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Cambridge IGCSE 0580/21, May/June 2025: Worked Solutions, Questions 15 to 23

Sir Faraz Hassan

22 Sept 2026

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Cambridge IGCSE Mathematics (0580)0580/21 - Extended - May/June 2025100 marks · 2 hours · Non-calculator

This is the rest of the paper. Questions 1 to 14, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Cambridge IGCSE Mathematics, Paper 0580/21 (Extended), May/June 2025 series, sat Friday 2 May 2025 in Cambridge administrative zone 2 – 100 marks, 2 hours, calculator not allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are Cambridge Assessment International Education's. This resource reproduces neither the exam paper nor the official mark scheme.

Cambridge publishes a timetable for each administrative zone, and the date above is the one published for Zone 2.

Cambridge has not yet published this paper on its public website. We will link the official question paper and mark scheme here as soon as they are released.

Try each question yourself first, then open the worked solution to check your method and see exactly where each method mark (M1) and independent mark (B1) is earned. The questions follow the same order as the original paper and carry the same marks.

Download printable PDF

All 23 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 15 to 23 of 23

(a) The expression below is to be expanded and simplified.
$(2 - \sqrt{5})(1 - 3\sqrt{5})$ [2 marks]

(b) The denominator of the fraction below is to be rationalised.
Give your answer in its simplest form.
$\dfrac{6}{\sqrt{10}}$ [2 marks]

(a)(b)

[Total 4 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

(2 - \sqrt{5})(1 - 3\sqrt{5})

, a product of two brackets, each holding a whole number and a multiple of

\sqrt{5}

\dfrac{6}{\sqrt{10}}

, a fraction with the surd

\sqrt{10}

in its denominator

There is no calculator, so every surd is kept exact and nothing is written as a decimal.

Find

(a)

(2 - \sqrt{5})(1 - 3\sqrt{5})

expanded and simplified (b)

\dfrac{6}{\sqrt{10}}

written with a rational denominator, in its simplest form

Plan the Solution

(a) Multiply each term of the first bracket by each term of the second, which gives four products. Then use $\sqrt{5} \times \sqrt{5} = 5$ , and collect the whole numbers and the multiples of $\sqrt{5}$ separately.
(b) Multiply the numerator and the denominator by $\sqrt{10}$ , so that the denominator becomes $\sqrt{10} \times \sqrt{10} = 10$ . Then cancel the common factor of the whole numbers.

Worked Solution [4 marks]

Rule - Surds:

\sqrt{a} \times \sqrt{a} = a

. Brackets holding surds are expanded term by term, exactly like brackets holding letters, and a denominator

\sqrt{a}

is made rational by multiplying the fraction by

\dfrac{\sqrt{a}}{\sqrt{a}}

, which is equal to

1

Step 1: (a) Multiply each term of the first bracket by each term of the second

2 \times 1 = 2

2 \times (-3\sqrt{5}) = -6\sqrt{5}

(-\sqrt{5}) \times 1 = -\sqrt{5}

(-\sqrt{5}) \times (-3\sqrt{5}) = 3 \times \sqrt{5} \times \sqrt{5}

(Reason: The first bracket holds the terms

2

and

-\sqrt{5}

, and the second holds

1

and

-3\sqrt{5}

, so there are four products, and each term keeps the sign in front of it. The last product is a negative times a negative, so it is positive, and the numbers in front of the two roots,

1

and

3

, multiply to give

3

Step 2: (a) Write the four terms as one expression

(2 - \sqrt{5})(1 - 3\sqrt{5}) = 2 - 6\sqrt{5} - \sqrt{5} + 3 \times \sqrt{5} \times \sqrt{5}

(Reason: This is the expansion before anything is simplified. It is the line the mark scheme's B1 is for: any

3

of these four terms written correctly earn that mark.)

Step 3: (a) Simplify

\sqrt{5} \times \sqrt{5}

3 \times \sqrt{5} \times \sqrt{5} = 3 \times 5 = 15

(2 - \sqrt{5})(1 - 3\sqrt{5}) = 2 - 6\sqrt{5} - \sqrt{5} + 15

(Reason: A square root multiplied by itself gives back the number under the root, so

\sqrt{5} \times \sqrt{5} = 5

. The last term is therefore a whole number, and it can join the

2

in the next step.)

Step 4: (a) Collect like terms

2 + 15 = 17

-6\sqrt{5} - \sqrt{5} = -7\sqrt{5}

(2 - \sqrt{5})(1 - 3\sqrt{5}) = 17 - 7\sqrt{5}

(Reason: Whole numbers are collected with whole numbers, and multiples of

\sqrt{5}

with multiples of

\sqrt{5}

, just as

x

terms would be:

6

lots of

\sqrt{5}

taken away and then

1

more taken away is

7

lots taken away.

17

and

7\sqrt{5}

cannot be combined, because one is a whole number and the other a multiple of

\sqrt{5}

, so the expression is fully simplified.)

Step 5: (b) Multiply the numerator and the denominator by

\sqrt{10}

\dfrac{6}{\sqrt{10}} = \dfrac{6}{\sqrt{10}} \times \dfrac{\sqrt{10}}{\sqrt{10}}

\sqrt{10} \times \sqrt{10} = 10

\dfrac{6}{\sqrt{10}} \times \dfrac{\sqrt{10}}{\sqrt{10}} = \dfrac{6\sqrt{10}}{10}

(Reason: Multiplying by

\dfrac{\sqrt{10}}{\sqrt{10}}

is multiplying by

1

, so the value of the fraction does not change. The denominator does:

\sqrt{10}

times itself is

10

, a whole number, and a whole-number denominator is what rationalising means. The multiplication written out in the first line is what the M1 is for.)

Step 6: (b) Cancel to the simplest form

\dfrac{6}{10} = \dfrac{3}{5}

\dfrac{6\sqrt{10}}{10} = \dfrac{3\sqrt{10}}{5}

(Reason: The whole numbers

6

and

10

share a factor of

2

, so both are divided by

2

, and the

\sqrt{10}

is left as it is. The root cannot be simplified either, because

10 = 2 \times 5

has no square factor other than

1

. So

\dfrac{3\sqrt{10}}{5}

is the simplest form the question asks for.)

(a)

17 - 7\sqrt{5}

(b)

\dfrac{3\sqrt{10}}{5}

Verification

Check 1 - (a) expanded the other way round: Multiply the whole first bracket by each term of the second instead:

(2 - \sqrt{5}) \times 1 = 2 - \sqrt{5}

, and

(2 - \sqrt{5}) \times (-3\sqrt{5}) = -6\sqrt{5} + 15

. Adding the two should give the same answer.

2 - \sqrt{5} - 6\sqrt{5} + 15 = 17 - 7\sqrt{5}

Check 2 - (a) with the partner brackets: Changing the sign of every

\sqrt{5}

in Steps 1 to 4 leaves the working true, so

(2 + \sqrt{5})(1 + 3\sqrt{5}) = 17 + 7\sqrt{5}

too. Multiply the two results together, using

(p - q\sqrt{5})(p + q\sqrt{5}) = p^2 - 5q^2

on each pair of partners. The brackets give

(2^2 - 5 \times 1^2)(1^2 - 5 \times 3^2)

, which is

(-1) \times (-44) = 44

, so the answer side must give

44

as well.

