Question bank · 18 min read
Compulsory Mathematics: Model Question Sets and Predicted Questions
This chapter does three things. It sets out the pattern of the SEE Compulsory Mathematics paper and shows how the grid spreads 75 marks over seven content areas. It then gives two complete model papers, each of 16 questions, 75 marks and three hours, followed by a full solution to every part. Last, it ranks practice priorities: question forms worth rehearsing first, with the reason for each position.
Nobody outside the examination board knows what will be set, and this chapter does not pretend to. A ranking here is a judgement from two public facts: the CDC specification grid (areas, number of questions and marks) and how standard a question form is. These are not forecasts. Treat the list as an order in which to practise, never as a promise that a question will appear. Every question in the model papers is an original practice item.
Seventy-five marks in 180 minutes is 2.4 minutes a mark. Plan it: Sets, Arithmetic and Mensuration (7 questions, 32 marks), 75 minutes; Algebra (3 questions, 15 marks), 35 minutes; Geometry (3 questions, 13 marks), 30 minutes; Statistics, Probability and Trigonometry (3 questions, 15 marks), 25 minutes; and 15 minutes to check. Bank the first one or two marks of every question first.
7.1 Pattern and weighting of the paper
7.1.1 The grid
The CDC grid for Compulsory Mathematics has 16 questions made of 49 items, for 75 marks. The areas and their weights are:
| Area | Hours | Questions | Marks |
|---|---|---|---|
| Sets | 12 | 1 | 6 |
| Arithmetic | 28 | 3 | 13 |
| Mensuration | 28 | 3 | 13 |
| Algebra | 32 | 3 | 15 |
| Geometry | 28 | 3 | 13 |
| Statistics and Probability | 24 | 2 | 11 |
| Trigonometry | 8 | 1 | 4 |
| Total | 160 | 16 | 75 |
7.1.2 The thinking levels
| Level | Marks |
|---|---|
| Knowledge | 12 |
| Understanding | 18 |
| Application | 30 |
| Higher ability | 15 |
| Total | 75 |
The questions in the model papers below follow this idea: most questions begin with a one-mark knowledge or understanding part (write a formula, draw a diagram, state a theorem) and end with a one- or two-mark higher-ability part. How the parts split is our design, not the board's.
7.1.3 How the model papers use the grid
Both papers give each area the marks the grid gives it:
| Question | Area | Marks |
|---|---|---|
| 1 | Sets | 6 |
| 2–4 | Arithmetic (tax, compound interest, growth or depreciation) | 4, 5, 4 |
| 5–7 | Mensuration (plane surfaces, cylinder and sphere, prism and pyramid) | 4, 5, 4 |
| 8–10 | Algebra (HCF and LCM, surds, indices, fractions, equations) | 5, 5, 5 |
| 11–13 | Geometry (circle, construction, area theorems) | 5, 4, 4 |
| 14–15 | Statistics (6) and Probability (5) | 6, 5 |
| 16 | Trigonometry | 4 |
| Total | 75 |
7.2 Model Question Set 1
Time: 3 hours. Full marks: 75. Attempt all questions. Use \pi=\frac{22}{7} where needed.
