Compulsory Mathematics: Unit-wise Objective Question Bank

The SEE Compulsory Mathematics paper has no multiple-choice group. Its 16 questions are built from items worth 1 mark upward, and the first items of most questions test knowledge: a formula, a definition, a one-line value. This chapter drills those objective items chapter by chapter, in multiple-choice form so that you can check each one in seconds. Each section opens with a facts line, gives six items, and ends with a key that has a one-line reason for every answer.

5.1 Sets

Facts. n(A\cup B)=n(A)+n(B)-n(A\cap B); (A\cup B)'=A'\cap B'; (A\cap B)'=A'\cup B'; n(U)=n(A\cup B)+n(A\cup B)'.

1. If n(A)=20, n(B)=15 and n(A\cap B)=5, then n(A\cup B) is [1]a40b35c30d25
2. If n(U)=50 and n(A\cup B)=40, then n(A\cup B)' is [1]a10b90c40d20
3. (A\cup B)' equals [1]aA'\cup B'bA'\cap B'cA\cap BdA\cup B'
4. If n(A)=25 and n(A\cap B)=8, the number of elements only in A is [1]a33b25c8d17
5. In a class of 40, every student likes football or cricket; 25 like football and 20 like cricket. The number liking both is [1]a5b10c15d45
6. If A\subset B, then A\cap B equals [1]aBb\emptysetcAdA\cup B

Answers

1. (c) — 20+15-5=30.

2. (a) — 50-40=10.

3. (b) — De Morgan's law.

4. (d) — 25-8=17.

5. (a) — 25+20-40=5.

6. (c) — every element of A is already in B.

5.2 Tax and Money Exchange

Facts. VAT at 13%: bill = price \times1.13. With excise duty e\% and VAT: price \times(1+\frac{e}{100})\times1.13. Currency: amount in one currency = amount in the other \times rate.

1. VAT at 13% on a price of Rs 2000 is [1]aRs 13bRs 26cRs 130dRs 260
2. A bill is Rs 11,300 including 13% VAT. The price without VAT is [1]aRs 9,830bRs 10,000cRs 11,287dRs 12,769
3. A shopkeeper sells goods worth Rs 8000 plus 13% VAT. The bill is [1]aRs 8,013bRs 8,104cRs 9,040dRs 8,130
4. If USD 1 = Rs 130, USD 200 is [1]aRs 650bRs 2,600cRs 13,000dRs 26,000
5. If Rs 160 = INR 100, then INR 5000 is [1]aRs 3,125bRs 8,000cRs 80,000dRs 5,160
6. A factory price of Rs 1000 carries 10% excise duty and then 13% VAT. The final price is [1]aRs 1,230bRs 1,243cRs 1,243.30dRs 1,100

Answers

1. (d) — 0.13\times2000=260.

2. (b) — 11300\div1.13=10000.

3. (c) — 8000\times1.13=9040.

4. (d) — 200\times130=26000.

5. (b) — 5000\times\frac{160}{100}=8000.

6. (b) — 1000\times1.10\times1.13=1243; VAT applies to price plus excise.

5.3 Compound Interest

Facts. A=P\left(1+\frac{R}{100}\right)^{T}; CI=A-P; half-yearly: A=P\left(1+\frac{R}{200}\right)^{2T}. For 2 years, CI-SI=P\left(\frac{R}{100}\right)^{2}.

1. The amount on Rs 10,000 for 2 years at 10% compound interest per annum is [1]aRs 12,000bRs 12,100cRs 12,200dRs 2,100
2. The compound interest in item 1 is [1]aRs 2,000bRs 2,100cRs 12,100dRs 1,000
3. Rs 4000 at 10% p.a. compounded half-yearly for 1 year amounts to [1]aRs 4,400bRs 4,410cRs 4,420dRs 4,200
4. For Rs 8000 at 5% for 2 years, the difference between CI and SI is [1]aRs 8bRs 20cRs 40dRs 800
5. A principal becomes Rs 1331 in 3 years at 10% compound interest. The principal is [1]aRs 1,000bRs 1,100cRs 900dRs 1,210
6. In half-yearly compounding the rate used per period is [1]aRb2Rc\frac{R}{2}d\frac{R}{4}

Answers

1. (b) — 10000(1.1)^2=12100.

