MYP 3 · Maths
ALGEBRAIC OPERATIONS
75 questions across 7 sub-topics
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Algebraic notation
The language of mathematics
Collecting like terms
Generalising arithmetic
Algebraic substitution
Algebraic products
Algebraic fractions
Algebraic notation 11 questions
QUESTION 1
4 marks
Criterion A
Medium
Write an algebraic expression for each of the following:
a.
5 more than twice a number $x$
[1] b.
The product of $a$ and $b$, decreased by 7
[1] c.
Half of the sum of $p$ and $q$
[1] d.
3 less than the square of $n$
[1] Show complete worked solution
(a)
$$2x + 5$$
(b)
$$ab - 7$$
(c)
$$\frac{p+q}{2}$$
(d)
$$n^2 - 3$$
QUESTION 2
3 marks
Criterion A
Medium
A number $y$ is tripled, then 8 is subtracted, and the whole result is then halved.
a.
Write an algebraic expression for the final result, using brackets where needed.
[2] b.
Evaluate your expression when $y=10$.
[1] Show complete worked solution
(a)
$$\frac{3y-8}{2}$$
(b)
$$\frac{3(10)-8}{2}=\frac{22}{2}=11$$
QUESTION 3
6 marks
Criterion B
Medium
Investigate how to represent 3 consecutive integers algebraically, and the pattern in their sum.
a.
If the first of 3 consecutive integers is $n$, write expressions for the second and third integers.
[2] b.
Write and simplify an expression for the sum of the 3 consecutive integers.
[2] c.
Test your formula with $n=5,6,7$ (sum should be 18) and $n=10,11,12$ (sum should be 33). Does it work? What do you notice about $3n+3$?
[2] Show complete worked solution
(a)
Second: $n+1$. Third: $n+2$.
(b)
$$n+(n+1)+(n+2) = 3n+3$$
(c)
For $n=5$: $3(5)+3=18$. For $n=10$: $3(10)+3=33$. Notice $3n+3=3(n+1)$ — the sum of 3 consecutive integers is always 3 times the MIDDLE integer.
QUESTION 4
4 marks
Criterion C
Medium
A student writes '2x' to mean 'the digit 2 followed by the digit x', like writing the number 25 for $x=5$.
a.
Explain what '2x' actually means in algebra, and evaluate it correctly for $x=5$.
[2] b.
Explain why the student's interpretation would cause serious problems if $x$ could be a decimal or negative number (e.g. $x=0.5$ or $x=-3$).
[2] Show complete worked solution
(a)
In algebra, placing a number directly next to a variable means MULTIPLICATION — '2x' means '2 times x'. For $x=5$: $2x=2\times5=10$, not 25.
(b)
The 'digit-joining' interpretation only makes sense for single positive whole-number digits — it breaks down completely for $x=0.5$ (there's no digit '0.5') or $x=-3$ (you can't 'write a negative digit' next to the 2). The multiplication interpretation works consistently for ANY value of $x$.
QUESTION 5
5 marks
Criterion D
Medium
A taxi company charges a \$4.50 flat fee plus \$2.20 per kilometre travelled.
a.
Write an algebraic expression for the total cost $C$ of a trip of $k$ kilometres.
[2] b.
Use your expression to find the cost of an 8 km trip.
[1] c.
A customer has exactly \$30. Find the maximum whole number of kilometres they can travel.
[2] Show complete worked solution
(a)
$$C = 4.50 + 2.20k$$
(b)
$$C=4.50+2.20(8)=4.50+17.60=\$22.10$$
(c)
$4.50+2.20k \le 30 \Rightarrow 2.20k\le25.5 \Rightarrow k\le11.59$, so the maximum whole number of km is 11.
QUESTION 6
4 marks
Criterion A
Hard
Write an algebraic expression for each, using $x$ for the unknown number:
a.
7 less than three times the square of $x$
[1] b.
The sum of $x$ and 4, all divided by twice $x$
[1] c.
5 more than $x$, then the whole result tripled, then 2 subtracted
[2] Show complete worked solution
(a)
$$3x^2-7$$
(b)
$$\frac{x+4}{2x}$$
(c)
$$3(x+5)-2$$
QUESTION 7
4 marks
Criterion A
Hard
A gym membership costs a \$60 joining fee plus \$35 per month. A student membership gets a 20\% discount on the MONTHLY fee only (not the joining fee).
a.
Write an expression for the total cost of a standard membership for $m$ months.
[2] b.
Write a SEPARATE, simplified expression for the total cost of a STUDENT membership for $m$ months.
[2] Show complete worked solution
(a)
$$60+35m$$
(b)
Monthly student rate: $35\times0.8=28$. $$60+28m$$
QUESTION 8
6 marks
Criterion B
Hard
Investigate how to represent 3 consecutive EVEN integers algebraically, and find a pattern in their sum.
a.
If the smallest of 3 consecutive even integers is $2n$, write expressions for the other two.
[2] b.
Write an expression for the SUM of the three consecutive even integers, and simplify it.
[2] c.
Factor your simplified sum, and explain what this factored form tells you about consecutive-even-integer sums (i.e. what number must they always be a multiple of).
[2] Show complete worked solution
(a)
Second: $2n+2$. Third: $2n+4$.
(b)
$$2n+(2n+2)+(2n+4)=6n+6$$
(c)
$6n+6=6(n+1)$ — this shows the sum of any 3 consecutive even integers is always a multiple of 6.
QUESTION 9
3 marks
Criterion C
Hard
A student translates 'half of $x$ plus 3' as $\frac{x+3}{2}$, but their classmate writes $\frac{x}{2}+3$.
a.
Determine which student's expression correctly matches the ORIGINAL phrase, and explain the difference in meaning using bracket placement.
[3] Show complete worked solution
(a)
The classmate ($\frac{x}{2}+3$) is correct. 'Half of $x$' means $\frac{x}{2}$, and THEN '3' is added afterward, giving $\frac{x}{2}+3$. The first student's version, $\frac{x+3}{2}$, would actually match a DIFFERENT phrase: 'half of ($x$ plus 3)' — the position of the addition relative to the division changes the meaning entirely.
QUESTION 10
6 marks
Criterion D
Hard
A rideshare app charges a base fare of \$3.50 plus \$1.20 per km, plus a \$0.15 per-minute charge for time spent in traffic.
a.
Write an expression for the total fare, using $k$ for kilometres travelled and $t$ for minutes spent in traffic.
[2] b.
A trip covers 8km and includes 6 minutes in traffic. Find the total fare, then find how much of the SAME total fare could instead have been achieved with ZERO traffic time but MORE kilometres travelled (find the equivalent extra distance, to 1dp).
[4] Show complete worked solution
(a)
$$3.50+1.20k+0.15t$$
(b)
Actual fare: $3.50+1.20(8)+0.15(6)=3.50+9.60+0.90=\$14$. Traffic charge portion: $\$0.90$. Extra distance equivalent: $0.90\div1.20=0.75$km — so travelling $0.75$km further (at 0 traffic minutes) would cost the same as the 6 minutes of traffic did.
QUESTION 11
5 marks
Criterion D
Hard
A phone data plan costs \$25 base, plus \$4 per GB for usage beyond the included 5GB.
a.
