Course Outline Year 7 Term 3
© XJS Coaching School Page 3 2016 version 1
Term 3
Week 1 Week 2 Week 3 Week 4 Week 5 Week 6 Week 7 Week 8 Week 9
Sat 16-Jul 23-Jul 30-Jul 6-Aug 13-Aug 20-Aug 27-Aug 3-Sep 10-Sep
Sun 17-Jul 24-Jul 31-Jul 7-Aug 14-Aug 21-Aug 28-Aug 4-Sep 11-Sep
Week Content Exercise
1
Algebra 1 (page86 - 89)
Key Words: Pronumerals, coefficient,
constant, Terms
Writing Expressions
‘Like Terms/Unlike Terms’
Summary of Index Laws (p. 88 – 89)
Show examples of Expressions and ask students
to find terms, pronumerals, coefficients and
constants,
E.g 2x 2 + 4x – 5xy + 9
Level 1 p.105 Qs.1, 2
Level 2 p. 106 Q.1
Work Sheet 1: Introductory
Algebra
2
Algebra 2 (pages 86 – 89, pages 94 – 97)
Go through Algebra, +, - , ×, ÷ of
Algebra
a) 2x + 3x = 5x, 5a – 4a = a
b) 5m × 2n = 10mn
c) 2
m ÷
6
n=
n
m
n
m 36
2
Algebraic Fractions:
a) 6
x +
6
2x =
6
3x
b) 4
5a +
3
4 a
= 12
17
12
)4(4)5(3
aaa
× and ÷ of Algebraic fractions.
a) 3
x ×
2
a =
6
ax
b) 25
aa =
5
2
c) 83
2 xx =
12
2x
d)5
3
6
aa
= 18
5
3
5
6
a
a
Level 1 p. 105 Qs. 3, 4, (left hand
side only)
L2 p. 106 Qs. 4, 5, 6(a, b, c)
L3 p. 107 Qs. 4, 5, 6
Work Sheet 2 – Algebra I
Course Outline Year 7 Term 3
© XJS Coaching School Page 4 2016 version 1
3
Algebra 3 (pages 88 – 89, 90 – 91, pages 98-
100)
Go through the 6 rules of Indices (p. 88 – 89)
Expanding values – Distributive Law
(p. 90 – 91)
2(m + 3) = 2m + 6 Do binomial expansion(p. 98 – 100)
(a + 4)(a – 3) = a2
+ a - 12
Level 1 p. 105 Qs. 5, 6
L 2 p. 106 Qs. 2, 7
L3 p. 107 Qs. 3, 7
L4 p. 108 Qs. 2, 5
Work Sheet 3 – Algebra II
Start revising Practice Exam 1
4
Algebra – Factorisation (page 92 – 93, page
102 – 104)
i) Taking out H.C.F
12a + 24 = 12(a + 2)(p. 92 – 93)
Grouping
am – 3a + 5m – 15 = (am -3a) + (5m –
15) = a(m – 3) + 5(m – 3) =(a + 5)(m –
3) D.O.P.S
a 2 - b 2 = (a + b)(a – b)
Trinomials
x2
- 2x – 8 = (x – 4)(x + 2)
Level 1 p.105 Qs. 7
L2 p. 106 Q. 3
L3 p. 107 Q. 5
L4 p. 108 Q.3
L6 p. 110 Q.4
Level 5 : Choose some relevant
worded problems.
Work Sheet 4 - Factorisation
5
Equations. (pages 121 – 129)
Define equation
One step Equation: x + 9 = 30, x = 21
Multiple step Equation: 6m – 10 = 20
6m = 20 + 10 = 30, m = 5
Equations with 2 unknowns: 9a – 2 = 6a
+ 16
9a – 6a = 16 + 2 = 18, 3a = 18, a = 6
Bracket Equations: 3(x –5) =24
3x – 15 = 24, 3x = 39, x = 13
Fraction Equation: 75
3
x
3x = 35, x = 7
Ability to write equations
Level 1 p. 134 Qs. 4, 6, 7
L2 p. 135 Qs. 5, 6, 8
L3 p. 136 Qs. 3, 4, 5, 6(LHS)
Work Sheet 5 – Equation 1
6
Equations (pages 128 – 132)
Substitution into Formulae, e.g V = u + at
where u = 7, a = 4, t = 3
V = 7 + 4 x 3 = 7 + 12= 19
Using Equations to solve problems(p.
128)
Level 2 p. 135 Q. 9
L3 p. 137 Qs. 7, 8
L6 p. 139 Qs. 5, 6, 7
Work Sheet 6 – Equation II
7 Do Practice Exam 2 in class Work Sheet 7 – Worded
Algebraic Problem
8 Term 3 Exam
9 Go through Term 3 Exam. Do holiday homework
End of term 3
Worksheet 1 Introductory Algebra Year 7 Term 3
© XJS Coaching School Page 5 2016 version 1
Name: ________________________________ Date: _______
Before you start, understand these key words:
a) pronumeral: the variable, usually represented by a letter (e.g., x, k, mn, x2, etc--)
b) coefficient: the number in front of the pronumeral
c) constant: the number by itself in an expression
d) term: all the values contained in an expression and includes a) or b) or c) or both
a) and b)
Example: 2x + 4y + 6 + m is an expression
a) there are 4 terms
b) the coefficient of x is 2 (also coefficient of the first term) and the coefficient of y is
4
c) the constant is 6
Q1. For each of the following expressions, write the coefficient of x.