- In a survey of 80 tourists in Pokhara, 45 visited Phewa Lake, 40 visited Sarangkot and 10 visited neither.(a) Write the cardinality notation for those who visited both places, using P and S.(b) Show the information in a Venn diagram.(c) How many tourists visited both places?(d) Find the ratio of those who visited only Phewa Lake to those who visited only Sarangkot. [1+1+3+1]
- The marked price of a refrigerator is Rs 60,000. A shopkeeper gives a 10% discount and charges 13% VAT on the discounted price.(a) Find the price after the discount.(b) Find the VAT amount.(c) Find the bill. [1+1+2]
- Rs 25,000 is deposited for 3 years at 8% per annum compound interest, compounded yearly.(a) Write the formula for the compound amount.(b) Find the amount.(c) Find the compound interest.(d) Find the simple interest on the same sum at the same rate and by how much the compound interest exceeds it. [1+2+1+1]
- The population of a town was 40,000 two years ago. It grew by 5% last year but by 8% this year.(a) Write the multiplying factor for a growth of R\%.(b) Find the present population.(c) If it falls by 10% next year, find the population then. [1+2+1]
- The sides of a triangle are 9 cm, 10 cm and 17 cm.(a) Write Heron's formula.(b) Find the area.(c) Find the height on the side of 10 cm. [1+2+1]
- A closed cylinder has radius 7 cm and height 10 cm.(a) Write the formula for the curved surface area.(b) Find the curved surface area.(c) Find the total surface area.(d) Find the volume.(e) Express the volume in litres. [1+1+1+1+1]
- A square pyramid has a base of side 16 cm and a height of 6 cm. Find(a) the slant height,(b) the lateral surface area,(c) the total surface area,(d) the volume. [1+1+1+1]
- If x=5+2\sqrt{6}, find(a) \frac1x in surd form,(b) x+\frac1x,(c) x^2+\frac{1}{x^2}. [2+1+2]
- (a) Simplify \frac{x^2-16}{x^2+8x+16}.(b) Prove that \frac{1}{x-3}-\frac{1}{x+3}-\frac{6}{x^2-9}=0. [2+3]
- The length of a rectangular garden is 5 m more than its width and its area is 84 m².(a) Form an equation.(b) Solve it.(c) Find the length and width.(d) Find the perimeter. [1+2+1+1]
- (a) Prove that the opposite angles of a cyclic quadrilateral are supplementary.(b) If one angle of a cyclic quadrilateral is 105°, find the opposite angle. [4+1]
- Construct a quadrilateral ABCD with AB = 5 cm, BC = 4 cm, CD = 4.5 cm, \angle B=90^\circ and \angle C=100^\circ, and then construct a triangle equal in area to it. Write the steps. [4]
- ABCD is a parallelogram and E is any point on side DC.(a) Prove that \triangle ABE=\frac12 parallelogram ABCD.(b) If the area of ABCD is 48 cm², find the area of \triangle ABE. [3+1]
- The marks of 20 students are in the table. Find the quartile deviation and its coefficient. [2+2+1+1]
| Marks | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| Students | 2 | 4 | 6 | 5 | 3 |
- A bag contains 5 red and 3 blue balls. Two balls are drawn one after another without replacement.(a) Draw a tree diagram.(b) Find the probability that both balls are of the same colour.(c) Find the probability that at least one ball is blue. [2+2+1]
- From a point on the ground 20\sqrt{3} m from the foot of a vertical pole, the angle of elevation of its top is 30^\circ.(a) Draw a figure.(b) Which ratio will you use?(c) Find the height of the pole. [1+1+2]
7.3 Solutions to Model Set 1
1. (a) n(P\cap S). (b) A Venn diagram with two overlapping circles P and S inside a rectangle U (80) and 10 outside both. (c) n(P\cup S)=80-10=70. n(P\cap S)=45+40-70= 15. (d) Only P=45-15=30; only S=40-15=25. Ratio =30:25= 6 : 5. Check: 30+15+25+10=80.
2. (a) Discount =10\% of 60000 =6000, so the price is Rs 54,000. (b) VAT =13\% of 54000 = Rs 7,020. (c) Bill =54000+7020= Rs 61,020.
3. (a) A=P\left(1+\frac{R}{100}\right)^T. (b) A=25000(1.08)^3=25000\times1.259712= Rs 31,492.80. (c) CI=31492.80-25000= Rs 6,492.80. (d) SI=\frac{25000\times8\times3}{100}=6000. The compound interest exceeds it by 6492.80-6000= Rs 492.80.
4. (a) \left(1+\frac{R}{100}\right). (b) P=40000\times1.05\times1.08= 45,360. (c) 45360\times0.9= 40,824.
5. (a) A=\sqrt{s(s-a)(s-b)(s-c)} with s=\frac{a+b+c}{2}. (b) s=\frac{9+10+17}{2}=18; A=\sqrt{18\times9\times8\times1}=\sqrt{1296}= 36 cm². (c) \frac12\times10\times h=36\Rightarrow h= 7.2 cm.
6. (a) CSA =2\pi rh. (b) 2\times\frac{22}{7}\times7\times10= 440 cm². (c) \text{TSA}=2\pi r(r+h)=440+2\times\frac{22}{7}\times49=440+308= 748 cm². (d) V=\frac{22}{7}\times49\times10= 1540 cm³. (e) 1540\text{ cm}^3=1.54 litres (1000 cm³ = 1 litre).
7. (a) Half the base =8 cm; l=\sqrt{6^2+8^2}= 10 cm. (b) Lateral area =\frac12\times64\times10= 320 cm². (c) \text{TSA}=320+16^2= 576 cm². (d) V=\frac13\times256\times6= 512 cm³.
