SPM 2021 K2 - Comprehensive Notes (Bullet-Point Summary)

1. Perimeter and area of a square

Given: Perimeter of a square is $P = 4x + 4$ .
For a square with side length $a$ , the perimeter is $P = 4a$ .
Express side in terms of x:
- $a = \dfrac{P}{4} = \dfrac{4x+4}{4} = x+1$
Area of a square is $A = a^2$ .
Therefore in terms of x:
- $A = (x+1)^2 = x^2 + 2x + 1$

2. Five cards arranged in a sequence (Diagram 1)

Diagram 1 shows five numbered cards arranged according to a sequence.
(a) Task: Nyatakan nilai J dan nilai K. / State the value of J and of K.
(b) Task: Seterusnya, nyatakan pola bagi jujukan itu. / Hence, state the pattern of the sequence.
Note: Specific values of J, K and the pattern depend on Diagram 1 (not provided in text). To solve, examine the given five cards and identify the rule (e.g., arithmetic, geometric, alternating, or a positional pattern) and apply it to determine J and K.

3. Staircase cross-section – gradient and handrail length

Diagram 2 shows cross-section of a staircase with labeled parts: Stair, Handrail, Floor, etc.
Given: The gradient of the staircase is $m = 0.25$ .
(a) Find the angle between the staircase and the floor.
- For a line with gradient $m$ , angle $\theta$ with the horizontal satisfies $\tan\theta = m$ .
- Therefore $\theta = \arctan(0.25)$ (approximately $14.0^{\circ}$ ).
(b) Find the length of the handrail (in metres).
- If vertical rise is $1.5\text{ m}$ (from the diagram annotation), then horizontal run is
  $\text{run} = \frac{\text{rise}}{m} = \frac{1.5}{0.25} = 6.0\text{ m}.$
- Handrail length (hypotenuse) is
  $L = \sqrt{(\text{rise})^2 + (\text{run})^2} = \sqrt{1.5^2 + 6^2} = \sqrt{38.25} \approx 6.19\text{ m}.$

4. Sofa prices in two bases (Diagram 3)

Diagram 3 shows RM prices for Sofa M and Sofa N in two different number bases.
Prices given: Sofa M = RM 547 (in some base), Sofa N = RM 24416 (in another base).
Goal: Determine which sofa is cheaper.
Approach:
- Convert each price to base 10 using its given base:
- If Sofa M is in base $bM$ : price in base-10 = $5\cdot bM^2 + 4\cdot b_M + 7$ .
- If Sofa N is in base $bN$ : price in base-10 = $2\cdot bN^4 + 4\cdot bN^3 + 4\cdot bN^2 + 1\cdot b_N + 6$ .
- Compare the resulting base-10 prices to decide which sofa is cheaper.
Note: Bases are not specified in the text excerpt; actual numeric comparison requires the base values.

5. Direct and inverse variation – headscarves sold

Relationship: Number sold $N$ varies directly as advertising budget $D$ and inversely as price per scarf $P$ .
Therefore: $N = \dfrac{k D}{P}$ for some constant k.
Given data: When $D = RM\,5000$ and $P = RM80$ , the number sold is $N = 480$ .
(a) Express $N$ in terms of $D$ and $P$ .
- $N = \dfrac{k D}{P}$ where $k$ is determined from data.
Solve for k using the provided values:
- $480 = \dfrac{k \cdot 5000}{80} \Rightarrow k = \dfrac{480 \cdot 80}{5000} = 7.68.$
- Hence, $N = 7.68 \dfrac{D}{P}$ .
(b) If the advertising budget is $D = RM75{,}000$ and the price per scarf remains the same ( $P = RM80$ ), find the number sold:
- $N = 7.68 \dfrac{75000}{80} = 7.68 \cdot 937.5 = 7200.$

