1998 - JAMB Mathematics Past Questions and Answers - page 4

x + 60o + 50o = 180o
x = 110o = 180o
x = 180o - 110o
= 70o
Users' Answers & Comments
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
Users' Answers & Comments
< TWR = < QTR = 40o (alternate segment)
< TWR = < TXR = 40o(Angles in the same segments)
< YXR = 40o + 32o = 72o
< YXR + < YTR = 180o(Supplementary)
72o + < YTR = 180o
< YTR = 180o - 72o
= 108o
Users' Answers & Comments
(\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
Users' Answers & Comments
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 = 360o - (120o + 90o)
= 360o - 210o = 150o
Angle of housing in a pie-chart = (\frac{2}{3}) x 150 = 100
(\frac{100}{360}) x 7200 = N2,000
Users' Answers & Comments
(\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%
Users' Answers & Comments10112 + x7 = 2510 = 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 2o
= 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
Users' Answers & Comments[(\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
Users' Answers & Commentsc(\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})
Users' Answers & Comments