1998 - JAMB Mathematics Past Questions and Answers - page 4

x + 60^o + 50^o = 180^o

x = 110^o = 180^o

x = 180^o - 110^o

= 70^o

A_{\(\bigtriangleup\)} = \(\frac{1}{2}\) x 8 x h = 20

= \(\frac{1}{2}\) x 8 x h = 4h

h = \(\frac{20}{4}\)

= 5cm

A_{\(\bigtriangleup\)}(PQTS) = L x H

A_{\(\bigtriangleup\)PQRT} = A_{\(\bigtriangleup\)QSR} + A_{\(\bigtriangleup\)PQTS}

20 + 50 = 70cm²

ALTERNATIVE METHOD

A_{\(\bigtriangleup\)PQRT} = \(\frac{1}{2}\) x 5 x 28

= 70cm²

< TWR = < QTR = 40^o (alternate segment)

< TWR = < TXR = 40^o(Angles in the same segments)
< YXR = 40^o + 32^o = 72^o

< YXR + < YTR = 180^o(Supplementary)

72^o + < YTR = 180^o

< YTR = 180^o - 72^o

= 108^o

\(\frac{10}{\sin 30^o} = \frac{15}{\sin x} = \frac{10}{0.5} = \frac{15}{\sin x}\)

\(\frac{15}{20} = \sin x\)

sin x = \(\frac{15}{20} = \frac{3}{4}\)

N.B x = < PRQ

Let the monthly expenditure on school fees be x let the monthly expenditure on housing be 2x; angle of housing and school fee in a pie chart = 360^o - (120^o + 90^o)

= 360^o - 210^o = 150^o

Angle of housing in a pie-chart = \(\frac{2}{3}\) x 150 = 100

\(\frac{100}{360}\) x 7200 = N2,000

\(\begin{array}{c|c} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \ \hline f & 1 & 9 & 4 & 7 & 10 & 8 & 7 & 9 & 8 & 2 & 1\end{array}\)
no pupils who failed the test = 1 + 3 + 4 + 7 + 10

= 25

5 of pupils who fail = \(\frac{25}{60}\) x 100%

= 41.70%

1011₂ + x₇ = 25₁₀ = 1011₂ = 1 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2^o

= 8 + 0 + 2 + 1

= 11₁₀

x₇ = 25₁₀ - 11₁₀

= 14₁₀

\(\begin{array}{c|c}
7 & 14 \ 7 & 2 R 0 \ & 0 R 2
\end{array}\)

X = 20₇

[\(\frac{1}{0.03}\) + \(\frac{1}{0.024}\)]

= [\(\frac{1}{0.03 \times 0.024}\)]^-1

= [\(\frac{0.024}{0.003}\)]^-1

= \(\frac{0.03}{0.024}\)

= \(\frac{30}{24}\) = 1.25

c^{\(\frac{1}{3}\)} = a^{\(\frac{1}{2}\)}b

= a^{\(\frac{1}{2}\)}b x a^-2

= a^{-\(\frac{3}{2}\)}

= (c^{\(\frac{1}{3}\)})³

= (a^{-\(\frac{3}{2}\)})^{\(\frac{1}{3}\)}

c = a^{-\(\frac{1}{2}\)}