6. Tax reliefs and medical cost (Kamil and Athar)

(a) Kamil’s chargeable income (2020):
- Gross annual salary: $RM\,56{,}000$ .
- Donations: $RM\,200$ to the state library.
- Tax reliefs (from Table 1 in text):
- Individual relief: $RM\,9{,}000$
- Life insurance and EPF (up to RM7{,}000): $RM\,1{,}800$
- Medical insurance (up to RM3{,}000): $RM\,3{,}120$
- Assuming donation is eligible relief as well (RM200).
- Total relief = $9{,}000 + 1{,}800 + 3{,}120 + 200 = RM\,14{,}120$ .
- Chargeable income = $56{,}000 - 14{,}120 = RM\,41{,}880.$
(b) Athar’s hospital costs with insurance (90/10 co-insurance; deductible RM2000):
- Total bill = RM 27,000.
- After deductible: RM 27,000 − RM 2,000 = RM 25,000.
- Co-insurance (Athar pays 10%): 0.10 × 25,000 = RM 2,500.
- Total amount borne by Athar = deductible + co-insurance = RM 2,000 + RM 2,500 = RM 4,500.

7. Cartesian plane geometry – distances and parallel lines

(a) Points: Liza’s house, Nina’s house, and Recreational Park with coordinates shown in Diagram 4; (h, 3) for Liza’s house, (2, 8) for Recreational Park, (2, 3) for Nina’s house, on a Cartesian plane.
(a)(i) Equation of straight line joining Liza’s house and Nina’s house.
- Given coordinates: Liza’s house at (h, 3); Nina’s house at (2, 3).
- Slope between (h,3) and (2,3) is $m = \dfrac{3-3}{h-2} = 0$ (horizontal line).
- Therefore equation is $y = 3$ .
(b) Liza cycles from her house to the Recreational Park, distance 13 km.
- (i) Find the value of $h$ .
- Using distance formula between Liza’s house (h,3) and Recreational Park (2,8): $\text{Distance}^2 = (h-2)^2 + (3-8)^2 = 13^2 = 169$
 - Compute: $(h-2)^2 + (-5)^2 = 169 \Rightarrow (h-2)^2 + 25 = 169 \Rightarrow (h-2)^2 = 144 \Rightarrow h-2 = \pm 12$ .
 - So $h = 14$ or $h = -10$ (taking both possibilities from distance formula).
- (ii) The road from Nina’s house to the school is parallel to the road from Liza’s house to Recreational Park.
- If the line from Liza’s house to Recreational Park has slope $m_{LP}$ , then the line from Nina’s house to the school must have the same slope.
- Using Nina’s house (2,3) and school unknown; The equation of the line with slope equal to $m{LP} = \dfrac{8-3}{2-h} = \dfrac{5}{2-h}$ (using Liza’s house as (h,3)) can be used to determine the school’s line once the school’s coordinates are provided. In general: find the line through Nina’s house with slope $m{LP} = \dfrac{8-3}{2-h}$ ; then express the equation of the line representing the road from Nina’s house to the school.

8. Table 2 – probability and relationships between pass/fail

Table 2 shows the outcome of Mathematics test for Class Yakin (Boys and Girls) with Pass/Fail counts across four groups.
Given: Probability that a student fails is $\frac{1}{2}$ (per context) and number of girls who pass is twice the number of boys who pass.
Tasks:
- Find the values of $x$ (the missing number of male/female counts in the table) and $y$ (an associated variable in Table 2) that satisfy:
- Total fails probability = $\tfrac{1}{2}$ .
- Girls who pass = 2 × Boys who pass.
Note: Specific numbers require the full Table 2 values; use the given probability and the stated relation to solve for the unknowns.

9. Congruent polygons and transformations (Diagram 5)

(a) List two pairs of congruent polygons.
(b) Draw image of triangle CDG under a clockwise rotation by 90° about point D.
(c) Triangle BCJ is the image of triangle EFJ under the combined transformation PQ.
- Describe, in full, the transformation:
- (i) Q: identify the first transformation associated with Q (e.g., rotation/reflection/translation) and its center/angle/axis.
- (ii) P: identify the second transformation after Q and its parameters.
Note: Exact polygons and coordinates depend on Diagram 5; the general approach is to identify congruent pairs and apply the specified transformations step-by-step.