(a) 7x (b) 6x2 – 7x (c) 4
x (d)
3
2x + 7
7 -7 ¼ -2/3
Q2. For each expression:
(i) state the number of terms
(ii) state the coefficient of the first term
(iii) state the constant.
(a) 597 2 yy (b) 73412 22 pqpq
i) 3 i) 4
ii) -7 ii) 12
iii) 5 iii) 7
(c) cabbba 35243 2222 (d) 5c2d 3c + 7d 2 + 8cd2 7cd
i) 5 i) 6
ii) 3 ii) 5
iii) 2 iii) -2
Q3. Write expressions for the following, where x and y represent numbers:
(a) 10 less than y (b) 10 minus 6 times x
y - 10 10 – 6x
(c) the sum of 3 times x and 5 times y
3x + 5y
Q4. David has 10 vintage cars. Write the number of cars he will have left if he sells:
Key Words: Pronumerals, coefficient, constant, Terms
Writing Expressions
Worksheet 1 Introductory Algebra Year 7 Term 3
© XJS Coaching School Page 6 2016 version 1
(a) 2 (b) p
8 10 – p
(c) Q (d) x on the first day and y the next day
10 – Q 10 – x – y
Q5. A bus leaves its depot with p passengers. Write an expression for the number of
passengers on the bus:
(a) on the first stop, 3 passengers get on p + 3
(b) on the 2nd stop, q passengers get off p + 3 – q
(c) next, r passengers get on and s passengers get off p + 3 – q + r – s
(d) finally, t passengers get on, u passengers get off p + 3 – q + r – s + t – u
Q6. Mark each of the following statements as either true (T) or false (F).
(a) A packet of Minties contains p Minties and they are divided amongst 7 people.
The number of Minties each person receives is 7/p. False (p/7)
(b) Amanda wants to buy 4 pairs of shoes each costing q dollars. The total cost is 4q
dollars. True
Q7. Kelly has a 60 cm-long string.
(a) If she cuts off x cm, how much string is remaining?
60 – x
(b) If she cuts off 1/5 of the remaining string, how much string was cut off?
)60(5
1x
(c) How much string is now remaining?
)60(5
4x
Q8. Shelley drives her car for p days over a 30-day period. Her sister Helen uses it for 6
1 of
the remaining days.
(a) For how long does Helen use the car?
)30(6
1p
(b) For how long do Shelley and Helen use the car over a 30-day period?
p + )30(6
1p or 5
6
5p
Worksheet 2 – Algebra I Year 7 Term 3
© XJS Coaching School Page 7 2016 version 1
Name: ________________________________ Date: _______
Q1. For each of the following terms, circle the like terms listed in brackets.
(a) 3ab ( 4a, 6b, –7ab, 3p, 2ab, ab)
(b) –7x2 ( –7x, –7, x2, –x2, 6x2)
(c) 3x2y ( 3x2, 2x2y, 3xy2, 2x, 3y)
(d) 5a2b ( 5ab2, 2a2b, 3ba2, 7
2ba, 3ab)
Q2. Simplify the following expressions by collecting like terms.
(a) 22 1520 yy (b) 35711 2 xxx
= 35 y 2. = 11𝑥2 + 2𝑥 + 3
(c) 76108 2 ggg (d) 12 7a2b 5 3ba2
= -8g2 – 4g + 7 = 7 – 10a2b.
Q3. Simplify the following expressions by collecting like terms.
(a) 222 387 pqqpqp (b) xxxx 25106 22
= 15p2q – 3pq2 = - 4x2 + 7x
(c) 22222 71139 abbaba (d) 3m2n + 5n 5m2n + 8n 11
= 2a2b2 + 4 - ab2 = - 2m2n + 13n – 11
Q4. Simplify the following expressions.
(a) pp 24 (b) pqq 63
= 8p2. = - 18 pq2.
(c) xyxy 410 (d) 6ab b 4a
= 40 x2y2 = - 24 a2b2.
Q5. Simplify the following expressions.
(a) 3x × 2x × 5y = (b) 11p × -3q2p × p =
= 30 x2y = - 33 p3q2.
(c) 3a2 4ab2 6a3b = (d) 5de 3d2 2e 10d4e3
= 72 a6b3 = - 300 d 7 e5.
Q6. Simplify the following expressions.
(a) 2
10d (b)
y24
12 (c) qq 3060 (d) 8x2y2 15xy
= 5d = y2
1 = 2
15
8xy
Key Words: Like Terms/Unlike Terms, Algebraic Fractions, Expanding
values
Worksheet 2 – Algebra I Year 7 Term 3
© XJS Coaching School Page 8 2016 version 1
Q7. Simplify the following expressions.
(a) y
xy
9
3 (b)
qr
pqr
12
40
(c) abcabc 306 (d)
22
423
24
30
yzx
zyx
= 3
x =
3
10
p =
5
1
=
4
5 2xyz
Q8. Evaluate the following:
(a) 6
5
4
3 (b)
3
2
10
9 (c)
6
5
3
2
= 12
1
12
109
=
5
3 =
5
4
5
6
3
2
Q9. Simplify the following expressions:
(a) 55
3 mm (b)
43
xx (c)
25
2 yy
= 5
4m =
12
7
12
34 xxx
=
10
9
10
54 yyy
(d) 77
4 nn (e)
82
qq (f)
83
2 rr
= 7
3n =
8
3
8
4 qqq
=
24
13
24
316 rrr
Q10. Simplify the following expressions:
(a) y
y 10
5 (b)
n
n
4
2
20
2 (c)
m
qr
q
m
8
316
= 2 = 20
1 = 6 r
Q11. Simplify the following expressions.