Write an expression for total cost, using $g$ for GB used BEYOND the included 5GB (assume $g\ge0$).
[2] b.
A customer's bill was \$53. Using your expression, determine how many GB beyond the included amount they used, and hence their TOTAL data usage for the month (including the included 5GB).
[3] Show complete worked solution
(a)
$$25+4g$$
(b)
$25+4g=53 \Rightarrow 4g=28 \Rightarrow g=7$GB beyond the included amount. Total usage: $5+7=12$GB.
The language of mathematics 10 questions
QUESTION 1
3 marks
Criterion A
Medium
Translate each phrase into an algebraic expression:
a.
The sum of $x$ and 3, multiplied by 4
[1] b.
The quotient of $y$ and 6, increased by 2
[1] c.
9 decreased by the product of 3 and $z$
[1] Show complete worked solution
(a)
$$4(x+3)$$
(b)
$$\frac{y}{6}+2$$
(c)
$$9-3z$$
QUESTION 2
3 marks
Criterion A
Medium
Translate: 'The square of a number, minus 5 times the number.'
a.
Write this as an algebraic expression, using $n$ for the number.
[1] b.
Evaluate the expression for $n=6$ and for $n=-2$.
[2] Show complete worked solution
(a)
$$n^2-5n$$
(b)
$n=6$: $36-30=6$. $n=-2$: $4-(-10)=14$.
QUESTION 3
5 marks
Criterion B
Medium
Investigate different word phrases that can translate to the SAME algebraic expression, $3x+2$.
a.
Write two different word phrases that would both translate to $3x+2$.
[2] b.
Now write a word phrase for $3(x+2)$ (a DIFFERENT expression), and explain what makes it different from $3x+2$ despite looking similar in words.
[3] Show complete worked solution
(a)
Example phrases: 'triple a number, then add 2' and '2 more than 3 times a number' — both give $3x+2$.
(b)
Phrase: 'a number increased by 2, then tripled'. The key difference is where the brackets fall: $3(x+2)$ means the ADDITION happens first, then multiplication — giving $3x+6$ when expanded, which is different from $3x+2$.
QUESTION 4
5 marks
Criterion C
Medium
A classmate translates 'a number decreased by 5' and '5 decreased by a number' as the same expression, $x-5$.
a.
Explain the classmate's error, writing the correct expression for each phrase.
[3] b.
Verify the difference by evaluating both expressions at $x=8$.
[2] Show complete worked solution
(a)
'A number decreased by 5' means starting with $x$ and subtracting 5: $x-5$. But '5 decreased by a number' means starting with 5 and subtracting $x$: $5-x$. These are different expressions (in fact, $5-x=-(x-5)$).
(b)
$x-5=8-5=3$. $5-x=5-8=-3$. Confirmed — they give opposite (different) results.
QUESTION 5
4 marks
Criterion D
Medium
A phone plan is described as: 'a base cost of \$20, plus \$0.15 for every text message sent beyond the first 100 (which are free).'
a.
Write a formula for the monthly cost $C$ if $t$ texts are sent beyond the free 100 (i.e. $t\ge0$).
[2] b.
A customer sent 340 texts in total this month. Find their bill.
[2] Show complete worked solution
(a)
$$C = 20 + 0.15t$$
(b)
Texts beyond free 100: $340-100=240$. $$C=20+0.15(240)=20+36=\$56$$
QUESTION 6
5 marks
Criterion A
Hard
Translate each phrase into an algebraic expression, then simplify where possible:
a.
The square of the sum of $x$ and 3, minus the square of $x$
[3] b.
Twice the difference between $y$ and 5, increased by the product of $y$ and 3
[2] Show complete worked solution
(a)
$$(x+3)^2 - x^2 = x^2+6x+9-x^2 = 6x+9$$
(b)
$$2(y-5)+3y = 2y-10+3y=5y-10$$
QUESTION 7
7 marks
Criterion B
Hard
Investigate whether the phrase '3 less than double a number' ALWAYS means the same thing as 'double the amount that is 3 less than a number'.
a.
Translate '3 less than double a number $x$' into an algebraic expression.
[2] b.
Translate 'double the amount that is 3 less than a number $x$' into a different algebraic expression.
[2] c.
Test both expressions at $x=10$, and explain why the two phrases (despite using similar words) give different results.
[3] Show complete worked solution
(a)
$$2x-3$$
(b)
$$2(x-3)$$
(c)
First: $2(10)-3=17$. Second: $2(10-3)=14$. They differ because of WHEN the subtraction happens relative to the doubling — 'less than double' means subtract AFTER doubling, while 'double the amount that is less' means subtract BEFORE doubling. Word order and structure in English can be genuinely ambiguous, which is exactly why precise mathematical notation (using brackets) is needed to remove ambiguity.
QUESTION 8
5 marks
Criterion B
Hard
Investigate how many DIFFERENT English phrases can describe the SAME expression, $5(x+2)$.
a.
Write THREE different English phrases that would all correctly translate to $5(x+2)$.
[3] b.
Explain why, despite using very different WORDING, all three phrases must translate to the EXACT same algebraic expression.
[2] Show complete worked solution
(a)
Examples: 'five times the sum of a number and 2'; '2 more than a number, all multiplied by 5'; 'a number increased by 2, then the result quintupled'. (Any three phrases that correctly capture 'add 2 first, then multiply by 5' are valid.)
(b)
All three phrases describe the SAME sequence of mathematical operations in the SAME order (add 2 to the number, THEN multiply the result by 5) — different words can describe an identical process, and since the underlying mathematical operations and their order are what determines the expression, the wording variation doesn't change the algebra.
QUESTION 9
3 marks
Criterion C
Hard
A student translates 'the difference between $x$ and $y$' as $x-y$, always, without exception.
a.
Explain why this translation can be problematic if $y > x$ (giving a negative result), and discuss whether 'difference' in everyday mathematical language usually implies a POSITIVE result (a distance/magnitude) rather than a signed value.
[3] Show complete worked solution
(a)
If $y>x$, then $x-y$ gives a NEGATIVE number, but the everyday/common meaning of 'difference' (e.g. 'the difference in our heights') usually implies a positive gap or distance, regardless of which quantity is larger. A more universally safe translation of 'difference between $x$ and $y$' would use absolute value, $|x-y|$, which is guaranteed non-negative regardless of which value is bigger — though in ALGEBRA specifically, 'difference' can also legitimately mean the signed subtraction $x-y$, so context matters.
QUESTION 10
5 marks
Criterion D
Hard
An insurance company calculates a premium using the description: 'a base rate of \$400, reduced by \$25 for every year of a driver's claim-free history, down to a minimum of \$150.'
a.
Write an algebraic expression for the premium based on $y$ claim-free years, IGNORING the minimum for now.
[2] b.
Find the number of claim-free years at which the premium would reach exactly the \$150 minimum, and explain why the algebraic expression $400-25y$ alone is NOT a complete/accurate model of the real premium for drivers with MORE claim-free years than this.
[3] Show complete worked solution
(a)
$$400-25y$$
(b)
$400-25y=150 \Rightarrow 25y=250 \Rightarrow y=10$ years. Beyond 10 claim-free years, the plain formula $400-25y$ would predict a premium BELOW \$150 (even negative for very large $y$), which doesn't match the real policy — the true model needs a MINIMUM condition applied (e.g. premium $=\max(400-25y, 150)$), since a real insurance premium can never keep decreasing indefinitely.
