Chapter 13 Volume of prisms

80 mm
Chapter 13 Volume of prisms
Activity 13.1
1. Discussion
2. Yes, it is possible to calculate volume using area of
cross-section × length, without separating the shape
into its basic shapes. The problem, however, is
deciding on which measurement to use for “length”.
In some complex shapes, it can be difficult to
identify the correct measurement for length.
3. In the example, we could find the area of the
composite cross-section and multiply by the length
of the prism, which in this case is clearly 1 cm. In
order to find the area of the composite cross-section
(the L-shape), however, we still need to divide
the shape into two basic shapes, a square and a
rectangle, to find the area.
Area of cross-section = 3 cm²; Length = 1 cm;
Volume = 3 cm³
Note: Choosing the wrong measurement for length may
result in an incorrect answer. Splitting the shape into
basic shapes and finding the sum of the volumes of all
the shapes, is a safer method to use.
Exercise 13.1
1. (a)
(c)
3 cm
3 cm
2 cm
2 cm
4 cm
3 cm
1 cm
2 cm
5 cm
(b)
8 m
2 m
(d) (i) Two rectangular prisms or cuboids
(ii) Cube and rectangular prism
(iii) Triangular prism and cuboid
2. (a)
(b)
1.8 m
(c)
1 cm
5 cm
1 cm
1 cm
2 cm
1 cm
25 mm
2 cm
2 cm
5 cm
32 m
1.5 m
1.8 m
1.5 m
32 m
4 m
4 m
2 m
3. (a) iv (b) ii (c) iii
4. (a) 1 250 cm³ (b) 84 cm³
(c) 173.61 cm³ (d) 157.5 m³
5. 132 m³ – __ 1 2 (42.39)
132 m³ – 21.195 m³ = 110.805 m³
Activity 13.2
Cylinder: 1 374.625 cm³
Rectangular prism: 575 cm³
Total: 1 949.625 cm³
Exercise 13.2
1. 33 m³ + 82.5 m³ = 115.5 m³
2. 8 cm³ – 1.3 cm³ = 6.7 cm³ for one nut and
3 350 cm³ for 500 nuts
3. Old volume: 2 358.75 m³ + 3 955 m³
= 6 333.75 m³
New volume: 18 870 m³ + 31 800 m³ = 50 670 m³
50 670 ÷ 6333.75 = 8
(a) Sewela is correct.
(b) The new volume is eight times bigger than the
old volume.
4. (a)
8 m
5 m
8 m
5 m
5 m
12.5 m
(b) 150 m³ + 500 m³ = 650 m³
5. (a) 1 440 m³ – 70.65 m³ = 1 369.35 m³
(b) 70.65 m³ is the volume of the tunnel now. If the
diameter changes to 3.5 m, the volume of the
tunnel will be 96.1625 m³.
Subtracting, we have 25.5125 m³ of granite
that will need to be cut away.
6. 27 m³ + 32.499 m³ = 59.499 m³ of grain
7. (i) (a) Surface area of 1 cylinder = 62.8
Therefore, surface area of 2 cylinders = 125.6 m 2
Volume = 75.36 m³
(b) Surface area = 105 m²
Volume = 43.875 m³
(c) Surface area = 132 m²
Volume = 72 m³
(ii) It seems as though the decision must be made
between (a) and (c), as (b) gives too small a
volume. (a) is the best option since it gives the
greater volume (space) and the smaller surface
area (material cost).
59

8. (a) 1 350 m³ – 42.39 m³ = 1 307.61 m³
of concrete
(b) 42.39 m³
(c) 9 000 m³ – 282.6 m³ = 8 717 m³ of concrete
Volume of water = 282.6 m³
9. (a) 4 m³ (b) 1 080 m³
(c) 270 bales of hay
10. (a) 9 450 m³ + 14 580 m³ + 11 340 m³
= 35 370 m³
(b) 35 370 m³ × 5 s/m³ = 176 850 s
Divide by 60 to get minutes: 2947.50 min
Divide by 60 to get hours: 49.125 h which is
49 h rounded to the nearest whole number.
Activity 13.3
1. (a) – (c)
2. Discussion.
Exercise 13.3
1.
2. (a) – (b)
60

