Learning Objectives
Convert length, area, volume and compound units correctly.
Express rounding and truncation information as intervals.
Find justified bounds for positive products and quotients.
Distinguish a rough estimate from a guaranteed bound.
Prerequisites: fractions, powers, inequalities and proportional scaling. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
A measurement gives a quantity in a unit, often rounded to a stated precision. This lesson explains how to calculate with such measurements without confusing a converted value, an estimate and a guaranteed range of possible values.
First we convert units, including squared and compound units. We then interpret rounded readings as intervals and combine those intervals to bound products and quotients. Finally, we distinguish a convenient estimate from an inequality that is guaranteed to contain the answer.
1. Unit Conversion: Keeping the Quantity Unchanged
A calculation involving distance, area or speed is meaningful only when the units match. Unit conversion changes the numerical representation of a quantity, not the quantity itself, so begin by identifying exactly what is being measured.
1.1 Length, area and volume: powers of a conversion factor
A measurement consists of a number and a unit. Converting the unit changes the number while preserving the quantity. Since 1 m = 100 cm, a square metre is a 100 cm by 100 cm square, so 1 m² = 10000 cm². Similarly, 1 m³ = 1000000 cm³.
The squared and cubed conversion factors come from converting each dimension. A square metre contains 100 rows of 100 square centimetres; this is why an area cannot be converted using only the length factor 100.
1.2 Compound units: converting numerator and denominator
A compound unit combines two or more units. In a speed such as kilometres per hour, the word per means division, so both the distance and the time units affect the numerical value.
Compound units require conversion of both parts. To turn kilometres per hour into metres per second, multiply by 1000 and divide by 3600. To convert back, multiply by 3600/1000. Keep units attached until the cancellation is clear.
Density is mass divided by volume, and pressure is force divided by area. Rates of pay and unit prices are also quotients. Before inserting values into a formula, make the units consistent.
Worked Example: Convert a Speed and an Area
Question: Convert 54 km/h to m/s and 250 cm² to m².
Method: Convert each part of the speed unit, and use the square of the length factor for the area conversion.
54 km/h = 54000 m/3600 s = 15 m/s. Check in one hour: 15 × 3600 = 54000 m = 54 km.
250 cm² = 250/10000 m² = 0.025 m². Dividing by 100 would convert a length, not an area.
2. Rounding and Truncation: Recovering Possible Original Values
A rounded reading does not identify one exact original measurement. It tells you a whole interval of values that would produce the same display. We need that interval before deciding what a calculation from the measurement can guarantee.
2.1 Nearest-unit rounding: halfway points form the boundaries
If a positive measurement is rounded to the nearest unit, the original is within half a unit of the displayed value. In this lesson exact halfway values round upward. Thus a displayed value 8 means 7.5 ≤ x < 8.5. The left endpoint rounds to 8; the right endpoint rounds to 9.
A filled endpoint on a number line means that boundary is allowed; an open endpoint means it is excluded. The convention for exact halfway values determines which end is included.
2.2 Precision and significant figures: finding the rounding step
For rounding to a step h > 0, a displayed value r gives r − h/2 ≤ x < r + h/2 under this convention. Nearest tenth uses h = 0.1; nearest hundred uses h = 100. Endpoint conventions should always be read from the question.
Significant figures begin at the first nonzero digit. Locate the place value of the last retained significant digit to find the rounding step. For instance, 420 to two significant figures has its last retained digit in the tens place, so its step is 10 rather than 1.
2.3 Truncation: discarding digits rather than rounding
Truncating a positive number simply discards the later digits. If truncating to one decimal place gives 3.4, then 3.4 ≤ x < 3.5. This is not the same interval as rounding to one decimal place, which gives 3.35 ≤ x < 3.45.
Figure 1. Values that round to 8 to the nearest integer, with exact halves rounded upward. A filled endpoint is included; an open endpoint is excluded.
Worked Example: Significant Figures
Question: A positive distance is reported as 420 m to two significant figures. State the error interval.
Method: Locate the tens place as the last retained significant digit and take half of that rounding step on either side.
The last retained significant digit is in the tens place, so the rounding step is 10 m and half a step is 5 m.
Therefore 415 m ≤ d < 425 m. The endpoint 425 m would round to 430 m. Saying only “420 m” without a precision would not determine this interval.
3. Bounds for Calculations: Choosing the Correct Extremes
Once each input is an interval, the next task is to find the range of possible outputs. The correct endpoint choices depend on whether increasing an input makes the result larger or smaller. Positivity and independence are conditions to check, not assumptions to leave implicit.
3.1 Positive products: both factors increase the result
For independent positive measurements, a product gets larger as either factor increases. Its smallest possible value uses the two lower endpoints; its upper bound uses the two upper endpoints. If an endpoint is excluded, distinguish a bound from a maximum actually attained.
3.2 Positive quotients: the denominator acts in the opposite direction
For a fixed positive distance, taking more time reduces the speed. This gives a concrete reason that a denominator must be treated differently from a numerator when finding bounds.
A positive quotient a/b gets larger as a increases and smaller as b increases. Its lower bound therefore uses the smallest numerator and largest denominator; its upper bound uses the largest numerator and smallest denominator. This rule requires a denominator bounded away from zero.
3.3 Independent inputs and attainable endpoints
If the quantities are linked, the separate extreme values may not occur together. For example, if a + b = 10, you cannot make both a and b approach 10 at once. Always check whether the question permits independent choices.
An upper bound is a value the result cannot exceed; a maximum is a largest value actually attained. An excluded endpoint may produce a sharp bound that can be approached as closely as desired but never reached. The strict inequality must preserve that distinction.