Collecting like terms 11 questions
QUESTION 1
3 marks
Criterion A
Medium
Simplify each expression by collecting like terms:
a.
$5x + 3y - 2x + 7y$
[1] b.
$8a - 3b - 5a + 9b - 2a$
[1] c.
$4p^2 + 3p - p^2 - 5p + 6$
[1] Show complete worked solution
(a)
$$3x+10y$$
(b)
$$a+6b$$
(c)
$$3p^2-2p+6$$
QUESTION 2
4 marks
Criterion A
Medium
Simplify: $7x - 4y + 3x^2 - 2x + 5y - x^2$
a.
Group the like terms together (constants, $x$ terms, $y$ terms, $x^2$ terms).
[2] b.
Simplify fully.
[2] Show complete worked solution
(a)
$x^2$ terms: $3x^2-x^2$. $x$ terms: $7x-2x$. $y$ terms: $-4y+5y$.
(b)
$$2x^2+5x+y$$
QUESTION 3
5 marks
Criterion B
Medium
Investigate whether simplifying an expression changes its value when a specific number is substituted.
a.
Evaluate $5x+3-2x+7$ (unsimplified) at $x=4$.
[2] b.
Simplify $5x+3-2x+7$, then evaluate the simplified version at $x=4$.
[2] c.
Compare your two answers, and explain why this makes sense.
[1] Show complete worked solution
(a)
$5(4)+3-2(4)+7=20+3-8+7=22$
(b)
Simplified: $3x+10$. At $x=4$: $3(4)+10=22$.
(c)
Both give 22 — simplifying doesn't change the VALUE of an expression, only how it's written, since collecting like terms is just applying the distributive law in reverse.
QUESTION 4
4 marks
Criterion C
Medium
A student simplifies $3x + 2y$ as $5xy$.
a.
Explain why this is incorrect, referring to what 'like terms' means.
[2] b.
Verify the student's error by evaluating both $3x+2y$ and $5xy$ at $x=2, y=3$.
[2] Show complete worked solution
(a)
Only LIKE terms (terms with exactly the same variable part) can be combined by addition. $3x$ and $2y$ have different variables ($x$ vs $y$), so they are NOT like terms and cannot be added together into a single term — $3x+2y$ is already in simplest form.
(b)
$3x+2y=3(2)+2(3)=6+6=12$. $5xy=5(2)(3)=30$. These are very different (12 vs 30), confirming $3x+2y\ne5xy$.
QUESTION 5
4 marks
Criterion D
Medium
A gardener buys $x$ packets of tomato seeds at \$3 each and $y$ packets of lettuce seeds at \$2 each. On a second visit, they buy 2 more packets of tomato seeds and 3 more packets of lettuce seeds, at the same prices.
a.
Write an expression for the total cost of the FIRST visit.
[1] b.
Write an expression for the total cost of the SECOND visit.
[1] c.
Write and simplify an expression for the TOTAL cost across both visits.
[2] Show complete worked solution
(a)
$$3x+2y$$
(b)
$$3(x+2)+2(y+3)$$
(c)
$$3x+2y+3(x+2)+2(y+3) = 3x+2y+3x+6+2y+6 = 6x+4y+12$$
QUESTION 6
5 marks
Criterion A
Hard
Simplify each expression fully:
a.
$5x+3y-2x+7y-4$
[2] b.
$3(2x-y)+2(x+3y)$
[3] Show complete worked solution
(a)
$$3x+10y-4$$
(b)
$$6x-3y+2x+6y = 8x+3y$$
QUESTION 7
5 marks
Criterion A
Hard
Simplify $3x^2-2x+5-x^2+4x-8$, then evaluate the simplified expression at $x=3$.
a.
Simplify by collecting like terms.
[2] b.
Evaluate at $x=3$, and verify by ALSO substituting $x=3$ into the ORIGINAL (unsimplified) expression, confirming both give the same result.
[3] Show complete worked solution
(a)
$$2x^2+2x-3$$
(b)
Simplified: $2(9)+2(3)-3=18+6-3=21$. Original: $3(9)-2(3)+5-9+4(3)-8=27-6+5-9+12-8=21$ — matches, confirming the simplification was correct.
QUESTION 8
5 marks
Criterion B
Hard
Investigate whether the ORDER in which you collect like terms (e.g. $x$ terms first vs constant terms first) ever affects the final simplified result.
a.
Simplify $7x+4-3x+9-2x$ by collecting the $x$ terms FIRST, then the constants.
[2] b.
Now simplify the SAME expression by collecting the constants FIRST, then the $x$ terms. Confirm you reach the same final answer, and explain why the order of collection never matters (referencing the commutative property of addition).
[3] Show complete worked solution
(a)
$x$ terms: $7x-3x-2x=2x$. Constants: $4+9=13$. Result: $2x+13$.
(b)
Constants first: $4+9=13$. $x$ terms: $7x-3x-2x=2x$. Result: $2x+13$ — identical. This is guaranteed by the COMMUTATIVE property of addition: since addition can be performed in any order without changing the total, grouping and combining like terms in any sequence always produces the same final simplified expression.
QUESTION 9
5 marks
Criterion C
Hard
A student simplifies $4x+3x^2$ as $7x^2$ (adding the coefficients and combining the different powers of $x$ into one term).
a.
Explain precisely why $4x$ and $3x^2$ are NOT like terms, and therefore cannot be combined into a single term.
[2] b.
Verify the error by evaluating both $4x+3x^2$ and the student's incorrect $7x^2$ at $x=2$.
[3] Show complete worked solution
(a)
Like terms must have the EXACT same variable part, including the same POWER of the variable. $4x$ has $x$ to the power 1, while $3x^2$ has $x$ to the power 2 — these are fundamentally different quantities (like trying to add a length to an area), so they cannot be combined; $4x+3x^2$ is already in its simplest form.
(b)
$4x+3x^2$ at $x=2$: $8+12=20$. Student's $7x^2$ at $x=2$: $7(4)=28$. These are different ($20\ne28$), confirming the terms cannot be combined.
QUESTION 10
4 marks
Criterion C
Hard
A classmate says: 'like terms must have the exact same NUMBER of variables multiplied together, so $xy$ and $x^2y$ can be combined since they both have $x$ and $y$ in them.'
a.
Explain why $xy$ and $x^2y$ are actually NOT like terms, correcting the classmate's misunderstanding about what makes terms 'alike'.
[2] b.
Give ONE example of a term that WOULD be a genuine like term with $xy$ (i.e. could be legitimately combined with it), and explain why it qualifies.
[2] Show complete worked solution
(a)
Like terms require the EXACT same combination of variables raised to the EXACT same powers — $xy$ means $x^1y^1$, while $x^2y$ means $x^2y^1$. Even though both terms involve $x$ and $y$, the DIFFERENT power of $x$ (1 vs 2) makes them fundamentally different terms, just like $4x$ and $3x^2$ from before.
(b)
Example: $5xy$ (or any multiple of $xy$, such as $-2xy$ or $\frac{1}{2}xy$) — it qualifies because it has the exact same variable parts with the exact same powers ($x^1y^1$), differing only in the numerical coefficient, which is precisely what makes terms 'like' and combinable.