3.
4.
5. (a) – (b)
61

Activity 13.4
Exercise 13.4
1.
2.
62

3.
4.
5.
63

Revision exercise
1. 300 cm³
2.
3. (a) 1 371.6 cm³ (b) 137 160 cm³
64

Chapter 14 Time, distance and
speed
Activity 14.1
1. Answers will vary.
2. In this case, there is no number (distance nor time)
that is the same for both people. It is therefore
difficult to work out mentally who was the fastest.
The correct answer is Sewela. There is a detailed
explanation in the Student’s Book.
Exercise 14.1
1. 50t coin = 0.2 m/s
25t coin = 0.17 m/s
2. 20 m/s
3. (a) Between A and B: ______ 30 km
= 1.5 km/min
20 min
Between B and C: 1.39 km/min
Between C and D: 1.33 km/min
(b) C and D
(c) A and B
4. (a) Kealaboga = 100 m/min or 1.67 m/s
(b) Kabelo 50 m/40 s = 1.25 m/s
(c) 0.42 metres faster per second
5. 0.31 km/min
6. (a) 6 km/h (b) 5.33 km/h
7. 23.3 km/h
8. (a) 70 km/h – Mr Koloi
72 km/h – Mrs Dube
Mr Koloi was going 10 km over the
speed limit.
Mrs Dube was going 12 km over the
speed limit.
(b) P160
9. (a) 125 km/h (b) 112.5 km/h
10. Set up the following equation:
720 – 32x = 480 + 32x
In this case, x stands for time in hours.
Solving for x, the length of time is 3.75 h.
This means that the aeroplane flies 720 km in
3.75 h (with the wind) and 480 km in 3.75 h
(against the wind).
Faster speed: 720 km ÷ 3.75 h = 192 km/h
Slower speed: 480 km ÷ 3.75 h = 128 km/h
Therefore in STILL air, we either add or subtract the
wind speed.
Faster speed: 192 km/h – 32 km/h = 160 km/h
Slower speed: 128 km/h + 32 km/h = 160 km/h
Speed in still air = 160 km/h
Exercise 14.2
1. 0.75 cm/min 2. 1.14 km/min
3. ________ 125 km
= 125 km/6.5 h = 19.2 km/h
6 h 30 min
4. Banyana ____ 5 km
1 h = 5 km/h
Kitso ______ 8 km
84 min = ____ 8 km = 5.7 km/h
1.4 h
Kitso walked faster.
5. _____ 580 m
= 20 m/s
29 s
6. 2 m/s
7. 30 km/ 5.5 h = 5.45 km/h (rounded to 2 d.p.)
8.
Distance (km) Speed (km/h) Time (h)
Bus d r 2
Taxi d + 30 2r – 40 2
r = __ d (Bus: Speed = _______ Distance
2 Time )
2r – 40 = _____ d + 30 (Taxi: Speed = _______
2
Distance
Time )
Substituting r = __ d we have:
2
2 __
( d
2 ) – 40 = _____ d + 30
2
[ d – 40 = _____ d + 30
2
[ 2d – 80 = d + 30
[ d = 110
Speed of bus = _______ Distance = ______ 110 km = 55 km/h
Time 2 h
Speed of taxi is twice the speed of the bus (55 × 2)
and 40 km slower.
= 110 km/h – 40 km/h = 70 km/h
Activity 14.2
Convert from Convert to Operation
km m × 1 000
m km ÷ 1 000
m cm × 100
cm m ÷ 100
h min × 60
min h ÷ 60
min s × 60
H s x 3 600
s h ÷ 3 600
Exercise 14.3
(Teachers, a useful online velocity converter can be
found at: www.translatorscafe.com)
1. (a) iii (b) i (c) iii
65

