Checkpoint Quiz 3

Maths Standard Year 11 · Module 2: Measurement

Lesson 11 — Rates Lesson 12 — Scale Drawings & Maps Lesson 13 — Errors & Limits of Accuracy Lesson 14 — Trig: Finding Unknown Sides Lesson 15 — Trig: Finding Unknown Angles Lesson 16 — Angles of Elevation & Depression Lesson 17 — Bearings & Navigation
10 MC 3 SA ~25 min
0/12
Multiple Choice Score
Answer questions to see your score.
Part A — Multiple Choice (1 mark each)
1 A train travels $450\text{ km}$ in $5$ hours. What is its average speed? L11
A $2250\text{ km/h}$
B $90\text{ km/h}$
C $45\text{ km/h}$
D $180\text{ km/h}$
B — $90\text{ km/h}$. $S = \dfrac{D}{T} = \dfrac{450}{5} = 90\text{ km/h}$.
2 A car uses $8\text{ L}$ per $100\text{ km}$. How much fuel is needed for a $350\text{ km}$ trip? L11
A $43.75\text{ L}$
B $35\text{ L}$
C $22.86\text{ L}$
D $28\text{ L}$
D — $28\text{ L}$. Fuel $= \dfrac{8}{100} \times 350 = 28\text{ L}$.
3 On a map, $1\text{ cm}$ represents $5\text{ km}$. Two towns are $8\text{ cm}$ apart on the map. What is the actual distance? L12
A $1.6\text{ km}$
B $40\text{ km}$
C $400\text{ km}$
D $13\text{ km}$
B — $40\text{ km}$. Actual distance $= 8 \times 5 = 40\text{ km}$.
4 A scale drawing uses a scale of $1:250$. A corridor measures $16\text{ cm}$ on the plan. What is the actual length? L12
A $6.4\text{ m}$
B $250\text{ m}$
C $400\text{ m}$
D $40\text{ m}$
D — $40\text{ m}$. Actual length $= 16 \times 250 = 4000\text{ cm} = 40\text{ m}$.
5 A measurement is recorded as $8.4\text{ cm}$ (rounded to the nearest millimetre). What is the absolute error? L13
A $0.1\text{ cm}$
B $0.5\text{ cm}$
C $0.05\text{ cm}$
D $0.01\text{ cm}$
C — $0.05\text{ cm}$. Absolute error $= \tfrac{1}{2} \times$ precision unit $= \tfrac{1}{2} \times 0.1\text{ cm} = 0.05\text{ cm}$.
6 A length is measured as $48\text{ cm}$ with an absolute error of $0.5\text{ cm}$. What is the percentage error, to 2 decimal places? L13
A $0.52\%$
B $1.04\%$
C $2.08\%$
D $0.48\%$
B — $1.04\%$. $\% \text{ error} = \dfrac{0.5}{48} \times 100 \approx 1.04\%$.
7 A ladder makes an angle of $70°$ with the horizontal ground. Its foot is $2.5\text{ m}$ from the base of the wall. How high up the wall does the ladder reach (to 2 d.p.)? L14
A $2.31\text{ m}$
B $0.86\text{ m}$
C $6.87\text{ m}$
D $7.31\text{ m}$
C — $6.87\text{ m}$. $\tan 70° = \dfrac{h}{2.5} \Rightarrow h = 2.5 \times \tan 70° \approx 2.5 \times 2.7475 \approx 6.87\text{ m}$.
8 From a point $30\text{ m}$ from the base of a tower, the angle of elevation to the top is $25°$. How tall is the tower (to 2 d.p.)? L14
A $71.29\text{ m}$
B $27.19\text{ m}$
C $12.68\text{ m}$
D $13.99\text{ m}$
D — $13.99\text{ m}$. $\tan 25° = \dfrac{h}{30} \Rightarrow h = 30 \times \tan 25° \approx 30 \times 0.4663 \approx 13.99\text{ m}$.
9 In a right-angled triangle, the side opposite angle $\theta$ is $7\text{ cm}$ and the adjacent side is $10\text{ cm}$. Find $\theta$ to the nearest degree. L15
A $44°$
B $35°$
C $55°$
D $46°$
B — $35°$. $\tan\theta = \dfrac{7}{10} = 0.7 \Rightarrow \theta = \tan^{-1}(0.7) \approx 34.99° \approx 35°$.