Worked Example: Speed from Rounded Measurements
Question: A distance is 100 m to the nearest metre and a time is 10 s to the nearest second. Bound the speed v = d/t, treating the measurements as independent.
Method: Write the distance and time intervals first, then maximise or minimise their positive quotient using opposite denominator endpoints.
99.5 ≤ d < 100.5 and 9.5 ≤ t < 10.5. All quantities are positive. Thus 99.5/10.5 < v < 100.5/9.5, or 199/21 < v < 201/19 m/s.
Both inequalities are strict: the lower limiting value needs t = 10.5, and the upper needs d = 100.5, each excluded. The nominal value 10 m/s lies between them. These are sharp bounds, not attained minimum and maximum speeds.
4. Estimation: Checking Size without Claiming Exactness
Bounds answer what must be true. An estimate instead helps you judge whether the size of a result is plausible. We finish by organising very large or small numbers in standard form and deciding when a rough calculation does or does not establish a bound.
4.1 Standard form: separating significant digits from scale
Standard form for a positive number is a × 10^n with 1 ≤ a < 10 and integer n. It separates size from significant digits. For example, 0.0048 = 4.8 × 10^(−3). Multiplication combines coefficients and adds exponents, followed by renormalising the coefficient if necessary.
The coefficient gives the significant digits, while the power of ten locates the decimal point. After multiplying coefficients, adjust the power of ten if the new coefficient is outside the interval from 1 up to, but not including, 10.
4.2 An estimate versus a proved bound
A rough estimate uses convenient nearby numbers to judge scale. It need not lie above or below the exact answer. A guaranteed bound needs inequalities whose directions you track. Rounding some factors up and others down does not by itself establish an upper bound.
To claim an upper or lower bound, justify the direction in which each replacement changes the result. An estimate obtained by a mixture of upward and downward rounding needs a separate argument before it can be called a bound.
Worked Example: An Estimate and a Bound
Question: Estimate 49.8 × 0.203 and also give simple guaranteed bounds.
Method: Use convenient rounded factors for a quick estimate, then use explicit inequalities to establish guaranteed lower and upper bounds.
For an estimate, use 50 × 0.2 = 10. For bounds, 49 < 49.8 < 50 and 0.20 < 0.203 < 0.21, so 9.8 < 49.8 × 0.203 < 10.5.
An independent exact calculation gives 10.1094, which satisfies the bounds. The estimate 10 is useful for checking scale, but its closeness alone was not a proof of an error limit.
Common Mistakes
Square or cube a length conversion factor when converting areas or volumes.
Do not assume the displayed rounded value is exact.
For a positive quotient, divide the upper numerator bound by the lower denominator bound to obtain the upper bound.
An excluded endpoint can be a sharp bound without being an attainable maximum.
5. Practice
The first questions check unit conversion and interpretation of precision. The later questions use those intervals in calculations and test whether an estimate has a justified direction. Include units and state clearly whether each endpoint is allowed.
1. Convert 72 km/h to m/s and 3.2 litres to cm³, given 1 litre = 1000 cm³.
2. A positive length rounds to 6.2 cm to the nearest tenth. State its interval.
3. A positive value truncates to 0.07 at two decimal places. State its interval.
4. A rectangle has independent lengths satisfying 3.5 ≤ l < 4.5 and 5.5 ≤ w < 6.5, in metres. Bound its area.
5. Independent positive measurements satisfy 8 ≤ a < 10 and 2 ≤ b < 4. Give sharp bounds for a/b and explain the endpoints.
6. Calculate (6 × 10^5)(4 × 10^(−3)) in standard form.
7. A student says: “Rounding 19.6 up to 20 and 0.49 down to 0.5 gives a lower bound for their product.” Identify both errors and find the exact product.
Worked Solutions
1. Convert the distance, time and volume units
72 × 1000/3600 = 20 m/s. Also 3.2 litres = 3200 cm³. Multiplying 20 by 3600 recovers 72000 m per hour.
2. Use half the stated rounding step
6.15 ≤ l < 6.25 cm under the halfway-up convention. The interval has width 0.1 cm and midpoint 6.2 cm.
3. Use the interval produced by truncation
0.07 ≤ x < 0.08. Truncation removes all digits after the hundredths place; it does not introduce half-step endpoints.
4. Multiply positive endpoint bounds
The product is smallest at both included lower endpoints: 3.5 × 5.5 = 19.25. The upper bound is 4.5 × 6.5 = 29.25, excluded. Thus 19.25 ≤ area < 29.25 m².
5. Choose opposite endpoints for a quotient
2 < a/b < 5. For the lower endpoint, a ≥ 8 and b < 4 imply a/b > 2; values approach 2 with a = 8 and b approaching 4. For the upper endpoint, a < 10 and b ≥ 2 imply a/b < 5; values approach 5 with b = 2 and a approaching 10. Neither endpoint is attained.
6. Renormalise the standard-form coefficient
The product is 24 × 10² = 2.4 × 10³. Check in ordinary notation: 600000 × 0.004 = 2400.
7. Check the rounding direction before claiming a bound
0.49 becomes 0.5 by rounding upward, not downward. Both positive factors have increased, so 20 × 0.5 = 10 is an upper bound, not a lower bound. The exact product is (20 − 0.4)(0.5 − 0.01) = 10 − 0.2 − 0.2 + 0.004 = 9.604.
6. Summary
- Convert the entire unit, including powers and denominators.
- Use the stated precision and rounding convention to find an interval.
- Track how each variable affects a product or quotient before choosing endpoints.
- Separate useful approximations from proved bounds and attainable extrema.