2. (b) — 12100-10000=2100.

3. (b) — rate per half-year 5%, two periods: 4000(1.05)^2=4410.

4. (b) — 8000\times(0.05)^2=20.

5. (a) — 1331\div(1.1)^3=1331\div1.331=1000.

6. (c) — the annual rate is halved and the periods doubled.

5.4 Population Growth and Depreciation

Facts. Growth: P_T=P_0\left(1+\frac{R}{100}\right)^{T}. Depreciation: V_T=V_0\left(1-\frac{R}{100}\right)^{T}. Net rate of growth = birth rate - death rate \pm net migration rate.

1. A population of 20,000 grows at 5% a year. After 2 years it is [1]a22,000b22,050c22,500d21,000
2. A population of 5000 decreases by 10% a year. After 2 years it is [1]a4,000b4,050c4,500d3,500
3. A town has 24,200 people now; the growth rate is 10% a year. Two years ago it had [1]a20,000b19,360c21,000d22,000
4. A machine of value Rs 80,000 depreciates 10% a year. After 2 years its value is [1]aRs 64,000bRs 64,800cRs 72,000dRs 60,000
5. Birth rate 3%, death rate 1% and out-migration 0.5% give a net growth rate of [1]a1.5%b2%c2.5%d4.5%
6. A quantity of 1000 grows 10% in year 1 and 20% in year 2. At the end it is [1]a1,300b1,320c1,330d1,440

Answers

1. (b) — 20000(1.05)^2=22050.

2. (b) — 5000(0.9)^2=4050.

3. (a) — 24200\div(1.1)^2=20000.

4. (b) — 80000(0.9)^2=64800.

5. (a) — 3-1-0.5=1.5.

6. (b) — 1000\times1.1\times1.2=1320.

5.5 Plane Surfaces

Facts. Heron: A=\sqrt{s(s-a)(s-b)(s-c)} with s=\frac{a+b+c}{2}; circle A=\pi r^2, C=2\pi r; ring A=\pi(R^2-r^2); sector A=\frac{\theta}{360}\pi r^2.

1. The area of a triangle with sides 13 cm, 14 cm and 15 cm is [1]a84 cm²b90 cm²c168 cm²d42 cm²
2. A ring has outer radius 14 cm and inner radius 7 cm. Its area is [1]a154 cm²b308 cm²c462 cm²d616 cm²
3. A path 1 m wide runs around the outside of a 20 m × 15 m garden. The area of the path is [1]a70 m²b74 m²c72 m²d76 m²
4. The number of 20 cm × 20 cm tiles needed for a 6 m × 4 m floor is [1]a60b120c600d6000
5. The area of a quadrant (90° sector) of radius 14 cm is [1]a154 cm²b308 cm²c616 cm²d77 cm²
6. The perimeter of a semicircle of diameter 14 cm is [1]a22 cmb36 cmc44 cmd58 cm

Answers

1. (a) — s=21; \sqrt{21\times8\times7\times6}=\sqrt{7056}=84.

2. (c) — \frac{22}{7}(196-49)=\frac{22}{7}\times147=462.

3. (b) — outer 22\times17=374; inner 20\times15=300; 374-300=74.

4. (c) — area 24 m² over 0.04 m² per tile gives 600.

5. (a) — \frac14\times\frac{22}{7}\times196=154.

6. (b) — arc \frac{22}{7}\times7=22, plus diameter 14, gives 36.

5.6 Cylinder and Sphere

Facts. Cylinder: curved surface area 2\pi rh, total 2\pi r(r+h), volume \pi r^2h. Sphere: surface area 4\pi r^2, volume \frac{4}{3}\pi r^3. Hemisphere: curved 2\pi r^2, total 3\pi r^2, volume \frac{2}{3}\pi r^3. 1 m³ = 1000 litres.

