4.1.1: Written Retrieval and Problem Solving Practice
- Page ID
- 183510
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Student Learning Objectives
4.1 Students will apply mental math and computational estimation techniques, such as using compatible numbers, properties, and rounding strategies, to solve mathematical problems efficiently and accurately.
4.2 Students will compare how basic and scientific calculators interpret and solve expressions, distinguishing between arithmetic logic (step-by-step entry) and algebraic logic (order of operations) while using calculator functions effectively (e.g., memory, exponent, and parentheses keys) to computer accurate results.
The mild assignment ๐ถ️ is a shorter and less challenging assignment. The medium assignment๐ถ️๐ถ️ is right in the middle in terms of length and level of difficulty. Finally, the spicy assignment ๐ถ️๐ถ️๐ถ️ is a longer assignment that is the most challenging.
(Spicy ๐ถ️๐ถ️๐ถ️ Problems are OPTIONAL, but go for it when you have the time to challenge yourself)
๐ถ️
Perform each of the following computations mentally and explain what technique you used to find the answer.
a) 40 + 160 + 29 + 31
b) 3670 − 474
c) 75 + 28
- Answer
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a) Step 1: Group numbers for easier addition Combine numbers that add up to round numbers: 40 + 160 = 200, 29 + 31 = 60, Step 2: Add results 200 + 60
b) First subtract 400 from 3679: 3670 − 400 = 3270
Then subtract the remaining 74: First do 3270 − 70 = 3200 Then 3200 − 4 = 3196
c) Step 1: Use compensation to simplify Round 28 up to 30 (easier to add): 75 + 30 = 105, Step 2: Subtract the extra 2 you added, 105 − 2 = 103
๐ถ️
Use compatible numbers to compute each of the following mentally.
a) 2 · 9 · 5 · 6
b) 82 + 37 + 18 + 13
- Answer
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a) (2 x 5) x (9 x 6) = 540
b) (82 + 18) + (37 + 13) = 150
๐ถ๐ถ️ ️
Calculate mentally using properties of operations, i.e. commutative, associative, distributive.
a) (37 + 25) + 43
b) 47 · 15 + 47 · 85
c) (4 x 13) x 25
d) 26 · 24 − 21 · 24.
- Answer
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a) (37 + 43) + 25 = 105
b) 47(15 + 85) = 4700
c) (4 x 25 x 13) = 1300
d) 24 ( 26 - 21) = 120
๐ถ️๐ถ️
Find each of the following differences using compensation method.
a) 43 − 17
b) 132 − 96
c) 250 − 167.
- Answer
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a) 46 - 20 = 26
b) 136 - 100 = 36
c) Count up: 33 + 50 = 83 OR (250-170+3) = 83
๐ถ๐ถ️ ️
Estimate using compatible number estimation.
a) 51 x 212
b) 3112 ÷ 62
c) 103 x 87.
- Answer
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a) 50 x 200 = 10000 b) 3000/60 = 50 c) 100 x 90 = 9000
๐ถ️
a) Find a range estimate for the sum 3741 + 1252.
b) Find a range estimate for the sum 289 x 12.
c) Find a range estimate for the sum 4787 ÷ 17.
- Answer
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a) The lower estimate or underestimate is 4000 and the upper estimate or overestimate is 6000. So the sum is in the range from 4000 to 6000.
b) The lower estimate or underestimate is (280 x 10 = 2800) and the upper estimate or overestimate is (300 x 15 = 4500).
c) The lower estimate or underestimate is (4600 / 20 = 230) and the upper estimate or overestimate is (4800/16=300)
๐ถ️
Round each of these to the position indicated.
a) 947 to the nearest hundred.
b) 27,462,312 to the nearest million.
c) 2561 to the nearest thousand.
- Answer
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a) 900 b) 27,000,000 c) 3,000
๐ถ๐ถ️ ️
Rounding to the left-most digit, calculate approximate values for each of the following:
a) 681 + 241
b) 678 − 431
c) 257 x 364
d) 28,329 ÷ 43.
- Answer
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a) 700 + 200 = 900 b) 700 - 400 = 300 c) 300 x 400 = 1200 d) 28,00 / 40 = 70
๐ถ️ ๐ถ️
Use estimation to tell whether the following calculator answers are reasonable. Explain why or why not.
(a) 657 + 542 + 707 = 193,346
(b) 26 x 47 = 1,222.
- Answer
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a) Not Reasonable: 700 + 500 + 700 = 1900 b) 25 x 50 = 1250, Reasonable
๐ถ️ ๐ถ️
Estimate each of the following using the methods: (i) range, (ii) one-column front end, (iii) two-column front end
- Answer
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i) Range: 700 + 0 + 0 = 700, 800 + 100 + 50 = 950, between 700 and 950 ii) 700 + 0 + 0 = 700 iii) 700 + 100 + 0 = 800
i) Range: 1700 + 1300 + 500 + 0 = 3500, 1800 + 1400 + 600 + 100 = 3900, between 3500 and 3900 ii) 2000 + 1400 + 500 + 0 = 3900 iii) 1700 + 1400 + 600 + 0 = 3700
๐ถ️
Simplify the following expressions using a calculator.
a) 135 − 7(48 − 33)
b) 32 · (50 - 7)2
c) \(\large \frac{1-2(3^{2}-4^{3})-1}{5+3\bullet 2}\)
- Answer
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a) 30 b) 59168 c) 10
๐ถ๐ถ️
Vanessa estimated 31 · 179 in the three ways shown below.
(i) 30 · 200 = 6000
(ii) 30 · 180 = 5400
(iii) 31 · 200 = 6200
Without finding the actual product, which estimate do you think is closer to the actual product? Explain.
- Answer
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ii) since it uses numbers closer to the original numbers
๐ถ️ ๐ถ️ ๐ถ️
There is a shortcut for multiplying a whole number by 99. For example, consider 15 x 99.
a) Why does 15 x 99 = (15 x 100) − (15 x 1)?
b) Compute 15 x 99 mentally, using the formula in part a)
c) Compute 95 x 99 mentally, using the same method. Then make up your own problems to try.
- Answer
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a) distributive property b) 1500 - 15 = 1485 c) 9500 - 95 = 9405
Problem Solving Practice
1) Develop a shortcut for multiplying by 25 mentally in a computation such as 24 x 25. Then Compute 44 x 25 using the same shortcut. Make sure to explain your method to someone and let them try a few problems.
2) Amari multiplied 3472 and 259 on their calculator and wrote down the answer of 89,248. How can you see immediately using estimation that they made an error? Can you see how ther error was made?
3) Conner tells you that multiplying by 5 is a lot like dividing by 2. For example, 48 x 5 = 240, but it is easier just to go 48 ÷ 2 = 24 and then affix a zero at the end. Will his method always work? Explain.
Directions: Using the digits 2 to 9 at most one time each, place a digit in each box to find the closest quotient to 250.
Directions: Using the digits 0 to 9 at most one time each, place a digit in each box to create a true equation with the greatest possible product.
Problem 4 and 5 Source: Open Middle