10. 3D prisms and cross-sections (Diagram 6)

Description: Two right prisms with rectangular bases ABCD and BCRNPQ lying on a horizontal plane. A semi-cylinder is removed from one prism. AFGKLPQB is the uniform cross-section.
Edges AF, DE, BK, and CJ are vertical. All equalities provided:
- AF = DE = EH = FG = JM = KL = QP = RN
- BK = CJ = HJ = GK
- Diameter of the semi-cylinder is 3 cm.
(a) Draw the elevation of the object on the plane parallel to AB and QP as seen from X.
(b) Draw the elevation of the object on the plane parallel to LM as seen from Y.
Notes:
- The task involves projecting the 3D object onto vertical planes, interpreting equal edges as height segments, and shading/constructing the two elevations accordingly.
- The diameter of the semi-cylinder is a key dimension for the solid’s cross-section.

11. (Diagram 5) Angles and shapes

The page shows an angle labeled 45° and other letters E, F, G, H, M, R, etc. Context suggests tasks may involve identifying right angles, angle measures, or applying congruence/transformations from Diagram 5.
Note: Specific questions depend on Diagram 5; use angle properties and the diagram’s given marks to answer.

12. Card sets and Venn diagrams (Table 4)

12(a) Cikgu Feeda has 12 numbered cards distributed among students as per: primes < 10; even numbers < 9; factors of 15.
(a) List all numbers received by Joe (J), Karl (K), and Liam (L) using set notation.
(b)
- (i) Draw a Venn diagram for the three sets (J, K, L).
- (ii) Represent the regions that contain numbers common to different sets using the symbols 'U' and/or 'n' (universal set and intersection notation).
Note: Exact numbers depend on the explicit sets:
- J = {2,3,5,7} (primes < 10)
- K = {2,4,6,8} (multiples of 2 < 9)
- L = {1,3,5} (factors of 15)
- Union = {1,2,3,4,5,6,7,8,15}; pairwise/intersection details follow from these definitions.

13. Vouchers, pizzas, and cupcakes (Matrix method)

Lisa buys 3 pizzas and 6 cupcakes with voucher 1 valued RM45: 3P + 6C = 45.
Lisa’s sister buys 7 pizzas with voucher 2 valued RM45, but the voucher is not enough; she must add cash equal to the price of one cupcake, so the shortfall is C: 7P − 45 = C.
(a) Using matrix method, solve for P and C (prices of pizza and cupcake):
- From equations: 3P + 6C = 45 and C = 7P − 45.
- Substituting gives 3P + 6(7P − 45) = 45 → 45P = 315 → P = 7; C = 7P − 45 = 49 − 45 = 4.
(b) Next week, they want to use the third voucher to buy 4 pizzas and 2 cupcakes. Using matrix multiplication only:
- Represent purchases as a row vector and prices as a column vector:
- Row: [4, 2] ; Prices: [P, C]^T = [7, 4]^T.
- Total cost = [4, 2] × [ [7], [4] ] = 4×7 + 2×4 = 28 + 8 = RM 36.
- Therefore, the third voucher must be at least RM36 to cover the cost.

