Compulsory Mathematics: Unit-wise Subjective Question Bank

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Compulsory Mathematics: Unit-wise Subjective Question Bank

The SEE Compulsory Mathematics paper is 16 questions, each worth about four or five marks and each broken into parts of 1 to 3 marks: 75 marks in all. There is no separate "short" and "long" group, so this chapter gives worked practice by chapter, in the sizes the paper uses: questions of 3 marks and 4 marks, with a complete model solution for each. Every numerical answer was recomputed with a program as well as by hand.

How to use this chapter
  • Attempt first. Allow about 11 minutes for a 4-mark question and 8 for a 3-mark one. Cover the solution while you work.
  • Write in steps. The mark scheme gives a mark for the formula, for the substitution, for the working and for the final answer with its unit. A correct final answer with no working loses marks.
  • Use the stated constants. Take \pi=\frac{22}{7} unless the question says otherwise, and give units (Rs, cm², m³, litres).
About years and sources

Every question is an original practice question written to the CDC pattern. No question is labelled with a year because none is claimed to have appeared in a specific paper.

6.1 Sets

  1. In a survey of 100 people, 60 like tea, 50 like coffee and 15 like neither.(a) Find the number who like both.(b) Draw a Venn diagram.(c) How many like only tea? [1+1+2]
  2. Verify (A\cup B)'=A'\cap B' for U=\{1,2,3,4,5,6,7,8\}, A=\{1,2,3,4\} and B=\{3,4,5,6\}. [3]
Answers

1. (a) n(T\cup C)=100-15=85. n(T\cap C)=n(T)+n(C)-n(T\cup C)=60+50-85=25. So 25 people like both.

(b) The Venn diagram has two overlapping circles inside a rectangle U (100): only tea =60-25=35; both =25; only coffee =50-25=25; neither =15 outside the circles. Check: 35+25+25+15=100.

(c) Only tea =60-25= 35 people.

2. A\cup B=\{1,2,3,4,5,6\}, so (A\cup B)'=\{7,8\}. A'=\{5,6,7,8\} and B'=\{1,2,7,8\}, so A'\cap B'=\{7,8\}. Both sides equal \{7,8\}, so (A\cup B)'=A'\cap B'. Verified.

6.2 Tax and Money Exchange

  1. A television has a marked price of Rs 40,000. The shopkeeper gives a 10% discount and then charges 13% VAT. Find the bill amount. [3]
  2. The factory price of a motorbike is Rs 80,000. Excise duty of 15% is charged on it, and VAT of 13% is then charged on the total. Find the price a customer pays. [4]
  3. Mr Karki changes USD 1000 into Rs at Rs 133.50 per dollar and then buys Indian rupees at Rs 160 per INR 100. How many INR does he get? [4]
  4. Income tax rates are: 1% on the first Rs 5,00,000, 10% on the next Rs 2,00,000 and 20% on the next Rs 3,00,000. Find the tax on a taxable income of Rs 8,00,000. [4]
Answers

1. Discount =10\% of 40000=4000; selling price =40000-4000=36000. VAT =13\% of 36000=4680. Bill =36000+4680= Rs 40,680.

2. Excise duty =15\% of 80000=12000. Price with excise =92000. VAT =13\% of 92000=11960. Price paid =92000+11960= Rs 1,03,960.

3. Rs from dollars =1000\times133.50=133500. Rs 160 buys INR 100, so Rs 1 buys INR \frac{100}{160} and Rs 133500 buys INR 133500\times\frac{100}{160}=83437.50. He gets INR 83,437.50.

4. First slab: 1\%\times500000=5000. Second slab: 10\%\times200000=20000. The remaining 800000-700000=100000 is in the third slab: 20\%\times100000=20000. Total tax =5000+20000+20000= Rs 45,000. (Slabs and rates are as stated in the question; actual rates change each year.)