8. (a) \frac1x=\frac{1}{5+2\sqrt{6}}\times\frac{5-2\sqrt{6}}{5-2\sqrt{6}}=\frac{5-2\sqrt{6}}{25-24}= 5-2\sqrt{6}. (b) x+\frac1x=(5+2\sqrt{6})+(5-2\sqrt{6})= 10. (c) x^2+\frac{1}{x^2}=10^2-2= 98.
9. (a) \frac{(x-4)(x+4)}{(x+4)^2}= \frac{x-4}{x+4}. (b) LCM =(x-3)(x+3)=x^2-9. LHS =\frac{(x+3)-(x-3)-6}{x^2-9}=\frac{x+3-x+3-6}{x^2-9}=\frac{0}{x^2-9}=0= RHS. Proved.
10. (a) Let the width be x m: x(x+5)=84. (b) x^2+5x-84=0\Rightarrow(x+12)(x-7)=0; x=-12 is rejected. So x=7. (c) Width 7 m, length 12 m. (d) Perimeter =2(12+7)= 38 m.
11. (a) Given: cyclic quadrilateral ABCD with centre O. To prove: \angle A+\angle C=180^\circ. Construction: join OB and OD. Proof: The arc BCD subtends \angle BOD (reflex) at the centre and \angle A at the circumference, so \angle A=\frac12\text{ reflex }\angle BOD. The arc BAD subtends \angle BOD (the non-reflex angle) at the centre and \angle C at the circumference, so \angle C=\frac12\angle BOD. Adding, \angle A+\angle C=\frac12(\angle BOD+\text{reflex }\angle BOD)=\frac12\times360^\circ=180^\circ. Proved. (b) 180^\circ-105^\circ= 75°.
12. Steps: (i) Draw AB = 5 cm. (ii) At B make \angle ABC=90^\circ and mark BC = 4 cm. (iii) At C make \angle BCD=100^\circ and mark CD = 4.5 cm. Join DA. (iv) Join the diagonal AC. (v) Through D draw a line parallel to AC (copy equal alternate angles). (vi) Extend BC to meet this line at E. (vii) Join AE; \triangle ABE is the required triangle. Justification: triangles ACD and ACE are on the base AC and between the parallels AC and DE, so they are equal in area; adding \triangle ABC to each gives quadrilateral ABCD =\triangle ABE.
13. (a) Given: parallelogram ABCD, E on DC. To prove: \triangle ABE=\frac12\text{ parallelogram ABCD}. Construction: join AC. Proof: AC bisects the area of the parallelogram, so \triangle ABC=\frac12 ABCD. Triangles ABC and ABE are on the same base AB and between the same parallels AB and DC, so they are equal in area. Hence \triangle ABE=\frac12 ABCD. (b) \frac12\times48= 24 cm².
14. Cumulative frequencies: 2, 6, 12, 17, 20; N=20. Q_1=\left(\frac{N+1}{4}\right)^{\text{th}} item =5.25th item. The cumulative frequency just above 5.25 is 6, so Q_1= 20. Q_3=3\left(\frac{N+1}{4}\right)^{\text{th}} item =15.75th item; the cumulative frequency just above is 17, so Q_3= 40. QD=\frac{40-20}{2}= 10. Coefficient of QD=\frac{40-20}{40+20}=\frac{20}{60}= \frac13 \approx0.33.
15. (a) The first branch splits R \left(\frac58\right) and B \left(\frac38\right); after R, the second splits R \left(\frac47\right) and B \left(\frac37\right); after B, it splits R \left(\frac57\right) and B \left(\frac27\right). (b) P(RR)=\frac58\times\frac47=\frac{20}{56}=\frac{5}{14}; P(BB)=\frac38\times\frac27=\frac{6}{56}=\frac{3}{28}. Same colour =\frac{10}{28}+\frac{3}{28}= \frac{13}{28}. (c) P(\text{at least one blue})=1-P(RR)=1-\frac{5}{14}= \frac{9}{14}.
16. (a) A right triangle with the pole as the vertical side, the ground distance as the base and 30^\circ at the point on the ground. (b) \tan30^\circ=\frac{\text{height}}{\text{base}}. (c) \tan30^\circ=\frac{h}{20\sqrt{3}}\Rightarrow\frac{1}{\sqrt{3}}=\frac{h}{20\sqrt{3}}\Rightarrow h= 20 m.
7.4 Model Question Set 2
Time: 3 hours. Full marks: 75. Attempt all questions. Use \pi=\frac{22}{7} where needed.
