Solution Method 1 1(ai)Kristian’s Amount corresponds to 3 = × $ ℎ 3+2 3 $72 = × $ ℎ 5 5 × $72 ℎ = 3 ℎ = $120 Stephanie=$120 − $72 = $48 Method 2 Kristian:Stephanie=3:2 $72: 3 $ℎ: 2 $72 × 2 ℎ = 3 = $24 × 2 = $48 aii) Price of computer game=45%$72 45 = × $72 100 162 = 5 = $32.4 iii) Total money spent=$8.40 + $32.4 = $40.8 Money Left=$72 − $40.8 = $31.2 $31.2 $72 31.2 = 72 312 = 720 104 = 240 13 = 30 iv) Original price:100% Reduced sale price :(100-20)%=80% Original price:100% $19.20: 80% Original price= $19.20×100% $1920 = 80 80% = $24 Method 2 Let be the original price Then 20% of will be 0.2 − 0.2 = $19.20 0.8 = $19.20 $19.20 = 0.8 $192 = 8 = $24 b) Amount = Principal +Simple Interest Simple interest = ×× 100 $550 × 10 × 2 = 100 = $110 Amount=$550 + $110 = $660 c) Amount = (1 + %) = $550(1 + 1.9%)10 = $550(1.019%)10 = $550 × 1.207096 = $663.9 d) Amount=$638.30 (1 + %) = $638.30 $550(1 + %)10 = $638.30 $638.30 10 (1 + %) = $550 (1 + %)10 = 1.160545454 1 (1.160545454)10 1 + % = 1 + % = 1.015 % = 1.015 − 1 = 0.015 100 = 0.015 × 100 = 1.5 2a) (−1, −2)-----------(4, −4) (−3, −2)------------(2,-4) (−3, −3)---------(2,-5) . . . T’’ . . . T’ (x,2k-y) ii. (−1, −2)-----------(−1,4) (−3, −2)------------(-3,4) (−3, −3)---------(-3,5) iii. Enlargement, scale factor 3, centre of enlargement (-6,-5) b)i 1 2 1 3 + =( )+( ) 3 4 0 6 To find the sum of two matrices just add the corresponding entries 1+1 2+3 + =( ) 3+0 4+6 2 5 =( ) 3 10 b)ii 1 3 1 2 = ( )( ) 0 6 3 4 Multiply the rows in the first matrix by the columns in the second 1×1+3×3 1×2+3×4 = ( ) 0×1+6×3 0×2+6×4 1+9 = ( 0 + 18 10 = ( 18 2 + 12 ) 0 + 24 14 ) 24 b) iii || = || The determinant of a 2 × 2 matrix say ( ) is given by − . 1×4−2×3=4×−3×1 4 − 6 = 4 − 3 1 = 4 1 = 4 c) If we transform any arbitrary point 0 −1 2 (, ) in by the matrix ( ) 1 0 we obtain ′( ′ , ′ ). ′ 0 −1 i.e ( ) () = ( ) ′ 1 0 0 − ′ ( )=( ) − 0 ′ − ′ ( )=( ) ′ Hence the given matrix maps (, ) onto ′(−, ). This is an anti-clockwise rotation through angle of 90° about the origin. Ciii If the point (, ) is reflected in the line = , we obtain ′ ( ′ , ′ ) = ′(, ) i.e ′ = ′ = This is the same as; ′ = 0 + --------(1) ′ = + 0-------(2) Equations (1) and (2) are systems of linear equations in and . We can transform this into a 2 × 2 matrix. 