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Estimation:
Rounding to a Specified Degree of Accuracy
Rounding helps simplify numbers. You can round to:
A specific place value (like nearest 10, 100, 1000)
A specific number of decimal places (d.p.)
A specific number of significant figures (s.f.)
Place Value Rounding:
To round, look at the digit after the one you’re rounding to:
If it’s 5 or more → round up
If it’s less than 5 → round down
Example: round 5,764 to:
Nearest 10 → 5,760
Nearest 100 → 5,800
Nearest 1,000 → 6,000
Decimal Places (d.p.)
Round to a certain number of digits after the decimal.
Example: round 4.6789 to:
1 d.p. → 4.7
2 d.p. → 4.68
Significant Figures (s.f.)
Start counting from the first non-zero digit, even if it's after a decimal point. Note that "100" alone is 1 s.f. while "100." is 3 since there's a point.
Example: round 0.007864 to:
1 s.f. → 0.008
2 s.f. → 0.0079
Practice Problem:
Round the following to 2 significant figures:
a) 3846
b) 0.006718
Answer: a) 3800 b) 0.0067
Estimating Calculations
When estimating, we round each number to 1 significant figure to make the math easier.
Example: estimate the value of (9.79×0.765)/41.3
Step 1: Round each number to 1 s.f.
9.79 ≈ 10, 0.765 ≈ 0.8, and 41.3 ≈ 40
Step 2: Substitute into expression
(10×0.8)/40 = 0.2
Practice Problem:
Estimate the value of 3.14×7.92 to 1 s.f.
Answer: 3.14 ≈ 3.1 and 7.92 ≈ 8. So, 3.1 × 8 = 24.8 ≈ 25
Rounding Answers in Context
In real-life problems, round your final answer to a reasonable level of accuracy that matches the context:
For money → 2 decimal places (e.g. $13.25)
For people/things → nearest whole number
For measurements → round to match the accuracy of original data
Example: you calculate that 57.893 people attended a show. Your answer is 58 people (you can’t have 0.893 of a person)
Practice Problem:
A recipe needs 163.782 grams of sugar. You measure 164 grams.
How many significant figures is your measurement?
Answer: 3 s.f.
Limits of Accuracy
What Are Upper and Lower Bounds?
When a number is rounded, we don’t know its exact original value—only the range it came from.
So we define:
Lower bound = the smallest it could be
Upper bound = the biggest it could be (but not including)
Rule:
If a value is rounded to the nearest unit (like nearest 1 cm, 1 m, etc):
Practice Problem:
A value is 62 kg to the nearest kg.
Find the lower and upper bounds.
Answer: lower is 62 - 0.5 = 61.5 kg and upper is 62 + 0.5 = 62.5 kg
Bounds in Calculations
Key Idea:
When using rounded values in calculations, we also apply bounds to estimate maximum and minimum possible outcomes.
Example A: Perimeter of a rectangle
A rectangle is measured as length = 12 cm (nearest cm) and width = 8 cm (nearest cm)
Step 1: Find bounds
Length: lower :12−0.5=11.5 and upper :12+0.5=12.5
Width: lower :8−0.5=7.5 and upper: 8+0.5=8.5
Step 2: Find upper bound of perimeter
𝑃= 2(𝐿+𝑊) = 2(12.5+8.5)=2×21=42 cm
Step 3: Find lower bound of perimeter
𝑃= 2(𝐿+𝑊) = 2(11.5+7.5)= 2×19 = 38 cm
Example B: Area of a Square Using Bounds
If a square’s side is 6 cm (nearest cm): Lower = 5.5 and Upper = 6.5
Upper bound of area = 6.5 x 6.5 = 42.25
Lower bound = 5.5 x 5.5 = 30.25
Example C: Speed Using Bounds
A car travels: distance = 120 km (nearest 10 km) and time = 2 hours (nearest hour)
Step 1: Find bounds
Distance lower = 115 and upper = 125 since its to the nearest 10 km
Time lower = 1.5 and upper = 2.5
Step 2: For maximum speed (distance/time) in km/h:
Use largest distance and smallest time:
125/1.5 = 83.33 km/h
Step 3: For minimum speed:
Use smallest distance and largest time:
115/2.5 = 46 km/h
Important Notices:
Why we used different bounds to get minimum and maximum?
For maximum speed, you want to see the maximum amount of distance you could've gone in the shortest amount of time (when someone arrives at school in 10 minutes, they're said to be faster than whoever arrived in 15 if they covered the same distance)
For minimum speed you want to see the smallest distance covered in the largest time (The slower the speed, the smaller the distance and bigger the time, just like your sleepy classmate who left early but still showed up after you)
Nearest 1 km vs 10 km:
When we say nearest km, it means the actual value could be half a km more or less. This is more precise. You’re saying the road is accurate within ±0.5 km.
Now, suppose the same road is measured to the nearest 10 km. That means the value is accurate to the nearest 10, so we’re dealing with ±5. This is less precise since the actual value could be up to 5 km off in either direction.
Practice Problem:
A runner completes a 100 m sprint in 12 s (both to nearest unit).
Estimate the maximum possible speed.
Answer: max speed = max distance/min time
Max distance: 100 + 0.5 =100.5 while minimum time is 12-0.5 = 11.5 so max speed = 8.7 m/s
NOTES DONE BY FARIDA SABET
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