Algebra & Surds Challenge (Non-calculator)

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Algebra & Surds Challenge (Non-calculator)

Time allowed: 90 minutesTotal: 64 marks9 questions

Candidate: Hana

Instructions

Answer all questions.
Write your answer to each part on the answer line. Use the maths keys that appear at the bottom of the screen for √, powers and other symbols.
Show all your working. You can earn marks for correct method even if your final answer is wrong.
You must not use a calculator.
Give answers in exact form — fractions or surds — unless the question says otherwise.
The number of marks for each question or part question is shown in brackets [ ].

Question 1

(a)

Solve the simultaneous equations. Show all your working.

\frac{x}{2} + \frac{y}{3} = 4

\frac{x}{4} - \frac{y}{6} = 0

x =

y =

[4]

Question 2

A theatre sells adult tickets for

a

dollars each and child tickets for

c

dollars each. On Monday, 40 adult and 25 child tickets were sold for a total of $1020. On Tuesday, 35 adult and 50 child tickets were sold for a total of $1230.

(a)

Show that

8 a + 5 c = 204

[2]

(b)

Find the cost of an adult ticket and the cost of a child ticket.

Adult $

Child $

[3]

(c)

On Wednesday, 30 adult tickets and some child tickets were sold for a total of $780. How many child tickets were sold?

[2]

Question 3

(a)

By substituting one equation into the other, find the coordinates of the two points where these graphs meet.

y = x + 1

y = x^{2} - 5

and

[5]

Question 4

(a)

Solve

6 x^{2} - x - 12 = 0

x =

x =

[3]

(b)

Solve

(x + 2) (x - 3) = 14

x =

x =

[3]

(c)

Solve

\frac{6}{x} + x = 5

x =

x =

[3]

Question 5

(a)

Write

3 x^{2} - 5 x + 2

in the form

a (x + b)^{2} + c

a =

b =

c =

[3]

(b)

x^{2} + 6 x + k

can be written as

(x + p)^{2} + 7

. Find the values of

p

and

k

p =

k =

[2]

(c)

x^{2} + b x + 11

can be written as

(x - 4)^{2} + q

. Find the values of

b

and

q

b =

q =

[2]

Question 6

(a)

By completing the square, solve

3 x^{2} - 12 x + 5 = 0

. Leave your answers in surd form.

x =

x =

[4]

Question 7

(a)

Show that the discriminant of

x^{2} + (k + 2) x + 2 k = 0

can be written as

(k - 2)^{2}

. Hence explain why the equation has real roots for every value of

k

[3]

(b)

Write down the value of

k

for which

x^{2} + (k + 2) x + 2 k = 0

has one repeated root.

k =

[1]

(c)

The equation

2 x^{2} + 3 x + c = 0

has two different real roots. Find the largest possible integer value of

c

c =

[3]

(d)

The equation

k x^{2} + 4 x + 1 = 0

has one repeated root. Find the value of

k

k =

[2]

Question 8

(a)

Simplify

12 + 75 - 27

[2]

(b)

Expand and simplify

(23 - 2)^{2}

[2]

(c)

Expand and simplify

(5 + 3) (5 - 3)

[2]

(d)

A square has area

(11 + 62)

^{2}

. Show that the length of one side is

(3 + 2)

cm.

[2]

Question 9

(a)

Solve

x 2 = 3 + x

. Give your answer in the form

a + b 2

x =

[3]

(b)

Write

\frac{3 + 1}{3 - 1}

in the form

a + b 3

a =

b =

[3]

(c)

Simplify

\frac{1}{5 - 2} + \frac{1}{5 + 2}

[3]

(d)

x = 2 + 3

. Show that

x + \frac{1}{x}

is an integer.