(a) 186
xx (b)
303
qpq (c)
10
2
6
7
5
2 xxx
= 3 = 10 p = 3
7x
Worksheet – Algebra II Year 7 Term 3
© XJS Coaching School Page 9 2016 version 1
Name: ________________________________ Date: _______
Q1. State the base and power (index) for each of the following: The first one has been done
for you.
(a) 45
Base : 4 Power: 5
(b) y9
Base: y, power: 9
(c) 23
Base: 2 Power: 3
(d) z7
Base: z Power: 7
Q2. Write the following in index form: The first one has been done for you.
(a) y y y (b) 7 7 7 7 7 (c) 5
3
5
3
5
3
5
3
5
3
5
3
=3y =
57 = 6)
5
3(
(d) 6 6 6 6 6 6 6 (e) 5 8 x x x x x y y y
= 76 = 40 35 yx
(f) 12 e 3 f f f f 2 g g g g
=72 44 gef
Q3. Write each of the following as a basic numeral: The first one has been done for you.
(a) 25 (b) 43. (c) (9)3
=2 x 2x 2 x 2 x 2 = 4 x 4 x4 -9 x -9x -9
= 32 = 64 = -729
(d) (2e)4 (e)
5
2
3
= 16 e4
= 32
243
2
35
5
Key Words: Rules of Indices/Expanding Values / Distributive Laws /
Simple and Binomial Expansion
Worksheet – Algebra II Year 7 Term 3
© XJS Coaching School Page 10 2016 version 1
Q4. Simplify each of the following:
(a) h2 h h4 (b) p3 q2 p q5 (c) 7a2c3 3a4c4 2a3c5
= 7h =
74qp = 42 129ca
Q5. Simplify each of the following:
(a) q10 q2 (b) d16 d 2 d 5 (c) 3
15
16
12
a
a
= 8q =
19d = 4
3 12a
(d) 10
69
24
48
p
pp (e)
7
5
10
9 6
7
5
l
k
l
k
= 3
4 5p =
3
4
42
5
l
k
Q6. Simplify each of the following:
(a) 2 0 (b) y 0 (c) (2y) 0 (d) 2 0 y (e) 2x 0 y
= 1 = 1 = 1 = y = 2y
Q7. Simplify each of the following:
(a) (y4)7 (b) (p5q0)11 (c) (4r8)3 (2r3)2
= 28y =
55p = 3062243 25624 rrr
(d) (a5)9 (a6)4 (e) 32
43
)(
)(
x
x
= 212445 aaa =
6
6
12
xx
x
Q8. Expand the following expressions.
(a) 2∙(y 3) (b) (q – 11) (c) 4∙(6x + 9)
= 2y – 6 = - q + 11 = - 24x – 36
(d) m∙(9 + m) (e) y∙(6y – 3) (f) 5x∙(7 2x)
= 9m + 2m = -6y
2+ 3y = - 35x + 10x
2
Worksheet – Algebra II Year 7 Term 3
© XJS Coaching School Page 11 2016 version 1
Q9. Expand and simplify the following expressions.
(a) 2(3x + 4y) + 2(4x – 3y) (b) 4(c – 4d) – 6(3c – 5d)
= - 6x + 8y + 8x – 6y = 2x + 2y = 4c – 16d – 18c + 30d
= 2(x + y) = -14c +14d = -14(c – 1)
(c) )2()3( yxyx (d) )44(3)43(2 ppp
= xy + 3x + xy – 2x = 6p 2 + 8 – 12p + 12
= 2xy + x = 6p2
- 12p + 20
Q10. Expand and simplify (binomial Expansion):
(a) )3)(14( aa (b) )67)(24( pp (c) )29)(29( xx
3134
3124
2
2
aa
aaa
123828
12142428
2
2
pp
ppp
2
2
481
4181881
x
xxx
(d) )2)(5( yxyx (e) )5)(3( baba
22
22
7110
2510
yxyx
yxyxyx
22
22
152
1535
baba
bababa
Worksheet 4 - Factorisation Year 7 Term 3
© XJS Coaching School Page 13 2016 version 1
Name: ________________________________ Date: _______
Q1. List all the factors of 48, then pick out the Highest Factor :-
Factors = 2, 3, 4, 6, 8, 12, 24, 48 Highest Factor = 48
Q2. (a) Find the highest common factor (HCF) of 42 and 24.
Factors of 42 = 2, 3, 6, 7, 21
Factors of 24 = 2, 3, 4, 6, 8, 12
HCF = 6 (b) Find the highest common factor (HCF) of 35 and 63.
Factors of 35 = 5, 7, 35
Factors of 63 = 3, 7, 9, 63
HCF = 7 Q3. Factorise by taking out the H.C.F:-
(a) 7x + 21. (b) 56 24x (c) 4xy + 20
= 7(x + 3) = 8(7 – 3x) = 4xy(1 + 5)
(d) 15m 2 + 25 (e) 12x 3 y + 21xy (f) 3c2d 3 + 9c3d 2 – 12cd 2
= 5m (3m +5) = 3xy (4x2
+7) = 3c d 2 (cd + 3c2
- 4)
Q4. Factorise by taking out the common factors in the brackets:-
(a) 3(x + y) – a(x + y). (b) 6x (b – c) + x2(b – c). (c) 8x (p – 2q) + 4x(p - 2q).
