NASS 2026 MA-O S4&5 Prelim P2 (MR)

Ngee Ann Secondary School Secondary 4&5 Mathematics-O 2026 Prelim Examinations Paper 2 Markers’ Report 1(a) 312 270 42BAC        2 2 2 418 680 2 418 680 cos 42BC     214661.0296BC  463.315 463.3 m (to 1 d.p.) This part of the question was generally well done. 1. Candidates need to understand that, for a “show” question where the final answer is given to a required accuracy, they must show an intermediate answer to at least two more decimal places/significant figures, as appropriate, than the required accuracy. Quite a number of candidates did not do so. 2. Candidates need to know the importance of reading the question carefully and following the specified accuracy requirements. 1(b) Let the drone be at point S. Let the point above C and at eye level of surveyor be point E. tan16.5 680 SE 680 tan16.5SE   201.425... Height of drone above C 201.425 1.74  203.165 203.2 m (1 d.p.) This part of the question was generally well done. 1. Candidates need to ensure that their presentation is clear (e.g. they should not use the letter O if it may be mistaken for the number 0). 2. Again, candidates should ensure that they read the question carefully, as it was specifically stated that the answer should be given to 1 decimal place.

2 NAS/2026/Prelim/MA-O/P2 1(c) This part of the question was generally well done. 1. Candidates who did not obtain full credit either did not attempt this part of the question or drew the perpendicular/angle bisectors inaccurately, with some construction arcs from the relevant points missing. 2. Candidates should also ensure that point T is clearly labelled on the diagram. 3. A number of candidates did not fully extend the bisectors and/or used dotted lines when solid lines should have been drawn. 1(d) sin sin 42 418 463.315 BCA   418sin 42sin 463.315BCA   1 418sin 42sin 463.315BCA        37.1343  Bearing of T from C 37.134390 2   071.43284  071.4  Let the foot of perpendicular from T to BC be D. 463.315 2CD 231.6575 37.1343 2DCT   18.56715 

3 NAS/2026/Prelim/MA-O/P2 231.6575cos18.56715 CT 244.377CT  244 m (3 s.f.) This part of the question was generally poorly done. Quite a number of candidates did not attempt this part of the question at all. 1. Among the candidates who attempted this part of the question, many did not realise that (x) represents a bearing. Hence, it should be expressed as a three-digit bearing and given to 1 decimal place. 2. Candidates should try to draw a diagram to help them visualise what is required by the question. The correct construction of perpendicular and angle bisectors would also help them solve this part of the question. 2(a)(i) 8 5 3 12x    5 5 15x   3 1 x   1. This part of the question was generally well done. 2. Candidates who split the compound inequality and solved it separately should use the word “and” when combining the inequalities. Some candidates wrote “or”, which is incorrect. 3. Quite a number of candidates did not take into account the reversal of the inequality sign when dividing by a negative number. Some candidates also made careless numerical errors when solving the inequalities. 4. A number of candidates ended up with inequalities that were mathematically impossible. E.g. 1 3 x  2(a)(ii) 1. This part of the question was generally well done. 2. A number of candidates did not draw the number line for the final solution set. Instead, they presented only the intermediate number lines. 3. Candidates need to understand when to use open circles and when to use closed circles on a number line. 4. Candidates should also write inequalities with the smaller value on the left and the larger value on the right. 2(b) 4 4 25 1 5 5 pq p q  1 This part of the question was generally poorly attempted. 1. There were quite a number of misconceptions such as 1 125 25p p   and/or   4 4 1 1 55 qq   2(c)

4 NAS/2026/Prelim/MA-O/P2   17 7 6 7 r r   or   27 7 6 7 7 r r    7 This part of the question was generally poorly attempted. 1. Candidates did not realise that they should first factorise out the common factor, in this case,, 7r . 2. There were quite a number of misconceptions such as   1 16 7 42r r  3. A number of candidates also misunderstood the question and attempted to find the value of (r) instead. 2(d) 2 2 3 5 8 2 4 4 3 a b a b a ab b       3 5 8 2 2 2 3 a b a b a b a b             3 2 3 5 8 2 2 3 a b a b a b a b         6 9 5 8 2 2 3 a b a b a b a b        2 2 3 a b a b a b    This part of the question was generally poorly attempted. 1. A majority of candidates did not realise that they should first factorise the denominator of the second fraction. Instead, they combined the two denominators immediately, resulting in unnecessarily complicated expressions in both the numerator and denominator. 2. A number of candidates did not use brackets when combining the numerators, which led to errors when simplifying the resulting expression. 3. Candidates should also leave their denominators in factorised form.

5 NAS/2026/Prelim/MA-O/P2 3(a)(i) 70 Okay 3(a)(ii) Most students didn’t realsise that this is a graph of reciporcal function and joined the points (-1,-20) to (1,-20). Smooth curves generally obtained. Marks were given so long points are connected to get a “smooth” curve. 3(a)(iii) Estimated gradient 98 72 5 0.4   5.65 (3 s.f.) Line of tangent is mostly drawn at the correct point. Revised range for gradient: 4.8 to 6.67. Students are advised to draw tangent better. No gap between line and curve and avoid “overlapping” with the curve (like “ multiple points of intersection” to avoid) 3(b)(i) At/After 3.5 h of charging, Power Bank A’s rate of charging is 5.65% per hour. Okay. Stronger students understand to make reference to (ii) and is able to conclude that it is the % per hour. Weaker students were not able to connect to (i) and give answers irrelevant to the question. 3(b)(ii) 3 2 5 12 0x x   poorly done, students did not know how to approach the question. Theu end up solving the cubic equation to obtain values of x. 2 125 0x x   2 12010 50 0x x   2 12010 50 100x x   Draw 10 50y x 

6 NAS/2026/Prelim/MA-O/P2 Since the battery percentage after 4h is higher with Power Bank A (92.5% > 90%), Carl should charge his phone using Power Bank A. 4(a) 33 12QR        3 2 PQ   Since 3 2QR PQ     , QR // PQ, and they have a common point Q, the points P, Q and R are collinear. Not well done. Weaker students unable to form the vector equation (did not show understanding of the meaning of the required equation) Many did not conclude that QR//PQ first, only state that Q is the common point. 4(b) PQ OQ OP        OP OQ PQ        0 3 1 1            3 0     Since the y-coordinate of P is 0, P lies on the x-axis. A number of students has difficulty in finding the position vector of P correctly with some who did not know how to approach the question. 4(c) A small number did not conclude at all. 3 3 3 1 1 2             31 12      1 2 PQ   So, 1 2PR PQ     , that means 1 2 PR PQ      : 1: 2PR PQ      n = 2 Poorly done. (No attempt by a number of students) Many is not aware that the ratio can be found using the vector equation, with some them mistakenly putting n as -2. A number of students did not use the exact values of PR = √2.5 and PQ = √10 to find the ratio (use estimated 5 s.f. values) A number made the mistake by squaring the ratio √2.5: √10 instead of dividi