RI 2024 Year 5 H2 Math Timed Practice (Modified) (Soln with comments)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 Page 1 of 14 2024 H2 Math Year 5 Timed Practice (Modified) Solutions with Comments 1 A smoothie shop sells three different types of smoothie blends, Berrylicious, Tropicality and Pumptein. The smoothie blends are dispensed from machines and each serving is prepared by blending protein powder, mixed fruits and milk. Relative to a serving of Berrylicious, a serving of Tropicality uses 50% the amount of protein powder, 30% more the amount of mixed fruits and 75% the amount of milk. Relative to a serving of Berrylicious, a serving of Pumptein uses 50% more the amount of protein powder, 60% the amount of mixed fruits and 25% more the amount of milk. A serving of Berrylicious, Tropicality and Pumptein weighs 360 grams, 350 grams and 355 grams respectively. Find the amount of protein powder, mixed fruits and milk required to make a serving of Berrylicious, in grams. [4] Solutions Comments Let the amount of protein powder, mixed fruits and milk required to make a serving of Berrylicious be x, y and z respectively. Total mass of a serving of Tropicality: 0.5 1.3 0.75 350x y z   -- (1) Total mass of a serving of Pumptein: 1.5 0.6 1.25 355x y z   -- (2) 360x y z   -- (3) Solving equations (1), (2) and (3) with the GC, we have 10, 150x y  and 200z Very often the last line in the SLE question gives a clue to the unknowns to define.

2024 H2 Math Year 5 Timed Practice (modified) Solutions with Comments __________________________________________________________________________________ Page 2 of 14 2 A graphic calculator is not to be used in answering this question. Given that 3 4i is a root of the equation 3 25 25 0,z az bz    find the values of the real numbers a and b and the remaining roots of the equation. [4] Solutions Comments Since the coefficients are real, 3 4iz  is another root of the equation. Method 1        2 2 2 2 3 4i 3 4i 3 4i 6 9 16 6 25 z z z z z z z               So,    3 2 25 25 6 25 5 1z az bz z z z       (By inspection) Comparing coefficients of 2z , 1 30 29a   Comparing coefficients of z, 125 6 119b   The other roots are 3 4iz  and 1 5z . Method 2 (Not recommended) Substitute 3 4iz  ( or 3 4iz  ) into the given eqn,                       3 2 5 3 4i 3 4i 3 4i 25 0 5 27 108i 144+64i 9 24i 16 3 4i 25 0 5 117 44i 7 24i 3 4i 25 0 560 7 3 4 55 6 i 0 a b a b a b a b a b                                 Comparing the real parts, 7 3 560a b  --- (1) Comparing the imaginary parts, 6 55a b  ---- (2) (1) + (2)  3: 7 18 560 55 3 25 725 29 a a a a       From (2): 55 29 6 119b    29a  , 119b        2 2 2 23 4i 3 4i 3 4i 6 9 16 6 25z z z z z z z              3 225 119 29 5 0z z z       2 6 25 5 1 0z z z    The other roots are 3 4iz  and 1 5z . Most students use Method 1, which is the recommended method, to do this question. As graphic calculator is not allowed in answering this question, students are to expand out   3 3 4i and   2 3 4i

2024 H2 Math Year 5 Timed Practice (modified) Solutions with Comments __________________________________________________________________________________ Page 3 of 14 3 With reference to the origin O, the points A and C have position vectors a and c respectively, where a and c are non-zero and non -parallel vectors. The point E lies on AC such that :AE AC = 2 : , where 2.    (a) Find the area of triangle OAE in terms of  , a and c. [3] (b) Given that the area of triangle OAE is 3 10 a c , solve for the value of  . [1] (c) Given that a is a unit vector, give the geometrical interpretation of (i) a.c , [1] (ii) .a c [1] (d) If 90OAC   , explain why a .c is positive. [1] Solutions Comments (a) [3] Using ratio theorem,  2 + 2OE    a c      1Area of Δ 2 2 + 21 2 21= since = 2 1 2 2 2 since 22 a ca ca a a 0 a c a c OAE OA OE                        Use a diagram to interpret the ratio : 2 :AE AC    The vector product gives a vector. This is the zero vector. (b) [1] 2 3Area of Δ 2 10OAE      a c a c By comparison, 2 3 2 10 10 20 6 4 20 5             (c)(i) [1] a .c gives the length of projection of c onto a. Note that a is a unit vector, so ˆcosOC AOCa.c . (c)(ii) [1] a c gives the perpendicular distance from point C to OA. Note that a is a unit vector, so ˆsinOC AOC a c . A E C ? O

2024 H2 Math Year 5 Timed Practice (modified) Solutions with Comments __________________________________________________________________________________ Page 4 of 14 Alternatively, a c gives area of parallelogram with OA and OC being the adjacent sides. Or, a c is equivalent to twice the area of triangle OAC. The description must accurately convey that it is the distance from a point to a line. (d) [1] Since 90OAC   ,   2 0 0 | | which is positive since is a nonzero ve ctor OA AC     . a. c a a.c a a     Alternatively, since 90OAC   , the angle between a and c, is less than 90 ,AOC  i.e., it is acute. Hence a .c is positive. You can draw a diagram to explain the relationship between the angles. O A C

2024 H2 Math Year 5 Timed Practice (modified) Solutions with Comments __________________________________________________________________________________ Page 5 of 14 4 The line 1l contains the points A and B with coordinates ( 1, 3, 7) and ( , 0, 2)a  respectively, where a is a constant. The line 2l has equation 7 10 , 10.2 3 x y z    It is given that 1l and 2l cross at the point C. (a) Find the value of a and the coordinates of C. [4] (b) The point P has coordinates ( 2,9,8). Find the point on 2l which is closest to P and hence find the exact perpendicular distance between P and 2l . [4] Solutions Comments (a) [4] 1 1 1 : 3 3 , 7 9 a l                      r  and 2 7 10: , 102 3 x yl z     i.e., 2 7 2 : 10 3 , 10 0 l                    r  When 1l and 2l intersect/cross, 1 1 3 3 7 9 a                    7 2 10 3 10 0                  1 2 6 ----(1) 3 3 7 ----(2) 9 3 ----(3) a              Solving (3), 1 3  . Then solving (2), 1 3 7 2     Solving (1),  1 1 4 6 53 a a      Hence C has coordinates ( 3, 4,10). Alternative method: 1 1 3 3 7 9 a                    10 x y         1 1 ----(1) 3 3 ----(2) 7 9 10 ----(3) a x y             Revise on your skill in converting from cartesian form to vector form. Careless mistakes can be costly as wrong equations will impact on working for part (b). Vector equation of a line is incomplete without “ r ”. Vectors should also be properly presented with the right notations such as “~” like a . Question asked to give your answer in coordinates, not as a position vector.

2024 H2 Math Year 5 Timed Practice (modified) Solutions with Comments __________________________________________________________________________________ Page 6 of 14 Solving (3), 1 3  . Then solving (2), 4y  7 10Sub into , we get 32 3 x y x    Solving (1),  11 1 3 53 a a      Hence C has coordinates ( 3, 4,10). (b) [4] Let the point which is closest to P be F. 7 2 10 3 10 0 OF                       , for some  PF OF OP        5 2 1 3 2            Since 2PF l 