2.
Convert from
Convert to
10 m/s 36 km/h
120 km/h 33.33 m/s
30 m/min 0.5 m/s
100 cm/min 60 m/h
3. It should have been ______ 3 600 s not _____ 360 s
1 h 1 h .
The correct answer is 36 km/h.
4. 22.22 m/s 5. 9 km/h 6. 126 km/h
7. 10.5 km/h
2.92 m/s
8. 120 km/h
Activity 14.3
1. 1 h 2. 2 h
3. 16:00 4. 6 h
5. 10:00 and 11:00. The graph is the steepest between
these two times.
6. A mark made at 14:30.
Exercise 14.4
1. (a) 08:20 (b) 20 min (c) 0.1 km/min
(d) Faster (e) 08:50
2. (a) 3 km (b) 1__
2 h
(c) Perhaps her friend was not home.
(d) 1 h
(e) 2.67 km/h (rounded to 2 d.p.)
3. (a) 140 m (b) 06:03 (c) 80 s
(d) 2 m/s (e) 0.78 m/s
4. (a) Team Silver Stars (b) 250 km
(c) Team Hot Shots (d) 3 h
(e) 150 km
(f) 100 km/h
5. (a) 5 km
(b) Skateboarder A stopped. Perhaps he fell and
hurt himself.
(c) 10 km/h (d) 07:15
(e) 37.5 min
(f) Skateboarder A won by 7.5min.
6. Answers will vary.
Exercise 14.5
1. (a)
Time of
day
Distance
from the start
Time from
the start
Comments
05:00 0 km 0 min Fetched sheep
06:30 3 km 1.5 h Arrived at grazing
area
07:30 3 km 2.5 h Stopped
09:00 0 km 4 h Arrived back at the
farm
(b)
Distance (km)
5
4
3
2
1
Shepherd’s journey
0
05:00 05:30 06:00 06:30 07:00 07:30 08:00 08:30 09:00
Time of day
(c) 6 km/ 4 h = 1.5 m/h average speed
2. (a)
Tiro and father’s journey
Distance (km)
5
4
3
2
1
0
09:00 09:30 10:00 10:30 11:00 11:30 12:00
Time of day
(b) 30 min
(c) 1 h
(d) ____ 8 km = 2.7 km/h (to 2 d.p.)
3 h
(e) 12:00
3. (a)
The journey of the bee
Distance (km)
8
7
6
5
4
3
2
1
0
0 5 10 15 20 25 30 35 40
Time (min)
(b) 15:00
(c) 10 min
(d) 0.5 km/min or 30 km/h
66

4. (a)
Distance (km)
300
250
200
150
100
Ambulances A and B’s journey
A
50
B
0
0 1 2 3 4 5 6 7 8
Time
Revision exercise
1. 116.67 km/h
2. (a) 0.51 km/min (b) 30.67 km/h
3. 6.25 m/s
4. (a) 07:30 (b) 1 h
(c) 4 km
(d) 2 km/h
5. 2.5 h
6. 15 km + 24.75 km = 39.75 km
5. (a)
(b) 100 km; 1 h (c) 2 h
(d) 4 h
The fishing boat’s journey
50
Distance (km)
40
30
20
10
0
0 1 2 3 4 5 6 7 8
Time (hours)
(b) 10 km/h (c) 6:00 (d) 12:00
Exercise 14.6
1. 200 km
2.
100 m practice sprints
Day Time Speed
Monday 12 s 8.3333 m/s
Tuesday 11 s 9.0909 m/s
Wednesday 10.5 s 9.523 m/s
Thursday 13 s 7.692 m/s
Friday 10.5 s 9.523 m/s
3. 4 h 4. 7 700 km
5. 3 h 45 min
6. He receives P105 from his mother for the
21 km walk.
He receives P73.50 from his grandfather.
He raises P178.50 in total.
7. (a) 90 km/h (b) 11__
2 h
(c) 45 km/h
8. No, the tree is 26.25 m high.
9. 52.5 s
67

