\]
So, \(667 + 295 = 962\) using direct addition.

Part b: Strategy 2 - Compensation (Rounding and Adjusting)

Step1: Round 295 to 300

Since \(295\) is close to \(300\), we can first add \(667 + 300\). This is easier because adding a multiple of 100 is straightforward.
\(667 + 300 = 967\)

Step2: Adjust for over - addition

We added \(5\) extra (because \(300 - 295 = 5\)), so we need to subtract that extra \(5\) from the result.
\(967-5 = 962\)

Calculation:

\(667 + 295=667+(300 - 5)=(667 + 300)-5 = 967 - 5=962\)

Part c: Preference of Strategy
Brief Explanations

I prefer the compensation strategy (Strategy 2). The reason is that rounding the number (295 to 300) makes the addition step (\(667 + 300\)) very simple and quick. The adjustment step (subtracting 5) is also easy to do mentally. It reduces the complexity of dealing with carrying over in the traditional place - value addition, especially when one of the numbers is close to a multiple of 10, 100, etc.

Answer:

a. Using direct addition: \(667+295 = 962\) (shown by place - value addition)
b. Using compensation: \(667 + 295=962\) (by rounding 295 to 300, adding and then adjusting)
c. Prefer compensation as it simplifies addition with mental math.