10 A right-angled triangle has a hypotenuse of $15\text{ cm}$ and an adjacent side of $9\text{ cm}$ for angle $\theta$. Find $\theta$ in degrees and minutes. L15
A $36°52'$
B $53°8'$
C $53°46'$
D $36°6'$
B — $53°8'$. $\cos\theta = \dfrac{9}{15} = 0.6 \Rightarrow \theta = \cos^{-1}(0.6) \approx 53.130° = 53° + 0.130 \times 60' \approx 53°8'$.
11 From a point on horizontal ground, the angle of elevation of the top of a building is $32°$. The horizontal distance from the point to the base of the building is $45\text{ m}$. What is the height of the building, to 1 decimal place? L16
A $24.0\text{ m}$
B $28.1\text{ m}$
C $38.2\text{ m}$
D $76.3\text{ m}$
B — $28.1\text{ m}$. $h = 45\tan 32° \approx 45 \times 0.6249 \approx 28.1\text{ m}$. Option A uses $\sin 32°$ instead of $\tan$. Option D divides instead of multiplying.
12 A ship sails on a bearing of $135°$ for $80\text{ km}$. How far south of its starting point has the ship travelled, to 1 decimal place? L17
A $56.6\text{ km}$
B $65.5\text{ km}$
C $56.6\text{ km}$
D $113.1\text{ km}$
A — $56.6\text{ km}$. A bearing of $135°$ is SE; the angle from South = $45°$. Southward distance $= 80\cos 45° \approx 80 \times 0.7071 \approx 56.6\text{ km}$. (Alternatively: southward $= 80\sin(180°-135°) = 80\sin 45°$.) Option B uses $\sin 35°$ — a labelling error.
Part B — Short Answer (show all working)
1 L11
A car travels $180\text{ km}$ at $60\text{ km/h}$, then $120\text{ km}$ at $80\text{ km/h}$.
(a) Find the time taken for each leg of the trip.
(b) Find the average speed for the entire trip, to 1 decimal place.
(a) Leg 1: $T = \dfrac{180}{60} = 3\text{ h}$  |  Leg 2: $T = \dfrac{120}{80} = 1.5\text{ h}$
(b) Total distance $= 180 + 120 = 300\text{ km}$; Total time $= 3 + 1.5 = 4.5\text{ h}$
Average speed $= \dfrac{300}{4.5} \approx 66.7\text{ km/h}$
Note: the average speed is NOT the average of 60 and 80, because different times are spent at each speed.
2 L12
A floor plan shows a rectangular room as $8.4\text{ cm} \times 5.6\text{ cm}$ using a scale of $1\,:\,50$.
(a) Find the actual dimensions of the room in metres.
(b) Find the actual area of the room in $\text{m}^2$.
(a) Length: $8.4 \times 50 = 420\text{ cm} = 4.2\text{ m}$
Width: $5.6 \times 50 = 280\text{ cm} = 2.8\text{ m}$
(b) Area $= 4.2 \times 2.8 = 11.76\text{ m}^2$
Check: on plan, area $= 8.4 \times 5.6 = 47.04\text{ cm}^2$. Scale factor for area $= 50^2 = 2500$. So $47.04 \times 2500 = 117\,600\text{ cm}^2 = 11.76\text{ m}^2$. ✓
3 L14 & L15
A vertical fence post of height $6\text{ m}$ casts a horizontal shadow of $8\text{ m}$.
(a) Find the angle of elevation of the sun from the tip of the shadow, to the nearest degree.
(b) Find the straight-line distance from the sun's position (tip of shadow to top of post), to 2 decimal places.
(a) $\tan\theta = \dfrac{6}{8} = 0.75 \Rightarrow \theta = \tan^{-1}(0.75) \approx 36.87° \approx 37°$
(b) $d = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ m}$   (or use $\dfrac{6}{\sin 37°} \approx 10.00\text{ m}$)

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