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1) Radio signals travel at a speed of 3 × 108 metres per second. A radio signal from Earth to a space probe takes 8 hours. What is the distance from Earth to the probe? Give your answer in scientific notation. Soln d s d t 8 hours = 8 hrs × 60 mins × 60 secs = 28800 seconds = s × t = ( 3 × 108 ) metres/sec × ( 8 × 60 × 60 ) secs = 3 × 108 × 8 × 60 × 60 = 3 × 108 × 28800 = 3 × 108 × 2.88 × 104 = 3 × 2.88 × 108 × 104 = 8.64 × 10(8 +4) = 8.64 × 1012 metres Change to km ÷ by 1000 = 103 = 8.64 × 1012 103 = 8.64 × 10(12 − 3) = 8.64 × 109 kilometres 2) A tank which holds 100 litres of water has a leak. After 150 minutes, there is no water left in the tank. v 100 litres 0 minutes t 150 The above graph represents the volume of water ( v litres ) against time ( t minutes ). (a) Find the equation of the line in terms of v and t. (b) How many minutes does it take for the container to lose 30 litres of water? Soln (a) The equation of a line is y = mx + c gradient m (0, 100) (150, 0) ÷ by 50 ( top and bottom) y = mx + c v = mt + c = y2 − y1 x2 − x1 = 0 − 100 150 − 0 = − 100 150 = −2 3 y intercept (0, 100) c = 100 ( from graph above y = v and x = t ) v = mt + c v = −2 t + 100 3 (b) ÷ 100 × 30 litres minutes 100 150 1 1.5 30 45 ÷ 100 × 30 It takes 45 minutes for the container to lose 30 litres of water 3) Bottles of juice should contain 50 millilitres. The contents of 7 bottles are checked in a random sample. The actual volumes in millilitres are as shown below. 52, 50, 51, 49, 52, 53, 50 Calculate the mean and standard deviation of the sample. Soln Mean Σx n Mean = 52 + 50 + 51 + 49 + 52 + 53 + 50 7 = 357 7 = 51 Standard Deviation √ x2 − ( x )2 / n (n−1) x2 = 522 + 502 + 512 + 492 + 522 + 532 + 502 x2 = 18219 n = 7 (n−1) = 6 = √ √ = √2 = 18219 − ( 127449 / 7 ) 6 12 6 Standard Deviation = 1.41 (2 d.p.) x = 357 ( x )2 = (357)2 ( x )2 = 127449 4) 250 milligrams of a drug are given to a patient at 12 noon. The amount of the drug in the bloodstream decreases by 20% every hour. How many milligrams of the drug are in the bloodstream at 3pm. Soln 12 noon 250 losing 20% each hour 1pm 250 × 0.8 2pm 250 × 0.8 × 0.8 left with 80% of the previous hour 2 = 250 × (0.8) 3pm 250 × 0.8 × 0.8 × 0.8 = 250 × (0.8)3 = 128 milligrams 80% = 80 = 0.8 100 5) A helicopter, at point H, hovers between two aircraft carriers at points A and B which are 1500 metres apart. H 500 A 550 1500 m B From carrier A, the angle of elevation of the helicopter is 500. From carrier B, the angle of elevation of the helicopter is 550. Calculate the distance from the helicopter to the nearest carrier. Soln H Shortest side is opposite the smallest angle 750 xm A 500 550 1500 m B Sine Rule x sin 50 = 1500 sin 75 ( x )( sin 75 ) = ( 1500 )( sin 50 ) ( ÷ sin 75 both sides ) x = ( 1500 )( sin 50 ) ( sin 75 ) x = 1189.6 m ( 1 d.p. ) The distance to the nearest carrier is 1189.6 metres. 