Chapter 15 Working with variables
Activity 15.1
1. (x – 5)²
= (x – 5)(x – 5)
= x² – 5x – 5x + 25
= x² – 10x + 25
2. They are both 5x.
3. Find the product of the two terms in the expression,
and double the product.
Exercise 15.1
1. (a) 3y + 18 (b) 4r² – 4r
(c) 6e + 6f
(d) 3a² + 3ab
(e) m²n – mn² (f) 2u² – 2uv
(g) 16k – 8jk
(h) 4pr² + 28p²r
(i) 30i + 6i²
(j) 8yk + 8y + 8yx
2. (a) 10a + 4 (b) 4tr + 42t + 2r
(c) 7m² + 2m
(d) 8u² – 13u
(e) 2rs + 4rt + 4ru (f) 12y² – 3y – 12
(g) 3b² + 26ab – 10a² (h) –3e² + 13e – 28
(i) –3t² + 5t (j) 4xm²n – 14x²
3. (a) 6r (b) 24t (c) –12ab (d) –24yz
4. (a) a² + 2a + 1 (b) b² + 6b + 9
(c) x² + 2xy + y² (d) 9d² + 12de + 4e²
(e) f² – 12f + 36 (f) g² – 2gh + h²
(g) 9d² – 12de + 4e² (h) 49p² –28pq + 4q²
(i) 25t² + 60t + 36 (j) 144r² + 72rs + 9s²
5. 49p 4 + 14p²q + q²
6. (a) (7x + 3)² (Any variable can be used.)
(b) 49x² + 42x + 9
7. (a) x² + 2x + 1 (b) y² – 6y + 9
(c) 4x² + 12x + 9 (d) 16x² + 8xy + y²
(e) 9x² + 18xy + 9y² (f) 64x² – 32xy + 4y²
(g) 16y² – 8xy + x²
(h) 121x² + 264xy + 144y²
(i) 64x² + 112xy + 49y² (j) 25y² + 90xy + 81x²
8. She has not doubled the product of the two terms,
to get the middle term. She has also forgotten to
square the 9 in the first term.
9. Correct. Since the binomial is being squared, the
first and last terms are also squared. Squaring a
positive or a negative number results in a positive
number.
10. (2x – 3y)² = 4x² – 12xy + 9y²
Exercise 15.2
1. (a) 6 (b) 2x (c) m (d) x
(e) 9w (f) 16ac (g) 4uv (h) 3m
2. P = 2(l + w)
3. (a) 12m(m – 2) (b) 4vr(1 + 2v)
(c) 2ab(5a + b) (d) 3g(2h²g – 6h + 1)
(e) 88xyz + 77xy² = 11xy(8z + 7y)
4. (a) b(4b + 1) (b) a(12 + 3b – 4c)
(c) t²(5 + 6r)
(d) 5(x² – 2y²)
(e) 3(6xy – 3x + y) (f) 2r(r + 1 + s)
(g) 3ab(4a + b) (h) 2y(z + 4)
(i) 5y(1 – 5y) (j) q(12 + 8r + q)
5. (a) (y + 1)(7 – a) (b) (r² – 1)(4 + t)
(c) (x – ½)(5 + w) (d) (m + n)(9 – i)
(e) (x – 4)(3x + 7)
(f) (b – 1)(4a + 2a²) = (b – 1)2a(2 + a)
6. (a) 18yp(4y – p)
(b) 3(x² – x + 3y – 4)
(c) 5a(4b² + b – 2a + 3ab)
(d) 2i(–6h + h² – 8i)
(e) 6a(b + 2bc – 3a)
(f) (h – 1)(g² + t + r²)
(g) 11efg(2e – fg + 11 – 9g)
(h) 27y²(3 – 4x)
(i) 15(3utv + ut² – 2u²v² + 4tv²)
(j) 6st(6s – 3 + 2st + 8t)
Extension activity
1. 2b(2a + b) 2. 3xy(1 – 3x)
3. r(12r – 4t + 3t²) 4. 5(2x + y)
5. 11hf(g + 4h) 6. 20pq(5p – 4q)
7. ab(2 – 2b + a) 8. –2i(4i + 9j)
9. 3r(1 + 2r – 4m) 10. uvw(17 + 4uvw)
Activity 15.2
1. (y + 2)(x + 1) 2. (b + 1)(x + 1)
3. (x – y)(m + 3) 4. (y – 2)(a + 3)
Exercise 15.3
1. (a) 10 (b) –28 (c) 18 (d) 36
(e) –8 (f) 16 (g) 30 (h) 4
2. (a) 7 (b) –4 (c) 40 (d) 1
(e) 66 (f) 28
3. 460 m 4. 40.5 kg
5. (a) 8a + 12b + 3c
(b) P6.40 + P13.20 + P8.70 = P28.30
6. (a) 156 (b) 2 022 (c) 1 017
(d) –32 (e) –112
7. (a) (x + x + 2 + 2x – 1) cm
= (4x + 1) cm
(b) 73 cm
8. P33.20 9. 720
10. (a) 5a + 3 where a is Sewela’s age.
(b) 63 years old
68

