2004 - JAMB Mathematics Past Questions and Answers - page 5

5000 : 5000 * 500

5000 : 2500000

5 : 2500

1 : 500

Tan θ = ^5√3/₅

Tan θ = √3

θ = 60^o

Coordinates of the line are (x₁ y₁)(0,2) respectively and (x₂ y₂)(2,0) respectively

Gradient (m) of the line = (y₂ - y₁) / (x₂ - x₁) = (0-6) / (2-0) = ^-6/₂

= -3

Equation of the line y-y₁ = m(x-x₁)

y - 6 = -3(x - 0)

y - 6 = -3x

y + 3x = 6

inequality; from the y axis, the shaded region is below the line

∴ y + 3x ∠ 6 and line is a broken line

= 4 x 30

= 120^o

i.e ∠PQR = 120^o ∴ RQD = ¹²⁰/₂ = 60^o also

∴ Δ QOR is equilateral Δ ∴ QO = 7cm

∴ PO = QO = RO ...... = 7cm radius of the circumference of the circle = 2πr

= 2 x ²²/₇ x ⁷/₁

= 44 cm

when x = 0, y = 0 (\to) (0, 6)

when y = 0, x = 2 (\to) (2, 0)

y + 3x < 6

< PRS = 20^o + 35^o = 55^o

< POS = 2 < PRS

2 x 55^o = 10^o

< OPS = < PSO = x

In (\bigtriangleup)POS

OP = OS(radii)

(\bigtriangleup)POS is an isosceles

2x + 110^o = 180^o

2x = 180^o - 110^o = 70^o

x = (\frac{70^o}{2})

= 35^o

beans angle = 360^o - (60^o + 90^o + 150^o)

360^o - 300^o = 60^o

150^o - 3000 tonnes

60^o - (\frac{60}{150} \times 3000) tonnes

= 1,200 tonnes

(\begin{array}{c|c} \text{class-marks(x)} & \text{(freq. (f)} & \text{cum. freq.}\hline 9 & &3.....3\14 & & 6....3 + 6 = 9\ 19 & & 9.....9 + 9 = 18\ 24 & & 3.....18 + 3 = 21\ 29 & & 2.....21 + 2 = 23\ 34 & & 2.....23 + 2 = 25\ & & 25\end{array})

% of students scoring more than 20 marks

= (\frac{7}{25}) x 100% = 28%