6.3 Compound Interest

  1. Find the amount and compound interest on Rs 50,000 for 3 years at 10% per annum compounded yearly. [4]
  2. Find the compound interest on Rs 20,000 for 1\frac12 years at 8% per annum compounded half-yearly. [4]
  3. Show that the difference between the compound interest and the simple interest on Rs 5000 for 2 years at 12% per annum is Rs 72. [3]
  4. The difference between the compound interest and the simple interest on a sum for 2 years at 10% per annum is Rs 90. Find the sum. [3]
Answers

1. A=P\left(1+\frac{R}{100}\right)^T=50000(1.1)^3=50000\times1.331=66550. CI=66550-50000= Rs 16,550; amount Rs 66,550.

2. Half-yearly: rate per period =\frac82=4\%, periods =1.5\times2=3. A=20000(1.04)^3=20000\times1.124864=22497.28. CI=22497.28-20000= Rs 2,497.28.

3. CI=5000(1.12)^2-5000=5000\times1.2544-5000=6272-5000=1272. SI=\frac{5000\times12\times2}{100}=1200. Difference =1272-1200=72. Shown.

4. For 2 years, CI-SI=P\left(\frac{R}{100}\right)^2. So 90=P\times(0.1)^2=0.01P, hence P=\frac{90}{0.01}= Rs 9,000.

6.4 Population Growth and Depreciation

  1. The population of a town is 60,000 and it grows at 3% per annum. Find its population after 2 years. [3]
  2. The population of a village was 20,000 two years ago and is 22,050 now. If it grew at the same rate every year, find the rate. [4]
  3. A car costs Rs 12,00,000. Its value depreciates by 10% in the first year and 8% in the second year. Find its value after two years. [3]
  4. A city has 80,000 people. The birth rate is 3.5%, the death rate 1.5% and 0.5% migrate out each year. Find the population after one year. [4]
Answers

1. P=60000(1.03)^2=60000\times1.0609= 63,654.

2. Let the rate be R\%. 22050=20000\left(1+\frac{R}{100}\right)^2\Rightarrow\left(1+\frac{R}{100}\right)^2=\frac{22050}{20000}=1.1025. So 1+\frac{R}{100}=1.05 (since 1.05^2=1.1025), giving R= 5%.

3. Value =1200000\times\left(1-\frac{10}{100}\right)\left(1-\frac{8}{100}\right)=1200000\times0.9\times0.92= Rs 9,93,600.

4. Net rate =3.5-1.5-0.5=1.5\%. Population after 1 year =80000\times1.015= 81,200. (Increase =1200.)

6.5 Plane Surfaces

  1. Find the area of a triangle with sides 7 cm, 15 cm and 20 cm using Heron's formula. [3]
  2. A rectangular garden 30 m by 20 m has a path 2 m wide inside it along all four sides. Find the area of the path and the cost of paving at Rs 50 per m². [4]
  3. A figure is a rectangle 20 m by 14 m with a semicircle drawn on one 14 m side. Find its area and perimeter. [4]
Answers

1. s=\frac{7+15+20}{2}=21. A=\sqrt{21(21-7)(21-15)(21-20)}=\sqrt{21\times14\times6\times1}=\sqrt{1764}= 42 cm².

2. Inner rectangle: (30-4)\times(20-4)=26\times16=416 m². Outer: 30\times20=600 m². Path area =600-416= 184 m². Cost =184\times50= Rs 9,200.

3. Rectangle area =20\times14=280 m². Semicircle radius 7 m: area =\frac12\pi r^2=\frac12\times\frac{22}{7}\times49=77 m². Total area =280+77= 357 m². Perimeter: three sides of the rectangle 20+14+20=54 m plus the arc \pi r=\frac{22}{7}\times7=22 m, so 76 m.

6.6 Cylinder and Sphere

  1. A closed cylindrical tin has radius 7 cm and height 15 cm. Find its total surface area and volume. [4]
  2. A metal sphere of radius 21 cm is melted and recast into small spheres of radius 3.5 cm. How many small spheres are made? [3]
  3. A cylindrical water tank has diameter 2.8 m and depth 5 m. Find its capacity in litres. [3]
  4. A hollow iron pipe is 7 m long. Its external radius is 5 cm and internal radius 4 cm. Find the volume of iron used. [4]
Answers

1. \text{TSA}=2\pi r(r+h)=2\times\frac{22}{7}\times7\times(7+15)=44\times22= 968 cm². V=\pi r^2h=\frac{22}{7}\times49\times15= 2310 cm³.

2. Number =\frac{\text{volume of big sphere}}{\text{volume of small sphere}}=\frac{\frac43\pi(21)^3}{\frac43\pi(3.5)^3}=\left(\frac{21}{3.5}\right)^3=6^3= 216.