1. The curved surface area of a cylinder with r=7 cm and h=10 cm is [1]a220 cm²b440 cm²c1540 cm²d308 cm²
2. The volume of the same cylinder is [1]a440 cm³b770 cm³c1540 cm³d3080 cm³
3. The surface area of a sphere of radius 7 cm is [1]a154 cm²b308 cm²c616 cm²d1437 cm²
4. The total surface area of a solid hemisphere of radius 7 cm is [1]a308 cm²b462 cm²c616 cm²d154 cm²
5. The volume of a sphere of radius 10.5 cm is [1]a2425 cm³b4851 cm³c4620 cm³d1617 cm³
6. A cylindrical tank has radius 0.7 m and height 2 m. Its capacity is [1]a308 litresb1540 litresc3080 litresd30 800 litres

Answers

1. (b) — 2\times\frac{22}{7}\times7\times10=440.

2. (c) — \frac{22}{7}\times49\times10=1540.

3. (c) — 4\times\frac{22}{7}\times49=616.

4. (b) — 3\pi r^2=3\times\frac{22}{7}\times49=462.

5. (b) — \frac43\times\frac{22}{7}\times(10.5)^3=4851.

6. (c) — \frac{22}{7}\times0.49\times2=3.08 m³ =3080 litres.

5.7 Prism and Pyramid

Facts. Prism: volume = base area \times height (length); lateral surface area = perimeter of base \times height; total = lateral +\ 2\times base area. Square pyramid: volume =\frac{1}{3}\times base area \times height; lateral area =\frac{1}{2}\times perimeter \times slant height; l^2=h^2+\left(\frac{b}{2}\right)^2. 1 m³ = 1000 litres.

1. A triangular prism has a base area of 6 cm² and length 10 cm. Its volume is [1]a16 cm³b30 cm³c60 cm³d120 cm³
2. A right triangular prism has a base with sides 3 cm, 4 cm, 5 cm and length 10 cm. Its lateral surface area is [1]a60 cm²b120 cm²c132 cm²d12 cm²
3. The total surface area of the prism in item 2 is [1]a120 cm²b126 cm²c132 cm²d144 cm²
4. A square pyramid has base side 6 cm and height 4 cm. Its slant height is [1]a4 cmb5 cmc6 cmd7 cm
5. The volume of the pyramid in item 4 is [1]a24 cm³b48 cm³c72 cm³d144 cm³
6. The area of the four walls of a room 5 m × 4 m × 3 m is [1]a27 m²b54 m²c60 m²d94 m²

Answers

1. (c) — 6\times10=60.

2. (b) — perimeter 12\times10=120.

3. (c) — base area \frac12\times3\times4=6, so 120+2(6)=132.

4. (b) — l=\sqrt{4^2+3^2}=5.

5. (b) — \frac13\times36\times4=48.

6. (b) — 2(5+4)\times3=54.

5.8 Highest Common Factor and Lowest Common Multiple

Facts. Factorise completely first. HCF is the product of the common factors of least power; LCM is the product of all factors of greatest power. \text{HCF}\times\text{LCM}= product of the two expressions.

1. The HCF of x^2-y^2 and x+y is [1]ax-ybx+ycx^2-y^2d1
2. The LCM of 3x^2y and 6xy^2 is [1]a3xyb6xyc6x^2y^2d18x^3y^3
3. The HCF of 12a^2b and 18ab^2 is [1]a36a^2b^2b6abc6a^2b^2d3ab
4. The LCM of x^2-4 and x-2 is [1]ax-2bx+2cx^2-4d(x-2)^2
5. The HCF of x^2+5x+6 and x^2+3x+2 is [1]ax+1bx+2cx+3dx-2
6. If one expression is (x+2)(x-3), the HCF is x+2 and the LCM is (x+2)(x-3)(x+1), the other expression is [1]ax+1bx-3c(x+2)(x+1)d(x-3)(x+1)

Answers

1. (b) — x^2-y^2=(x-y)(x+y), so the common factor is x+y.

2. (c) — LCM of 3 and 6 is 6, and the highest powers are x^2 and y^2.

3. (b) — HCF of 12 and 18 is 6, lowest powers a and b.

