Renee's Extra Practice — Sec 2 Mathematics

26 questions · 124 marks · full step-by-step answer key
Prepared by Miss Clarissa Ng
www.clartutors.com
Questions 1 to 26 · 124 marks
Q1. Simplify (9a4b−6)½ × (2a−1b2)3, leaving your answer in positive index form. [3]
Q2. A pen and a notebook cost $15 together. Mrs Lim bought some notebooks for $57. If she had bought pens instead, she would have 4 more pens by paying $2 less. The cost of a notebook is $n.

(a) Write down an expression, in terms of n, for

(i) the number of notebooks that was bought with $57, [1]

(ii) the number of pens she could have bought by paying $2 less. [1]

(b) Form an equation in n and show that it reduces to 4n2 + 52n − 855 = 0. [3]

(c) Solve the equation 4n2 + 52n − 855 = 0 and find the value of n. [2]

(d) How many pens could Mrs Lim buy with $55? [1]

Q3.

(a) Express 5x2 + 20x + 6 in the form a(x + p)2 + q. [3]

(b) Hence write down the coordinates of the turning point of the graph of y = 5x2 + 20x + 6. [1]

(c) State the equation of the line of symmetry of the same graph. [1]

Q4. Solve the equation 8x−2 ÷ 14x+1 = 23x−5. [3]
Q5. Without the use of a calculator, evaluate

27125−⅓ ÷ √(1 79) × 49½

Show all your working clearly. [4]
Q6. The pair of simultaneous equations

5x + ky = 8    and    10x − 6y = 15

has no solution. Find the value of k. [2]
Q7.

(a) Factorise mn − 5m + 2n − 10. [2]

(b) Hence, express (4m − 12)(m2 − m − 6)  ÷  (m − 5)[(m + 2)(n − 5)] as a single fraction in its simplest form. [3]

Q8. Express 36(x − 5)2 − x(5 − x) as a single fraction. [3]
Q9. By completing the square, express the equation y = −2x2 + 4x + 1.5 in the form y = a(x + b)2 + c, where a ≠ 0 and b, c are rational numbers. [3]
Q10. Solve the simultaneous equations

y − x = −4

3y2 − 2xy = 0

Q11. Leaving your answer in positive index form, simplify the expression

(−8p6q−3)⅓ × 3p2q06p−½q¼

Q12. A bakery stamps cookies. Machine A takes t seconds to stamp one cookie.

(a) Write an expression, in terms of t, for the number of cookies machine A stamps in one second. [1]

(b) Machine B takes 15 seconds less than machine A to stamp one cookie. Write an expression, in terms of t, for the number of cookies stamped by machine B in one second. [1]

(c) One morning, both machines were used for 1 hour. They stamped a total of 200 cookies. Write down an equation in t to represent this information and show that it reduces to t2 − 51t + 270 = 0. [3]

(d) Solve the equation t2 − 51t + 270 = 0. [2]

(e) Find the number of cookies machine B stamps in one hour. [1]

Q13. Simplify the following, giving your answers in positive index form.

(a) 5ab0√(25a4) [2]

(b) bx3(xy)3  ×  3x−1y−4 [3]

(c) (−ab½)3 × a4/3a−2b [3]

Q14. Simplify (16x6y−4)½ × (3x−2y3)2, leaving your answer in positive index form. [3]
Q15.

(a) Express 3x2 − 18x + 8 in the form a(x + p)2 + q. [3]

(b) Hence write down the coordinates of the turning point of the graph of y = 3x2 − 18x + 8. [1]

(c) State the equation of the line of symmetry of the same graph. [1]

Q16. A marker and a ruler cost $13 together. Mrs Tan bought some markers for $48. If she had bought rulers instead, she would have 3 more rulers by paying $3 less. The cost of a marker is $m.

(a) Write down an expression, in terms of m, for

(i) the number of markers that was bought with $48, [1]

(ii) the number of rulers she could have bought by paying $3 less. [1]

(b) Form an equation in m and show that it reduces to 3m2 + 54m − 624 = 0. [3]

(c) Solve the equation 3m2 + 54m − 624 = 0 and find the value of m. [2]

(d) How many rulers could Mrs Tan buy with $45? [1]

Q17. Solve the equation 9x+1 ÷ 13x−2 = 34x−3. [3]
Q18. Without the use of a calculator, evaluate

2764−⅔ ÷ √(1 916) × 2516½

Show all your working clearly. [4]
Q19. The pair of simultaneous equations

3x − ky = 4    and    9x + 15y = 2

has no solution. Find the value of k. [2]
Q20.