17^2 - 5 \times 7^2 = 289 - 245 = 44

Check 3 - (b) by squaring: Both fractions are positive, so they are equal if their squares are. Squaring a root gives the number under it, so

\left(\dfrac{6}{\sqrt{10}}\right)^2 = \dfrac{36}{10}

and

\left(\dfrac{3\sqrt{10}}{5}\right)^2 = \dfrac{9 \times 10}{25} = \dfrac{90}{25}

\dfrac{36}{10} = \dfrac{18}{5}

and

\dfrac{90}{25} = \dfrac{18}{5}

Check 4 - (b) multiplying back: If

\dfrac{3\sqrt{10}}{5}

is equal to

\dfrac{6}{\sqrt{10}}

, then multiplying it by

\sqrt{10}

must give back the numerator

6

\dfrac{3\sqrt{10}}{5} \times \sqrt{10} = \dfrac{3 \times 10}{5} = 6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The expansion, simplified	Note	The Answer column gives $17 - 7\sqrt{5}$ final answer, and the Marks column gives $2$ for it.	✓
(a) Three of the four terms of the expansion	B1	For $3$ correct terms from $2 - 6\sqrt{5} - \sqrt{5} + 3 \times \sqrt{5} \times \sqrt{5}$ oe.	✓
(b) The fraction with a rational denominator	Note	The Answer column gives $\dfrac{3\sqrt{10}}{5}$ cao, and the Marks column gives $2$ for it.	✓
(b) Multiplying top and bottom by $\sqrt{10}$	M1	For $\dfrac{6}{\sqrt{10}} \times \dfrac{\sqrt{10}}{\sqrt{10}}$ oe.	✓
(b) What cao and oe mean	Note	cao is the scheme's abbreviation for correct answer only, and oe for or equivalent. The answer asked for is the simplest form, $\dfrac{3\sqrt{10}}{5}$ ; the line $\dfrac{6\sqrt{10}}{10}$ before the cancelling is equal to it, but it is not yet in its simplest form.	✓

Full marks: 4/4

The expression below is to be expanded and simplified.
$(x + 4)(x - 3)(3x + 2)$ [3 marks]

[Total 3 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

(x + 4)(x - 3)(3x + 2)

, a product of three brackets, each holding a multiple of

x

and a number

There is no calculator, so every product is worked out by hand with whole numbers.

Find

The product multiplied out and simplified. Three brackets that each hold an

x

term multiply to give an

x^3

term, so expect terms in

x^3

x^2

and

x

, and a number.

Plan the Solution

Expand the first two brackets, $(x + 4)(x - 3)$ , and collect like terms. That gives a quadratic with three terms.
Multiply each of those three terms by each term of $(3x + 2)$ . That gives $3 \times 2 = 6$ terms.
Collect like terms: the two $x^2$ terms together, and the two $x$ terms together.

Worked Solution [3 marks]

Rule - Expanding three brackets: multiply two of the brackets together and simplify, then multiply every term of that result by every term of the third bracket, and collect like terms.

Step 1: Expand the first two brackets

(x + 4)(x - 3) = x^2 - 3x + 4x - 12

-3x + 4x = x

(x + 4)(x - 3) = x^2 + x - 12

(Reason: Each term of

(x + 4)

multiplies each term of

(x - 3)

x \times x = x^2

x \times (-3) = -3x

4 \times x = 4x

and

4 \times (-3) = -12

. Those four terms are one pair of brackets expanded, which is the line the mark scheme's B1 describes, with at least

3

of the

4

correct. Then

-3x

and

4x

are like terms, and together they make

x

Step 2: Multiply each term of

x^2 + x - 12

(3x + 2)

x^2(3x + 2) = 3x^3 + 2x^2

x(3x + 2) = 3x^2 + 2x

-12(3x + 2) = -36x - 24

(Reason: The quadratic has three terms and

(3x + 2)

has two, so there are six products. Each term keeps the sign in front of it. The last term of the quadratic is

-12

, so

-12 \times 3x = -36x

and

-12 \times 2 = -24

Step 3: Write the six terms as one expression

(x + 4)(x - 3)(3x + 2) = 3x^3 + 2x^2 + 3x^2 + 2x - 36x - 24

(Reason: This is the product fully multiplied out, before anything is collected. It is a correct expansion unsimplified, the first of the two forms the mark scheme's B2 accepts.)

Step 4: Collect like terms

2x^2 + 3x^2 = 5x^2

2x - 36x = -34x

(x + 4)(x - 3)(3x + 2) = 3x^3 + 5x^2 - 34x - 24

(Reason: The two

x^2

terms add to

5x^2

, and

2x

take away

36x

leaves

-34x

. The

x^3

term and the number have nothing to join with, so they stay as they are. No two of the four terms left are alike, so the expression is fully simplified, with the powers of

x

in descending order.)

3x^3 + 5x^2 - 34x - 24

Verification

Check 1 - put $x = 1$ into both: An expansion is equal to the product it came from for every value of

x

, so the brackets and the answer must give the same number when

x = 1

. The brackets give

(1 + 4) \times (1 - 3) \times (3 \times 1 + 2) = 5 \times (-2) \times 5 = -50

3 \times 1^3 + 5 \times 1^2 - 34 \times 1 - 24 = -50

Check 2 - the brackets in a different order: Expand

(x - 3)(3x + 2)

first instead:

3x^2 + 2x - 9x - 6 = 3x^2 - 7x - 6

. Then multiply by

(x + 4)

x(3x^2 - 7x - 6) = 3x^3 - 7x^2 - 6x

and

4(3x^2 - 7x - 6) = 12x^2 - 28x - 24

. Adding the two should give the same answer.

3x^3 - 7x^2 + 12x^2 - 6x - 28x - 24 = 3x^3 + 5x^2 - 34x - 24

Check 3 - the answer is $0$ when $x = 3$ : When

x = 3

the bracket

(x - 3)

0

, so the whole product is

0

. The answer must be

0

there as well.

3 \times 3^3 + 5 \times 3^2 - 34 \times 3 - 24 = 81 + 45 - 102 - 24 = 0

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The expansion, simplified	Note	The Answer column gives $3x^3 + 5x^2 - 34x - 24$ final answer, and the Marks column gives $3$ for it.	✓
The full expansion	B2	For correct expansion unsimplified, or simplified four-term expression of correct form with $3$ terms correct.	✓
One pair of brackets expanded	B1	Or for one pair of brackets expanded with at least $3$ terms out of $4$ correct.	✓

Full marks: 3/3

(a) In a jar there are $6$ red beads, $3$ green beads and $1$ blue bead, and two of them are taken at random with replacement.
What is the probability that the two beads taken are both green? [2 marks]

(b) In a second jar there are $4$ red tokens and $2$ yellow tokens, and two of them are taken at random without replacement.

(i) Write the missing probabilities on the tree diagram. [2 marks]

(ii) What is the probability that one of the two tokens is yellow? [3 marks]

(a)(b)(ii)

[Total 7 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

First jar:

6

red,

3

green and

1

blue bead, so

10

beads in all. Two are taken with replacement.

Second jar:

4

red and

2

yellow tokens, so

6

tokens in all. Two are taken without replacement.

The tree already shows

\dfrac{4}{6}

for a red first token,

\dfrac{2}{6}

for a yellow first token, and

\dfrac{3}{5}

for red after red.

Find

(a) The probability that both beads are green. (b)(i) The three probabilities missing from the tree. (b)(ii) The probability that one of the two tokens is yellow.