QUESTION 11
4 marks
Criterion D
Hard
A landscaper charges a callout fee of \$50, plus \$15 per hour of labour ($h$ hours), plus \$8 per bag of mulch ($b$ bags). A customer orders TWO separate jobs on the same day: Job 1 needs 3 hours and 5 bags; Job 2 needs 2 hours and 3 bags.
a.
Write a simplified SINGLE expression for the TOTAL cost of both jobs combined, in terms of a single callout fee (charged once) plus combined labour and mulch costs.
[4] Show complete worked solution
(a)
Total labour hours: $3+2=5$. Total mulch bags: $5+3=8$. Total cost (one callout fee): $$50+15(5)+8(8)=50+75+64=189$$
Generalising arithmetic 11 questions
QUESTION 1
4 marks
Criterion A
Medium
Consider the pattern: $1\times3=3$, $2\times4=8$, $3\times5=15$, $4\times6=24$.
a.
Describe the pattern in words (what is being multiplied each time?).
[1] b.
Write a general algebraic expression for the $n$th term of this pattern.
[1] c.
Use your expression to find the 10th term of the pattern, then verify by direct multiplication.
[2] Show complete worked solution
(a)
Each time, a number is multiplied by the number 2 more than itself.
(b)
$$n(n+2)$$
(c)
$10(10+2)=10\times12=120$. Direct check: $10\times12=120$.
QUESTION 2
4 marks
Criterion A
Medium
Prove, using algebra, that the sum of any two consecutive even numbers is always a multiple of 4... or is it?
a.
Let the two consecutive even numbers be $2n$ and $2n+2$. Write and simplify an expression for their sum.
[2] b.
Is $4n+2$ always a multiple of 4? Test with $n=1$ (giving even numbers 2 and 4) to check.
[2] Show complete worked solution
(a)
$$2n+(2n+2) = 4n+2$$
(b)
$n=1$: sum $=2+4=6$. Is 6 a multiple of 4? No. So the original claim in the question title is FALSE — $4n+2$ is actually always 2 more than a multiple of 4, not a multiple of 4 itself.
QUESTION 3
5 marks
Criterion B
Medium
Investigate the claim: 'the sum of any 5 consecutive integers is always a multiple of 5.'
a.
Let the 5 consecutive integers be $n, n+1, n+2, n+3, n+4$. Write and simplify an expression for their sum.
[2] b.
Factor your expression to show it is always a multiple of 5.
[1] c.
Test the claim with the integers 6,7,8,9,10 (sum should be a multiple of 5).
[2] Show complete worked solution
(a)
$$n+(n+1)+(n+2)+(n+3)+(n+4) = 5n+10$$
(b)
$$5n+10 = 5(n+2)$$
(c)
$6+7+8+9+10=40=5\times8$, confirming the claim (and matching $5(n+2)$ with $n=6$: $5(8)=40$).
QUESTION 4
5 marks
Criterion C
Medium
A classmate says: 'I checked that $n^2+n+41$ gives a prime number for $n=0,1,2,3,4$, so it must give a prime number for EVERY value of $n$.'
a.
Verify the classmate's claim for $n=0,1,2$ by calculating $n^2+n+41$ in each case.
[2] b.
Explain why checking a few cases is NOT enough to prove a general algebraic claim, even though this particular formula happens to work for many small values of $n$ (this is a famous example in mathematics!).
[3] Show complete worked solution
(a)
$n=0$: $0+0+41=41$ (prime). $n=1$: $1+1+41=43$ (prime). $n=2$: $4+2+41=47$ (prime).
(b)
Testing specific numbers can only show a pattern HOLDS for those cases — it can never prove the pattern holds for ALL possible values, since there are infinitely many numbers to check. A single counterexample (and one does exist for this formula, at $n=40$, where $n^2+n+41=41^2$, which is NOT prime) is enough to disprove a general claim, no matter how many earlier cases worked.
QUESTION 5
4 marks
Criterion D
Medium
A conveyor belt at a factory moves at a constant speed. It was observed that after 3 seconds, a box has moved 1.5 m, and after 7 seconds, the same box has moved 3.5 m.
a.
Assuming the distance moved is directly proportional to time, write a general formula for distance $d$ (in m) after $t$ seconds, in the form $d=kt$.
[2] b.
Use your formula to predict how far the box will have moved after 1 minute (60 seconds), and comment on whether this seems like a reasonable real-world prediction.
[2] Show complete worked solution
(a)
Using either data point: $k=1.5/3=0.5$ (checking: $0.5\times7=3.5$). So $$d=0.5t$$
(b)
$d=0.5(60)=30$ m. This seems reasonable for a factory conveyor belt moving at a slow, steady pace over a full minute.
QUESTION 6
6 marks
Criterion A
Hard
Prove, using algebra, that the sum of the squares of any 2 consecutive ODD numbers is always 2 more than a multiple of 8.
a.
Let the two consecutive odd numbers be $2n+1$ and $2n+3$. Expand and simplify $(2n+1)^2+(2n+3)^2$.
[3] b.
Rewrite $8n^2+16n+10$ in the form $8(\ldots)+2$, to show it is always 2 more than a multiple of 8.
[2] c.
Verify this holds for the pair 5 and 7.
[1] Show complete worked solution
(a)
$$(2n+1)^2+(2n+3)^2 = 4n^2+4n+1+4n^2+12n+9 = 8n^2+16n+10$$
(b)
$$8n^2+16n+10 = 8(n^2+2n+1)+2 = 8(n+1)^2+2$$
(c)
$5^2+7^2=25+49=74=8(9)+2$.
QUESTION 7
5 marks
Criterion B
Hard
Investigate the pattern in $n^2-n$ for various whole numbers $n$, and what this reveals about divisibility.
a.
Calculate $n^2-n$ for $n=2,3,4,5,6$.
[2] b.
Every result (2,6,12,20,30) is EVEN. Factor $n^2-n$ algebraically, and use the factored form to explain WHY $n^2-n$ is always even, for any whole number $n$.
[3] Show complete worked solution
(a)
$n=2$: $2$. $n=3$: $6$. $n=4$: $12$. $n=5$: $20$. $n=6$: $30$.
(b)
$n^2-n=n(n-1)$ — this is a product of TWO CONSECUTIVE integers. Since any two consecutive integers must include exactly one even number, their product must always be even (an even number times anything is always even).
QUESTION 8
6 marks
Criterion B
Hard
Investigate whether the claim 'the product of any 3 consecutive integers is always divisible by 6' is true.
a.
Test the claim with 4,5,6 and with 7,8,9.
[2] b.
Explain WHY this must always be true, by arguing separately that the product of 3 consecutive integers must always be divisible by 2 (since one of any 2 consecutive integers is even), AND always divisible by 3 (since one of any 3 consecutive integers is a multiple of 3), and hence divisible by $2\times3=6$.
[4] Show complete worked solution
(a)
$4\times5\times6=120=6\times20$. $7\times8\times9=504=6\times84$.
(b)
Among any 3 consecutive integers, AT LEAST one must be even (divisibility by 2 guaranteed), and exactly one out of every 3 consecutive integers must be a multiple of 3 (since multiples of 3 occur every third number, guaranteeing divisibility by 3). Since the product is guaranteed divisible by both 2 AND 3 (which share no common factor), it must be divisible by their product, $2\times3=6$.