14. Roller coaster – linear inequalities and graphing

A roller coaster has 12 rows and 3 seats per row (max 36 passengers).
Rule: A child must be accompanied by at least one adult.
Let x = number of children, y = number of adults.
(a) Write two linear inequalities (besides x > 0 and y ≥ 0) representing the situation with maximum capacity:
- Capacity constraint: $x + y \le 36$
- Accompaniment constraint (at least one adult per child): $x \le y$
(b) On graph paper (scale 2 cm = 5 persons on both axes), draw and shade the feasible region defined by the inequalities in (a).
(c) From the graph, determine whether 14 children and 10 adults can ride together:
- Check (i) x + y ≤ 36: 14 + 10 = 24 ≤ 36 → satisfied.
- Check (ii) x ≤ y: 14 ≤ 10 → not satisfied.
- Conclusion: 14 children and 10 adults cannot ride together under these rules.
(d) After 4 rows are out of service, maximum capacity changes to 24 passengers.
- (i) New inequality: $x + y \le 24$
- (ii) Draw the line for this inequality on the same graph to shade the new feasible region.

15. Reza’s distance-time graph, diet, and smashes

(a) Reza runs 10 km every weekend. Diagram 7 shows distance-time graph.
- (i) State the duration of rest (time when the graph is horizontal).
- (ii) Compute the average speed over the run: average speed = total distance / total time.
(b) Diet: total daily calories = 3000. Diagram 8 shows a pie chart of nutrition categories.
- Protein intake = 250 calories; angle for protein = (250/3000) × 360 = 30°.
- The difference between the angle for sodium and calcium is 36°; Fat angle is 11°.
- Find the angle for calcium intake (subject to the full chart: fat, protein, sodium, calcium, carbohydrates sum to 360°).
- Approach: Let angle for calcium be θC, angle for sodium be θN. Then:
- Protein angle = 30°, Fat angle = 11°.
- If sodium angle is larger than calcium by 36°, then θN = θC + 36°; total: 30° + 11° + θC + (θC + 36°) + θCarb = 360° → 3θC + θ_Carb = 283°.
- If calcium is larger, θN = θC − 36°; total: 30° + 11° + θC + (θC − 36°) + θCarb = 360° → 2θC + θ_Carb = 355°.
- Therefore, θC cannot be determined uniquely without the value of θCarb (the carbohydrate angle) or which of sodium/calcium is larger; use the diagram to resolve.
(c) Pukulan junam drill: Distances RMN = 38°; MN = 4 m; ∆height due to jumps at R: 0.25 m and 0.15 m.
- Given RM = RN (isosceles), and MN = 4 m, use Law of Sines to find RM = RN:
- Base angles at M and N are each 71° (since apex angle at R is 38°).
- RM / sin(71°) = MN / sin(38°) ⇒ RM = RN = MN · sin(71°) / sin(38°).
- Numerically, RM ≈ RN ≈ 4 × sin(71°)/sin(38°) ≈ 6.14 m.
- The difference between the angle of depression from P to M and to N depends on the vertical height at the moment of observation and the horizontal distances; with the stated vertical jumps, compute using right-triangle trigonometry considering the height change and horizontal distance to M and N. (The exact numeric value requires the precise geometry from Diagram 9 and the height values given.)
(d) Racket choice – speeds for smashes
- From Table 5, speeds (km/h) for S: 254, 251, 256, 260, 253; and T: 252, 255, 258, 254, 255.
- Compute means:
- Mean(S) = (254+251+256+260+253)/5 = 1274/5 = 254.8 km/h.
- Mean(T) = (252+255+258+254+255)/5 = 1274/5 = 254.8 km/h.
- Conclusion: Both racquets give the same average shuttle speed in the given trials; additional factors (consistency, maximum speed, control) may be considered, but based on data, neither is superior by average speed alone.

16. Lisa and Reza – nutrition angles and ratio

(a) Rezza is to compute the angle for calcium intake using the given pie chart data (protein and fat angles given, sodium-calcium difference).
(b) Given: protein angle corresponds to 250 calories; total calories = 3000; protein angle = 30° as shown earlier. Difference between sodium and calcium angles is 36°. Fat angle is 11°. Find the calcium angle.
(c) The remaining portion would be carbohydrates; use the total sum to verify consistency with the diagram.
Note: The exact numeric value for calcium angle depends on which sector is larger (sodium vs calcium) and the carbohydrate angle; use the given diagram values to compute the precise angle.