= (x + y)(3 – a) = (b – c)(6x + x2
) = (p – 2q)(8x + 4)
Q5. Factorise by grouping:
(a) ax + 2x + 2ay + 4y. (b) 7a2 – 14ab + 10ax – 20bx.
= x (a + 2) + 2y (a + 2) = 7a (a – 2b) + 10x (a – 2b)
= (x + 2y)(a + 2) = (a – 2b)(7a + 10x)
(c) 8x2 – 12xy –2px + 3py.
= 4x (2x – 3y) – p(2x + 3y)
= (4x – p)(2x – 3y)
Q6. Factorise using the Difference Of Perfect Square rule. (D.O.P.S):-
(a) 22 99100 (b) x2 – 6
2 (c)
22 nm
= (100 + 99)(100 – 99) = (x + 6)(x – 6) = (m – n)(m + n)
= 199
(d) x2 – 49. (e) 2281 y (f) 10025 2 m
= (x – 7)(x + 7) = (81 – y)(81 + y) = (5m – 10)(5m + 10)
(g) 4a2 9b2
= (2a – 3b)(2a + 3b)
Key Words: H.C.F/Grouping/D.O.P.S/Factorise
Worksheet 4 - Factorisation Year 7 Term 3
© XJS Coaching School Page 14 2016 version 1
Q7 (a) 25a2 – 16b2 (b) 4 – 9x2 y2 (c) x3 – 49x y2
(5a – 4b) (5a + 4b) (2 – 3xy) (2 + 3xy) x∙(x – 7y) (x +7y)
(d) 1 – 36m2 (e) 2ax – ay + 2bx - by (f) 3x(22a – b) + 7y(22a – b)
(1 – 6m) (1 + 6m) (2x – y) (a + b) (3x + 7y) (22a – b)
(g) 3a2 – 75 (h) 20m2n – 5n3 (i) 𝑚 2 −1
4
3(a – 5) (a + 5) 5n(2m – n) (2m + n) (𝑚 − 1
2)(𝑚 +
1
2)
(j) 1
9−
𝑎2
25 (k)
𝑥2−1
3𝑥+3 (l)
(3𝑥2−27)(𝑎+𝑦)
𝑎𝑥+𝑦𝑥−3𝑎−3𝑦
(1
3−
𝑎
5) (
1
3+
𝑎
5)
𝑥−1
3 3(x + 3)
BONUS QUESTION
Q8 Fully factorise 3𝑥4 + 243
625
3𝑥4 + 192
625
3(𝑥4 + 64
625)
3 [𝑥4 + 64
625+
16𝑥2
25−
16𝑥2
25]
3 [(𝑥2 +8
25)
2
−16𝑥2
25]
3 [(𝑥2 +8
25+
4𝑥
5) (𝑥2 +
8
25−
4𝑥
5)]
3
625(25𝑥2 + 20𝑥 + 8)(25𝑥2 − 20𝑥 + 8)
Worksheet 5 - Equation I Year 7 Term 3
© XJS Coaching School Page 15 2016 version 1
Name: ________________________________ Date: _______
Q1. Solve the following one step equations:
(a) x + 2 = 4 (b) m – 4 = 8 (c) 4x = 24.
x = 4 – 2 = 2 m = 12 x = 6
(d) 3
x = 5. (e) - x + 5 = 3 (f) m – 3 = -7
x = 15 -x = - 2, x = 2 m = - 4
(g) -3x = 32 (h) 55
y
x = -103
2 y = -25
Q2. Solve the following two step equations:
(a) 2x – 1 = 13. (b) 3x + 3 = 9 (c) 6 – 5x = 14.
2x = 14, x = 7 3x = -12, x = - 4 - 5x = - 20, x = 4
(d) 85
4
x (e) 5
7
2
x (f) 10
4
11
x
.
4x = 40, x = 10 2x = 35, x = 14.5 x – 11 = 40, x = 51
(g) 32
312
x (h) 4
5
6
x
12 – 3x = 6, -3x = - 6 - 6x = 20,
x = 2 x = - 3 2/3
Q3. Solve the following multiple-step equations:
(a) 3
12 x = 4. (b) 5
4
35
x (c) 3
7
)4(2
y
2x – 1 = -12, 5 – 3x = 20, 2y + 8 = 21
2x = -11 -3x = 15 2y = 13
x = - 5.5 x = - 5 y = 6.5
Q4. Solve the equation with pronumerals on both sides:
(a) 2x + 4 = x + 6 (b) 12x + 1 = 3x + 19 (c) 7x - 8 = 12x – 28
2x – x = 6 – 4, 9x = 18 -5x = -20
x = 2 x = 2 x = 4
(d) 3x + 1 = 12x + 10. (e) 14y + 3 = 10y – 14 (f) 2x + 11 = 13 9x
-9x = 9, x = -1 4y = -17, y = 4 ¼ 11x = 2, x = 2/11
Key Words: One step / Multiple / bracket / Equations with two unknowns /
Fractional Equations / fraction / Equations / Substitutions
Worksheet 5 - Equation I Year 7 Term 3
© XJS Coaching School Page 16 2016 version 1
Q5. Solve the following bracket equations:
(a) 3(x + 2) = 10. (b) 5(2x – 7) = 5 (c) 3.5(7.5x + 2.2) = 0
3x + 6 = 10, 3x = 4 10x – 35 = 5, 10x = 40 26.25x + 7.7 = 0,
x = 4/3 x = 4 x = -7.7/26.25
(d) – (3x + 4) = 13 (e) - 4(y – 5) = -40 (f) -5(m + 7) = 50
3x – 12 = 13, -3x = 25 - 4y + 20 = -40, = - 60 - 5m – 35 = 50,
x = -25/3 y = 15 -5m = 85, m = - 17
Q6. Solve the equation by first inverting the equation or by cross- multiplying.
