2002 - JAMB Mathematics Past Questions and Answers - page 5

o

30 - x + x + 40 - x = 80 - 20
70 - x = 60
- x = 60 - 70
- x = - 10
∴ x = 10
Music only = 30 - x
= 30 - 10
20
P(music only) = 20/80
= 1/4
= 0.25

In Δ PQS, ∠PSQ = X(base ∠s of isoc Δ PQS)
In Δ QRS, ∠RQS = ∠PSQ + X(Extr ∠ = sum of two intr. opp ∠s)
∴ ∠RQS = X + X
= 2X
Also ∠QRS = 2X(base ∠s of isoc Δ QRS in Δ PRS,
72 = ∠RPS + ∠PRS(Extr, ∠ = sum of two intr. opp ∠s)
∴ 72 = x + 2x
72 = 3x
x = ⁷²/₃
x = 24^o

r = \(\frac{2}{3}\)R

∴R = \(\frac{3}{3}\)R

Area of small circle = πr²

= π(\(\frac{2R}{3}\))²

Area of the big circle πr² = π\(\frac{(3R)^2}{3}\)

Area of shaded portion = π(\(\frac{3R}{3}\))² - π(\(\frac{2R}{3}\))²

= π[(\(\frac{3R}{3}\))² - (\(\frac{2R}{3}\))²]

= π[(\(\frac{3R}{3}) + (\frac{2R}{3}) - (\frac{3R}{3}\)) - (\(\frac{2R}{3}\))]

= π[(\(\frac{5R}{3}\)) (\(\frac{R}{3}\))]

= π x \(\frac{5R}{3}\) x \(\frac{R}{3}\)

= ⁵/₉πR²

Volume of cylinder = πr²h
= π x 3² x 20
= π x 9 x 20
= 180 πcm³
Volume of hemisphere = ²/_3πr³
= ²/₃ x π x 3²
= 2 x π x 3²
= 2 x π x 9
= 18π
= Volume of the solid = 180π x 18π
= 198πcm³

∠PQR = 180 - 128 (∠s on a straight line)
= 52^o
∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52^o
∴ΔPQR is an isosceles triangle

20 + X + 50 + 40 + 2X + 60 = 260
3X + 170 = 260
3X = 260 - 170
3x = 90
x = 30
Absent for at least 4 days

4	5	6
40	2x	60

i.e 40 + 2x + 60 = 40 + (2 x 30) + 60
= 40 + 60 + 60
= 160

80 = (30 - x) + x + 40 - x + 20

80 = 90 - x, x = 90 - 80 = 10

N (music only) = 30 - x = 30 - 10 = 20

P (music only) = \(\frac{20}{80} = \frac{1}{4}\)

= 0.25

m = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{0 - (4)} = \frac{2}{4} = \frac{1}{2}\)

\(\frac{y_2 - y_1}{x_2 - x_1} \geq m\)

\(\frac{y - 0}{x + 4} \geq \frac{1}{2}\)

2y \(\geq\) x + 4, -x + 2y \(\geq\) 4 = px + qy \(\geq\) 4

p = -1, q = 2

< PRQ = 180^o - 128^o = 52^o = < R

< P + < R + < Q = 180^o

76^o + 52^o + < Q = 180^o

128^o + < Q = 180^o

< Q = 180^o - 128^o

= 52^o

\(\bigtriangleup\)PQR is an isosceles