Plan the Solution

(a) With replacement the jar is the same for both picks, so green has probability $\dfrac{3}{10}$ each time. Multiply the two.
(b)(i) Without replacement the second pick is from one token fewer. Count what is left after each colour of first token.
(b)(ii) One yellow can happen in two ways: red then yellow, or yellow then red. Multiply along each path, then add the two paths.

Worked Solution [7 marks]

Rule - Probability tree: multiply the probabilities along a path, then add the paths that give the outcome you want.

(a) Green on one pick

\dfrac{3}{6 + 3 + 1} = \dfrac{3}{10}

(Reason: The bead goes back after the first pick, so each pick is from all

10

beads, and

3

of them are green.)

(a) Green on both picks

\dfrac{3}{10} \times \dfrac{3}{10} = \dfrac{9}{100}

(Reason: Putting the bead back makes the second pick independent of the first, so the two probabilities multiply.)

(b)(i) After a red first token

\dfrac{2}{6 - 1} = \dfrac{2}{5}

\dfrac{3}{5} + \dfrac{2}{5} = 1

(Reason: Taking a red token leaves

5

tokens:

3

red and both yellow ones. So the Red to Yellow branch is

\dfrac{2}{5}

, and with the printed

\dfrac{3}{5}

it makes

1

, as the two branches from one point must.)

(b)(i) After a yellow first token

\dfrac{4}{6 - 1} = \dfrac{4}{5}

\dfrac{2 - 1}{6 - 1} = \dfrac{1}{5}

\dfrac{4}{5} + \dfrac{1}{5} = 1

(Reason: Taking a yellow token leaves

5

tokens: all

4

red and only

1

yellow. These go on the Yellow to Red and Yellow to Yellow branches.)

(b)(ii) Red, then yellow

\dfrac{4}{6} \times \dfrac{2}{5} = \dfrac{8}{30}

(Reason: Multiply along the path: a red first token, then a yellow one from the tokens that are left.)

(b)(ii) Yellow, then red

\dfrac{2}{6} \times \dfrac{4}{5} = \dfrac{8}{30}

(Reason: Multiply along the other path: a yellow first token, then a red one from the tokens that are left.)

(b)(ii) Add the two paths

\dfrac{8}{30} + \dfrac{8}{30} = \dfrac{16}{30}

(Reason: The two paths cannot both happen, so their probabilities add.

\dfrac{16}{30}

is the form the mark scheme gives, and it accepts an equivalent such as

\dfrac{8}{15}

(a)

\dfrac{9}{100}

(b)(i) Red then Yellow

\dfrac{2}{5}

, Yellow then Red

\dfrac{4}{5}

, Yellow then Yellow

\dfrac{1}{5}

(b)(ii)

\dfrac{16}{30}

Verification

Check 1 - (a): Count ordered picks instead: there are

10 \times 10 = 100

equally likely pairs, and

3 \times 3 = 9

of them are green then green.

\dfrac{9}{100}

again

Check 2 - (b)(i): The two branches leaving any one point of the tree must add to

1

\dfrac{3}{5} + \dfrac{2}{5} = 1

and

\dfrac{4}{5} + \dfrac{1}{5} = 1

Check 3 - (b)(ii): Take the two same-colour paths away from

1

. Both red is

\dfrac{4}{6} \times \dfrac{3}{5} = \dfrac{12}{30}

and both yellow is

\dfrac{2}{6} \times \dfrac{1}{5} = \dfrac{2}{30}

1 - \dfrac{12}{30} - \dfrac{2}{30} = \dfrac{16}{30}

Check 4 - (b)(ii): Count ordered pairs of different tokens:

6 \times 5 = 30

in all, and

4 \times 2 + 2 \times 4 = 16

of them hold exactly one yellow token.

\dfrac{16}{30}

again

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Both beads green	Note	The Answer column gives $\dfrac{9}{100}$ oe, and the Marks column gives $2$ for it.	✓
(a) The method	M1	For $\dfrac{3}{10} \times \dfrac{3}{10}$ .	✓
(b)(i) The three missing probabilities	Note	The Answer column gives $\dfrac{2}{5}$ , $\dfrac{4}{5}$ and $\dfrac{1}{5}$ , and the Marks column gives $2$ for it.	✓
(b)(i) Partial Marks	B1	For $\dfrac{2}{5}$ .	✓
(b)(ii) One of the two tokens yellow	Note	The Answer column gives $\dfrac{16}{30}$ oe, and the Marks column gives $3$ for it.	✓
(b)(ii) Full marks FT their tree diagram	3FT	Full marks FT their tree diagram dep on probabilities $< 1$ .	✓
(b)(ii) Without the 3FT: both paths	M2FT	Otherwise, for $\dfrac{4}{6} \times \dfrac{2}{5} + \dfrac{2}{6} \times \dfrac{4}{5}$ .	✓
(b)(ii) Without the 3FT: one path	M1FT	Or for $\dfrac{4}{6} \times \dfrac{2}{5}$ or for $\dfrac{2}{6} \times \dfrac{4}{5}$ .	✓

Full marks: 7/7

On Saturday, Tanya paddles a kayak $12$ km across a lake at a speed of $x$ km/h.
On Sunday she hikes $10$ km at a speed of $(x - 4)$ km/h.

(a) In terms of $x$ , how long does Tanya spend paddling? [1 mark]

(b) In terms of $x$ , how long does she spend hiking? [1 mark]

(c) Tanya's hike takes $1$ hour longer than her paddle.
Form an equation in terms of $x$ from this fact, and show that it can be simplified to $x^2 - 2x - 48 = 0$ . [4 marks]

(d) Solve $x^2 - 2x - 48 = 0$ by factorising. [3 marks]

(e) How many hours does Tanya spend paddling? [1 mark]

(a) h(b) h(d) x =(d) or x =(e) h

[Total 10 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

Saturday:

12

km paddled at

x

km/h

Sunday:

10

km hiked at

(x - 4)

km/h

The hike takes

1

hour longer than the paddle

There is no calculator, so every step is done by hand and every value stays exact.

Find

(a) the paddling time, in terms of

x

(b) the hiking time, in terms of

x

x

, shown to simplify to

x^2 - 2x - 48 = 0

(d) the solutions of

x^2 - 2x - 48 = 0

; it is a quadratic, so expect two (e) the paddling time in hours

Plan the Solution

(a) and (b) Time is distance divided by speed.
(c) The hiking time minus the paddling time is $1$ . Multiply every term by $x(x - 4)$ to clear both fractions, multiply out the brackets, then collect every term on one side.
(d) Find two numbers that multiply to $-48$ and add to $-2$ , split the middle term with them, factorise, and set each bracket equal to zero.
(e) A speed cannot be negative, so keep the positive solution and put it into the answer to (a).

Worked Solution [10 marks]

Rule - Speed, then a quadratic:

\text{time} = \dfrac{\text{distance}}{\text{speed}}

; an equation with algebraic fractions is cleared by multiplying every term by a common denominator; and a quadratic equal to zero is solved by factorising and setting each bracket equal to zero.

Step 1: (a) The time spent paddling

\text{time} = \dfrac{\text{distance}}{\text{speed}}

\text{time paddling} = \dfrac{12}{x}

(Reason: Distance is speed multiplied by time, so time is distance divided by speed. Tanya paddles

12

km at

x

km/h, so the time is

12

divided by

x

, written as a fraction. The speed is in kilometres per hour and the distance in kilometres, so the time is in hours, the unit on the answer line.)