Exercise 15.4
1. (a) __ x
2
(b)
(e) 6__
a²
(f)
3__
y
1__
2t
2. (a) ___ 2t²
y (b) ____ 1
3um 5
y²
(d) –2y²z 4 (e) ____
(g) _____
( –1q
3y 2 rs ) (h) ___ 5h²
2 k ³
3. (a) 4__
y (b) ____ 3
4hg 2
(d) _____ 5m 5 n²
(e) ____ 3
3u 2ab 4
(g) ___ 4s² (h) ____
–5q 4
3r
3p²r 5
Exercise 15.5
1. (c)
2. (a) 7__
5 + 2y (b) __ x
(d) ______ 1
2(j – 2)
2mj
(g) ______
m – 13
3. (a) _____ 2
5x + 6
2p
(d) ____
p – 2
y
(g) ____
y + 2
4. (a) ____ 6
3 + x
(d) ______ 1
3xy – 2
(g) ____ h
h + 9
5. ______ 2x
3x² + 1
(e)
(c) __ 2z
p
(g) ___ a²
2b²
(c) ____ 3i²j³
k
___
h³
10r 4 (f) 3u²
(c) __ 2b
3a c
(f) ____ –3x²
2
3 + 2a (c) __ 2z
2z – 3
____ 2wv (f) _______ b
5 + v 10(2 – a)
(h) ______ 1
5(r + 6)
(b) ____ 2
y – 1
(e) _____ 1
3a – 1
(h) ____ a²
a + b
(b) _____ 2
7b + 1
(e) 6
(h) ______ 4m²n
2m – n
(c)
y _____
y² + 2
(f) ____ 2
x – y
(c) _____ 4t²
6 + 5t
(f) ____ –5
5 + r
(d) – __ u
2v
(h) ___ 6n
m²i
Revision exercise
1. (a) 5x² + 5xy (b) 4r²p – 24rp + 4rp²
(c) –2t² + 14t
(d) 4r²s + 2r²t
(e) x² + 6x + 9 (f) 16 – 8a + a²
(g) 4w² + 12wr + 9r² (h) 25y² – 30xy + 9x²
2. (a) b(b + 1) (b) 7e(e + 2)
(c) 5(y² – 2y + 3) (d) ab(4 + 2a – b)
(e) 3uv(2uv – 1 + 6v – 3u)
(f) 2(y – 1)(3 + 4x) (g) (15 + a)(a + b)
3. (a) 0 (b) –25 (c) –5 (d) –2
(e) –113 (f) 26
4. (a) __ x
6y
(e) ______ 15
3 + 10y
(b) ___ 2h
gi²
(f) ______ 7r²
3(r + 3)
(c)
3v ____
2uw
(d)
2x ____
4 – x
69

