2002 - JAMB Mathematics Past Questions and Answers - page 5

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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

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