Learning Objectives
Choose between exact calculation, proportional modelling and bounds.
Combine surds, prime factors and percentage multipliers in unfamiliar questions.
Explain why an answer is valid and why a competing argument fails.
Identify a specific skill to revisit using worked solutions.
Prerequisites: the four preceding Section 1A lessons. This review combines their methods without introducing new topic requirements. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
This review connects the four preceding lessons on exact arithmetic, indices and surds, proportion, and measurement bounds. Its purpose is to help you recognise which method a mixed problem needs, rather than practise one rule in isolation.
We first choose a useful representation of the quantities. We then check the conditions that make a calculation valid and combine percentage multipliers with intervals. The final practice asks for both answers and reasons, including explanations of why some tempting alternatives fail.
1. Choosing a Representation for a Mixed Problem
In a mixed question, identifying the kind of quantity is part of the work. A recurring decimal, a surd length and a rounded measurement each suggest a different representation. Choosing the form first often reveals the calculation you actually need.
1.1 Exact fractions, prime factors and multipliers: matching form to purpose
A long-looking expression often becomes simpler after you choose its form. Use fractions when repeating decimals or cancellations appear, prime factors when divisibility or squares matter, and percentage multipliers when a quantity changes repeatedly. Use an interval, rather than a single value, when a measurement has been rounded.
1.2 The requested quantity: value, ratio, range or claim
First identify the requested quantity. A question may ask for a ratio, a range of possible values, or whether an argument is always valid. Solving for an unnecessary intermediate value can obscure the structure. Record domain and sign restrictions before manipulating the expressions.
For instance, two lengths can both contain √2 while their ratio is rational because that common nonzero factor cancels. If the question asks for the ratio, approximate decimal lengths only make the task less clear.
Worked Example: A Surd Ratio
Question: A rectangle has sides √50 cm and √8 cm. Find its perimeter and the ratio of its longer side to its shorter side.
Method: Simplify both surd lengths into multiples of √2, then use addition for the perimeter and cancellation for their ratio.
Simplify the sides to 5√2 and 2√2. The perimeter is 2(5√2 + 2√2) = 14√2 cm. The ratio is 5√2:2√2 = 5:2.
The common factor √2 cancels in the ratio because it is positive and nonzero. As a check, the area is (5√2)(2√2) = 20 cm², which is rational even though the side lengths are not.
2. Checking Conditions and Testing Claims
A calculation is justified only when the operations used are valid for the stated inputs. This section revisits the restrictions from the earlier lessons, then distinguishes a numerical check from a proof about every allowed number.
2.1 Conditions on a method: nonzero, nonnegative and similar
The answer alone does not establish that a method was valid. A cancellation needs a nonzero factor, a root needs an allowed radicand, a quotient bound needs a denominator away from zero, and a similarity argument needs genuinely similar shapes.
State these conditions alongside the method instead of adding them after an answer has already been selected. A formula that works for positive inputs may need a different conclusion when negative inputs are allowed.
2.2 Counterexamples and boundary cases: challenging an always claim
When assessing a universal claim, test boundary cases deliberately: zero when permitted, a negative value, an endpoint or a special prime such as 2. A counterexample must satisfy every assumption in the claim. If the claim survives testing, it still needs an argument covering all allowed cases.
A useful counterexample meets the assumptions of the statement but breaks its conclusion. Explain both parts so the reader can see why that particular example settles the claim.
Worked Example: A False Root Argument
Question: A learner claims that √(x²) = x for every real x because “squaring and square-rooting undo each other”. Assess the claim.
Method: Test a permitted negative input, then use the definition of the principal square root to identify the missing condition.
At x = −3, √(x²) = √9 = 3, whereas x = −3. The claim is false. The correct identity is √(x²) = |x|.
The original identity becomes valid with the extra assumption x ≥ 0. This identifies the missing condition, rather than merely spotting one incorrect substitution.
3. Combining Percentage Changes with Measurement Bounds
A percentage calculation usually begins with an exact starting or finishing value. If that value was rounded instead, the same multiplier must act on an interval. This brings together proportion from Lesson 03 and bounds from Lesson 04.
3.1 Scaling an interval by a positive multiplier
When the transformation is multiplication by a positive constant, both endpoints of an interval are multiplied by that constant and their inclusion status is preserved. This joins percentage reasoning to measurement bounds.
For a positive factor, larger inputs still give larger outputs, so the order of the endpoints is preserved. A strict upper endpoint stays strict because multiplying cannot turn an excluded input into an included one.
3.2 Reversing a rounded result: an interval of original values
If a displayed price or measurement has been rounded, reversing a change generally gives an interval of possible original values. The single value obtained by dividing the display is only a nominal estimate.
Start with the interval for the displayed final value, then divide both endpoints by the positive change multiplier. This gives all possible originals, whereas dividing the display alone gives just one representative value.
Worked Example: A Rounded Discounted Price
Question: After a 20% discount, a positive price is reported as £48 to the nearest pound, with halves rounded upward. Find the interval of possible original prices P, treating prices as real amounts.
Method: Convert the rounded discounted price into an interval before reversing the positive discount multiplier.
The actual discounted price D satisfies 47.5 ≤ D < 48.5, and D = (4/5)P. Multiply the interval by 5/4 to obtain 59.375 ≤ P < 60.625, in pounds.
The nominal original price £60 works, since its discounted price is £48. The lower endpoint also works, giving £47.50; the upper endpoint is excluded because £48.50 rounds to £49.
Common Mistakes
An answer option that matches an unchecked squared equation may be an extraneous solution.
Do not replace an interval by its midpoint when the question asks what must be true.
A counterexample outside the allowed domain does not disprove a claim.