- In a hostel of 60 students, 35 like football, 28 like volleyball and 8 like neither.(a) Write in set notation the students who like football but not volleyball.(b) Show the information in a Venn diagram.(c) How many students like both games?(d) Find the ratio of those who like only football to those who like only volleyball. [1+1+3+1]
- The factory price of a refrigerator is Rs 45,000. Excise duty of 10% is charged on it, and VAT of 13% is then charged on the price including excise duty.(a) Find the excise duty.(b) Find the price including excise duty.(c) Find the VAT.(d) Find the price the customer pays. [1+1+1+1]
- Rs 40,000 is invested for 2 years at 10% per annum, compounded half-yearly.(a) State the rate per period and the number of periods.(b) Find the amount.(c) Find the compound interest.(d) By how much does it exceed the compound interest if the interest were compounded yearly? [1+2+1+1]
- A machine bought for Rs 5,00,000 depreciates by 10% every year.(a) Write the formula for the value after T years.(b) Find its value after 2 years.(c) Find the total loss in value over the 2 years. [1+2+1]
- A circular garden of radius 14 m has a path 7 m wide all around it outside.(a) Write the outer radius.(b) Find the area of the path.(c) Find the cost of paving it at Rs 60 per m². [1+2+1]
- A solid iron sphere of radius 10.5 cm is melted and recast into a solid cylinder of radius 7 cm.(a) Find the volume of the sphere.(b) Equate volumes.(c) Find the height of the cylinder. [2+1+2]
- A tent is in the shape of a square pyramid with base side 6 m and slant height 5 m.(a) Find its height.(b) Find the area of canvas needed.(c) Find the cost of the canvas at Rs 150 per m².(d) Find the volume of air inside. [1+1+1+1]
- Consider the expressions x^3-x and x^2+x-2.(a) Find the HCF.(b) Find the LCM.(c) Verify that HCF \times LCM = product of the expressions. [2+2+1]
- (a) Simplify 8^{\frac23}\times\left(\frac14\right)^{-\frac12}.(b) Solve 2^{x+1}+2^{x+2}=24. [2+3]
- Three pens and two notebooks cost Rs 130, while two pens and three notebooks cost Rs 120.(a) Form two equations.(b) Solve them.(c) Find the cost of one pen and one notebook together. [2+2+1]
- (a) Prove that angles in the same segment of a circle are equal.(b) If \angle ACB=35^\circ in the same segment as \angle ADB, find \angle ADB. [4+1]
- Construct a triangle PQR with PQ = 6 cm, QR = 5 cm and \angle Q=60^\circ. Then construct a rectangle equal in area to it. Write the steps. [4]
- In \triangle ABC, P is the midpoint of BC.(a) Prove that the area of \triangle ABP equals the area of \triangle ACP.(b) If the area of \triangle ABC is 54 cm², find the area of \triangle ABP. [3+1]
- The scores of a batter in six matches are 3, 4, 6, 7, 10 and 18.(a) Find the mean.(b) Find the standard deviation.(c) Find the coefficient of variation. [1+3+2]
- A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that it is(a) a heart,(b) a king,(c) a king or a heart,(d) neither a king nor a heart. [1+1+2+1]
- If \tan\theta=\frac34 and \theta is acute, find(a) \sin\theta,(b) \cos\theta,(c) the value of (\sin\theta+\cos\theta)^2. [1+1+2]
7.5 Solutions to Model Set 2
1. (a) F-V (or n(F\cap V')). (b) Two overlapping circles F and V in a rectangle of 60, with 8 outside. (c) n(F\cup V)=60-8=52; both =35+28-52= 11. (d) Only football =35-11=24; only volleyball =28-11=17. Ratio = 24 : 17.
2. (a) 10\% of 45000 = Rs 4,500. (b) 45000+4500= Rs 49,500. (c) VAT =13\% of 49500 = Rs 6,435. (d) 49500+6435= Rs 55,935.
3. (a) Rate per period =\frac{10}{2}=5\%; periods =2\times2=4. (b) A=40000(1.05)^4=40000\times1.21550625= Rs 48,620.25. (c) CI=48620.25-40000= Rs 8,620.25. (d) Yearly: A=40000(1.1)^2=48400, CI=8400. The half-yearly interest exceeds it by 8620.25-8400= Rs 220.25.
4. (a) V=V_0\left(1-\frac{R}{100}\right)^T. (b) V=500000(0.9)^2=500000\times0.81= Rs 4,05,000. (c) Loss =500000-405000= Rs 95,000.