3. r=1.4 m, h=5 m. V=\frac{22}{7}\times(1.4)^2\times5=\frac{22}{7}\times1.96\times5=30.8 m³. Since 1 m³ =1000 litres, capacity = 30,800 litres.

4. Length =7 m =700 cm. Volume of iron =\pi(R^2-r^2)l=\frac{22}{7}(25-16)\times700=\frac{22}{7}\times9\times700= 19,800 cm³.

6.7 Prism and Pyramid

  1. The base of a right prism is a right-angled triangle with sides 6 cm, 8 cm and 10 cm, and the length of the prism is 15 cm. Find the total surface area and the volume. [4]
  2. A square pyramid has base side 10 cm and slant height 13 cm. Find its height, total surface area and volume. [4]
  3. A room is 6 m long, 5 m wide and 3.5 m high. Find the area of its four walls and the cost of painting them at Rs 40 per m². [3]
Answers

1. Base area =\frac12\times6\times8=24 cm². Perimeter of base =6+8+10=24 cm. Lateral area =24\times15=360 cm². \text{TSA}=360+2\times24= 408 cm². V=24\times15= 360 cm³.

2. Half the base =5 cm. h=\sqrt{l^2-5^2}=\sqrt{169-25}=\sqrt{144}= 12 cm. Lateral area =\frac12\times\text{perimeter}\times l=\frac12\times40\times13=260 cm². \text{TSA}=260+10^2= 360 cm². V=\frac13\times100\times12= 400 cm³.

3. Area of four walls =2(l+b)h=2(6+5)\times3.5=77 m². Cost =77\times40= Rs 3,080.

6.8 Highest Common Factor and Lowest Common Multiple

  1. Find the HCF and LCM of x^2-4x+3 and x^2-9. [3]
  2. The HCF of two expressions is x+1 and their LCM is (x+1)^2(x-1). If one expression is x^2+2x+1, find the other. [3]
  3. Find the HCF of x^2-1, x^2+2x+1 and x^2+x. [3]
Answers

1. x^2-4x+3=(x-1)(x-3) and x^2-9=(x-3)(x+3). HCF =x-3. LCM =(x-1)(x-3)(x+3).

2. \text{HCF}\times\text{LCM}=A\times B. Here A=x^2+2x+1=(x+1)^2. B=\frac{(x+1)\cdot(x+1)^2(x-1)}{(x+1)^2}=(x+1)(x-1)= x^2-1.

3. x^2-1=(x-1)(x+1); x^2+2x+1=(x+1)^2; x^2+x=x(x+1). The common factor is x+1, so HCF =x+1.

6.9 Radical and Surd

  1. Rationalise the denominator of \frac{3}{\sqrt{7}-\sqrt{5}}. [3]
  2. If x=3+2\sqrt{2}, find x+\frac1x, x^2+\frac{1}{x^2} and x^3+\frac{1}{x^3}. [4]
  3. Find the values of a and b if \frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}. [3]
Answers

1. Multiply by the conjugate: \frac{3}{\sqrt{7}-\sqrt{5}}\times\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{3(\sqrt{7}+\sqrt{5})}{7-5}= \frac{3(\sqrt{7}+\sqrt{5})}{2}.

2. \frac1x=\frac{1}{3+2\sqrt{2}}\times\frac{3-2\sqrt{2}}{3-2\sqrt{2}}=\frac{3-2\sqrt{2}}{9-8}=3-2\sqrt{2}. So x+\frac1x=(3+2\sqrt{2})+(3-2\sqrt{2})= 6. x^2+\frac{1}{x^2}=\left(x+\frac1x\right)^2-2=36-2= 34. x^3+\frac{1}{x^3}=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right)=216-18= 198.

3. \frac{3+\sqrt{2}}{3-\sqrt{2}}\times\frac{3+\sqrt{2}}{3+\sqrt{2}}=\frac{(3+\sqrt{2})^2}{9-2}=\frac{9+6\sqrt{2}+2}{7}=\frac{11}{7}+\frac{6}{7}\sqrt{2}. Hence a=\frac{11}{7} and b=\frac{6}{7}.