Step 2: (b) The time spent hiking

\text{time hiking} = \dfrac{10}{x - 4}

(Reason: The same rule, with the hike's

10

km and its speed of

x - 4

km/h. The whole of

x - 4

is the speed, so all of it goes underneath the

10

. Writing

\dfrac{10}{x} - 4

instead would divide by

x

alone and then take

4

away, which is a different expression.)

Step 3: (c) Turn the sentence into an equation

\text{time hiking} - \text{time paddling} = 1

\dfrac{10}{x - 4} - \dfrac{12}{x} = 1

(Reason: The hike takes

1

hour longer, so it is the longer of the two times, and taking the paddling time away from it leaves

1

. The two times are the expressions from (b) and (a). The other order, paddling time minus hiking time, would equal

-1

, not

1

Step 4: (c) Multiply every term by

x(x - 4)

x(x - 4) \times \dfrac{10}{x - 4} - x(x - 4) \times \dfrac{12}{x} = x(x - 4) \times 1

10x - 12(x - 4) = x(x - 4)

(Reason: Both denominators,

x - 4

and

x

, divide into

x(x - 4)

, so multiplying every term by it clears both fractions. Every term means the

1

on the right as well: an equation stays true only when both sides are multiplied by the same thing. In the first fraction the

x - 4

cancels and leaves

10x

; in the second the

x

cancels and leaves

12(x - 4)

, which stays in its bracket because the minus sign in front of it belongs to all of it.)

Step 5: (c) Multiply out the brackets

-12(x - 4) = -12x + 48

x(x - 4) = x^2 - 4x

10x - 12x + 48 = x^2 - 4x

(Reason: The

-12

multiplies both terms inside its bracket:

-12

times

x

-12x

, and

-12

times

-4

+48

, because two negatives multiply to give a positive. On the right,

x

times each term of

x - 4

gives

x^2 - 4x

Step 6: (c) Collect every term on one side

-2x + 48 = x^2 - 4x

0 = x^2 - 4x + 2x - 48

0 = x^2 - 2x - 48

(Reason: On the left,

10x - 12x = -2x

. Adding

2x

to both sides and subtracting

48

from both sides moves every term to the right, where the

x^2

term is already positive. Then

-4x + 2x = -2x

, and the equation reads

x^2 - 2x - 48 = 0

, as the question asked. In a show-that question the working is the answer, so every line is written down.)

Step 7: (d) Two numbers with product

-48

and sum

-2

6 \times (-8) = -48

6 + (-8) = -2

(Reason: To factorise

x^2 - 2x - 48

, look for two numbers that multiply to the constant term,

-48

, and add to the coefficient of

x

, which is

-2

. The product is negative, so one number is positive and the other negative; the sum is negative, so the negative one is the larger in size. The factor pairs of

48

are

1

and

48

2

and

24

3

and

16

4

and

12

, and

6

and

8

, and only

6

and

8

differ by

2

Step 8: (d) Split the middle term and factorise

x^2 - 2x - 48 = x^2 + 6x - 8x - 48

x^2 + 6x - 8x - 48 = x(x + 6) - 8(x + 6)

x(x + 6) - 8(x + 6) = (x + 6)(x - 8)

(Reason: The

-2x

is written as

+6x - 8x

, using the pair just found. Taking

x

out of the first two terms and

-8

out of the last two leaves the same bracket,

(x + 6)

, in both, and that common bracket is taken out as a factor. With practice the pair can be written straight into the brackets,

(x + 6)(x - 8)

Step 9: (d) Solve

(x + 6)(x - 8) = 0

x + 6 = 0 \text{ or } x - 8 = 0

x = -6 \text{ or } x = 8

(Reason: Two numbers multiply to give zero only when one of them is zero, so one bracket or the other is zero. Setting each bracket equal to zero gives one solution each.)

Step 10: (e) Keep the value that can be a speed, then find the time

x = 8

\dfrac{12}{8} = \dfrac{3}{2}

\dfrac{3}{2} = 1.5

(Reason: The paddling speed is

x

km/h and a speed cannot be negative, so

x = -6

is rejected: it would make the paddling time

\dfrac{12}{-6} = -2

hours. With

x = 8

the hiking speed is

8 - 4 = 4

km/h, positive as it must be. Putting

x = 8

into the answer to (a) gives

1.5

hours, which is

1\dfrac{1}{2}

hours, or

1

hour and

30

minutes.)

(a)

\dfrac{12}{x}

hours(b)

\dfrac{10}{x - 4}

hours(c)

x^2 - 2x - 48 = 0

, as required(d)

x = -6

x = 8

(e)

1.5

hours

Verification

Check 1 - (c) and (e) together: With

x = 8

, the paddle takes

\dfrac{12}{8} = 1.5

hours and the hike takes

\dfrac{10}{8 - 4} = 2.5

hours. The hike should take exactly

1

hour longer.

2.5 - 1.5 = 1

Check 2 - (d) multiplying the brackets back out: Multiply out

(x + 6)(x - 8)

term by term. It should give back the quadratic.

x^2 - 8x + 6x - 48 = x^2 - 2x - 48

Check 3 - (d) both solutions in the equation: Put each solution into

x^2 - 2x - 48

. Each should give

0

8^2 - 2 \times 8 - 48 = 0

and

(-6)^2 - 2 \times (-6) - 48 = 0

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The paddling time	Note	The Answer column gives $\dfrac{12}{x}$ , and the Marks column gives $1$ for it.	✓
(b) The hiking time	Note	The Answer column gives $\dfrac{10}{x - 4}$ , and the Marks column gives $1$ for it.	✓
(c) The equation from the two times	M1	For $\text{their } \dfrac{10}{x - 4} - \text{their } \dfrac{12}{x} = 1$ oe.	✓
(c) Multiplying out and clearing the fractions	M2	For $10x - 12x + 48 = x^2 - 4x$ : correctly multiplying their brackets and clearing algebraic fractions, e.g. $(x - 4)\left(\dfrac{12}{x} + 1\right) = 10$ leading to $12 - \dfrac{48}{x} + x - 4 = 10$ and then $12x - 48 + x^2 - 4x = 10x$ .	✓
(c) In place of the M2: clearing the fractions only	M1	Or for correctly clearing, or correctly collecting into a single fraction, two fractions both with different algebraic denominators, e.g. $10x - 12(x - 4) = x(x - 4)$ or $\dfrac{10x - 12(x - 4)}{x(x - 4)}\ [= 1]$ .	✓
(c) The quadratic	A1	For leading to $0 = x^2 - 2x - 48$ , with no errors or omissions seen, dep on M3.	✓
(d) The factorisation	M2	For $(x + 6)(x - 8)$ .	✓
(d) Without the M2: the grouped form or the pair of numbers	M1	Otherwise, for $x(x - 8) + 6(x - 8)$ or $x(x + 6) - 8(x + 6)$ or $(x + a)(x + b)$ where $ab = -48$ or $a + b = -2$ .	✓
(d) The two solutions	B1	For $-6$ , $8$ .	✓
(e) The paddling time in hours	Note	The Answer column gives $1.5$ or $1\dfrac{1}{2}$ , and the Marks column gives $1$ for it.	✓
(e) Follow-through	Note	FT $\dfrac{12}{\text{their } 8}$ .	✓

Full marks: 10/10

What is the value of $27^{-\dfrac{2}{3}}$ ? [2 marks]

[Total 2 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

The number to evaluate is

27^{-\dfrac{2}{3}}

: base

27

, index

-\dfrac{2}{3}

No calculator may be used on this paper, so every step is worked by hand.