(a) Factorise ab − 6a + 5b − 30. [2]

(b) Hence, express (5a − 15)(a2 + 2a − 15)  ÷  (a − 6)[(a + 5)(b − 6)] as a single fraction in its simplest form. [3]

Q21. Express 9(x − 2)2 − x(2 − x) as a single fraction. [3]
Q22. By completing the square, express the equation y = −4x2 + 8x + 2.5 in the form y = a(x + b)2 + c, where a ≠ 0 and b, c are rational numbers. [3]
Q23. Solve the simultaneous equations

y − 3x = −6

y2 − 2xy = 0

Q24. Leaving your answer in positive index form, simplify the expression

(−27a3b−6)⅓ × 4a4b08a−½b¾

Q25. A workshop prints keychains. Machine A takes t seconds to print one keychain.

(a) Write an expression, in terms of t, for the number of keychains machine A prints in one second. [1]

(b) Machine B takes 30 seconds less than machine A to print one keychain. Write an expression, in terms of t, for the number of keychains printed by machine B in one second. [1]

(c) One morning, both machines were used for 1 hour. The keychains printed came to 100 altogether. Write down an equation in t to represent this information and show that it reduces to t2 − 102t + 1080 = 0. [3]

(d) Solve the equation t2 − 102t + 1080 = 0. [2]

(e) Find the number of keychains machine B prints in one hour. [1]

Q26. Simplify the following, giving your answers in positive index form.

(a) 4xy0√(16x6) [2]

(b) cy2(yz)3  ×  2y−1z−3 [3]

(c) (−xy½)2 × x2/3x−3y [3]