6.10 Indices

  1. Prove that \left(\frac{x^a}{x^b}\right)^{a+b}\times\left(\frac{x^b}{x^c}\right)^{b+c}\times\left(\frac{x^c}{x^a}\right)^{c+a}=1. [4]
  2. Solve 3^{x+2}-3^x=72. [3]
  3. If 2^x=3^y=6^{-z}, show that \frac1x+\frac1y+\frac1z=0. [4]
Answers

1. The first factor is (x^{a-b})^{a+b}=x^{(a-b)(a+b)}=x^{a^2-b^2}. Similarly the others are x^{b^2-c^2} and x^{c^2-a^2}. The product is x^{(a^2-b^2)+(b^2-c^2)+(c^2-a^2)}=x^{0}=1. Proved.

2. 3^x\cdot3^2-3^x=72\Rightarrow3^x(9-1)=72\Rightarrow3^x=9=3^2. Equating powers, x=2.

3. Let 2^x=3^y=6^{-z}=k. Then 2=k^{1/x}, 3=k^{1/y} and 6=k^{-1/z}. Since 2\times3=6, k^{1/x}\cdot k^{1/y}=k^{-1/z}, so k^{\frac1x+\frac1y}=k^{-\frac1z}. Equating powers, \frac1x+\frac1y=-\frac1z, hence \frac1x+\frac1y+\frac1z=0. Shown.

6.11 Algebraic Fraction

  1. Simplify \frac{a^2-b^2}{a^2+2ab+b^2}\times\frac{a+b}{a-b}. [3]
  2. Simplify \frac{1}{x-2}-\frac{1}{x+2}. [3]
  3. Simplify \frac{x^2-5x+6}{x^2-4}\div\frac{x^2-9}{x^2+5x+6}. [4]
Answers

1. \frac{(a-b)(a+b)}{(a+b)^2}\times\frac{a+b}{a-b}=\frac{(a-b)(a+b)(a+b)}{(a+b)^2(a-b)}= 1.

2. LCM =(x-2)(x+2). \frac{(x+2)-(x-2)}{(x-2)(x+2)}=\frac{4}{x^2-4}.

3. Factorise: \frac{(x-2)(x-3)}{(x-2)(x+2)}\div\frac{(x-3)(x+3)}{(x+2)(x+3)}. Simplify each fraction: \frac{x-3}{x+2}\div\frac{x-3}{x+2}= 1.

6.12 Equations

  1. Solve 2x+3y=12 and 3x-y=7. [3]
  2. The sum of the ages of Anil and Bina is 40 years. Five years ago, Anil was twice as old as Bina. Find their present ages. [4]
  3. The perimeter of a rectangle is 28 m and its area is 48 m². Find its length and breadth. [4]
  4. The product of two consecutive positive integers is 132. Find them. [3]
Answers

1. From the second equation, y=3x-7. Substitute: 2x+3(3x-7)=12\Rightarrow11x-21=12\Rightarrow x=3. Then y=9-7=2. x=3, y=2. Check: 2(3)+3(2)=12 ✓.

2. Let Anil be a and Bina be b years. a+b=40 and a-5=2(b-5), i.e. a=2b-5. Then 2b-5+b=40\Rightarrow3b=45\Rightarrow b=15, a=25. Anil 25 years, Bina 15 years. Check: five years ago 20 and 10 ✓.

3. Let the length be l; l+b=\frac{28}{2}=14, so b=14-l. l(14-l)=48\Rightarrow l^2-14l+48=0\Rightarrow(l-6)(l-8)=0. Length =8 m and breadth =6 m (the other root gives the same rectangle). Check: 8\times6=48 ✓, 2(8+6)=28 ✓.

4. Let the integers be x and x+1: x(x+1)=132\Rightarrow x^2+x-132=0\Rightarrow(x+12)(x-11)=0. x=-12 is rejected (positive integers), so x=11: the integers are 11 and 12.