Chapter 16 Formulae
Exercise 16.1
1. (a) –7 (b) –2 (c) 18
(d) 1 (e) 4
2. (a) 11__
2
(d) 6
3.
(b) 21__
4
(e) 71__
2
(c) 3
3__
4
Side length a Side length b Side length c
3 cm 4 cm 5 cm
12 cm 16 cm 20 cm
6 mm 8 mm 10 mm
1 m 2 m 2.236 m
4. (a) 10 cm (b) 13.9 cm
(c) 20 cm
(d) 29.2 cm
5. (a) 10.83 cm² (b) 62.35 mm²
(c) 27.71 m²
(d) 18.29 cm²
6. 10 cm 7. P1 450
8. (a) P16.80 (b) P17.80
(c) P19.80 (d) P22.80
9. √ ____
200 cm 2 = 14.14 cm 2
Exercise 16.2
1. (a) (iii) (b) (iii) (c) (i)
2. (a) y = z – x (b) d = st
(c) t = ____ y + h
r (d) t = s
(e) p = ____ c + q
5
3. (a) b = ____ t – 5
3
(c) d = w – 1 _____
7
(e) x =
12y – 4u ______
m
______
6(1 + r)
(f) b = √ ______
c² – a²
(b) d = __ g 6
(d) c = f – 1__
____ 2
12
Or c = _____ 2f – 1
24
(f) s =
3
√ _____
x + 7
___
4. (a) r = √
__ A
πh
(b) radius = 3 cm; diameter = 6 cm
5. (a) m = __ W g
(b) g = __ W m
__
6. a = √
__ A
6
= 0.75 m
7. x = ( y __
2 ) 2 + 8 or x =
8. r = D – 6a ______
(2π + 3)
y² + 32
______
4
_____
9. (a) c = √
____ a + b
T
(b) 6
10. (a) y = –5 + 2x (b) y = –5 + q + p
(c) y = 9 – x
(d) y = __ p 2
(e) y = ______ –15 + x (f) y = _____ x + 11
6
13
(g) y = ___ 3xa
(h) y = __ 2b
5b
7a
(i) y = ___ 5a²c
(j) y = __ 2a
3b
rt
Exercise 16.3
1. (a) h = ____ Y
(b) Hourly rate: P23.60
2080
2. (a) d = F – P2.50 ________
2
(b) 14 km
3. (a) 150 ug (b) m = A – 30 _____
8
(c) 21.25 kg
4. (a) 38 white onions
(c) 6 red onions
(b) r = _____ w – 2
3
5. (a) n = ___ C
0.3
(b) 600 gallons
6. (a)
Polygon Square Pentagon
Hexagon
Heptagon
Octagon
Sides (n) 4 5 6 7 8
Diagonals (d) 2 5 9 14 20
(b) 170
7. (a) P4 000 (b) P = __
rt
I (c) r = __ I
Pt
(d) t = __ I (e) 3 years
Pr
8. (a) 22 pieces (b) f = ______ 50 – O
2
(c) 16 players
9. 90 km
Revision exercise
1. (a) –1 (b) 4 (c) 0
(d) 3.94 (2 d.p.)
2. (a) b = √ ______
c² – a² (b) 4 cm
(c) a = √ ______
c² – b² (d) 36 m
3. (a) 39 (b) 48 (c) 4
4. (a) a = 2 – b (b) a = ____ 7 + b
(c) a = 7 – 3b
2
(d) a = ____ 2 – b
6
3
√ __
___ b
5
(e) a = (b – 5) ² (f) a =
5. (a) h = n + 3 _____
2
(b) (i) 30 (ii) 168 (iii) 51
70