QUESTION 9
5 marks
Criterion C
Hard
A student claims: 'I tested $n^2+n+1$ for $n=1,2,3,4,5$ and got 3,7,13,21,31 — all of these look prime, so this formula must always produce primes.'
a.
Verify the student's 5 calculated values are correct, then TEST $n=4$ specifically to see if 21 is actually prime.
[3] b.
Explain the broader lesson: why must a claim like 'this formula always gives primes' be checked EXTREMELY carefully (ideally proven, not just tested), even when early results look promising?
[2] Show complete worked solution
(a)
$n=1$:3, $n=2$:7, $n=3$:13, $n=4$:21, $n=5$:31 — all calculated correctly. However, $21=3\times7$ is NOT prime — the student's claim that 'all of these look prime' was actually incorrect even for the tested values, showing they didn't check carefully enough.
(b)
Testing only a handful of small cases can create a false sense of confidence — as shown here, even the student's OWN test cases contained a counterexample they missed. A pattern appearing to hold for several values is never a substitute for genuine proof; formulas that produce primes for many small inputs (like the famous $n^2+n+41$) eventually fail for larger $n$, which is why rigorous verification of every single claimed case matters, not just a quick glance.
QUESTION 10
7 marks
Criterion D
Hard
A company's revenue model claims: 'Revenue (in thousands) after $n$ months of operation follows $R=2n^2+5n+3$.'
a.
Find the revenue after 6 months, and after 12 months.
[3] b.
The company wants to predict revenue after 5 YEARS (60 months) using the same formula, but a financial analyst warns this could be highly unreliable. Calculate the predicted 60-month revenue, and explain, referencing the SHAPE of a quadratic function, why extrapolating this formula so far into the future is risky.
[4] Show complete worked solution
(a)
$R(6)=2(36)+30+3=72+30+3=105$ thousand. $R(12)=2(144)+60+3=288+60+3=351$ thousand.
(b)
$R(60)=2(3600)+300+3=7200+300+3=7503$ thousand (over \$7.5 million). This is risky because a quadratic model grows increasingly steeply (the growth RATE itself keeps increasing) — real business revenue growth typically slows or plateaus over time due to market saturation, competition, or capacity limits, so a model fitted to early data is very unlikely to remain accurate when extrapolated many years into the future; the quadratic's ever-accelerating growth is not realistic indefinitely.
QUESTION 11
7 marks
Criterion D
Hard
A staircase is built so that step $n$ has a total rise, measured from the ground, of $R(n)=180n$ mm, for $n=1,2,3,\ldots$
a.
Find the rise of step 15.
[2] b.
Building regulations require the TOTAL rise of a staircase (top step's rise) to be between 2500mm and 3200mm for a standard single flight. Using your formula, find the RANGE of step numbers $n$ that would satisfy this regulation, and comment on whether a 170-step staircase (unrealistically large) would ever actually be needed in practice, even though the formula could technically calculate its rise.
[5] Show complete worked solution
(a)
$$R(15)=180(15)=\boxed{2700\text{ mm}}.$$
(b)
$$2500\le180n\le3200$$
$$13.888\ldots\le n\le17.777\ldots$$
Since $n$ is a whole number, $\boxed{14\le n\le17}$. A 170-step flight would rise $180(170)=30{,}600$ mm, far above the stated range.
Algebraic substitution 10 questions
QUESTION 1
3 marks
Criterion A
Medium
Given $a=4$, $b=-3$, $c=2$, evaluate each expression:
a.
$2a+3b$
[1] b.
$a^2-bc$
[1] c.
$\dfrac{a+b}{c}$
[1] Show complete worked solution
(a)
$$2(4)+3(-3)=8-9=-1$$
(b)
$$4^2-(-3)(2)=16+6=22$$
(c)
$$\frac{4+(-3)}{2}=\frac{1}{2}$$
QUESTION 2
4 marks
Criterion A
Medium
The formula for the area of a trapezium is $A=\frac{1}{2}(a+b)h$, where $a,b$ are the parallel side lengths and $h$ is the height.
a.
Find the area when $a=8$ cm, $b=12$ cm, and $h=5$ cm.
[2] b.
A second trapezium has $a=6$ cm, $h=4$ cm, and an area of 32 cm$^2$. Find $b$.
[2] Show complete worked solution
(a)
$$A=\frac{1}{2}(8+12)(5)=\frac{1}{2}(20)(5)=50 \text{ cm}^2$$
(b)
$32=\frac{1}{2}(6+b)(4) \Rightarrow 32=2(6+b) \Rightarrow 16=6+b \Rightarrow b=10$ cm.
QUESTION 3
6 marks
Criterion B
Medium
Investigate how the value of $2x^2$ changes as $x$ increases, compared to how $(2x)^2$ changes.
a.
Evaluate $2x^2$ and $(2x)^2$ for $x=1,2,3$.
[3] b.
Notice that $(2x)^2$ is always exactly double $2x^2$ in each case. Explain algebraically why this happens.
[3] Show complete worked solution
(a)
$x=1$: $2x^2=2$, $(2x)^2=4$. $x=2$: $2x^2=8$, $(2x)^2=16$. $x=3$: $2x^2=18$, $(2x)^2=36$.
(b)
$(2x)^2 = 2^2 \times x^2 = 4x^2$, while $2x^2$ is just $2\times x^2$. Since $4x^2 = 2\times(2x^2)$, $(2x)^2$ will always be exactly double $2x^2$, for any value of $x$.
QUESTION 4
4 marks
Criterion C
Medium
A student substitutes $x=3$ into $5-2x$ and writes: $5-2\times3 = 3\times3=9$.
a.
Identify the error in the student's working, and give the correct answer.
[2] b.
Explain, using the order of operations rules, exactly why the multiplication must be done before the subtraction here.
[2] Show complete worked solution
(a)
The student incorrectly combined $5-2$ into $3$ before multiplying — but order of operations requires multiplication before subtraction. Correct working: $5-2\times3=5-6=-1$.
(b)
In the expression $5-2\times3$, there are no brackets, so standard order of operations applies: multiplication and division are always performed before addition and subtraction, regardless of the order they're written left to right.
QUESTION 5
5 marks
Criterion D
Medium
The speed of a moving object is given by $v = u + at$, where $u$ is initial speed (m/s), $a$ is acceleration (m/s$^2$), and $t$ is time (s).
a.
A car starts at $u=5$ m/s and accelerates at $a=2.5$ m/s$^2$. Find its speed after $t=8$ seconds.
[2] b.
The car's speed limit is 90 km/h. Convert this to m/s (1 km/h $\approx$ 0.278 m/s), and determine after how many seconds (from the same start) the car would reach this limit.
[3] Show complete worked solution
(a)
$$v=5+2.5(8)=5+20=25 \text{ m/s}$$
(b)
$90 \times 0.278 \approx 25.0$ m/s. Since $v=25$ m/s was already reached at $t=8$ s (from part a), the car reaches the speed limit at approximately $t=8$ seconds.
QUESTION 6
6 marks
Criterion A
Hard
Given $a=-3$, $b=4$, $c=-2$, evaluate each expression:
a.