Find

The value of

27^{-\dfrac{2}{3}}

Plan the Solution

Each part of the index $-\dfrac{2}{3}$ has its own meaning. The $3$ underneath asks for a cube root, the $2$ on top asks for a square, and the minus sign turns the result upside down.
Take the cube root before squaring. $27$ is a cube number, so its cube root is a small whole number and the square that follows stays small too.
Deal with the minus sign last, by writing one over the positive power.

Worked Solution [2 marks]

Rule - Fractional and negative indices:

a^{\dfrac{m}{n}} = \left(\sqrt[n]{a}\right)^{m}

and

a^{-k} = \dfrac{1}{a^{k}}

Cube root, for the

3

in the denominator

27 = 3 \times 3 \times 3

27^{\dfrac{1}{3}} = \sqrt[3]{27} = 3

(Reason: A power of

\dfrac{1}{3}

is a cube root, and

3 \times 3 \times 3 = 27

, so the cube root of

27

3

Square, for the

2

in the numerator

27^{\dfrac{2}{3}} = \left(27^{\dfrac{1}{3}}\right)^{2} = 3^{2} = 9

(Reason: A power of

\dfrac{2}{3}

means the cube root, then squared. Squaring the

3

from Step 1 gives

9

Reciprocal, for the minus sign

27^{-\dfrac{2}{3}} = \left(27^{\dfrac{2}{3}}\right)^{-1} = 9^{-1}

9^{-1} = \dfrac{1}{9}

(Reason: A negative power is one over the matching positive power, so the value is the reciprocal of

9

. Reaching

9^{-1}

\dfrac{1}{3^{2}}

on the way is the working the method mark rewards.)

\dfrac{1}{9}

Verification

Check 1 - $27$ as a power of $3$ : Replace

27

with

3^{3}

. A power of a power multiplies the indices, and

3 \times \left(-\dfrac{2}{3}\right) = -2

\left(3^{3}\right)^{-\dfrac{2}{3}} = 3^{-2} = \dfrac{1}{3^{2}} = \dfrac{1}{9}

, which agrees.

Check 2 - cube the answer: Cubing multiplies the index by

3

, so the cube of

27^{-\dfrac{2}{3}}

27^{-2}

. The cube of

\dfrac{1}{9}

must therefore be one over

27^{2}

\left(\dfrac{1}{9}\right)^{3} = \dfrac{1}{729} = \dfrac{1}{27^{2}}

, as it should be.

Check 3 - square before the cube root: Doing the square first gives

27^{2} = 729

. Since

9^{3} = 729

, the cube root of

729

9

\dfrac{1}{\sqrt[3]{729}} = \dfrac{1}{9}

, the same value.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The answer	Note	The Answer column gives $\dfrac{1}{9}$ , and the Marks column gives $2$ for it.	✓
Method for the value	M1	For $9^{-1}$ or for $\dfrac{1}{3^{2}}$ or $\dfrac{1}{\sqrt[3]{729}}$ .	✓

Full marks: 2/2

What is the exact value of $x$ ? [4 marks]

x =

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

A right-angled triangle, with the angle

30^\circ

at one end of the side

x

cm and the right angle at the other end

The side of

6

cm, across the triangle from the

30^\circ

angle

The figure is not to scale, and no calculator may be used

Find

The exact value of

x

, so a square root stays as a root instead of becoming a rounded decimal

Plan the Solution

Name each side from the $30^\circ$ angle. The side of $6$ cm is opposite it and the side $x$ cm is adjacent to it, so tan is the ratio that links them.
Find $\tan 30^\circ$ exactly, by hand, from half of an equilateral triangle.
Rearrange to make $x$ the subject, then divide by the exact value and simplify.

Worked Solution [4 marks]

Rule - The tangent ratio:

\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}

, with the exact value

\tan 30^\circ = \dfrac{1}{\sqrt{3}}

from half of an equilateral triangle.

Name the sides from the

30^\circ

angle

\tan 30^\circ = \dfrac{\text{opposite}}{\text{adjacent}}

\tan 30^\circ = \dfrac{6}{x}

(Reason: The

6

cm side is across the triangle from the

30^\circ

angle, so it is the opposite side. The

x

cm side runs from the

30^\circ

angle to the right angle, so it is the adjacent side. The hypotenuse is neither given nor asked for, so the ratio to use is tan, the one that links the opposite and adjacent sides.)

The exact value of

\tan 30^\circ

2^2 - 1^2 = 4 - 1 = 3

\tan 30^\circ = \dfrac{1}{\sqrt{3}}

(Reason: No calculator is allowed, so the value comes from an equilateral triangle with sides of

2

, cut in half down its line of symmetry. Each half is a right-angled triangle with angles of

30^\circ

60^\circ

and

90^\circ

, a hypotenuse of

2

and a side of

1

opposite the

30^\circ

. Pythagoras gives the side next to the

30^\circ

\sqrt{3}

, so

\tan 30^\circ

1

over

\sqrt{3}

. This exact value, or the equal form

\dfrac{\sqrt{3}}{3}

, is what the B1 is for.)

Make

x

the subject

6 = x \tan 30^\circ

x = \dfrac{6}{\tan 30^\circ}

(Reason: Multiply both sides of

\tan 30^\circ = \dfrac{6}{x}

x

, then divide both sides by

\tan 30^\circ

. Reaching

\dfrac{6}{\tan 30^\circ}

is the working the M2 is for. The equation

\dfrac{6}{x} = \tan 30^\circ

on its own earns the M1 instead.)

Divide by

\dfrac{1}{\sqrt{3}}

\dfrac{1}{\tan 30^\circ} = \sqrt{3}

x = 6 \times \sqrt{3} = 6\sqrt{3}

(Reason: Dividing by a fraction is the same as multiplying by its reciprocal, and the reciprocal of

\dfrac{1}{\sqrt{3}}

\sqrt{3}

. So dividing

6

\tan 30^\circ

is the same as multiplying

6

\sqrt{3}

. The root cannot be simplified, because

3

has no square factor other than

1

, so

6\sqrt{3}

is the exact value.)

x = 6\sqrt{3}

Verification

Check 1 - through the hypotenuse: The

6

cm side is opposite the

30^\circ

and

\sin 30^\circ = \dfrac{1}{2}

, so the hypotenuse is twice as long,

12

cm. Pythagoras then gives the square of the third side, which must equal the square of the answer.

12^2 - 6^2 = 144 - 36 = 108

and

(6\sqrt{3})^2 = 36 \times 3 = 108

, which agree.

Check 2 - the sine rule: The third angle is

180 - 90 - 30 = 60

degrees, and it is opposite

x

. The sine rule pairs each side with the sine of the angle opposite it, which is the scheme's second form of the M2.

x = \dfrac{6 \sin 60^\circ}{\sin 30^\circ} = 6 \times \dfrac{\sqrt{3}}{2} \times 2 = 6\sqrt{3}

, the same value.