4. (c) — x^2-4=(x-2)(x+2) already contains x-2.

5. (b) — (x+2)(x+3) and (x+1)(x+2) share x+2.

6. (c) — other =\frac{\text{HCF}\times\text{LCM}}{\text{given}}=\frac{(x+2)\cdot(x+2)(x-3)(x+1)}{(x+2)(x-3)}=(x+2)(x+1).

5.9 Radical and Surd

Facts. \sqrt{ab}=\sqrt{a}\sqrt{b}; the conjugate of a+\sqrt{b} is a-\sqrt{b} and (a+\sqrt{b})(a-\sqrt{b})=a^2-b. Rationalise by multiplying by the conjugate. If x+\frac1x=k then x^2+\frac{1}{x^2}=k^2-2.

1. \sqrt{12} in simplest form is [1]a3\sqrt{2}b2\sqrt{3}c4\sqrt{3}d6
2. \sqrt{18}+\sqrt{8} equals [1]a\sqrt{26}b5\sqrt{2}c6\sqrt{2}d13
3. (2+\sqrt{3})(2-\sqrt{3}) equals [1]a1b7c4-\sqrt{3}d2
4. The rationalised form of \frac{1}{\sqrt{3}-1} is [1]a\sqrt{3}-1b\frac{\sqrt{3}+1}{2}c\frac{\sqrt{3}-1}{2}d\sqrt{3}+1
5. If x=2+\sqrt{3}, then \frac1x is [1]a2+\sqrt{3}b\sqrt{3}-2c2-\sqrt{3}d\frac{1}{2}
6. If x=2+\sqrt{3}, then x^2+\frac{1}{x^2} equals [1]a4b14c16d18

Answers

1. (b) — \sqrt{4\times3}=2\sqrt{3}.

2. (b) — 3\sqrt{2}+2\sqrt{2}=5\sqrt{2}.

3. (a) — 4-3=1.

4. (b) — \frac{\sqrt{3}+1}{(\sqrt{3})^2-1}=\frac{\sqrt{3}+1}{2}.

5. (c) — (2+\sqrt{3})(2-\sqrt{3})=1, so the reciprocal is 2-\sqrt{3}.

6. (b) — x+\frac1x=4, so x^2+\frac{1}{x^2}=16-2=14.

5.10 Indices

Facts. a^m\times a^n=a^{m+n}; \frac{a^m}{a^n}=a^{m-n}; (a^m)^n=a^{mn}; a^0=1; a^{-n}=\frac{1}{a^n}; a^{\frac{m}{n}}=\sqrt[n]{a^m}. If a^x=a^y then x=y.

1. 2^3\times2^4 equals [1]a2^{12}b4^7c2^7d2^{1}
2. The value of 64^{\frac23} is [1]a4b8c16d32
3. The value of 2^{-3} is [1]a-8b-6c\frac{1}{6}d\frac{1}{8}
4. If 3^x=81, then x is [1]a3b4c27d5
5. \frac{(x^2)^3}{x^4} equals [1]ax^2bx^3cx^{10}dx^{-2}
6. The value of (25)^0+5^{-1}\times5 is [1]a1b2c5d6

Answers

1. (c) — add the powers: 2^{3+4}=2^7.

2. (c) — 64^{2/3}=(4^3)^{2/3}=4^2=16.

3. (d) — 2^{-3}=\frac{1}{2^3}.

4. (b) — 81=3^4.

5. (a) — \frac{x^6}{x^4}=x^2.

6. (b) — 1+\frac15\times5=1+1=2.

5.11 Algebraic Fraction

Facts. Factorise, then cancel common factors. Add or subtract over the LCM of the denominators. \frac ab\div\frac cd=\frac{ad}{bc}. The difference of squares: a^2-b^2=(a-b)(a+b).