(a) 9
2 =
x
5 (b)
5
25
y (c) -
8
63
m (d)
5
34
x
-2x = 45 2y = 25 6m = - 24 3x = -20
x = - 22.5 y = 12.5 m = - 4 x = -20/3
Worded Problems: Write an equation first and show working out in solving the number.
Q7. (a) If 3 less than 6 times a certain number is equal to 11 more than 4 times that certain
number, find the number.
6x – 3 = 4x + 11, 2x = 14,
x = 7 (b) Find a certain number which when added to 34 is equal to 200 less twice the
certain number.
x + 34 = 2x – 200, - x = - 200 – 34 = - 234
x = 234
(c) A school is given special funding according to the formula that there will be $300
given for each student along with a single $4000 grant.
A certain school received $124 900. How many students did it have?
Fund without the grant = 124900 – 4000 = 120900
Number of students = 120900/300 = 403
(d) The difference of 40 and twice a certain number is equal to 20. Find the number.
2x – 40 = 20, 2x = 60
x = 30
(e) Half of a number is one quarter. What is the number?
½ x = ¼ , x = ¼ × 2
x = ½
It is possible that 40 – 2x = 20
x = 10
Worksheet 6 - Equation II Year 7 Term 3
© XJS Coaching School Page 17 2016 version 1
Name: ________________________________ Date: _______
Q1. Substitutions: Find the pronumeral stated.
(a)
T
DS , find S if D = 80 and T = 2
(b) 2mcE , find m if E = 40 and c =
2
S = 80/2 = 40 m = 40/4 = 10
(c) d = b2 – 4ac , find d if b = 3, a = 2
and c = 4
(d) m = x 2 - y , find x if m = 51 and y
= 13
D = 9 – 24, D = - 15 x = ym
x = 8641351
x = 8
Q2. If a = 21 and b =
52 , evaluate:
(a) 3a – 2b (b) a
b
3
2
3 × ½ - 2 × 2/5 = 3/2 – 4/5
= 7/10
(2/5 × 2/5) /3/2 = 4/25 × 2/3
= 8/75
Q3. If 2
2
1atuts , evaluate s when:
(a) u = 0, a = 9.8 and t = 10 (b) u = –3, a = 10 and t = 5
s = 0 + ½ × 9.8 × 10 × 10
s = 490
s = -3 x 5 + ½ x 10 x 5 x 5
s = -15 + 125
s = 110
Q4. If the formula for the conversion of temperature from Celsius (°C) to Fahrenheit (°F) is
given by 325
9 CF . Find:
(a) F when C = 450 (b) C when F = 59
0
F = 9/5 x 45 + 32 = 81 + 32
F = 113° F
9/5 C = F – 32
= 9/5C = 59 – 32 = 27
= C = 27 x 5/9 = 15°C
Key Words: Formulae / Substitutions / Inequations
Worksheet 6 - Equation II Year 7 Term 3
© XJS Coaching School Page 18 2016 version 1
Q5. (a) Use the formula 6x = 2y 3 to find x when y = 11.
6x = 22 – 3 = 19
x = 19/6 = 3 𝟏
𝟔
(b) Use the formula in (a) to find y when x = 12
2y = 6x + 3
2y = 72 + 3 = 75
y = 75/2 = 37.5
Q6. The sum of n terms of an arithmetic sequence is given by the formula ][2
Lan
S
where a is the first term and L is the last term. Find the first term in a sequence which
has n = 5, L = 7 and S = 10.
a = S x 2/n – L,
a = 10 x 2/5 – 7, a = 4 – 7
a = -3
Q7. (a) Construct a formula for P and n where P is the profit a company makes after
selling n electric fans at $70 each, while the cost of manufacturing of each fan is
$25 and the fixed cost per month (which includes rent, electricity, and so on) is
$729.
Profit for each fan = $70 - $25 = $45
Formula: P = 45n - $729
(b) Find the profit made by the company if 2000 electric fans were sold in one month.
P = $45 x 2 000 - $729 = $90 000 - $ 729 = $89 271
(c) Find the minimum number of fans that need to be sold for the company to not
make a loss.
45n - $729 = 0
45n = $729
n = $729 / 45 = 16.2
Minimum number of fans that need to be sold = 17 fans
Worksheet 6 - Worded Algebraic Problems Year 7 Term 3
© XJS Coaching School Page 19 2016 version 1
Name: ________________________________ Date: _______
Q1. The sides of a square table top have dimensions of x2 metres.
(a) Write an expression for its perimeter.