Answer Key — Renee's Extra Practice

Total: 124 marks · 26 questions · step-by-step working for every question. Marks are for a correct method even if the final answer is wrong; the last line of each entry is the final answer.
Q1 · Indices — 24b3a  [3 marks]
(9a4b−6)½ = 3a2b−3
(2a−1b2)3 = 8a−3b6
Product = 24a2−3b−3+6 = 24a−1b3
Positive index form: 24b3 / a
Q2 · Quadratic word problem — (a)(i) 57n   (ii) 5515 − n   (c) n = 9.5   (d) 10  [8 marks]
(a)(i) each notebook costs $n, so 57n
(ii) $2 less than $57 is $55, and a pen costs $(15 − n), so 5515 − n
(b) pens − notebooks = 4: 5515 − n − 57n = 4
55n − 57(15 − n) = 4n(15 − n) → 55n − 855 + 57n = 60n − 4n2
4n2 + 52n − 855 = 0  □
(c) n = −52 ± √(522 + 4×4×855) ÷ 8 = (−52 ± 128) ÷ 8
n = 768 = 9.5 or n = −1808 = −22.5 (rejected — a price cannot be negative)
(d) a pen costs 15 − 9.5 = $5.50, so 55 ÷ 5.5 = 10 pens
Q3 · Completing the square — (a) 5(x + 2)2 − 14   (b) (−2, −14)   (c) x = −2  [5 marks]
5x2 + 20x + 6 = 5(x2 + 4x) + 6
= 5[(x + 2)2 − 22] + 6
= 5(x + 2)2 − 20 + 6 = 5(x + 2)2 − 14
(b) turning point (−2, −14)  —  the bracket is zero there, giving the least value of y
(c) line of symmetry x = −2
Q4 · Indicial equation — x = −0.5  [3 marks]
8 = 23 and 14 = 2−2, so 23(x−2) ÷ 2−2(x+1) = 23x−5
23x−6 ÷ 2−2x−2 = 23x−5 → 2(3x−6) + (2x+2) = 23x−5
5x − 4 = 3x − 5 → 2x = −1
x = −0.5
Q5 · Indices and surds, no calculator — 56  [4 marks]
27125−⅓ = 35−1 = 53   (because 27125 = 353)
√(1 7⁄9) = √169 = 43
49½ = 23
53 ÷ 43 × 23 = 53 × 34 × 23
= 3036 = 56
Q6 · No solution / parallel lines — k = −3  [2 marks]
No solution means the lines are parallel: the same gradient, different intercepts.
5x + ky = 8 → y = 8 − 5xk, gradient −5k
10x − 6y = 15 → y = 10x − 156, gradient 53
−5k = 53 → 5k = −15 → k = −3
Q7 · Factorising by grouping + algebraic fractions — (a) (m + 2)(n − 5)   (b) 4(n − 5)m − 5  [5 marks]
(a) mn − 5m + 2n − 10 = m(n − 5) + 2(n − 5)
= (m + 2)(n − 5)
(b) 4m − 12 = 4(m − 3) and m2 − m − 6 = (m − 3)(m + 2)
so (4m − 12)(m2 − m − 6) = 4(m − 3)[(m − 3)(m + 2)] = 4m + 2
invert the second fraction: (m + 2)(n − 5)m − 5
4m + 2 × (m + 2)(n − 5)m − 5 = 4(n − 5)m − 5
Q8 · Algebraic fractions — x2 − 5x + 36(x − 5)2  [3 marks]
5 − x = −(x − 5), so x5 − x = −xx − 5
36(x − 5)2 − x5 − x = 36(x − 5)2 + xx − 5
common denominator (x − 5)2: = 36 + x(x − 5)(x − 5)2
= x2 − 5x + 36(x − 5)2
Q9 · Completing the square (rational) — y = −2(x − 1)2 + 3.5  [3 marks]
−2x2 + 4x + 1.5 = −2(x2 − 2x) + 1.5
= −2[(x − 1)2 − 12] + 1.5
= −2(x − 1)2 + 2 + 1.5
= −2(x − 1)2 + 3.5
Q10 · Simultaneous: linear + quadratic — (4, 0) and (12, 8)  [6 marks]
y = x − 4
3(x − 4)2 − 2x(x − 4) = 0
3(x2 − 8x + 16) − 2x2 + 8x = 0 → x2 − 16x + 48 = 0
(x − 4)(x − 12) = 0 → x = 4 or x = 12
x = 4 → y = 0;   x = 12 → y = 8
(4, 0) and (12, 8)
Q11 · Positive index form (algebraic) — −p9/2q5/4  [4 marks]
(−8p6q−3)⅓ = −2p2q−1
numerator: −2p2q−1 × 3p2 = −6p4q−1
−6p4q−1 ÷ 6p−½q¼ = −p4−(−½)q−1−¼
= −p9/2q−5/4 = −3p9/2q5/4
Q12 · Rate word problem — (a) 1t   (b) 1t − 15   (d) t = 45   (e) 120  [8 marks]
(a) 1t
(b) 1t − 15
(c) 1 hour = 3600 s, so 3600t + 3600t − 15 = 200
3600(t − 15) + 3600t = 200t(t − 15) → 7200t − 54000 = 200t2 − 3000t
200t2 − 10200t + 54000 = 0 → t2 − 51t + 270 = 0  □
(d) (t − 45)(t − 6) = 0 → t = 45 or t = 6
t = 6 is rejected: machine B would take 6 − 15 = −9 seconds, which is impossible, so t = 45
(e) machine B takes 45 − 15 = 30 s per cookie, so 3600 ÷ 30 = 120 cookies
Q13 · Indices, three parts — (a) 1a   (b) 3byx   (c) −a19/3b½  [8 marks]