6) The diagram below shows a spotlight at point S, mounted 10 metres directly above a point P at the front edge of a stage. The spotlight swings 450 from the vertical to illuminate another point Q, also at the front point of the stage. S 450 10 m P Q B Through how many more degrees must the spotlight swing to illuminate a point B, where Q is the mid-point of PB? Soln Diag. 1 S (a) 0 10 m P y0 (a) 10 m 45 x (o) Q Diag. 2 S P 20 m ( 2x ) (o) tan 450 = o = x a 10 0 tan 45 = x 10 ( × 10 both sides ) tan y0 = o = 20 a 10 0 tan y = 20 10 0 tan y = 2 ( 10 )( tan 450 ) = x x = 10 y0 = 63.40 Spotlight must swing through 63.40 − 450 = 18.40 more to illuminate a point at B. B 7) A square trapdoor of side 80 centimetres is held open by a rod as shown. Trapdoor Rod 80 cm 760 40 cm 40 cm The rod is attached to a corner of the trapdoor and placed 40 centimetres along the edge of the opening. The angle between the trapdoor and the opening is 760. Calculate the length of the rod to 2 significant figures. Soln Rod x cm 80 cm 760 40 cm Cosine Rule x2 = 802 + 402 − (2)(80)(40)(cos 760) x2 = 8000 −(6400)(cos 760) x2 = 6451.699…… x = √ 6451.699…… x = 80.322…… x = 80 ( 2 s.f. ) 8) The curved part of a doorway is an artic circle with radius 500 millimetres and centre C. The height of the doorway to the top of the arc is 2000 millimetres. The vertical edge of the doorway is 1800 millimetres. C 500 mm 1800 mm 2000 mm Calculate the width of the doorway. Soln 200 mm x mm 300 mm C 1800 mm x mm (a) 300 mm (b) 500 mm (h) 500 mm Pythagoras 2000 mm a2 x2 x2 x2 x x = = = = = = h2 − b2 5002 − 3002 250000 − 90000 160000 √ 160000 400 mm The width of the doorway is 2x mm = (2)(400) mm = 800 mm 9) A gift box, 8 centimetres high, is prism shaped. A B O 8 cm 10 cm The uniform cross-section is a regular pentagon. Each vertex of the pentagon is 10 centimetres from the centre O. Calculate the volume of the box. Soln A A 10 cm B 0 72 O O 10 cm Area of 3600 = 720 5 720 B 10 cm AOB = 1 × 10 × 10 × sin 720 2 = 47.55 cm2 ( 2 d.p. ) Area of Pentagon = 5 × 47.55 = 237.75 cm2 Volume of the box = Area of Pentagon × Height of Box = 237.75 cm2 × 8 cm = 1902 cm3 10) Solve algebraically the equation 4 sin xo + 1 = −2 0 ≤ x ≤ 360. Soln 4 sin xo + 1 = −2 ( − 1 both sides ) 4 sin xo = −3 ( ÷ 4 both sides ) sin xo = −3 4 xo = (180 + 48.6)0 or xo = (360 − 48.6)0 xo = 228.60 or xo = 311.40 180 0 S A T C sin yo = 3 4 o y = 48.60 0/3600 11) A rectangular lawn has a path, 1 metre wide, on 3 sides as shown. xm 1m The breadth of the lawn is x metres. The length of the lawn is three times its breadth. The area of the lawn equals the area of the path. (a) Show that 3x2 − 5x − 2 = 0. (b) Hence find the length of the lawn. Soln 3x m (a) 1m (x + 1) m (x + 1) m xm 1m Area of Lawn 3x m = = (3x)(x) 3x2 Area of lawn = Area of path 3x2 = 5x + 2 3x2 − 5x − 2 = 0 1m Area of Path = = = = 2(1)(x + 1) + (3x)(1) 2(x + 1) + 3x 2x + 2 + 3x 5x + 2 (b) 3x2 − 5x − 2 = 0 (3x + 1)(x − 2) = 0 (3x + 1) = 0 or (x − 2) = 0 3x = −1 x=2 or x = −1 3 ( NOT possible x is a length ) The length of the garden is 3x = (3)(2) = 6 metres