Check 3 - the size of the answer: Without a calculator,

1.7^2 = 2.89

and

1.8^2 = 3.24

, so

\sqrt{3}

lies between

1.7

and

1.8

. Multiplying by

6

gives

6 \times 1.7 = 10.2

and

6 \times 1.8 = 10.8

10.2 < 6\sqrt{3} < 10.8

: shorter than the

12

cm hypotenuse, and longer than the

6

cm side because it faces the larger angle,

60^\circ

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The exact value of $x$	Note	The Answer column gives $6\sqrt{3}$ oe, and the Marks column gives $4$ for it.	✓
An exact trigonometric value	B1	For $\tan 30 = \dfrac{1}{\sqrt{3}}$ or $\dfrac{\sqrt{3}}{3}$ , or for $\sin 60 = \dfrac{\sqrt{3}}{2}$ and $\sin 30 = \dfrac{1}{2}$ .	✓
A calculation for $x$	M2	For $\dfrac{6}{\tan 30}$ or $\dfrac{6\sin(60)}{\sin(30)}$ .	✓
An equation for $x$	M1	Or for $\dfrac{6}{x} = \tan 30$ or $\dfrac{\sin 60}{x} = \dfrac{\sin 30}{6}$ .	✓

Full marks: 4/4

In the diagram, $OABC$ is a rhombus whose vertex $O$ is at the origin, and $P$ is the point where its two diagonals cross.
It is given that $\overrightarrow{AP} = \mathbf{n}$ and $\overrightarrow{OP} = 2\mathbf{m}$ .

(a) Express each of these vectors in terms of $\mathbf{m}$ and $\mathbf{n}$ , in its simplest form.
(i) $\overrightarrow{OA}$ [1 mark]

(ii) $\overrightarrow{OC}$ [1 mark]

(b) The point $D$ is placed so that $\overrightarrow{AD} = 10\mathbf{m} - 3\mathbf{n}$ .
The quadrilateral $OADC$ is a trapezium. Show that this is true. [3 marks]

(a)(i) OA =(a)(ii) OC =

[Total 5 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

A rhombus

OABC

with its vertex

O

at the origin; its diagonals cross at

P

\overrightarrow{OP} = 2\mathbf{m}

and

\overrightarrow{AP} = \mathbf{n}

For (b):

\overrightarrow{AD} = 10\mathbf{m} - 3\mathbf{n}

The figure is not to scale, and no calculator may be used

Find

(a)

\overrightarrow{OA}

and

\overrightarrow{OC}

in terms of

\mathbf{m}

and

\mathbf{n}

, each in its simplest form (b) a reason why

OADC

is a trapezium, which means showing that two of its opposite sides are parallel

Plan the Solution

(a)(i) There is no given vector from $O$ to $A$ , so go the long way round through $P$ : out along $\overrightarrow{OP}$ , then back along $\overrightarrow{AP}$ .
(a)(ii) The diagonals of a rhombus bisect each other, so $P$ is the midpoint of $AC$ and $\overrightarrow{PC}$ is the same vector as $\overrightarrow{AP}$ . Then go from $O$ to $P$ to $C$ .
(b) Find $\overrightarrow{CD}$ by the route $C$ to $O$ to $A$ to $D$ , simplify it, and compare it with $\overrightarrow{OA}$ . If one is a number times the other, the two sides are parallel.

Worked Solution [5 marks]

Rule - Vector routes: the vector from one point to another is the sum of the vectors along any route between them; going backwards along a vector changes its sign, so

\overrightarrow{PA} = -\overrightarrow{AP}

; and if

\overrightarrow{CD} = k\overrightarrow{OA}

for a number

k

, then

CD

is parallel to

OA

Step 1: (a)(i) The route from

O

A

through

P

\overrightarrow{OA} = \overrightarrow{OP} + \overrightarrow{PA}

\overrightarrow{PA} = -\overrightarrow{AP} = -\mathbf{n}

\overrightarrow{OA} = 2\mathbf{m} + (-\mathbf{n}) = 2\mathbf{m} - \mathbf{n}

(Reason: Both given vectors end at

P

, so the journey from

O

A

goes through it: first along

\overrightarrow{OP}

, then from

P

A

. The arrow labelled

\mathbf{n}

points from

A

P

, and this part of the journey goes the other way, so it is

-\mathbf{n}

. The two terms are different vectors and cannot be combined, so

2\mathbf{m} - \mathbf{n}

is already in its simplest form.)

Step 2: (a)(ii)

P

is the midpoint of

AC

\overrightarrow{PC} = \overrightarrow{AP}

\overrightarrow{PC} = \mathbf{n}

(Reason: The diagonals of a rhombus bisect each other: they cut each other exactly in half. So

P

is halfway along

AC

, and the step from

P

C

carries on in the same direction as the step from

A

P

, for the same distance. Two steps with the same length and the same direction are the same vector.)

Step 3: (a)(ii) The route from

O

C

through

P

\overrightarrow{OC} = \overrightarrow{OP} + \overrightarrow{PC}

\overrightarrow{OC} = 2\mathbf{m} + \mathbf{n}

(Reason: Out along

\overrightarrow{OP}

as before, but this time on to

C

, which is forwards along

\mathbf{n}

rather than backwards. Compare (a)(i): the only change is the sign of

\mathbf{n}

, because

A

and

C

lie on opposite sides of

P

along the same diagonal.)

Step 4: (b) A route from

C

D

\overrightarrow{CD} = \overrightarrow{CO} + \overrightarrow{OA} + \overrightarrow{AD}

\overrightarrow{CO} = -\overrightarrow{OC} = -(2\mathbf{m} + \mathbf{n})

\overrightarrow{CD} = -(2\mathbf{m} + \mathbf{n}) + 2\mathbf{m} - \mathbf{n} + 10\mathbf{m} - 3\mathbf{n}

(Reason: The vector

\overrightarrow{AD}

is given, and the answers to (a) give the rest of a route: from

C

back to the origin, out to

A

, then on to

D

. Going from

C

O

is backwards along

\overrightarrow{OC}

, so the whole of

2\mathbf{m} + \mathbf{n}

changes sign, and it stays in a bracket until that minus sign is dealt with.)

Step 5: (b) Expand the bracket and collect like terms

\overrightarrow{CD} = -2\mathbf{m} - \mathbf{n} + 2\mathbf{m} - \mathbf{n} + 10\mathbf{m} - 3\mathbf{n}

\overrightarrow{CD} = 10\mathbf{m} - 5\mathbf{n}

(Reason: The minus sign in front of the bracket multiplies both terms inside it, so

-(2\mathbf{m} + \mathbf{n})

becomes

-2\mathbf{m} - \mathbf{n}

. Then the

\mathbf{m}

terms and the

\mathbf{n}

terms are collected separately: the numbers in front of

\mathbf{m}

give

-2 + 2 + 10 = 10

, and the numbers in front of

\mathbf{n}

give

-1 - 1 - 3 = -5

Step 6: (b) Write

\overrightarrow{CD}

as a multiple of

\overrightarrow{OA}

\overrightarrow{CD} = 5(2\mathbf{m} - \mathbf{n})

\overrightarrow{CD} = 5\overrightarrow{OA}

(Reason: Both

10

and

5

are multiples of

5

, so

5

is a common factor. Taking it out leaves

2\mathbf{m} - \mathbf{n}

, which is

\overrightarrow{OA}

from (a)(i). When one vector is a number times another, the lines they run along are parallel. Here the number,

5

, is positive, so the two vectors also point the same way.)

Step 7: (b) Why this makes

OADC

a trapezium

CD \parallel OA

(Reason: Going round

OADC

in order, its sides are

OA

AD

DC

and

CO

, so

OA

and

DC

are opposite sides. A quadrilateral with a pair of parallel opposite sides is a trapezium. The factor

5

also shows that

CD

is five times as long as

OA

, so the two parallel sides are different lengths and

OADC

is not a parallelogram. In a show-that question the working is the answer, so every line is written down.)