[2]

End of paper

Mark scheme

Algebra & Surds Challenge (Non-calculator) · 64 marks

Question	Answer	Marks	Partial marks
1(a)	$x = 4$ , $y = 6$	4	M1 for clearing the fractions in one equation, e.g. $3 x + 2 y = 24$ or $3 x - 2 y = 0$ M1 for correctly eliminating one variable A1 for $x = 4$ A1 for $y = 6$
2(a)	$40 a + 25 c = 1020$ , then divide by 5	2	M1 for $40 a + 25 c = 1020$ A1 for dividing by 5 to reach the given equation
2(b)	Adult $18, child $12	3	M1 for $35 a + 50 c = 1230$ (or $7 a + 10 c = 246$ ) used with $8 a + 5 c = 204$ to eliminate one variable A1 for $a = 18$ A1 for $c = 12$
2(c)	20	2	M1 for $780 - 30 \times 18$ or $(780 - 540) \div 12$ (FT their values)
3(a)	$(3, 4)$ and $(- 2, - 1)$	5	M1 for $x^{2} - 5 = x + 1$ M1 for $x^{2} - x - 6 = 0$ M1 for $(x - 3) (x + 2)$ A1 for $x = 3$ and $x = - 2$ A1 for both correct points
4(a)	$- \frac{4}{3}$ and $\frac{3}{2}$	3	M2 for $(3 x + 4) (2 x - 3)$ or M1 for a pair of brackets that expands to give two of the three terms correctly
4(b)	$5$ and $- 4$	3	M1 for $x^{2} - x - 6 = 14$ M1 for $x^{2} - x - 20 = 0$ and $(x - 5) (x + 4)$
4(c)	$2$ and $3$	3	M1 for $6 + x^{2} = 5 x$ M1 for $x^{2} - 5 x + 6 = 0$ and $(x - 2) (x - 3)$
5(a)	$3 (x - \frac{5}{6})^{2} - \frac{1}{12}$ oe	3	B1 for $a = 3$ B1 for $b = - \frac{5}{6}$ B1 for $c = - \frac{1}{12}$
5(b)	$p = 3$ , $k = 16$	2	B1 for $p = 3$ B1 for $k = 16$
5(c)	$b = - 8$ , $q = - 5$	2	B1 for $b = - 8$ B1 for $q = - 5$
6(a)	$\frac{6 + 21}{3}$ and $\frac{6 - 21}{3}$ oe	4	M1 for $x^{2} - 4 x + \frac{5}{3} = 0$ or $3 (x^{2} - 4 x) + 5$ M1 for $(x - 2)^{2} - 4$ seen M1 for $(x - 2)^{2} = \frac{7}{3}$ oe A1 for both answers, e.g. $2 \pm \frac{7}{3}$
7(a)	$(k + 2)^{2} - 4 \times 1 \times 2 k = k^{2} - 4 k + 4 = (k - 2)^{2}$ ; a square is never negative, so $b^{2} - 4 a c \geq 0$	3	M1 for $(k + 2)^{2} - 8 k$ A1 for $k^{2} - 4 k + 4 = (k - 2)^{2}$ B1 for explaining that $(k - 2)^{2} \geq 0$ so the roots are real
7(b)	2	1
7(c)	1	3	M1 for $3^{2} - 4 \times 2 \times c > 0$ A1 for $c < \frac{9}{8}$
7(d)	4	2	M1 for $4^{2} - 4 \times k \times 1 = 0$
8(a)	$43$	2	M1 for two of $23$ , $53$ , $33$ seen
8(b)	$14 - 46$	2	M1 for $12 - 26 - 26 + 2$ oe
8(c)	2	2	M1 for $5 - 15 + 15 - 3$ or $5 - 3$
8(d)	$(3 + 2)^{2} = 9 + 62 + 2 = 11 + 62$	2	M1 for $(3 + 2)^{2} = 9 + 32 + 32 + 2$ A1 for reaching $11 + 62$ with no errors
9(a)	$3 + 32$	3	M1 for $x (2 - 1) = 3$ M1 for $x = \frac{3}{2 - 1}$ multiplied by $\frac{2 + 1}{2 + 1}$
9(b)	$2 + 3$	3	M1 for multiplying by $\frac{3 + 1}{3 + 1}$ M1 for $\frac{4 + 2 3}{2}$
9(c)	$25$	3	M1 for $5 + 2$ or $5 - 2$ from rationalising either fraction M1 for common denominator $5 - 4 = 1$
9(d)	$\frac{1}{2 + 3} = 2 - 3$ , so $x + \frac{1}{x} = 4$	2	M1 for $\frac{1}{x} = 2 - 3$ A1 for $= 4$

M method mark · A accuracy mark (depends on the M mark) · B mark for a correct result or statement, independent of method · oe or equivalent · cao correct answer only · FT follow through after an earlier error · isw ignore subsequent working

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