$2a^2-3bc$
[2] b.
$\dfrac{a-c}{b+c}$
[2] c.
$(a+b)^2-c^2$
[2] Show complete worked solution
(a)
$$2(-3)^2-3(4)(-2) = 18+24=42$$
(b)
$$\frac{-3-(-2)}{4+(-2)}=\frac{-1}{2}$$
(c)
$$(-3+4)^2-(-2)^2=1-4=-3$$
QUESTION 7
6 marks
Criterion A
Hard
The formula for the distance between two points is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
a.
Find the distance between the points $(1,4)$ and $(3,7)$.
[2] b.
A third point, $(x, 10)$, is exactly $\sqrt{13}$ units from $(3,7)$ as well, with $x>3$. Set up and solve an equation to find $x$.
[4] Show complete worked solution
(a)
$$d=\sqrt{(3-1)^2+(7-4)^2}=\sqrt{4+9}=\sqrt{13}\approx3.61$$
(b)
$\sqrt{(x-3)^2+(10-7)^2}=\sqrt{13} \Rightarrow (x-3)^2+9=13 \Rightarrow (x-3)^2=4 \Rightarrow x-3=\pm2$. Since $x>3$: $x=5$.
QUESTION 8
6 marks
Criterion B
Hard
Investigate how the value of $a^3+2a^2b-b^2$ changes when $a$ and $b$ swap their VALUES (not their positions in the formula).
a.
Evaluate $a^3+2a^2b-b^2$ at $a=2,b=-1$, then again at $a=-1,b=2$ (values swapped).
[3] b.
Interesting — both give $-1$! Test ONE more swap-pair, e.g. $a=3,b=1$ then $a=1,b=3$, to see if this 'swap gives the same result' pattern continues, or if the first case was a coincidence.
[3] Show complete worked solution
(a)
At $a=2,b=-1$: $8+2(4)(-1)-1=8-8-1=-1$. At $a=-1,b=2$: $-1+2(1)(2)-4=-1+4-4=-1$.
(b)
$a=3,b=1$: $27+2(9)(1)-1=27+18-1=44$. $a=1,b=3$: $1+2(1)(3)-9=1+6-9=-2$. These are DIFFERENT ($44\ne-2$) — so the earlier match was just a coincidence for those SPECIFIC numbers, not a general property of this expression.
QUESTION 9
5 marks
Criterion C
Hard
A student substitutes $x=-2$ into $x^2$ and gets $-4$ (calculating $-2\times2$ instead of $(-2)\times(-2)$).
a.
Explain the student's error precisely, and give the correct value of $x^2$ when $x=-2$.
[3] b.
Contrast this with $-x^2$ (note: no brackets around the base) when $x=2$. Explain why $-x^2$ and $(-x)^2$ can give DIFFERENT results, using this example.
[2] Show complete worked solution
(a)
$x^2$ means $x\times x$ — substituting $x=-2$ gives $(-2)\times(-2)$, NOT $-2\times2$. The student seems to have only used ONE negative sign instead of squaring the entire negative value. Correct: $(-2)^2=(-2)\times(-2)=4$.
(b)
$-x^2$ at $x=2$ means $-(x^2)=-(4)=-4$ (square FIRST, then negate). $(-x)^2$ at $x=2$ means $(-2)^2=4$ (negate first, THEN square). These differ because of where the negative sign sits relative to the squaring operation — without brackets, exponents apply before the leading negative sign, following order of operations.
QUESTION 10
6 marks
Criterion D
Hard
A projectile's height (in metres) after $t$ seconds is modelled by $h=20t-5t^2$.
a.
Find the height after 1 second and after 3 seconds.
[2] b.
Interesting — both give 15m! Find the height after 2 seconds, and use ALL THREE results (at $t=1,2,3$) to describe the general SHAPE of the projectile's flight path, and estimate approximately when it reaches its highest point.
[4] Show complete worked solution
(a)
$h(1)=20-5=15$m. $h(3)=60-45=15$m.
(b)
$h(2)=40-20=20$m — higher than both $t=1$ and $t=3$. This confirms the height rises then falls (a parabolic arc), with the peak somewhere between $t=1$ and $t=3$ — and since $h(1)=h(3)=15$m exactly (equal heights on either side), the peak must occur exactly halfway between them, at $t=2$ seconds, matching where we found the maximum height of 20m.
Algebraic products 11 questions
QUESTION 1
4 marks
Criterion A
Medium
Simplify each product:
a.
$3x \times 4y$
[1] b.
$5a \times (-2a)$
[1] c.
$(3x)^2$
[1] d.
$2p \times 3q \times 4p$
[1] Show complete worked solution
(a)
$$12xy$$
(b)
$$-10a^2$$
(c)
$$9x^2$$
(d)
$$24p^2q$$
QUESTION 2
4 marks
Criterion A
Medium
A rectangle has width $4x$ and length $7x$.
a.
Write and simplify an expression for the area of the rectangle.
[2] b.
If $x=3$ cm, find the numerical area, and verify it matches substituting $x=3$ directly into your simplified expression.
[2] Show complete worked solution
(a)
$$\text{Area} = 4x \times 7x = 28x^2$$
(b)
Width $=12$ cm, length $=21$ cm, area $=12\times21=252$ cm$^2$. Check via formula: $28(3)^2=28\times9=252$ cm$^2$.
QUESTION 3
4 marks
Criterion B
Medium
Investigate the pattern when multiplying powers of the same variable, e.g. $x^2 \times x^3$.
a.
Expand $x^2 \times x^3$ fully (writing out all the individual $x$ factors) to find the simplified result.
[2] b.
Repeat for $x^3\times x^4$ and $x^2\times x^5$, and state the pattern connecting the original exponents to the final exponent.
[2] Show complete worked solution
(a)
$x^2\times x^3 = (x\times x)\times(x\times x\times x) = x\times x\times x\times x\times x = x^5$
(b)
$x^3\times x^4=x^7$. $x^2\times x^5=x^7$. Pattern: when multiplying powers of the same base, ADD the exponents: $x^m\times x^n=x^{m+n}$.
QUESTION 4
4 marks
Criterion C
Medium
A student simplifies $(2x)^2$ as $2x^2$.
a.
Explain the student's error, and give the correct simplification.
[2] b.
Verify the difference by evaluating both $(2x)^2$ and $2x^2$ at $x=3$.
[2] Show complete worked solution
(a)
$(2x)^2$ means $(2x)\times(2x)$, where BOTH the 2 and the $x$ get squared: $=4x^2$, not $2x^2$. The student forgot that the exponent applies to everything inside the brackets, including the coefficient 2.
(b)
$(2x)^2=(6)^2=36$. $2x^2=2(9)=18$. These are different (36 vs 18), confirming the error.
QUESTION 5
4 marks
Criterion D
Medium
A square sheet of metal has side length $5x$ cm. A smaller square with side length $2x$ cm is cut from one corner.
a.
Write and simplify an expression for the area of the original large square.
[1] b.
Write and simplify an expression for the area of the small square cut out.
[1] c.
Write a simplified expression for the remaining area of metal, and evaluate it when $x=4$ cm.