1. \frac{x^2-9}{x-3} simplifies to [1]ax-3bx+3cx^2-3d3
2. \frac1x+\frac1y equals [1]a\frac{2}{x+y}b\frac{1}{xy}c\frac{x+y}{xy}d\frac{xy}{x+y}
3. \frac{a}{b}\div\frac{c}{d} equals [1]a\frac{ac}{bd}b\frac{bc}{ad}c\frac{ad}{bc}d\frac{a}{bc}
4. \frac{1}{x-1}-\frac{1}{x+1} equals [1]a0b\frac{2}{x^2-1}c\frac{2x}{x^2-1}d\frac{1}{x^2-1}
5. \frac{x^2+3x+2}{x^2+x} simplifies to [1]a\frac{x+2}{x}b\frac{x+1}{x}c\frac{x+2}{x+1}d2
6. \frac{2}{x}+\frac{3}{x} equals [1]a\frac{5}{2x}b\frac{6}{x^2}c\frac{5}{x}d\frac{5}{x^2}

Answers

1. (b) — \frac{(x-3)(x+3)}{x-3}=x+3.

2. (c) — common denominator xy.

3. (c) — invert the divisor and multiply.

4. (b) — \frac{(x+1)-(x-1)}{x^2-1}=\frac{2}{x^2-1}.

5. (a) — \frac{(x+1)(x+2)}{x(x+1)}=\frac{x+2}{x}.

6. (c) — same denominator, add numerators.

5.12 Equations

Facts. Elimination or substitution for two linear equations. Quadratic by factorising or x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. Check roots in the original problem and reject any that make no sense (such as a negative length).

1. If x+y=7 and x-y=1, then x is [1]a3b4c5d6
2. The roots of x^2-5x+6=0 are [1]a1 and 6b-2 and -3c2 and 3d-1 and -6
3. The solutions of x^2=49 are [1]a7 onlyb\pm7c\pm 24.5d-7 only
4. The equation x^2+2x+1=0 has [1]atwo distinct rootsbno real rootcone repeated root x=-1droot x=1
5. Two numbers have sum 10 and difference 2. The larger is [1]a4b5c6d8
6. The length of a rectangle is 3 m more than its breadth and its area is 40 m². The breadth is [1]a4 mb5 mc8 md10 m

Answers

1. (b) — adding gives 2x=8.

2. (c) — (x-2)(x-3)=0.

3. (b) — both 7 and -7 square to 49.

4. (c) — (x+1)^2=0.

5. (c) — a+b=10, a-b=2 gives a=6.

6. (b) — b(b+3)=40\Rightarrow(b+8)(b-5)=0; reject b=-8.

5.13 Area of Triangles and Quadrilaterals

Facts. Triangle =\frac12\times base \times height. Parallelograms on the same base and between the same parallels are equal in area. A triangle and a parallelogram on the same base and between the same parallels: triangle =\frac12 parallelogram. A median divides a triangle into two triangles of equal area.

1. The area of a triangle of base 10 cm and height 6 cm is [1]a16 cm²b30 cm²c60 cm²d36 cm²
2. A triangle and a parallelogram stand on the same base and between the same parallels. If the parallelogram has area 40 cm², the triangle has area [1]a80 cm²b40 cm²c20 cm²d10 cm²
3. A median divides a triangle of area 36 cm² into two triangles each of area [1]a9 cm²b12 cm²c18 cm²d36 cm²
4. The area of a rhombus with diagonals 10 cm and 24 cm is [1]a240 cm²b120 cm²c34 cm²d60 cm²
5. The area of a trapezium with parallel sides 8 cm and 12 cm and height 5 cm is [1]a40 cm²b50 cm²c100 cm²d60 cm²
6. Parallelograms on the same base and between the same parallels are [1]acongruentbequal in areacsimilar onlydequal in perimeter

Answers

1. (b) — \frac12\times10\times6=30.

2. (c) — triangle is half the parallelogram.

3. (c) — a median bisects the area: 36\div2=18.

4. (b) — \frac12\times10\times24=120.

5. (b) — \frac12(8+12)\times5=50.

6. (b) — equal in area, though not necessarily congruent.

5.14 Construction

Facts. To construct a triangle equal in area to a quadrilateral, join a diagonal and draw a line through the opposite vertex parallel to that diagonal. Triangles on the same base and between the same parallels are equal in area. A rectangle equal to a triangle has the same base and half the height.