4 × 2x = 8x
(b) Write an expression for its area.
2x × 2x = 4x2 .
(c) If all the sides are decreased by 1 m, what are the new dimensions?
New side = 2x – 1
(d) Write an expression for the new area and expand.
(2x – 1)(2x – 1) = 4x2 -2x – 2x + 1 = 4x2 - 4x + 1
Q2. A rectangular door is 180 cm long and 50 cm wide. (Area of rectangle = length x
width)
(a) Find the area of the door.
Area = 180 × 50 = 9000 cm2.
(b) If the length of the door is increased by y cm, write an expression for the new
length.
New length = 180 + y
(c) If the width of the door is increased by z cm, write an expression for the new
width.
New width = 50 + z
(d) Write an expression for the new area of the door and expand.
New Area = (180 + y)(50 + z) = 9000 + 180z + 50y + yz cm2.
(e) Find the area of the door if y = 2 cm and z = 3 cm.
Area = 9000 + 180 × 3 + 50 × 2 + 2 × 3 = 9646 cm2.
Q3. If a rectangular table top has a length of x3 cm and a width of x cm, an expression for
its perimeter is (circle the correct answer):
A 4x cm
B 3x2 cm2
C 4x2 cm2
D 8x cm
E 8x2 cm2
Worded Algebraic Problems
Worksheet 6 - Worded Algebraic Problems Year 7 Term 3
© XJS Coaching School Page 20 2016 version 1
Q4. A square has a side length of 3y cm.
(a) Write an expression for its perimeter.
Perimeter = 4 × 3y = 12 y
(b) Write an expression for its area.
Area = 3y × 3y = 9y2.
(c) If the side length is decreased by z cm, write an expression for the new
perimeter.
New side = 3y – z, New Perimeter = 4(3y – z) = 12y – 4z
(d) Find the perimeter when y = 5 cm and z = 6 cm.
P = 12 × 5 – 4 × 6 = 60 – 24 = 36
Q5. A rectangular pond has dimensions 5 m by 3 m. A pathway around the edge of the pond
is y m wide on each side.
(a) Sketch and label the pond and the pathway around the pond.
(b) Find the area of the pond.
Area of Pond = 3 × 5 = 15 m2.
(c) What are the lengths of the “pond + pathway” and the widths of the “pond +
pathway”?
Length of pond + pathway = 5 + 2y, Width of pond + pathway = 3 + 2y
(d) Find the area of the pond and the pathways.
Area of pond + pathways = (5 + 2y)∙(3 + 2y) = 15 + 10y + 6y + 4y2
= 15 + 16y + 4y2
(e) Find the area of the pathways.(Area = Area of (d) – Area of (b))
Area of pathways = 15 + 16y + 4y2 - 15 = 16y + 4y2.
3 m
5 m
y m
y m
Practice Test 1 Year 7 Term 3
© XJS Coaching School Page 21 2016 version 1
Name: ________________________________ Date: _______
Section A: Multiple Choice Questions.
Please circle the correct answer.
Q1. The coefficient of x in the expression
4x2 + 3
2x +
x3
1 is:
A 4
B 2
C 3
2
D 3
E 3
1
Q2. Brad buys licorice by the metre. He
cuts off y cm. The amount of licorice
left is:
A 1 – y
B 100 – y
C 100 + y
D 100
y
E y
Q3. What is 5xy2 – 4 – 3xy2 + 3 simplified?
A 2xy2 – 1
B 8xy2 + 7
C 2xy2 + 7
D 8xy2 – 1
E. 2x2y
Q4. What is 20b 30b simplified?
A 30
20b
B 3
2
C 30
20
D 3
2b
E b3
2
Q5. If y = 2 + 6x 3x2, the value of y when
x = 3 is:
A 5
B 7
C 3
D 2
E 5
Q6. What does 6
2 p
p equal?
A 2
B 2
1
C 12p
D 12p2
E 2
12
p
Q7. What is 29
6 xx simplified?
A 18
7x
B 18
21x
C x18
21
D x18
7
E x6
7
Q8. What is x3 in expanded form?
A 3x
B x x x
C 3 x x x
D x + x + x
E x x x
Q9. The answer to n3 n2 is:
A n5
B n6
C n1
D n32
E n23
Q10. What is y7 y5?
A y12
B y2
C y35
D y57
E y75
Practice Test 1 Year 7 Term 3
© XJS Coaching School Page 22 2016 version 1
Q11. zzyyy 42 in index form is:
A. 23 42 zxy
B zy 38
C 58yz
D 238 zy
E. 23 24 zy
Q12. What is equivalent to )43(5 yy ?
A yy 2015 2
B 18 2 y
C 18 2 y
D 115 2 y
E 2015 2 y
Q13. What is the equivalent of
5)42(36 yyy ?
A 566 2 yy
B 5618 2 yy
C 512 y
D 566 2 yy
E 512 y
Q14. The equivalent of )2)(1( xx is:
A 22 xx
B 22 xx
C 22 xx
D 32 xx
E 22 yx
Section B: Short Answer Questions: Please show workings.