An arithmetic check supports a calculation; a universal claim needs a general argument.
4. Practice
Try these questions without first looking at the solution headings. For each, identify the representation or model you chose and check its conditions. Questions 1–6 include alternatives; explain why your choice works. The later questions ask for bounds, justification and combined methods.
1. Evaluate (0.363636…)/(2/3). A: 2/11. B: 6/11. C: 4/9. D: 11/6.
2. Find the least positive integer k such that 48k is a square. A: 2. B: 3. C: 6. D: 12.
3. Simplify 1/(√3 − √2). A: √3 − √2. B: 1/√5. C: √3 + √2. D: √5.
4. A positive quantity rises by 10% and then falls by 10%. What percentage increase restores the original from the final value? A: 1%. B: (100/99)%. C: 10%. D: (1/99)%.
5. A:B = 3:4 and B:C = 6:5. What fraction of A + B + C is C? A: 5/12. B: 5/21. C: 10/31. D: 5/9.
6. For positive x, y = k/x² with k > 0. If x is multiplied by 3/2, what is the multiplier for y? A: 2/3. B: 4/9. C: 9/4. D: 3/2.
7. A positive mass rounds to 2.0 kg to the nearest tenth and an independent positive volume rounds to 0.5 litres to the nearest tenth. Give sharp bounds for density in kg/litre. Use halves rounded upward.
8. An integer n has the property that n² is divisible by 12. Must n be divisible by 6? Explain.
9. Solve √(3x + 4) = x, giving all real solutions.
10. Two similar containers have surface-area ratio 9:16. The smaller holds 81 litres. Find the larger capacity.
11. Assess the claim: if positive a and b satisfy a < b, then 1/a < 1/b. Give a counterexample and the correct relation.
12. A journey consists of 60 km at 30 km/h followed by 60 km at 60 km/h, with no stops. Find the average speed over the whole journey and explain why the arithmetic mean of the two speeds fails.
Worked Solutions
1. Convert the recurring decimal before dividing
B. Set t = 0.363636…. Then 100t − t = 36, so t = 36/99 = 4/11. Dividing by 2/3 gives (4/11)(3/2) = 6/11. Multiplying 6/11 by 2/3 recovers 4/11. A loses a factor of 3, and D is the reciprocal of the correct answer.
2. Pair prime factors to make a square
B. 48 = 2⁴ × 3, so k must supply an odd exponent of 3 and even exponents of other primes. The least is 3, giving 144 = 12². Although 12 also works, it is not least; 2 and 6 leave an odd exponent of 2.
3. Rationalise using a conjugate
C. Multiply numerator and denominator by √3 + √2. The denominator is 3 − 2 = 1. Check: (√3 + √2)(√3 − √2) = 1. B and D incorrectly combine unlike surds.
4. Reverse the combined percentage multiplier
B. The multiplier after both changes is (11/10)(9/10) = 99/100. Restore with 100/99, an increase of 1/99 of the final value. Expressed as a percentage, this is (100/99)%. A confuses 1% of the original with the needed fraction of the reduced value; D omits the conversion to percent.
5. Link ratios before taking a share of the total
C. Match B at 12 parts: A:B:C = 9:12:10. C is 10/31 of the whole. Check: 9/12 = 3/4 and 12/10 = 6/5. Ratios using just two parts do not give the share of the total.
6. Apply inverse-square scaling
B. y changes by 1/(3/2)² = 4/9. For example, x = 2 becomes 3, while y changes from k/4 to k/9. Their ratio is 4/9; inverse square scaling requires both inversion and squaring.
7. Bound the density using measurement intervals
The intervals are 1.95 ≤ m < 2.05 and 0.45 ≤ V < 0.55. Density m/V satisfies 1.95/0.55 < m/V < 2.05/0.45, hence 39/11 < density < 41/9 kg/litre. Both endpoints are excluded: the lower needs V = 0.55 and the upper needs m = 2.05. The nominal density 4 lies between these bounds.
8. Use prime factors to justify divisibility
Yes. If n = 0 the conclusion holds. Otherwise, prime factorisation of |n| shows that a factor of 2 in n² requires a factor of 2 in n, and a factor of 3 in n² requires a factor of 3 in n. Thus n is divisible by both 2 and 3, hence by 6. Conversely, n = 6t gives n² = 36t², which is divisible by 12.
9. Check candidates after squaring
Any solution has x ≥ 0. Squaring gives x² − 3x − 4 = 0, or (x − 4)(x + 1) = 0. The candidate 4 works since √16 = 4. The candidate −1 fails since √1 ≠ −1. Only x = 4.
10. Convert the area ratio to a volume ratio
The length ratio is 3:4, so the volume ratio is 27:64. One volume part is 81/27 = 3 litres; the larger capacity is 64 × 3 = 192 litres. Check: 81/192 = 27/64.
11. Test and prove the reciprocal comparison
The claim is false: a = 1, b = 2 gives 1/a = 1 > 1/2 = 1/b. In general, divide a < b by the positive product ab to obtain 1/b < 1/a. Thus taking reciprocals reverses order for positive numbers.
12. Use total distance divided by total time
Total distance is 120 km. Times are 60/30 = 2 hours and 60/60 = 1 hour, so average speed is 120/3 = 40 km/h. The arithmetic mean 45 km/h would apply for equal times, but the traveller spends twice as long at the slower speed. Check: 40 km/h for 3 hours covers 120 km.
5. Summary
- Choose a form that exposes structure before calculating.
- Carry domain, sign, unit and endpoint conditions into the final answer.
- Use a second calculation, substitution or counterexample to test reasoning.
- Revisit exact arithmetic for questions 1–2 and 8, surds for 3 and 9, proportion for 4–6 and 10–12, and bounds for 7.