1998 - JAMB Mathematics Past Questions and Answers - page 3

(\begin{array}{c|c} \text{Class intervals} & F & \text{Class boundary}\ \hline 5 - 7 & 17 & 4.5 - 9.5\ 10 - 14 & 32 & 9.5 - 14.5\ 15 - 19 & 25 & 14.5 - 19.5\ 20 - 24 & 24 & 19.5 - 24.5 \end{array})

mode = 9 + (\frac{D_1}{D_2 + D_1}) x C

= 9.5 + (\frac{5(32 - 17)}{2(32) - 17 - 25})

= 9.5 + (\frac{75}{27})

= 12.27

(\approx) 2.3

mean (x) = (\frac{\sum x}{N})

= k + k + 1 + k + 3

= (\frac{3k + 3}{3})

= k + 1

(\begin{array}{c|c} X & (X -X) & (X -X)^2\ \hline k & -1 & 1 \ k + 1 & 0 & 0\ k + 2 & 1 \ \hline & & 2\end{array})

Variance (52) = (\frac{\sum (x - x)^2}{N})

= (\frac{2}{3})

mean (x) = (\frac{1 + x + 1 + 2x + 1}{3})

= (\frac{3x + 3}{3})

= 1 + x

(\begin{array}{c|c} X & (X -X) & (X -X)^2\ \hline 1 & -x & x^2 \ x + 1 & 0 & 0\2x + 1 & x & x^2\ \hline & & 2x^2\end{array})

S.D = (\sqrt{\frac{\sum(x - 7)^2}{\sum f}})

= (\sqrt{(6)}^2)

= (\frac{2x^2}{3})

= 2x²

= 18

x² = 9

∴ x = (\pm) (\sqrt{9})

= (\pm)3

Number of red balls = 16,

Number of blue balls = 20

Let x represent the No of white balls to be added

∴ Total number of balls = 36 + x

2(36 + x) = 80

= 2x + 80 - 72

= 8

x = (\frac{8}{2})

= 4

p(x) = (\frac{2}{5}) p(y) = (\frac{1}{3})

p(x or y) = p(x ∪ y)

= p(x) + p(y)

= (\frac{2}{5}) + (\frac{1}{3})

= (\frac{11}{5})

(\begin{vmatrix}& \hline 1 & 2 & 3 & 4\hline 1 & 1 & 2& 3 & 4\2 & 2& 4 & 6 & 8\ 3& 3& 6& 9 & 12\4 & 4 & 8 & 12& 16\end{vmatrix})

p (product of x, and y > 6) = (\frac{6}{16})

= (\frac{3}{8})

y = 6 sin (2x - 25)^o

(\frac{dy}{dx}) = -12 cos(2x - 25^o)

(\frac{dy}{dx}) = 0

-12 cos x (2x - 25^o) = 0

= cos(2x - 25^o) = 0

= 2x - 25^o = cos^-1(0)

= 2x - 25^o

= 90^o

x = (\frac{115}{2})

m = (\frac{y_2 - y_1}{x_2 - x_2} = \frac{3 - 0}{0 - 2} = \frac{-3}{2})

= (\frac{y - y_1}{x- x_1})

m = (\frac{y - 3}{x}) (\geq) (\frac{-3}{2})

2(y - 3) (\geq) - 3x = 2y - 6 (\geq) - 3x

= 2y + 3x (\leq) 6 ; x (\leq) 0, y (\leq) 0

< PSR = (\frac{1}{2})(180^o) = 90^o (angle substended by a semi-circle)

∴ < TSR = 90^o

< SRT = 90^o - 30^o = 60^o

< PRS = 90^o - 35^o = 55^o

< PRQ = 180 - (60^o + 55^o) = 180 - 115^o = 65^o

< SQR = 35^o(angles in the same segment)

< QSR + < SRQ + < SQR = 180^o

< QSR + 120^o + 35^o = 180^o

< QSR + 155 = 180^o

< QSR = 180^o - 155^o

= 25^o