Q1. Answer the following for each expression.
i) State the number of terms.
ii) State the constant (if there is one)
iii) State the coefficient of x
(a) 10x2 + 9x + 5x2 + 7
i) State the number of terms. 3 (10x2 and 5x2 are like terms)
ii) State the constant (if there is one) 7
iii) State the coefficient of x 9
(b) 3y3 + 4y2 2y – 5
i) State the number of terms. 4
ii) State the constant (if there is one) - 5
iii) State the coefficient of x 0
Q2. A class of z students go to the Melbourne Zoo. One fifth of the class visit the Reptile
House while one sixth of the class go to the Butterfly House.
(a) Write an expression to represent the number of students who went to the Reptile
House.
1
5𝑧
(b) Write an expression to represent the number of students who went to the Butterfly
House.
1
6𝑧
(c) Write an expression to represent the number of students who went to both the
Practice Test 1 Year 7 Term 3
© XJS Coaching School Page 23 2016 version 1
Reptile and Butterfly Houses.
(1
5+
1
6) 𝑧 =
11
30𝑧
Q3. Simplify the following expressions by collecting like terms.
(a) 4x + 7x = 11x (b) 12y – 8y = 4y
(c) 3z 10z + 4z = - 3 z (d) 15x – 5y + 6 – 2x = 13x – 5y + 6
(e) 13 + ab – 4b2 + 8ab – 7b2 = 13 + 9ab – 11b2.
Q4. Expand the following expressions and simplify where possible.
(a) )77(5 q (b) )52(3 pp (c) ttt 4)38(612
- 35q + 35 6 pp 152 12 + 48t2
+14t
Q5. Expand and simplify each of the following.
(a) )1)(2( yy (b) )37)(37( xx
23
22
2
2
yy
yyy
2
2
94249
9212149
xx
xxx
(c) )7)(7( xx (d) )52)(52( dd
49
4977
2
2
x
xxx
254
2510104
2
2
d
ddd
Q6. Simplify the following expressions.
(a) 12
3
6
mm (b)
15
2
5
2
12
5
12
32 mmm
15
25
15
26
15
)2(32
qqqqq
Practice Test 1 Year 7 Term 3
© XJS Coaching School Page 24 2016 version 1
(c) 2
23
20
45
xx
20
2435
20
203045
20
)23(1045
x
xxxx
Practice Test 2 Year 7 Term 3
© XJS Coaching School Page 25 2016 version 1
Name: ________________________________ Date: _______
Section A: Multiple Choice Questions.
Please circle the correct answer.
Q1. The highest common factor of 75 and
100 is:
A 5
B 25
C 50
D 75
E 100
Q2. When factorised, 24x – 18y equals:
A 24(x – 2y)
B 3(8x – 6y)
C 18(6x – y)
D 6(4x – 3y)
E None of the above
Q3. The solution to the equation
7(x – 15) = 28 is:
A x = 11
B x = 19
C x = 20
D x = 6.14
E None of the above
Q4. When factorised, 2x(c + d) – 4y(c + d)
equals:
A (2x – 4y)(c + d)
B 2(x – 2y)(c + d)
C 2(2x – 4y)(c + d)
D 2x – 4y(c + d)
E None of the above
Q5. When factorised, m2 – 16n2 equals:
A (m – n)(m + n)
B (5m – 4n)(5m + 4n)
C (m – 4n)(m + 4n)
D (5m + n)(5m + n)
E (m – 4n)(m – 4n)
Q6. When factorised, what does x2 + x – 6
equal?
A (x – 3)(x + 2)
B (x – 6)(x + 1)
C (x – 2)(x + 3)
D (x – 1)(x + 6)
E None of the above
Q7. What is equivalent to )43(5 yy ?
F yy 2015 2
G 18 2 y
H 18 2 y
I 115 2 y
J 2015 2 y
Q8. What is the equivalent of
5)42(36 yyy ?
F 566 2 yy
G 5618 2 yy
H 512 y
I 566 2 yy
J 512 y
Q9. The equivalent of )2)(1( xx is:
F 22 xx
G 22 xx
H 22 xx
I 32 xx
J 22 yx
Q10. The solution to the equation 4
x = 5 is:
A x = 20
B x = 20
C x = 1.25
D x = 1
E x = 9
Q11. What is the solution to the equation
7 = 14 + x?
A x = 2
B x = 7
C x = 7
D x = 2
E None of the above
Q12. What is the solution to the equation
5x + 3 = 37?
A x = 8
B x = 8
C x = 6.8
D x = 106
E x = –106
Practice Test 2 Year 7 Term 3
© XJS Coaching School Page 26 2016 version 1
Section B: Short Answer Questions: Please show workings.
Q1. Factorise the following:
(a) 45x2 – 18x. (b) 5a2b3 + 15a3b2 – 25ab2.
3x(15x – 6) )53(5 22 aabab
(c) 36 – y2 (d) 25x2 – 16y2
(6 – y)(6 + y) (5x – 4y)(5x + 4y)
(e) x2 + 12x + 11
(x + 11)(x + 1)
Q2. Solve the following equations.
(a) a + 46 = 243 (b) b – 6.1 = 3.2
a =243 – 46 = 197 b = 3.2 + 6.1 = 9.3
(c) 5c = 49 (d) 8
x = 2.25
c = 5
49
5
49 x = 2.25 x 8 = 18.
Q3. Solve the following equations.