6.13 Area of Triangles and Quadrilaterals

  1. Prove that parallelograms on the same base and between the same parallels are equal in area. [4]
  2. A parallelogram has base 12 cm and height 5 cm. A triangle stands on the same base and between the same parallels. Find the area of each. [3]
  3. Prove that a triangle and a parallelogram on the same base and between the same parallels satisfy: area of triangle =\frac12 area of parallelogram. [4]
Answers

1. Given: parallelograms ABCD and ABEF on the same base AB and between the parallels AB and DF. To prove: area ABCD = area ABEF. Proof: In \triangle ADF and \triangle BCE: AD=BC (opposite sides of parallelogram ABCD), AF=BE (opposite sides of ABEF), and \angle DAF=\angle CBE (corresponding angles, as AD\parallel BC and AF\parallel BE). So \triangle ADF\cong\triangle BCE (SAS) and their areas are equal. Let the points D, F, C, E lie on the line in that order, so that ABCF is common to both. Area ABCD = area ABCF + area \triangle ADF and area ABEF = area ABCF + area \triangle BCE. Since \triangle ADF=\triangle BCE, area ABCD = area ABEF. Hence proved.

2. Area of the parallelogram =\text{base}\times\text{height}=12\times5= 60 cm². The triangle on the same base and between the same parallels has half that area: \frac12\times12\times5= 30 cm².

3. Given: triangle ABE and parallelogram ABCD on the same base AB and between the parallels AB and DC (E lies on DC). To prove: \triangle ABE=\frac{1}{2}\text{ parallelogram ABCD}. Construction: join AC. Proof: The diagonal AC divides the parallelogram into two congruent triangles, so \triangle ABC=\frac{1}{2}\text{ parallelogram ABCD}. Triangles ABC and ABE are on the same base AB and between the same parallels, so \triangle ABC=\triangle ABE. Hence \triangle ABE=\frac{1}{2}\text{ parallelogram ABCD}. Proved.

6.14 Construction

  1. Describe the steps to construct a triangle equal in area to a quadrilateral ABCD, and justify the construction. [4]
  2. Describe how to construct a parallelogram equal in area to a given triangle ABC with one angle equal to a given angle \theta. [4]
Answers

1. Steps: (i) Construct the quadrilateral ABCD with the given measurements. (ii) Join the diagonal AC. (iii) Through D draw a line parallel to AC (copy equal alternate angles with the compass). (iv) Extend the side BC to meet this parallel at E. (v) Join AE. Then triangle ABE is the required triangle. Proof: Triangles ACD and ACE are on the same base AC and between the same parallels AC and DE, so they are equal in area. Adding triangle ABC to each, quadrilateral ABCD =\triangle ABE. Leave all arcs visible.

2. Steps: (i) Construct triangle ABC. (ii) Bisect BC at D with a perpendicular bisector (arcs visible). (iii) At D construct \angle CDE=\theta (a given angle) on the same side as A. (iv) Through A draw a line parallel to BC to meet DE at E. (v) Through C draw a line parallel to DE to meet AE at F. Then DCFE is the required parallelogram. Proof: Join AD; triangle ABC is twice triangle ADC, since D bisects BC. Parallelogram DCFE and triangle ADC are on the same base DC and between the same parallels, so the parallelogram is twice triangle ADC. So the parallelogram equals triangle ABC in area.

6.15 Circle

  1. Prove that the angle at the centre of a circle is twice the angle at the circumference standing on the same arc. [4]
  2. ABCD is a cyclic quadrilateral with \angle A=(2x+10)^\circ and \angle C=(3x-20)^\circ. Find x and the angles A and C. [3]
  3. A tangent PT is drawn to a circle with centre O and radius 8 cm from a point P with OP = 17 cm. Find PT. [3]
Answers

1. Given: a circle with centre O; arc AB subtends \angle AOB at the centre and \angle ACB at the circumference. To prove: \angle AOB=2\angle ACB. Construction: join CO and produce it to D. Proof: OA=OC (radii), so \angle OAC=\angle OCA. The exterior angle \angle AOD=\angle OAC+\angle OCA=2\angle OCA. Similarly \angle BOD=2\angle OCB. Adding, \angle AOB=\angle AOD+\angle BOD=2(\angle OCA+\angle OCB)=2\angle ACB. Proved.

2. Opposite angles of a cyclic quadrilateral are supplementary: (2x+10)+(3x-20)=180\Rightarrow5x-10=180\Rightarrow x=38. \angle A=2(38)+10= 86° and \angle C=3(38)-20= 94°. Check: 86+94=180 ✓.

3. The tangent is perpendicular to the radius at the point of contact, so \angle OTP=90^\circ. By Pythagoras, PT^2=OP^2-OT^2=17^2-8^2=289-64=225, so PT= 15 cm.