Chapter 17 Equations
Activity 17.1
The answer is four bananas. Since one apple equals two
bananas, if we take away the apple and two bananas
from the first diagram, we are left with one watermelon
and four bananas.
Activity 17.2
1. (a) × by 3; a = 18 (b) × by 5; b = __ –5
2
(c) × by 4; c = –8
Exercise 17.1
1. (a) 4 (b) 6 (c) 12
(d) 12 (e) 20 (f) 18
2. (a) 10 (b) 10 (c) 100
(d) 100 (e) 100 (f) 1 000
3. (a) x = 2 (b) y = –1 (c) u = –1 __
2
(f) t = –12
(d) w = 3 (e) r = __ 9 2
4. (a) y = 12 (b) y = –16 (c) y = __ 3
2
(d) y = __ 8 (e) y = __
3
11 (f) y = 10
2
(g) y = 90 (h) y = –20 (i) y = 48
(j) y = 6 (k) y = __
77
6 (l) y = 12
5. (a) a = 5 (b) r = 2 (c) t = __ 3 5
(d) k = 2 (e) h = __
10
1 (f) w = –4.71 (2 d.p.)
(g) u = –1 (h) f = 2 (i) t = __ 1 3
(j) h = __ 1 (k) m = __ 38 or 7.6
5
5
Exercise 17.2
1. a = 0 2. a = 10 3. a = –9
4. a = –13 5 a = 3 6. a = –15
7. a = –9 8. a = 12 __
5
9. a = –2
10. a = __ –1
6
11. a = 36 12. a = __ –1
8
Exercise 17.3
1. x = 3, y = 2 2. x = 3, y = 9
3. x = 0, y = 2 4. a = 5, b = –1
5. p = –1, q = 2 6. x = 15, y = 9
7. x = –1, y = –5 8. r = –1, s = 4
9. v = __ –2
3 , w = ___ –19
3
10. x = __ 3 2 , y = 0
Exercise 17.4
1. (a) r = 1, t = 2 (b) x = 8, y = 4
(c) a = 2, b = 0 (d) x = 5, y = 1
(e) i = 2, h = 6 (f) h = __ –1
5 , g = __ –4
5
(g) a = –1, b = 1
2. At the point x = __ 5 2 , y = __ 3 2 .
3. x + 4y = 7
x – 3y = 0 (original equation was x = 3y)
y = 1 and x = 3
A pencil costs P3.00 and a sweet costs P1.00.
4. (a) d = –3, e = –4 (b) r = 5, s = 3
(c) a = 1, b = 4 (d) m = –3, n = –2
(e) x = –1, y = –5 (f) x = –3, y = –4
Exercise 17.5
1. (a) x = 1, y = 2 (b) x = 3, y = 2
(c) x = 2, y = 6 (d) x = 2, y = 5
(e) x = 6, y = 4 (f) x = 1, y = 2
(g) x = __ 20 , y = 51/7 (h) x = 1, y = –1
7
(i) x = ____ –111
27 , y = ___ –21
9
2. (a) 2l + 2w = 40
l × w = 51
(b) By trial and error or diagrammatically,
l = 17 cm and w = 3 cm.
Therefore, area = 17 cm × 3 cm = 51 cm²
3. (a) a = –3, b = –1 (b) s = –1, t = –5
(c) a = 11, b = –4 (d) t = 6, r = –1
(e) p = 52__
3 , q = __ –7 (f) e = __
3
1 2 , f = __ 3 2
4. The point (1,4).
Exercise 17.6
1. (a) x + y = 2 000, y = x + 660
(b) Smaller dose: 670 mg and bigger dose:
1 330 mg.
2. (a) c + g = 24, c = 2g – 3
(b) Clarabell: 15 l, Gertie: 9 l
3. (a) B gets 756 votes and A gets 704 votes.
(b) B wins by 52 votes.
4. Let Malebogo’s share be x and Banyana’s share be y.
1__
4 x = __ 1 y and x + y = 800
6
x = P320 and y = P480
5. (a) 2a + 3b = 385, 4a + 2b = 470
Speed 1: 80km/h, Speed 2: 75km/h
6. (a) x + y = 75, 4x = y
(b) x = 15min, y = 60 min
Clinic: 15 min, Hospital: 60 min
7. 8n + 2e = 340, 6n + 4e = 380
Normal rate of pay: P30 per hour
Overtime: P50 per hour
8. a + c = 2 200
4a + 1.5c = 5 050
Adults = 700, Children = 1 500
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9. (a) 13s + 4t = 487
6s + 2t = 232
(b) Shrubs cost P23 each and trees cost P47 each.
10. Challenge:
Thabo: 3a + 5b = 76, a + b = 20
Shot 1: 12 times
Shot 2: 8 times
Mpho: 3a + 5b = 82, a + b = 20
Shot 1: 9 times
Shot 2: 11 times
Activity 17.3
1.
2.
3.
4.
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5.
6.
7.
8.
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Activity 17.4
1.
2.
3.
4.
5.
6.
7.
8.
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Extension activity
f = 6
Revision exercise
1. (a) a = __ 1 (b) a = –3
3
(c) a = __ –4
(d) a = –9.6
9
(e) a = –11 (f) a = –2
2. (a) g = –1 (b) h = 1
(c) r = __ 4 (d) y = –1
5
(e) u = 2
(f) t = __ 5 3
3. (a) a = 2; b = 1 (b) d = 11; e = 4
4. (a) x = 1; y = 2 (b) a = 3; b = 1
(c) p = 1; q = –5
5.
6. 7s + 2m = 191
3s + 5m = 144 where s = sugar and m = milk
s = 23; m = 15
One bag of sugar costs P23.00 and two litres of milk
cost P15.00.
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