[2] Show complete worked solution
(a)
$$(5x)^2=25x^2 \text{ cm}^2$$
(b)
$$(2x)^2=4x^2 \text{ cm}^2$$
(c)
Remaining area $=25x^2-4x^2=21x^2$. At $x=4$: $21(16)=336$ cm$^2$.
QUESTION 6
6 marks
Criterion A
Hard
Simplify each product:
a.
$(2x)^3$
[2] b.
$3x^2 \times 4x^3$
[2] c.
$(-3x)^2 \times 2x$
[2] Show complete worked solution
(a)
$$8x^3$$
(b)
$$12x^5$$
(c)
$$9x^2 \times 2x = 18x^3$$
QUESTION 7
8 marks
Criterion B
Hard
Investigate the pattern in $(x+a)^2-(x-a)^2$ for different values of $a$.
a.
Fully expand $(x+5)^2-(x-5)^2$.
[3] b.
Repeat with $(x+3)^2-(x-3)^2$ and $(x+7)^2-(x-7)^2$.
[3] c.
State the general pattern: what is $(x+a)^2-(x-a)^2$ in terms of $x$ and $a$? Verify your general rule using one of your specific results.
[2] Show complete worked solution
(a)
$(x^2+10x+25)-(x^2-10x+25) = 20x$.
(b)
$(x+3)^2-(x-3)^2=12x$. $(x+7)^2-(x-7)^2=28x$.
(c)
$(x+a)^2-(x-a)^2 = 4ax$. Check with $a=5$: $4(5)x=20x$, matching part (a).
QUESTION 8
5 marks
Criterion B
Hard
Investigate whether $(x+y)^3$ equals $x^3+y^3$ (i.e. whether cubing 'distributes' over addition, the way some students assume).
a.
Test with $x=2, y=3$: calculate $(x+y)^3$ and separately $x^3+y^3$.
[2] b.
Fully expand $(x+y)^3=(x+y)(x+y)(x+y)$ step by step to find the CORRECT general expansion, confirming it does NOT simplify to just $x^3+y^3$.
[3] Show complete worked solution
(a)
$(2+3)^3=5^3=125$. $2^3+3^3=8+27=35$. These are very different ($125\ne35$).
(b)
$(x+y)(x+y)=x^2+2xy+y^2$. Then $(x^2+2xy+y^2)(x+y) = x^3+x^2y+2x^2y+2xy^2+xy^2+y^3 = x^3+3x^2y+3xy^2+y^3$. This is clearly NOT just $x^3+y^3$ — it has two extra 'cross terms' ($3x^2y$ and $3xy^2$) that a naive 'distribute the power' approach would miss.
QUESTION 9
5 marks
Criterion C
Hard
A student simplifies $2x \times 3x$ as $6x$ (adding the coefficients of $x$ together, treating multiplication like addition).
a.
Explain the student's error, distinguishing between how coefficients behave under MULTIPLICATION versus ADDITION of like terms.
[3] b.
Verify the correct answer ($6x^2$) using $x=2$, comparing to the student's incorrect $6x$.
[2] Show complete worked solution
(a)
The student is confusing two different rules: for ADDITION of like terms ($2x+3x$), you DO add the coefficients, giving $5x$. But for MULTIPLICATION ($2x\times3x$), you multiply the coefficients (2×3=6) AND apply the index law for the variable part ($x\times x=x^2$), giving $6x^2$, not $6x$.
(b)
$2x\times3x$ at $x=2$: $4\times6=24$. Correct formula $6x^2$ at $x=2$: $6(4)=24$. Student's $6x$ at $x=2$: $12\ne24$ — confirms the error.
QUESTION 10
5 marks
Criterion C
Hard
A classmate says: '$(x^2)^3$ and $x^2\times x^3$ both involve the numbers 2 and 3 with $x$, so they must be equal.'
a.
Calculate both expressions using the correct index laws, and determine whether the classmate is right.
[3] b.
Verify numerically at $x=2$: calculate $(2^2)^3$ and $2^2\times2^3$ directly.
[2] Show complete worked solution
(a)
$(x^2)^3=x^{2\times3}=x^6$ (power of a power: MULTIPLY exponents). $x^2\times x^3=x^{2+3}=x^5$ (product of powers: ADD exponents). These are NOT equal ($x^6\ne x^5$), so the classmate is wrong.
(b)
$(2^2)^3=4^3=64$. $2^2\times2^3=4\times8=32$. Different ($64\ne32$), confirming the two expressions are genuinely different.
QUESTION 11
6 marks
Criterion D
Hard
A rectangular garden bed's length is $3x$ metres and width is $2x$ metres. A second, similar bed has BOTH dimensions doubled (length $6x$, width $4x$).
a.
Find the area of each bed, in terms of $x$.
[3] b.
Find the RATIO of Bed 2's area to Bed 1's area, and explain why doubling BOTH dimensions doesn't just double the area (connect this to why the ratio isn't simply 2).
[3] Show complete worked solution
(a)
Bed 1: $3x\times2x=6x^2$. Bed 2: $6x\times4x=24x^2$.
(b)
Ratio $=\frac{24x^2}{6x^2}=4$. Doubling both dimensions doesn't just double the area because area depends on the PRODUCT of two dimensions — doubling each of two multiplied quantities multiplies the result by $2\times2=4$, not just 2. This is a general principle: scaling both dimensions of a 2D shape by a factor $k$ scales the area by $k^2$.
Algebraic fractions 11 questions
QUESTION 1
3 marks
Criterion A
Medium
Simplify each algebraic fraction fully:
a.
$\dfrac{6x}{9}$
[1] b.
$\dfrac{8ab}{12a}$
[1] c.
$\dfrac{15x^2}{5x}$
[1] Show complete worked solution
(a)
$$\frac{2x}{3}$$
(b)
For $a\ne0$, $$\frac{8ab}{12a}=\frac{2b}{3}.$$ The restriction $a\ne0$ is required because the original denominator is zero when $a=0$.
(c)
For $x\ne0$, $$\frac{15x^2}{5x}=3x.$$ The simplified expression agrees with the original fraction only on its original domain, $x\ne0$.
QUESTION 2
3 marks
Criterion A
Medium
Simplify $\dfrac{12x^2y}{18xy^2}$, showing full working.
a.
Find the HCF of the numerical coefficients (12 and 18), and simplify the numerical part.
[1] b.
Simplify the $x$ terms and the $y$ terms separately (using index laws), then combine everything into the final simplified fraction.
[2] Show complete worked solution
(a)
HCF$(12,18)=6$, so $\frac{12}{18}=\frac{2}{3}$
(b)
The original denominator $18xy^2$ requires $x\ne0$ and $y\ne0$. Then $$\frac{x^2}{x}=x,\qquad\frac{y}{y^2}=\frac1y,$$ so $$\frac{12x^2y}{18xy^2}=\boxed{\frac{2x}{3y}},\qquad x\ne0,\ y\ne0.$$
QUESTION 3
5 marks
Criterion B
Medium
Investigate what happens to an algebraic fraction when the SAME factor is present in both the numerator and denominator.
a.
Simplify $\dfrac{x(x+3)}{2(x+3)}$, explaining which factor cancels.
[2] b.
Now try to simplify $\dfrac{x+3}{2+3}$. Does anything cancel here? Explain the difference between this and part (a).