6.16 Trigonometry

  1. If \sin\theta=\frac{5}{13} and \theta is acute, find \cos\theta, \tan\theta and \sec\theta. [3]
  2. Prove that (\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1. [3]
  3. The angle of elevation of the top of a tower from a point 30 m from its foot is 60^\circ. Find the height of the tower. (\sqrt{3}=1.732) [4]
  4. Evaluate \tan45^\circ+\sin30^\circ\cos60^\circ. [3]
Answers

1. Take opposite =5, hypotenuse =13. Adjacent =\sqrt{13^2-5^2}=12. \cos\theta=\frac{12}{13}, \tan\theta=\frac{5}{12}, \sec\theta=\frac{13}{12}.

2. LHS =\sec^2\theta-\tan^2\theta=(1+\tan^2\theta)-\tan^2\theta=1= RHS. Proved.

3. Let the tower be AB with height h and the point be C with BC=30 m. In right triangle ABC: \tan60^\circ=\frac{h}{30}\Rightarrow h=30\sqrt{3}=30\times1.732= 51.96 m. (About 52 m.)

4. \tan45^\circ=1, \sin30^\circ=\frac12, \cos60^\circ=\frac12. Value =1+\frac12\times\frac12=1+\frac14= \frac54.

6.17 Statistics

  1. Find the quartile deviation and its coefficient for the data 5, 8, 10, 12, 15, 18, 20. [3]
  2. Find the mean and standard deviation of 10, 12, 14, 16, 18, 20, 22. [4]
  3. Team A has mean score 50 with standard deviation 5. Team B has mean 40 with standard deviation 6. Which team is more consistent? [3]
Answers

1. The data are in order, n=7. Q_1=\left(\frac{n+1}{4}\right)^{\text{th}} item =2nd item =8. Q_3=3\left(\frac{n+1}{4}\right)^{\text{th}}=6th item =18. QD=\frac{Q_3-Q_1}{2}=\frac{18-8}{2}= 5. Coefficient =\frac{Q_3-Q_1}{Q_3+Q_1}=\frac{10}{26}=\frac{5}{13}\approx0.385.

2. Mean \bar x=\frac{112}{7}=16. Deviations x-\bar x: -6,-4,-2,0,2,4,6; squares: 36,16,4,0,4,16,36, sum =112. \sigma=\sqrt{\frac{\sum(x-\bar x)^2}{n}}=\sqrt{\frac{112}{7}}=\sqrt{16}= 4.

3. CV_A=\frac{5}{50}\times100=10\%; CV_B=\frac{6}{40}\times100=15\%. The smaller coefficient of variation means more consistent, so Team A is more consistent.

6.18 Probability

  1. A bag contains 4 red, 3 blue and 2 green balls. Two balls are drawn one after another without replacement. Find the probability that(a) both are red,(b) one is red and one is blue. [2+2]
  2. Two fair dice are thrown. Find the probability that the sum is 7. [3]
  3. If P(A)=\frac35, P(B)=\frac12 and P(A\cup B)=\frac45, find P(A\cap B). [3]
Answers

1. Total balls =9. (a) P(RR)=\frac49\times\frac38=\frac{12}{72}= \frac16. (b) P(RB)+P(BR)=\frac49\times\frac38+\frac39\times\frac48=\frac{12}{72}+\frac{12}{72}=\frac{24}{72}= \frac13.

2. The sample space has 6\times6=36 outcomes. Sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1), that is 6 outcomes. P=\frac{6}{36}= \frac16.

3. P(A\cup B)=P(A)+P(B)-P(A\cap B), so P(A\cap B)=\frac35+\frac12-\frac45=\frac{6+5-8}{10}= \frac{3}{10}.

Chapter summary

  • The 16 questions of the paper break into parts of 1 to 3 marks; this chapter practises the 3- and 4-mark forms chapter by chapter.
  • Show the formula, the substitution, the working and the final answer with its unit; each earns a mark.
  • Proofs (parallelograms, circle theorems) are written as Given, To prove, Construction, Proof, with a reason on each line.
  • Constructions are written as numbered steps with a one-line justification and all arcs visible.
  • Fast checks: cross-check a fraction answer by substituting back; for money questions state the unit (Rs, INR, USD).