5. (a) R=14+7=21 m. (b) Area of path =\pi(R^2-r^2)=\frac{22}{7}(441-196)=\frac{22}{7}\times245= 770 m². (c) Cost =770\times60= Rs 46,200.
6. (a) V_s=\frac43\times\frac{22}{7}\times(10.5)^3= 4851 cm³. (b) Volume of cylinder = volume of sphere: \pi r^2h=4851. (c) \frac{22}{7}\times49\times h=4851\Rightarrow154h=4851\Rightarrow h= 31.5 cm.
7. (a) Half the base =3 m; h=\sqrt{5^2-3^2}= 4 m. (b) Canvas = lateral area =\frac12\times(4\times6)\times5= 60 m². (c) Cost =60\times150= Rs 9,000. (d) V=\frac13\times36\times4= 48 m³.
8. x^3-x=x(x-1)(x+1) and x^2+x-2=(x+2)(x-1). (a) HCF =x-1. (b) LCM =x(x-1)(x+1)(x+2). (c) HCF \times LCM =(x-1)\cdot x(x-1)(x+1)(x+2); the product of the expressions is x(x-1)(x+1)\cdot(x+2)(x-1). The two are equal. Verified.
9. (a) 8^{2/3}=(2^3)^{2/3}=2^2=4; \left(\frac14\right)^{-1/2}=4^{1/2}=2. Product =4\times2= 8. (b) 2^x\cdot2+2^x\cdot4=24\Rightarrow2^x(2+4)=24\Rightarrow2^x=4=2^2. So x=2.
10. (a) Let a pen cost Rs p and a notebook Rs n: 3p+2n=130 and 2p+3n=120. (b) Multiply the first by 3 and the second by 2: 9p+6n=390 and 4p+6n=240. Subtract: 5p=150\Rightarrow p=30. Then n=\frac{130-90}{2}=20. Pen Rs 30, notebook Rs 20. Check: 2(30)+3(20)=120 ✓. (c) 30+20= Rs 50.
11. (a) Given: a circle with centre O; \angle ACB and \angle ADB are in the same segment standing on chord AB. To prove: \angle ACB=\angle ADB. Construction: join OA and OB. Proof: The angle at the centre is twice the angle at the circumference on the same arc, so \angle AOB=2\angle ACB and \angle AOB=2\angle ADB. Hence 2\angle ACB=2\angle ADB, so \angle ACB=\angle ADB. Proved. (b) \angle ADB= 35°.
12. Steps: (i) Draw PQ = 6 cm. (ii) At Q make \angle PQR=60^\circ and mark QR = 5 cm; join PR. (iii) Draw the altitude from R to PQ and bisect it (the height of the rectangle is half the altitude). (iv) On PQ construct a rectangle PQST with height equal to half the altitude (draw perpendiculars at P and Q). Justification: area of triangle =\frac12\times PQ\times\text{altitude}=PQ\times\frac{\text{altitude}}{2}= area of the rectangle.
13. (a) Given: \triangle ABC with P the midpoint of BC. To prove: \triangle ABP=\triangle ACP. Construction: draw AM perpendicular to BC. Proof: \triangle ABP=\frac12\times BP\times AM and \triangle ACP=\frac12\times PC\times AM. Since BP=PC, the two areas are equal. (b) Each is half of 54: 27 cm².
14. (a) Mean =\frac{3+4+6+7+10+18}{6}=\frac{48}{6}= 8. (b) Deviations: -5,-4,-2,-1,2,10; squares: 25,16,4,1,4,100; sum =150. \sigma=\sqrt{\frac{150}{6}}=\sqrt{25}= 5. (c) CV=\frac{5}{8}\times100= 62.5%.
15. (a) P(\text{heart})=\frac{13}{52}=\frac14. (b) P(\text{king})=\frac{4}{52}=\frac{1}{13}. (c) P(\text{king or heart})=\frac14+\frac{1}{13}-\frac{1}{52}=\frac{13+4-1}{52}=\frac{16}{52}= \frac{4}{13}. (d) 1-\frac{4}{13}= \frac{9}{13}.
16. Take opposite =3, adjacent =4, hypotenuse =5. (a) \sin\theta= \frac35. (b) \cos\theta= \frac45. (c) (\sin\theta+\cos\theta)^2=\left(\frac35+\frac45\right)^2=\left(\frac75\right)^2= \frac{49}{25}. (Check: 1+2\sin\theta\cos\theta=1+\frac{24}{25}=\frac{49}{25}.)