17. Smash drill geometry – distances and angles

(a) R is the jump point; M and N are markers; MR = RN and MN = 4 m; MRN = 38° (apex angle at R) implying RM = RN by isosceles triangle.
(b) A vertical jump of 0.25 m and 0.15 m at R changes the height of hitting point P; compute the difference between the angle of depression to M and to N from P.
Steps:
- Compute RM = RN using Law of Sines as in 15(c).
- Compute horizontal distances to M and N from the projection, then use tan(θ) = opposite/adjacent, with the height adjustments, to find the two depression angles and their difference.
Note: The exact numeric value depends on the full geometry provided in Diagram 9.

18. Badminton racquet – data interpretation

Table 5 shows shuttlecock speeds (km/h) for smashes using racquets S and T across five trials.
Question: Which racquet should Reza buy? Justify your answer.
Analysis:
- Compute average speeds for each racquet:
- Mean(S) = 254.8 km/h (as computed in 15(d) step).
- Mean(T) = 254.8 km/h.
- Since the averages are equal, other factors (consistency across trials, variance, feel, control) should be considered; with given data, averages do not show a clear winner.

19. SMK field trip – costs and timings

(a) Group: 5 teachers and 35 students (total participants = 40).
(i) Jadual 6 lists student counts by Form and gender.
- (Form Four): Boys = 11, Girls = 8 -> total 19.
- (Form Five): Boys = 7, Girls = 9 -> total 16.
- Total students listed = 19 + 16 = 35 (as given).
- (i) State the ratio of teachers to students.
- Teachers: 5, Students: 35; Ratio = 5:35 = 1:7.
(ii) A student is chosen at random to be photographer. Probability the chosen student is a Form 5 boy.
- If selection is among students only: P(Form 5 boy) = 7/35 = 1/5 = 0.2.
- If selection is from all participants (teachers included): P = 7/40 = 0.175. Typically, since they say "murid" (student), the intended answer is 7/35 = 0.20.
(b) Total fees collected RM 3000. Each teacher pays RM 40 more than a student.
- Let student fee = s; teacher fee = s + 40.
- 5(s + 40) + 35s = 3000 ⇒ 5s + 200 + 35s = 3000 ⇒ 40s = 2800 ⇒ s = 70.
- Teacher fee = 110.
- Therefore: student pays RM 70; teacher pays RM 110.
(c) Jadual 7 – speed-time data for bus; (i) Compute total distance of the journey.
- Use area under speed-time graph or sum distance over time intervals, depending on provided data points.
(ii) Rajah 11 – complete the distance-time graph using part (i) results; describe bus motion from 8:15 a.m. to 10:20 a.m.

20. Table 3 – distribution and mean/variance (summary)

(a) Given mean = 55.6, find m (the frequency in 50–54 class).
(b) Compute variance of the distribution using grouped data.
Approach:
- Determine total N = 39 + m; compute mean via sum of (midpoint × frequency) / N equals 55.6; solve for m (as shown in previous worked example, m = 11).
- With m found, compute variance using the formula for grouped data:
- Var = [Σ fi (midi − mean)^2] / N (population variance) or divide by (N − 1) for sample variance, depending on convention.
Result (example): Using m = 11 and N = 50, the computed variance is approximately $\sigma^2 \approx 42.04$ (population variance).

21. Complex polygons, distances, and travel plans (Diagrams 11 and 12)

(a) Diagram 5 shows multiple polygons on a Cartesian plane.
- (i) List two pairs of congruent polygons.
(b) Rotate triangle CDG by 90° clockwise about D and draw its image.
(c) Triangle BCJ is the image of triangle EFJ under a combined transformation PQ.
- (i) Describe the transformation associated with Q.
- (ii) Describe the transformation associated with P.
(Notes): The exact pairs and transformations depend on Diagram 5 and the labeling PQ.
Diagram 12 provides an undirected graph of five attractions; timing labels indicate travel times between destinations.