(a) √(25a4) = 5a2, so 5a ÷ 5a2 = 1a
(b) (xy)3 = x3y3, so bx3x3y3 = by3
by3 × 3y4x = 3byx
(c) (−ab½)3 = −a3b3/2   and   a4/3 ÷ (a−2b) = a10/3b
−a3b3/2 × a10/3b = −a3 + 10/3b3/2 − 1 = −a19/3b½
Q14 · Indices — 36y4x  [3 marks]
(16x6y−4)½ = 4x3y−2
(3x−2y3)2 = 9x−4y6
Product = 36x3−4y−2+6 = 36x−1y4
Positive index form: 36y4 / x
Q15 · Completing the square — (a) 3(x − 3)2 − 19   (b) (3, −19)   (c) x = 3  [5 marks]
3x2 − 18x + 8 = 3(x2 − 6x) + 8
= 3[(x − 3)2 − 32] + 8
= 3(x − 3)2 − 27 + 8 = 3(x − 3)2 − 19
(b) turning point (3, −19)
(c) line of symmetry x = 3
Q16 · Quadratic word problem — (a)(i) 48m   (ii) 4513 − m   (c) m = 8   (d) 9  [8 marks]
(a)(i) 48m
(ii) $3 less than $48 is $45, and a ruler costs $(13 − m), so 4513 − m
(b) rulers − markers = 3: 4513 − m − 48m = 3
45m − 48(13 − m) = 3m(13 − m) → 45m − 624 + 48m = 39m − 3m2
3m2 + 54m − 624 = 0  □
(c) divide by 3: m2 + 18m − 208 = 0 → (m − 8)(m + 26) = 0
m = 8 or m = −26 (rejected — a price cannot be negative)
(d) a ruler costs 13 − 8 = $5, so 45 ÷ 5 = 9 rulers
Q17 · Indicial equation — x = 3  [3 marks]
9 = 32 and 13 = 3−1, so 32(x+1) ÷ 3−(x−2) = 34x−3
32x+2 ÷ 3−x+2 = 34x−3 → 2x + 2 + x − 2 = 4x − 3
3x = 4x − 3
x = 3
Q18 · Indices and surds, no calculator — 169  [4 marks]
2764−⅔ = 34−2 = 169   (because 2764 = 343)
√(1 9⁄16) = √2516 = 54
2516½ = 54
169 ÷ 54 × 54 = 169 × 45 × 54
= 169
Q19 · No solution / parallel lines — k = −5  [2 marks]
3x − ky = 4 → y = 3x − 4k, gradient 3k
9x + 15y = 2 → y = 2 − 9x15, gradient −35
3k = −35 → 3k = −15 → k = −5
Q20 · Factorising by grouping + algebraic fractions — (a) (a + 5)(b − 6)   (b) 5(b − 6)a − 6  [5 marks]
(a) ab − 6a + 5b − 30 = a(b − 6) + 5(b − 6)
= (a + 5)(b − 6)
(b) 5a − 15 = 5(a − 3) and a2 + 2a − 15 = (a + 5)(a − 3)
so (5a − 15)(a2 + 2a − 15) = 5(a − 3)[(a + 5)(a − 3)] = 5a + 5
invert the second fraction: (a + 5)(b − 6)a − 6
5a + 5 × (a + 5)(b − 6)a − 6 = 5(b − 6)a − 6
Q21 · Algebraic fractions — x2 − 2x + 9(x − 2)2  [3 marks]
2 − x = −(x − 2), so x2 − x = −xx − 2
9(x − 2)2 + xx − 2
common denominator (x − 2)2: = 9 + x(x − 2)(x − 2)2
= x2 − 2x + 9(x − 2)2
Q22 · Completing the square (rational) — y = −4(x − 1)2 + 6.5  [3 marks]
−4x2 + 8x + 2.5 = −4(x2 − 2x) + 2.5
= −4[(x − 1)2 − 1] + 2.5
= −4(x − 1)2 + 4 + 2.5
= −4(x − 1)2 + 6.5
Q23 · Simultaneous: linear + quadratic — (2, 0) and (6, 12)  [6 marks]
y = 3x − 6
(3x − 6)2 − 2x(3x − 6) = 0
9x2 − 36x + 36 − 6x2 + 12x = 0 → 3x2 − 24x + 36 = 0
x2 − 8x + 12 = 0 → (x − 2)(x − 6) = 0 → x = 2 or x = 6
x = 2 → y = 0;   x = 6 → y = 12
(2, 0) and (6, 12)
Q24 · Positive index form (algebraic) — −3a11/22b11/4  [4 marks]
(−27a3b−6)⅓ = −3ab−2
numerator: −3ab−2 × 4a4 = −12a5b−2
−12a5b−2 ÷ 8a−½b¾ = −128a5+½b−2−¾
= −32a11/2b−11/4 = −3a11/22b11/4
Q25 · Rate word problem — (a) 1t   (b) 1t − 30   (d) t = 90   (e) 60  [8 marks]
(a) 1t
(b) 1t − 30
(c) 1 hour = 3600 s, so 3600t + 3600t − 30 = 100
3600(t − 30) + 3600t = 100t(t − 30) → 7200t − 108000 = 100t2 − 3000t
100t2 − 10200t + 108000 = 0 → t2 − 102t + 1080 = 0  □
(d) (t − 90)(t − 12) = 0 → t = 90 or t = 12
t = 12 is rejected: machine B would take 12 − 30 = −18 seconds, which is impossible, so t = 90
(e) machine B takes 90 − 30 = 60 s per keychain, so 3600 ÷ 60 = 60 keychains
Q26 · Indices, three parts — (a) 1x2   (b) 2cy2   (c) x17/3  [8 marks]
(a) √(16x6) = 4x3, so 4x ÷ 4x3 = 1x2
(b) (yz)3 = y3z3, so cy2y3z3 = cyz3
cyz3 × 2z3y = 2cy2
(c) (−xy½)2 = x2y   and   x2/3 ÷ (x−3y) = x11/3y
x2y × x11/3y = x2 + 11/3 = x17/3
Miss Clarissa Ng | www.clartutors.com