(a)(i)

\overrightarrow{OA} = 2\mathbf{m} - \mathbf{n}

(a)(ii)

\overrightarrow{OC} = 2\mathbf{m} + \mathbf{n}

(b)

\overrightarrow{CD} = 10\mathbf{m} - 5\mathbf{n} = 5\overrightarrow{OA}

, so

CD

is parallel to

OA

and

OADC

is a trapezium

Verification

Check 1 - (a)(i) and (a)(ii) together: The long diagonal

OB

is twice

OP

, because

P

is its midpoint, so

\overrightarrow{OB} = 4\mathbf{m}

. Going from

O

A

and then along

AB

, which is the same vector as

\overrightarrow{OC}

, also reaches

B

, so the two answers must add up to

4\mathbf{m}

(2\mathbf{m} - \mathbf{n}) + (2\mathbf{m} + \mathbf{n}) = 4\mathbf{m}

Check 2 - (b) by a route through P: Go from

C

D

through

P

and

A

instead.

C

P

is back along

\mathbf{n}

P

A

is back along

\mathbf{n}

again, and then

A

D

is given. This route uses only the vectors in the question, not the answers to (a).

\overrightarrow{CD} = -\mathbf{n} - \mathbf{n} + 10\mathbf{m} - 3\mathbf{n} = 10\mathbf{m} - 5\mathbf{n}

, the same vector.

Check 3 - with numbers: Take a real rhombus:

O(0, 0)

A(3, 4)

B(8, 4)

and

C(5, 0)

, with every side

5

long. Its diagonals cross at their midpoint

P(4, 2)

, so

2\mathbf{m} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}

\mathbf{m} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}

and

\mathbf{n} = \overrightarrow{AP} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}

. Then

2\mathbf{m} - \mathbf{n} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}

, which is

\overrightarrow{OA}

, and

\overrightarrow{AD} = 10\mathbf{m} - 3\mathbf{n} = \begin{pmatrix} 17 \\ 16 \end{pmatrix}

, which puts

D

(20, 20)

\overrightarrow{CD} = \begin{pmatrix} 15 \\ 20 \end{pmatrix} = 5\begin{pmatrix} 3 \\ 4 \end{pmatrix} = 5\overrightarrow{OA}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The vector $\overrightarrow{OA}$	Note	The Answer column gives $2\mathbf{m} - \mathbf{n}$ , and the Marks column gives $1$ for it.	✓
(a)(ii) The vector $\overrightarrow{OC}$	Note	The Answer column gives $2\mathbf{m} + \mathbf{n}$ , and the Marks column gives $1$ for it.	✓
(b) The vector $\overrightarrow{CD}$	M2	For $\overrightarrow{CD} = 10\mathbf{m} - 5\mathbf{n}$ . Allow M2 for equivalents: $\overrightarrow{CD} = -2\mathbf{n} + 10\mathbf{m} - 3\mathbf{n}$ or $\overrightarrow{CD} = -(2\mathbf{m} + \mathbf{n}) + 2\mathbf{m} - \mathbf{n} + 10\mathbf{m} - 3\mathbf{n}$ . For M2, FT their (a), e.g. $\overrightarrow{CD} = \text{their } \overrightarrow{CO} + \text{their } \overrightarrow{OA} + 10\mathbf{m} - 3\mathbf{n}$ .	✓
(b) In place of the M2: a correct route for $\overrightarrow{CD}$	M1	Or for correct route for $\overrightarrow{CD}$ using the lines of the diagram with $\overrightarrow{AD}$ , e.g. $\overrightarrow{CD} = \overrightarrow{CA} + \overrightarrow{AD}$ oe.	✓
(b) $\overrightarrow{CD}$ as a multiple of $\overrightarrow{OA}$	A1	For $\overrightarrow{CD} = 5\overrightarrow{OA}$ leading to $CD$ is parallel to $OA$ [ $\therefore$ $OACD$ is a trapezium]. Dependent on M2.	✓

Full marks: 5/5

The equation of a curve is $y = x^n + qx^2 + 9x$ .
For this curve, $\dfrac{dy}{dx} = 3x^2 - 12x + 9$ .

(a) What are the values of $n$ and $q$ ? [2 marks]

(b) What are the coordinates of the curve's turning points? [4 marks]

(a) n =(a) q =(b)

[Total 6 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

The curve

y = x^n + qx^2 + 9x

, with two unknown constants,

n

and

q

Its gradient function,

\dfrac{dy}{dx} = 3x^2 - 12x + 9

There is no calculator, so every step is done by hand, and every value stays exact.

Find

(a) The values of

n

and

q

(b) The coordinates of the turning points. The gradient function is a quadratic, so expect two points

Plan the Solution

(a) Differentiate $y$ term by term, keeping $n$ and $q$ as letters, then match the result term by term with $3x^2 - 12x + 9$ .
(b) At a turning point the gradient is zero, so solve $3x^2 - 12x + 9 = 0$ for $x$ . Then put each value of $x$ into the equation of the curve, using the values from (a), to find $y$ .

Worked Solution [6 marks]

Rule - Differentiation and turning points: the derivative of

ax^k

kax^{k-1}

, and at a turning point the gradient is zero, so

\dfrac{dy}{dx} = 0

Step 1: (a) Differentiate

y = x^n + qx^2 + 9x

term by term

\dfrac{dy}{dx} = nx^{n-1} + 2qx + 9

(Reason: Every term follows the same rule: multiply by the power, then take

1

off the power. So

x^n

becomes

nx^{n-1}

. In

qx^2

the constant

q

simply stays as a multiplier, so that term becomes

2qx

, and

9x

becomes

9

Step 2: (a) Match the terms with the given gradient function

nx^{n-1} + 2qx + 9 = 3x^2 - 12x + 9

n - 1 = 2 \implies n = 3

2q = -12 \implies q = -6

(Reason: The two expressions are equal for every value of

x

, so their terms must match one by one. The only term that can match

3x^2

nx^{n-1}

: its power

n - 1

must be

2

, which gives

n = 3

, and its coefficient

n

must be

3

, which agrees. The

x

terms give

2q = -12

, so

q = -6

, and the constant

9

is already the same on both sides.)

Step 3: (b) Set

\dfrac{dy}{dx}

equal to zero

3x^2 - 12x + 9 = 0

x^2 - 4x + 3 = 0

(Reason: At a turning point the curve is momentarily flat, so its gradient is

0

. Every coefficient is a multiple of

3

, so dividing the whole equation by

3

gives a simpler quadratic with exactly the same solutions.)

Step 4: (b) Factorise and solve

(x - 1)(x - 3) = 0

x - 1 = 0 \implies x = 1

x - 3 = 0 \implies x = 3

(Reason: The two numbers that multiply to

3

and add to

-4

are

-1

and

-3

. A product is zero only when one of its factors is zero, which gives the two values of

x

. The quadratic formula on

3x^2 - 12x + 9 = 0

gives the same pair:

x = \dfrac{12 \pm \sqrt{144 - 108}}{6} = \dfrac{12 \pm 6}{6}

, which is

3

1

Step 5: (b) Find each

y

-coordinate from the equation of the curve

y = x^3 - 6x^2 + 9x

1^3 - 6 \times 1^2 + 9 \times 1 = 1 - 6 + 9 = 4

3^3 - 6 \times 3^2 + 9 \times 3 = 27 - 54 + 27 = 0

(Reason: With

n = 3

and

q = -6

from (a), the curve is

y = x^3 - 6x^2 + 9x

. The

y

-coordinates come from this equation, not from

\dfrac{dy}{dx}

, which is

0

at both points. Powers are worked out before multiplying:

3^3 = 27

and

6 \times 3^2 = 6 \times 9 = 54

. So the turning points are

(1, 4)

and

(3, 0)

(a)

n = 3

q = -6

(b)

(1, 4)

and

(3, 0)

Verification

Check 1 - (a) differentiate the finished curve: With

n = 3

and

q = -6

the curve is

y = x^3 - 6x^2 + 9x

. Differentiating it term by term should give back the gradient function the question states.