[3] Show complete worked solution
(a)
The factor $(x+3)$ appears in both the numerator and denominator, so it cancels (as long as $x\ne-3$): $$\frac{x(x+3)}{2(x+3)} = \frac{x}{2}$$
(b)
Nothing cancels here — $\frac{x+3}{2+3}=\frac{x+3}{5}$, which cannot be simplified further. The key difference is that in part (a), $(x+3)$ was a FACTOR (multiplied) in both parts, while here, 3 is just a TERM being added — you can only cancel common FACTORS, not individual terms within a sum.
QUESTION 4
4 marks
Criterion C
Medium
A student simplifies $\dfrac{x+4}{x}$ by 'cancelling the $x$', writing the answer as $4$.
a.
Explain why this cancellation is incorrect.
[2] b.
Verify the error by evaluating both $\dfrac{x+4}{x}$ and $4$ at $x=2$.
[2] Show complete worked solution
(a)
You can only cancel a factor that multiplies the ENTIRE numerator and the entire denominator. Here, $x$ is only part of a SUM in the numerator ($x+4$), not a factor of the whole numerator, so it cannot be cancelled with the $x$ in the denominator.
(b)
$\dfrac{x+4}{x}=\dfrac{6}{2}=3$. The claimed 'simplified' answer of 4 does not match (3 vs 4), confirming the cancellation was invalid.
QUESTION 5
4 marks
Criterion D
Medium
A recipe requires $\dfrac{3x}{4}$ cups of flour per batch, where $x$ represents the number of people being served.
a.
Simplify the expression for the amount of flour needed per person (i.e. divide by $x$), and interpret what this simplified fraction means.
[2] b.
A caterer is serving 24 people. Use the ORIGINAL expression (not the per-person one) to find the total flour needed.
[2] Show complete worked solution
(a)
$$\dfrac{3x}{4} \div x = \dfrac{3x}{4x} = \dfrac{3}{4}$$ This means each person requires $\frac{3}{4}$ cup of flour, regardless of how many people are being served — a constant amount per person.
(b)
$$\dfrac{3(24)}{4} = \dfrac{72}{4}=18 \text{ cups}$$
QUESTION 6
7 marks
Criterion A
Hard
Simplify each algebraic fraction, factoring first where needed:
a.
$\dfrac{x^2+5x}{x}$
[2] b.
$\dfrac{x^2-9}{x-3}$
[3] c.
$\dfrac{2x^2+6x}{4x}$
[2] Show complete worked solution
(a)
Factor the numerator and state the original restriction: $$\frac{x^2+5x}{x}=\frac{x(x+5)}{x}=\boxed{x+5},\qquad x\ne0.$$
(b)
The original denominator requires $x\ne3$. Then $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=\boxed{x+3},\qquad x\ne3.$$
(c)
The original denominator requires $x\ne0$. Then $$\frac{2x^2+6x}{4x}=\frac{2x(x+3)}{4x}=\boxed{\frac{x+3}{2}},\qquad x\ne0.$$
QUESTION 7
5 marks
Criterion B
Hard
Investigate what happens when you simplify $\dfrac{x^2-a^2}{x-a}$ for different values of $a$.
a.
Simplify $\dfrac{x^2-9}{x-3}$ (i.e. $a=3$) by factoring the numerator as a difference of squares.
[2] b.
Repeat for $\dfrac{x^2-16}{x-4}$ ($a=4$) and $\dfrac{x^2-49}{x-7}$ ($a=7$). State the general pattern for $\dfrac{x^2-a^2}{x-a}$.
[3] Show complete worked solution
(a)
The original expression is undefined at $x=3$. For $x\ne3$, $$\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=\boxed{x+3}.$$
(b)
$$\frac{x^2-16}{x-4}=x+4\quad(x\ne4),\qquad \frac{x^2-49}{x-7}=x+7\quad(x\ne7).$$ In general, $$\frac{x^2-a^2}{x-a}=\frac{(x-a)(x+a)}{x-a}=\boxed{x+a},\qquad x\ne a.$$
QUESTION 8
3 marks
Criterion C
Hard
A student simplifies $\dfrac{x+7}{7}$ by 'cancelling the 7', writing the answer as $x$.
a.
Explain why this cancellation is invalid, and verify using $x=14$ that the student's answer is wrong.
[3] Show complete worked solution
(a)
The 7 in the numerator is part of a SUM ($x+7$), not a FACTOR of the entire numerator — you can only cancel factors that multiply the WHOLE numerator and WHOLE denominator, not terms being added. Verify: $\dfrac{14+7}{7}=\dfrac{21}{7}=3$, but the student's 'simplified' answer would give $x=14$ — clearly different (3 vs 14), confirming the error.
QUESTION 9
5 marks
Criterion C
Hard
A student simplifies $\dfrac{2x+4}{2}$ as $x+4$ (only dividing the FIRST term by 2, forgetting the second).
a.
Explain the error, and give the correct simplification, making clear that BOTH terms in the numerator must be divided.
[3] b.
Verify the correct answer using $x=5$.
[2] Show complete worked solution
(a)
$\dfrac{2x+4}{2} = \dfrac{2x}{2}+\dfrac{4}{2} = x+2$ — EVERY term in the numerator must be divided by the denominator, not just the first one. The student's answer ($x+4$) incorrectly left the second term unchanged.
(b)
$\dfrac{2(5)+4}{2}=\dfrac{14}{2}=7$. Correct formula: $5+2=7$ (student's wrong version would give $5+4=9\ne7$).
QUESTION 10
6 marks
Criterion D
Hard
A water tank drains according to the formula: time to drain $=\dfrac{V}{r}$ minutes, where $V$ is volume in litres and $r$ is drain rate in litres/minute. A tank has volume $V=3x^2+9x$ litres and drains at rate $r=3x$ litres/minute.
a.
Simplify the expression for draining time, factoring the numerator first.
[3] b.
If $x=8$, find the actual draining time, and verify by calculating $V$ and $r$ separately first, then dividing directly.
[3] Show complete worked solution
(a)
Because $r=3x$ is a positive drain rate, $x>0$. Then $$\frac{3x^2+9x}{3x}=\frac{3x(x+3)}{3x}=\boxed{x+3\text{ minutes}},\qquad x>0.$$
(b)
Using simplified formula: $8+3=11$ minutes. Verify: $V=3(64)+9(8)=192+72=264$L. $r=3(8)=24$L/min. Direct: $264\div24=11$ minutes.
QUESTION 11
6 marks
Criterion D
Hard
A charity's total fundraising amount, in dollars, is $5x^2+15x$. The money is divided equally among $5x$ volunteer teams for their operating budgets.
a.
Simplify the expression for the amount each team receives.
[2] b.
If there are 40 volunteer teams (i.e. $5x=40$), find the amount each team receives, and the TOTAL fundraising amount that was originally divided.
[4] Show complete worked solution
(a)
The number of teams is $5x$, so $x>0$. Then $$\frac{5x^2+15x}{5x}=\frac{5x(x+3)}{5x}=\boxed{x+3\text{ dollars per team}},\qquad x>0.$$
(b)
$5x=40$, so $x=8$.
$$x+3=8+3=\boxed{\$11}$$
$$5x^2+15x=5(8)^2+15(8)=320+120=\boxed{\$440}.$$
This agrees with $40\times\$11=\$440$.