(a) 4p + 2.6 = 9 (b) 3 – 0.4s = 11
4p = 9 – 2.6 = 6.4 -0.4s = -11 -3 = -14
p = 6.4/ 4 = 1.6
35
4
140
4.0
14
s
(c) 2.0
t + 5 = 2.3 (d) 8(x – 5) = 30
54.02.07.2
7.253.22.0
xt
t
8x – 40 = -30
8x = -30 + 40 = 10
4
11
8
10x
Practice Test 2 Year 7 Term 3
© XJS Coaching School Page 27 2016 version 1
(e) 3(m – 7) = 4(m – 3)
3m – 21 = 4m – 12
3m – 4m = -12 + 21 = 9
- m = 9
m = - 9
Q4. Mary Jane wishes to hire a car for 1 day. She has a total of $65 to spend. The Acapulco
Gold Car Rental company charges a flat fee of $30 per day plus $0.23 per kilometre.
How far can she travel on her budget?
Equation: C = $30 + 0.23k
65 = 30 + 0.23k
0.23k = 65 -30 = 35
k =
152
23
3500
23.0
35
She can travel approximately 152 km.
Q5. The Columbian Car Rental Company charges a flat fee of $38 per day, but only $0.19
per kilometre. Comparing this rate with the Acapulco Gold company from the previous
question, for what distance will the total cost from the two companies be the same
Equation of Columbian Car Company is; C = $38 + 0.19k
Equation of Acapulco Gold Company is: C = $30 + 0.23k
If cost is the same, then, 38 + 0.19k = 30 + 0.23k
38 – 30 = 0.23k – 0.19k
8 = .04k k = 8/0.04 = 200
Cost will be the same after 200 kilometres
Holiday Homework Year 7 Term 2
© XJS Coaching School Page 29 2016 version 1
Name: ________________________________ Date: _______
Q1. Three consecutive odd numbers added up to 69. What is the first, second and third
number?
x + x+2 + x+4 = 69, 3x + 6 = 69
3x = 69 – 6 = 63, x = 21 .
Q2. Four consecutive odd numbers added up to 192. What are the four numbers?
x + x+2 + x+4 + x +6 = 192, 4x + 12 = 192
4x = 192 – 12 = 180, x = 180/4
x = 45
Q3. The side of an equilateral triangle is 2x + 3. If the perimeter of the triangle is 69 cm, find
x.
3(2x + 3) = 69, 2x +3 = 69/3 = 23
2x = 23 -3 = 20, x = 20/2
x = 10
Q4. Adam walks x km on the first day. On the 2nd day, he only walks half the distance of the
first day. On the 3rd day he walks a third of the distance of the first day. On the forth day
he walks a quarter of the distance and on the fifth day, he walks a fifth of the distance he
walked on the first day. Find out how far he walked on the first day (to the nearest whole
number) if his total journey was 100km?
x + 1/2x + 1/3 x + ¼ x + 1/5 x = 100
10060
137
60
1215203060
xxxxxx, 137x = 6000
x = 6000/137 = 43.79
x = 44km
Q5. A taxi charges $3 flag fall and then $0.80 per kilometre. Charles had $60 on him.
(a) How far can he travel in a taxi with the money he had?
60/.8 = 75 km
(b) If Charles only wanted to travel 30 kilometres, how much ( to the nearest cent)
must he pay the taxi?
Cost = 3 + 30 x 0.8 =$27
Q6. A plumber charges according to the formula: C = $70 + 60h where C represents Cost
and h represents the number of hours.
(a) If the plumber worked for 2 hours, how much will he be paid?
Cost = 70 + 60 x 2 = $190
(b) A person asking the plumber to repair his tap only had $100. How long can the
plumber worked on the tap?
(100 – 70 ) /60 = 0.5 hours
Holiday Homework Year 7 Term 2
© XJS Coaching School Page 30 2016 version 1
Q7. Mary wanted to buy a radio which cost $450. She already saved $70 and could save $20
per month. How long will it take her to save enough to buy her radio?
She has to save 450 – 70 = $380, No of months needed = 380/20 = 19
Q8. A school is given a lump sum of $2500 and another $250 per student. If the government
gave $72500, how many students are there in the school?
Amount of money minus the lump sum = 72500 – 2500 = $70000
No. of students in school = 70000/250 = 280
Q9 If y represents a certain number, write expressions for the following numbers.
(a) A number 12 more than y y + 12
(b) A number 5 less than y y - 5
(c) A number that is double y 2y
(d) The number that is formed when y is divided by 8 y/8
(e) The number that is formed when y is multiplied by 6
and 10 is added to the result 6y + 10
Q10 If tickets to a soccer match cost $25 for adults and $17 for children, write an
expression for the cost of:
(a) y adult tickets $ 25y
(b) d children tickets $ 17d
(c) y adult tickets and d children tickets $ (25y + 17d)
Q11 Find the value of the following expressions, if a = 3 and b = 5
(a) 6 a 18
(b) 6 b 30
(c) 3a + 5 b 34
(d) (b – 2) ÷ (2 – a) - 3
(e)
2
1
4
a
5
4
Q12 Tristan is now a years old.
(a) Write an expression for his age in 5 years’ time. a + 5
(b) How old was Tristan 7 years ago? a - 7
(c) Tristan’s mum is 27 years older than him. How old is
his mum?
a + 27
Q14 The formula for finding the perimeter (P) of a rectangle of length l and width w is
wlP 22 . Use this formula to find the perimeter of a 50 m swimming pool with a
width of 20 m
P = 2∙(50) + 2∙(20) = 100 + 40 = 140 m