22. City tour graph theory scenario (Cikgu Linda vs Cikgu Kumar)

Context: After 12:30 p.m., a plan to visit five attractions (National Science Centre, National Library, Twin Towers, Petronas, National Mosque, Istana Negara, Menara KL, etc.) is represented by an undirected graph with weights (times in minutes).
Task: Decide whose opinion is better about the next destination to visit first to save time, with justification.
Approach:
- Compare travel times along potential routes using the graph (shortest-path considerations, e.g., Dijkstra’s algorithm conceptually).
- Justify which sequence minimizes total travel time to visit the next destination(s).
Notes: The precise justification depends on Diagram 12 and its edge weights between attractions.
Overall study tips from this transcript:
- Translate word problems into standard math models (algebraic expressions, linear inequalities, matrices, probability, geometry, calculus-free trigonometry, etc.).
- Always identify what is given, what needs to be found, and which formula or principle applies (e.g., P = 4a for squares, tanθ = gradient, law of sines, mean/variance for grouped data, matrix representation for systems of equations, etc.).
- When bases or diagrams are involved, convert to a common basis (decimal) or read the diagram carefully to extract numeric values before arithmetic.
- For transformation questions, name the type of transformation (translation, rotation, reflection, enlargement) and specify centers/angles or lines of reflection as required; apply step-by-step to obtain images.
- In probability problems, decide whether you are computing unconditional probabilities or conditional probabilities (e.g., choosing a student vs. choosing from all participants).
- In graphing/LP problems, set up objective vs. constraint inequalities, identify feasible regions, and interpret shading and intercepts.
Key formulas (recap in LaTeX):
- Square: $P = 4a,\, A = a^2$ , hence $A = \left(\frac{P}{4}\right)^2$ .
- Gradient to angle: $\tan\theta = m\quad\Rightarrow\quad \theta = \arctan(m)$ .
- Hypotenuse: if rise R and run r, $L = \sqrt{R^2 + r^2}$ ; with gradient $m = \dfrac{R}{r} \Rightarrow r = \dfrac{R}{m}$ .
- Direct/inverse variation: if $N \propto \dfrac{D}{P}$ , then $N = k\dfrac{D}{P}$ with a constant $k$ determined from data.
- Means in grouped data (midpoints, frequencies):
- $\bar{x} = \dfrac{\sum fi \cdot mi}{\sum fi}$ , where $mi$ is the class midpoint.
- Variance for grouped data (population):
- $\sigma^2 = \dfrac{\sum fi (mi - \bar{x})^2}{\sum fi}$ ; for sample variance, divide by one less: $\frac{\sum fi (mi - \bar{x})^2}{\sum fi - 1}$ .
- Distance between two points: $d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2}$ .
- Distance-time and speed-time relationships involve areas under curves and slopes for average speed; speed = distance / time, and total distance is the integral/area under a speed-time curve for non-constant speed.
Optional practice prompts (from this transcript) for exam prep:
- Solve for unknowns in problems where a perimeter is given in terms of x and express area in x.
- Practice pattern recognition for sequences from diagram-based problems.
- Convert prices stated in different bases to base-10 and compare.
- Use the Law of Sines or Pythagoras in staircase problems where run and rise are deduced from gradient.
- Set up and solve matrix equations for word problems involving two equations with two unknowns (e.g., voucher problems).
- Model real-world constraints with linear inequalities and interpret feasible regions.
- Compute and interpret mean and variance for grouped data, including solving for missing frequencies.
- Apply set theory and Venn diagram reasoning to three-set problems.
- Use probability concepts carefully when the problem defines “a student” vs “any person.”
- Practice reading and interpreting diagrams from geometry/topology problems (rotations, reflections, parallelism).
- Be comfortable with basic trigonometry in practical contexts (angles from gradients, distances in isosceles triangles, etc.).