0 + ′ ( )=( ) + 0 ′ ′ 0 1 ( )=( ) () ′ 1 0 0 1 Hence the matrix is ( ) 1 0 3a) i. 1 Median = ∑ ℎ occurrence 2 1 = (200) = 100th occurrence 2 Tracing from graph,this corresponds to = 4003 ii. 1 The lower quartile is ∑ th 4 occurrence 1 = (200) = 50th occurrence 4 Tracing from graph,this corresponds to = 3503 iii. The inter-quartile range = upper quartile – lower quartile = 420 − 350 = 703 iv. From the graph 3003 corresponds to 30 students. The number of students who estimate that the volume is greater than 300 3 = 200 − 30 = 170 3b Area(2 ) Mid-Values() Frequency() 20 < ≤ 60 60 < ≤ 100 100 < ≤ 150 150 < ≤ 250 40 80 125 200 32 64 80 24 1280 5120 10,000 4800 ∑ = 445 ∑ = 200 ∑ = 21,200 ∑ ̅ = ∑ 21200 = 200 = 106 ii. Frequency density 0.8 1.6 1.6 0.24 iii. ( > 1002 ) 80 24 = + 200 200 104 = 200 If the students are two then event is mutually inclusive i.e the occurrence of the first event will affect the second. P(both student estimate 2 104 A>100 )= × 200 10712 = 39800 103 99 4. a. The volume, V, of a sphere with 4 radius r is = 3 3 Substitute = 15 4 = (15)3 3 = 45003 = 14137.173 = 141403 correct 4 s.f bi) Volume of cylinder, = 2 ℎ = (25)2 60 = 375003 = 117809.723 volume of water required to fill the tank= − ℎ = 117809.72 − 14140 = 103669.723 = 1037003 to 4 s.f ii. There is no change in volume of water, likewise the radius. (25)2 = 1037003 1037003 = 6252 = 52.81cm ci) The volume, V, of a cone with radius r 1 and height h is = 2 ℎ 3 The volume of the sphere is now equal to the volume of the cone. But this cone has height, ℎ = 54. 1 2 × 54 = 14140 3 14140 2 = 18 2 = 250.05 = √250.05 = 15.81 ii. Total surface area =base area + curved surface area . . = 2 + . 54 15.81 Using Pythagoras’ theorem we can find 2 = 542 + 15.812 = √3165.9561 = 56.2668 . . = ( × 15.812 + × 15.81 × 56.2668)2 = 1,139.532 = 3,579.952 = 35802 to 4 s.f. 5a. 20 ( ) = + , ≠ 0 −10 −8 −5 −2 −1.6 () −12 −10.5 −9 −12 −14.1 1.6 2 5 8 10 14.1 12 9 10.5 12 . C) from the graph = 2.4 8.5 d) for values of k within: −8.94 < < 8.94, () is has no solution. The prime numbers within this range are: K=2,3,5,7 e) We first determine the gradient function: 20 =− 2 +1 For gradient to be −4, then: = −4 20 − 2 + 1 = −4 2 = 4 = −2 2 Hence the other coordinate is: (−2, −12) f) ( ) = 2 20 + = 2 Multiplying through by ,we obtain: 20 + 2 = 3 3 − 2 − 20 = 0 Comparing coefficients to: 3 + 2 + = 0 = −1 = −20 f) = 2 −4 16 −2 4 0 0 2 4 4 16 ii)We plot and draw the graph as follows iii. from the graph the solution to 3 − 2 − 20 = 0 is = 3.2 iv From iii, = 3.2 since that is solution of 3 − 2 − 20 = 0 6. ai.The perimeter of the rectangle is given by, = 2 + 2 Let’s substitute = 80, = , = ⇒ = . ⇒ 80 = 2 + 2 ( ) Multiplying through by , we have, 2 2 − 80 + 2 = 0 Divide through by 2, ⇒ 2 − 40 + = 0 ii. When = 300, the equation becomes; 2 − 40 + 300 = 0 2 − 10 − 30 + 300 = 0 ( − 10) − 30( − 10) = 0 ( − 10)( − 30) = 0 − 10 = 0 − 30 = 0 = 10 0 = 30 iii. When = 200, the equation becomes 2 − 40 + 200 = 0 = 1, = −40, = 200 By the quadratic formula, − ± √ 2 − 4 = 2 −(−40) ± √(−40)2 − 4 × 1 × 200 = 2×1 40 ± √1600 − 800 = 2 40 ± √800 = 2 40 ± 20√2 = 2 = 20 ± 10√2 = 34.14 = 5.86 b) i Average speed = 200 = 200 = Also; 200 ( + 10) = 200 = ( + 10) 200 200 − = − ( + 10) − = − = − = 200−200(+10) (+10) 200−200+2000) (+10) 2000) (+10) ii. When = 80, We have; 2000) − = 80(80 + 10) 2000 = 7200 5 = ℎ 18 5 = × 60 18 50 = 3 2 = 16 + 3 2 = 16 + × 60 3 = 16 40 7ai. 1 ⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ 2 1 = ⃗⃗⃗⃗⃗ 2 1 = 2 ⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ . ⃗⃗⃗⃗⃗⃗ = 1 ⃗⃗⃗⃗⃗⃗ = + ⃗⃗⃗⃗⃗ 3 1 ⃗⃗⃗⃗⃗⃗ = + ⃗⃗⃗⃗⃗ 3 1 ⃗⃗⃗⃗⃗⃗ = − ⃗⃗⃗⃗⃗ 3 1 1 = − 2 3 . ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ 2 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = + ⃗⃗⃗⃗⃗ 3 2 =+ 3 From triangle ORU, ⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⃗ Since ∆ ≡ ∆ ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗⃗ 1 =++ 2 3 =+ 2 2 ⃗⃗⃗⃗⃗⃗ | = √180 c) ⃗⃗⃗⃗⃗⃗ = ( ) and | − ⃗⃗⃗⃗⃗⃗ |, Using the components of | ⃗⃗⃗⃗⃗⃗ | = √(2)2 + (−)2 | ⃗⃗⃗⃗⃗⃗ | = √5 2 | On substitution, √180 = √5 2 Squaring both sides, 180 = 5 2 2 = 36 Taking positive square root of both sides, = √36 =6 8. a ( ) = (1) 2 + 1 = 12 + 4 2 + 1 = 5 2 = 4 =2 b) ℎ(3) = 2(23 ) + 1 = 16 + 1 = 17 c) ( ) = 2 + 1 We know that: ( −1 ( )) = 2 −1 ( ) + 1 = 2 −1 ( ) = − 1 −1 −1 ( ) = 2 d) ( ) = (2 + 1)2 + 4 = (2 + 1)(2 + 1) + 4 = 4 2 + 2(2 ) + 1 + 4 = 4 2 + 4 + 5 c) ℎ( ) = 2 ℎ(ℎ−1 ( )) = ℎ−1 ( ) 2 = Taking logarithm of both sides to base ; ℎ−1 ( ) ln 2 = ln ln −1 ( ) ℎ = ln 2 Now back to the question, ℎ−1 ( ) = 0.5 ln = 0.5 ln 2 ln = 0.5 ln 2 ln = ln √2 = √2 f) 1 ℎ() = 2 1 ⟹ = 2 2 ⟹ 2− = 2 Since the base is the same, − = = −1 9) ai. Given (4,0) (0,2) We first of all find the gradient; ∆ = ∆ 2−0 = 0−4 1 =− 2 We know -intercept to be 2 1 =− +2 2 ii) 2 2 + =1 2 2 The point (4,0) lies on this curve, 42 02 ⟹ 2+ 2=1 16 0 ⟹ 2+ 2=1 16 ⟹ 2 +0=1 16 ⟹ 2=1 16 ⟹ = 2 1 2 = 16 The point (0,2) lies on this curve, 02 22 ⟹ 2+ 2=1 0 4 ⟹ 2+ 2=1 4 ⟹0+ 2 =1 4 ⟹ 2=1 4 ⟹ = 2 1 2 = 4 b) If (2, )or (2, −) lies on 2 2 + =1 16 4 Then it must satisfy it. (2)2 (−)2 + =1 16 4 4 2 + =1 16 4 4 + 4 2 = 16 4 2 = 12 2 = 3 ∴ = √3 ii. The point (2, √3) can be used to determine angle , √3 2 1 √3 tan ( < ) = 2 2 1 3 −1 √ < = tan ( ) 2 2 < = 2 tan −1 = 81.78 c) = Comparing: 2 2 + 2=1 2 to √3 ( ) 2 2 2 + =1 16 4 2 = 16, ⟹ = 4 2 = 4, ⟹ = 2 ∴ = × 4 × 2 = 8 ii. If the curve intersects the x-axis at (12,0) (−12,0) , then either point must satisfy the equation of the curve; 122 02 + 2=1 2 2 12 +0=1 2 2 = 122 = 12 By comparison, the length of = 12 is a magnification of = 4, by a scale factor of 3. The corresponding scale factor for the area is 32 = 9 Hence the new area= 9 × 8 = 72