1. The theorem used to construct a triangle equal in area to a given quadrilateral is [1]atriangles on the same base and between the same parallels are equal in areabangle in a semicircle is 90°cPythagoras' theoremdthe sine rule
2. A rectangle equal in area to a triangle of base 8 cm and height 6 cm has the same base and a height of [1]a6 cmb12 cmc3 cmd4 cm
3. A square equal in area to a rectangle of 9 cm × 4 cm has side [1]a5 cmb6 cmc6.5 cmd13 cm
4. The area of a triangle with base 8 cm and height 6 cm is [1]a14 cm²b48 cm²c24 cm²d12 cm²
5. To draw a line through a point parallel to a given line, you use [1]aequal alternate anglesbthe mediancthe centroiddthe perimeter
6. In the steps of construction, the compass arcs [1]ashould be rubbed outbshould be left visiblecare not requireddare replaced by a ruler line

Answers

1. (a) — shifting the vertex along a parallel keeps the area.

2. (c) — triangle area 24 cm²; with base 8 cm the rectangle height is 24\div8=3 cm, half the triangle's height.

3. (b) — area 36, side 6 cm.

4. (c) — \frac12\times8\times6=24.

5. (a) — copy an angle so the alternate angles are equal.

6. (b) — examiners credit visible construction arcs.

5.15 Circle

Facts. Angle at the centre =2\times angle at the circumference on the same arc. Angles in the same segment are equal. Angle in a semicircle =90^\circ. Opposite angles of a cyclic quadrilateral sum to 180^\circ. A tangent is perpendicular to the radius at the point of contact. A perpendicular from the centre bisects a chord.

1. An arc subtends 40° at the circumference. The angle at the centre is [1]a20°b40°c80°d160°
2. The angle in a semicircle is [1]a45°b90°c120°d180°
3. One angle of a cyclic quadrilateral is 110°. The opposite angle is [1]a70°b110°c90°d250°
4. A point is 13 cm from the centre of a circle of radius 5 cm. The length of a tangent from the point is [1]a8 cmb12 cmc18 cmd144 cm
5. Angles in the same segment standing on one chord are [1]acomplementarybsupplementarycequalddouble
6. A chord of length 16 cm lies in a circle of radius 10 cm. Its distance from the centre is [1]a4 cmb6 cmc8 cmd12 cm

Answers

1. (c) — the centre angle is twice: 2\times40=80.

2. (b) — a diameter subtends a right angle.

3. (a) — 180-110=70.

4. (b) — \sqrt{13^2-5^2}=\sqrt{144}=12.

5. (c) — equal angles in the same segment.

6. (b) — half-chord 8, so \sqrt{10^2-8^2}=6.

5.16 Trigonometry

Facts. \sin^2\theta+\cos^2\theta=1; 1+\tan^2\theta=\sec^2\theta. \sin30^\circ=\frac12, \cos30^\circ=\frac{\sqrt{3}}{2}, \tan45^\circ=1, \sin60^\circ=\frac{\sqrt{3}}{2}. \sin\theta=\cos(90^\circ-\theta).

1. The value of \sin30^\circ is [1]a\frac{\sqrt{3}}{2}b1c\frac12d\frac{1}{\sqrt{2}}
2. If \sin\theta=\frac{5}{13} and \theta is acute, then \cos\theta is [1]a\frac{12}{13}b\frac{8}{13}c\frac{5}{12}d\frac{13}{12}
3. \sin^2\theta+\cos^2\theta equals [1]a0b\sin2\thetac\tan^2\thetad1
4. \sin25^\circ equals [1]a\cos25^\circb\cos65^\circc\tan65^\circd\sin65^\circ
5. The angle of elevation of a tower top from a point 20 m away is 45^\circ. The height of the tower is [1]a10 mb20 mc20\sqrt{2} md40 m
6. The value of \sin^230^\circ+\cos^230^\circ is [1]a\frac12b\frac34c\frac{\sqrt{3}}{2}d1

Answers

1. (c) — standard value.

2. (a) — hypotenuse 13, opposite 5, adjacent \sqrt{169-25}=12.