\dfrac{dy}{dx} = 3x^2 - 2 \times 6x + 9 = 3x^2 - 12x + 9

Check 2 - (b) the gradient is zero at $x = 1$ and $x = 3$ : Put each value of

x

into

3x^2 - 12x + 9

. Both should give

0

3 \times 1^2 - 12 \times 1 + 9 = 0

and

3 \times 3^2 - 12 \times 3 + 9 = 0

Check 3 - (b) the curve in factorised form: Taking out the common factor

x

gives

y = x(x^2 - 6x + 9) = x(x - 3)^2

. This form finds each

y

-coordinate by a different calculation.

1 \times (1 - 3)^2 = 4

and

3 \times (3 - 3)^2 = 0

Check 4 - (b) both points really are turning points: The gradient should change sign at each turning point. At

x = 0

and

x = 4

, outside the two points,

3x^2 - 12x + 9

9

. At

x = 2

, between them, it should be negative.

3 \times 2^2 - 12 \times 2 + 9 = -3

: positive, negative, positive, so the curve rises to

(1, 4)

, falls to

(3, 0)

and rises again

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The values of $n$ and $q$	Note	The Answer column gives $[n =]\ 3$ , $[q =]\ -6$ , and the Marks column gives $2$ for it.	✓
(a) Each value, $n$ or $q$	B1	For each correct value.	✓
(b) The two turning points	Note	The Answer column gives $(1, 4)$ and $(3, 0)$ , and the Marks column gives $4$ for it.	✓
(b) One correct point, or both values of $x$	B3	For $(1, 4)$ or $(3, 0)$ , or for two correct values of $x$ .	✓
(b) In place of the B3: the equation factorised, or the quadratic formula	M2	Or for $[3](x - 1)(x - 3)\ [= 0]$ oe, or $x = \dfrac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 3 \times 9}}{2 \times 3}$ oe.	✓
(b) In place of the B3 or the M2: the gradient set to zero	M1	Or for writing $\dfrac{dy}{dx} = 0$ , or for $3x^2 - 12x + 9 = 0$ .	✓

Full marks: 6/6

The expression below is to be simplified.
$\dfrac{2x^2 + 10x}{x^2 - 25}$ [3 marks]

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 23 - Exam Solution

Understanding the Question

Given

\dfrac{2x^2 + 10x}{x^2 - 25}

, an algebraic fraction with a quadratic expression on the top and another on the bottom

No calculator is allowed, and none is needed: every step is factorising by hand.

Find

The fraction in its simplest form, where the top and the bottom have no factor in common.

Plan the Solution

Factorise the top by taking out the highest common factor of $2x^2$ and $10x$ .
Factorise the bottom as a difference of two squares, because $25 = 5^2$ .
Divide the top and the bottom by the bracket they share.

Worked Solution [3 marks]

Rule - Simplifying an algebraic fraction: factorise the top and the bottom fully, then cancel any factor they have in common. Only a factor of the whole top and of the whole bottom may be cancelled, never a single term.

Step 1: Factorise the numerator

2x^2 + 10x = 2x \times x + 2x \times 5

2x^2 + 10x = 2x(x + 5)

(Reason: The highest common factor of

2x^2

and

10x

2x

: the highest common factor of the numbers

2

and

10

2

, and both terms contain

x

. Dividing each term by

2x

gives

\dfrac{2x^2}{2x} = x

and

\dfrac{10x}{2x} = 5

, and those two go inside the bracket. This factorised top is what the first B1 on the mark scheme is for.)

Step 2: Factorise the denominator

x^2 - 25 = x^2 - 5^2

x^2 - 25 = (x - 5)(x + 5)

(Reason:

25

5^2

, so

x^2 - 25

is a difference of two squares, and

a^2 - b^2 = (a - b)(a + b)

with

a = x

and

b = 5

. Multiplying the brackets back out,

-5x

and

+5x

cancel and leave

x^2 - 25

. This factorised bottom is what the second B1 is for.)

Step 3: Cancel the common factor

\dfrac{2x^2 + 10x}{x^2 - 25} = \dfrac{2x(x + 5)}{(x - 5)(x + 5)}

\dfrac{2x(x + 5)}{(x - 5)(x + 5)} = \dfrac{2x}{x - 5}

(Reason: The bracket

(x + 5)

is a factor of the whole top and of the whole bottom, so both are divided by it. That is allowed: the fraction only has a value when

x^2 - 25

is not

0

, and then

x + 5

is not

0

either. What is left has no factor in common, so it is fully simplified. The

x

on the top does not cancel with the

x

x - 5

: there it is a term of the bottom, not a factor of it.)

\dfrac{2x}{x - 5}

Verification

Check 1 - put $x = 10$ into both: The simplified fraction must equal the original for every value of

x

where the original has a value. With

x = 10

the original gives

\dfrac{2 \times 10^2 + 10 \times 10}{10^2 - 25} = \dfrac{300}{75} = 4

\dfrac{2 \times 10}{10 - 5} = \dfrac{20}{5} = 4

Check 2 - put $x = 3$ into both: A second value, this time one that makes the bottom negative. With

x = 3

the original gives

\dfrac{2 \times 3^2 + 10 \times 3}{3^2 - 25} = \dfrac{48}{-16} = -3

\dfrac{2 \times 3}{3 - 5} = \dfrac{6}{-2} = -3

Check 3 - rebuild the original: Multiplying the top and the bottom of the answer by

(x + 5)

must give back the fraction the question started with:

2x(x + 5) = 2x^2 + 10x

on the top and

(x - 5)(x + 5) = x^2 - 25

on the bottom.

\dfrac{2x(x + 5)}{(x - 5)(x + 5)} = \dfrac{2x^2 + 10x}{x^2 - 25}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The simplified fraction	Note	The Answer column gives $\dfrac{2x}{x - 5}$ final answer, and the Marks column gives $3$ for it.	✓
The top factorised	B1	For $2x(x + 5)$ .	✓
The bottom factorised	B1	For $(x - 5)(x + 5)$ .	✓

Full marks: 3/3

Keep revising

That is the whole paper. Read what the IGCSE is and how it is graded, or compare Edexcel 4MA1 with Cambridge 0580 if you are still choosing a board. Check the IGCSE grade boundaries to set your target, and if the exam is close, the four-week IGCSE Maths revision plan sets out what to do week by week.

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Cambridge IGCSE 0580/21, May/June 2025: Worked Solutions, Questions 15 to 23

Worked solutions, questions 15 to 23 of 23

Question 15, Non-calculator

Question 15 - Exam Solution

Question 16, Non-calculator

Question 16 - Exam Solution

Question 17, Non-calculator

Question 17 - Exam Solution

Question 18, Non-calculator

Question 18 - Exam Solution

Question 19, Non-calculator

Question 19 - Exam Solution

Question 20, Non-calculator

Question 20 - Exam Solution

Question 21, Non-calculator

Question 21 - Exam Solution

Question 22, Non-calculator

Question 22 - Exam Solution

Question 23, Non-calculator

Question 23 - Exam Solution

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