7.6 Predicted questions: ranked practice priorities
These are practice priorities, ranked from the CDC grid and the syllabus. They are not forecasts. A form ranks higher when its area carries many marks, when it is a standard form with a clean answer, and when it has a one-mark first part you can always bank.
7.6.1 Tier A — rehearse until automatic
| Rank | Topic and likely form | Marks | Why it is ranked here |
|---|---|---|---|
| 1 | Arithmetic: tax with discount, excise and VAT; currency exchange | 4 | Arithmetic carries 13 marks across three questions, and tax questions have a fixed order of steps. |
| 2 | Arithmetic: compound interest (amount, CI, half-yearly, CI minus SI) | 4 or 5 | The same area; a formula with a clean answer. |
| 3 | Arithmetic: population growth and depreciation | 4 | Third arithmetic question; a one-line formula and changing rates. |
| 4 | Sets: survey with Venn diagram, both, only, ratio | 6 | A fixed single question worth 6 marks, with 1-mark parts to bank. |
| 5 | Algebra: simultaneous equations and quadratic word problems | 5 | Algebra is the heaviest area (15 marks) and equations give the longest parts. |
| 6 | Mensuration: cylinder and sphere, melted and recast, pyramid | 4 or 5 | 13 marks across three questions; the formulae are fixed. |
7.6.2 Tier B — second pass
| Rank | Topic and likely form | Marks | Why it is ranked here |
|---|---|---|---|
| 7 | Statistics: quartile deviation, standard deviation, coefficient of variation | 6 | Statistics and Probability carry 11 marks across two questions. |
| 8 | Probability: two draws, tree diagram, at least one | 5 | Standard tree-diagram steps; a complement shortcut helps. |
| 9 | Algebra: surds — rationalise, x+\frac1x, x^2+\frac{1}{x^2} | 5 | A reliable 2 + 1 + 2 structure. |
| 10 | Geometry: a circle theorem proof with reasons | 4 or 5 | A written proof gains marks for each given step and reason. |
| 11 | Geometry: area theorem or a numerical application | 4 | Short proof plus a numerical. |
| 12 | Mensuration: pyramid with slant height | 4 | A square pyramid has a clean right-triangle step. |
7.6.3 Tier C — worth a quick pass
| Rank | Topic and likely form | Marks | Why it is ranked here |
|---|---|---|---|
| 13 | Construction: triangle equal to a quadrilateral, parallelogram or rectangle equal to a triangle | 4 | Steps are marked, but the area is four marks. |
| 14 | Algebra: HCF and LCM; algebraic fractions; indices | 3–5 | Factorising skills decide the whole answer. |
| 15 | Trigonometry: ratios from one ratio; height and distance | 4 | The smallest area (8 hours, 4 marks), but a sure question. |
| 16 | Plane surfaces: Heron's formula, path around a garden | 4 | Direct substitution with a chance of a unit slip. |
Work Tier A until the method is automatic, then Tier B, and keep a day for Tier C. Practise the whole answer — formula, substitution, working and final statement with a unit — because every question is marked by steps. The first one or two marks of every question are the knowledge and understanding marks; take those first.
Chapter summary
- The paper is 16 questions and 75 marks: Sets 6, Arithmetic 13, Mensuration 13, Algebra 15, Geometry 13, Statistics and Probability 11, Trigonometry 4.
- Plan the three hours: 75 minutes for the first seven questions, 35 for Algebra, 30 for Geometry, 25 for Statistics, Probability and Trigonometry, 15 to check.
- Arithmetic formulae: bill =\text{price}\times(1+\frac{r}{100}); A=P\left(1+\frac{R}{100}\right)^T; half-yearly rate \frac{R}{2} and 2T periods.
- Mensuration: cylinder 2\pi r(r+h) and \pi r^2h; sphere 4\pi r^2 and \frac43\pi r^3; pyramid V=\frac13\times\text{base}\times h with l^2=h^2+\left(\frac b2\right)^2.
- Statistics: QD=\frac{Q_3-Q_1}{2}; \sigma=\sqrt{\frac{\sum(x-\bar x)^2}{n}}; CV=\frac{\sigma}{\bar x}\times100\%. Probability: P(A\cup B)=P(A)+P(B)-P(A\cap B).
- A ranking is an order of practice, never a forecast of what will be set.