3. (d) — the basic identity.

4. (b) — complementary angles: \sin\theta=\cos(90^\circ-\theta).

5. (b) — \tan45^\circ=1=\frac{h}{20}, so h=20 m.

6. (d) — \frac14+\frac34=1.

5.17 Statistics

Facts. Range =L-S; coefficient of range =\frac{L-S}{L+S}. Quartile deviation =\frac{Q_3-Q_1}{2}; its coefficient is \frac{Q_3-Q_1}{Q_3+Q_1}. Standard deviation \sigma=\sqrt{\frac{\sum(x-\bar x)^2}{n}}. Coefficient of variation =\frac{\sigma}{\bar x}\times100\%; the smaller it is, the more consistent the data.

1. The largest value is 48 and the smallest is 12. The range is [1]a30b36c60d4
2. The coefficient of range in item 1 is [1]a0.6b0.4c0.75d1.6
3. If Q_1=20 and Q_3=40, the quartile deviation is [1]a20b30c10d60
4. The coefficient of quartile deviation for Q_1=20 and Q_3=40 is [1]a\frac12b\frac13c\frac23d20
5. The standard deviation of 2, 4, 4, 4, 5, 5, 7, 9 is [1]a1b2c4d5
6. A set has mean 30 and standard deviation 6. The coefficient of variation is [1]a5%b18%c20%d180%

Answers

1. (b) — 48-12=36.

2. (a) — \frac{36}{60}=0.6.

3. (c) — \frac{40-20}{2}=10.

4. (b) — \frac{20}{60}=\frac13.

5. (b) — mean 5, squared deviations sum to 32, variance \frac{32}{8}=4, SD =2.

6. (c) — \frac{6}{30}\times100=20\%.

5.18 Probability

Facts. P(E)=\frac{\text{favourable outcomes}}{\text{total outcomes}}; P(E')=1-P(E); P(A\cup B)=P(A)+P(B)-P(A\cap B). Without replacement, the second draw has one fewer ball.

1. When a fair coin is tossed, P(\text{head}) is [1]a0b\frac12c1d\frac14
2. A die is thrown. The probability of a prime number is [1]a\frac13b\frac12c\frac23d\frac16
3. If P(A)=0.3, then P(A') is [1]a0.3b0.07c0.7d1.3
4. A bag has 3 red and 2 blue balls. P(\text{red}) for one draw is [1]a\frac23b\frac25c\frac35d\frac15
5. Two coins are tossed. P(\text{at least one head}) is [1]a\frac14b\frac12c\frac34d1
6. If P(A)=0.5, P(B)=0.4 and P(A\cap B)=0.2, then P(A\cup B) is [1]a0.9b0.7c0.1d0.4

Answers

1. (b) — one favourable outcome in two.

2. (b) — primes 2, 3, 5 give \frac36=\frac12.

3. (c) — 1-0.3.

4. (c) — \frac{3}{3+2}.

5. (c) — 1-P(\text{TT})=1-\frac14.

6. (b) — 0.5+0.4-0.2=0.7.

Chapter summary

• The paper has no MCQ group, but its first item in most questions is a one-mark knowledge item; this bank drills them.
• Key formulae: n(A\cup B)=n(A)+n(B)-n(A\cap B); A=P\left(1+\frac{R}{100}\right)^T; Heron \sqrt{s(s-a)(s-b)(s-c)}; cylinder 2\pi rh, \pi r^2h; sphere 4\pi r^2, \frac43\pi r^3.
• Algebra: (a+\sqrt{b})(a-\sqrt{b})=a^2-b; a^m a^n=a^{m+n}; a^{-n}=\frac{1}{a^n}; quadratic roots by factorising.
• Geometry: centre angle =2\times circumference angle; cyclic quadrilateral opposite angles sum to 180^\circ; tangent length =\sqrt{d^2-r^2}.
• Trigonometry and data: \sin^2\theta+\cos^2\theta=1; Q.D.=\frac{Q_3-Q_1}{2}; C.V.=\frac{\sigma}{\bar x}\times100\%; P(E')=1-P(E).