1998 - JAMB Mathematics Past Questions and Answers - page 4

31
In diagram, PQ || ST and < PQR = 120o, < RST = 130o, find the angle marked x
A
50
B
65
C
70
D
80
correct option: c

x + 60o + 50o = 180o

x = 110o = 180o

x = 180o - 110o

= 70o

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32
In the figure, PQST is a parallelogram and TSR is a straight line. If the area of \(\bigtriangleup\)QRS is 20cm2, find the area of the trapezium PQRT.
A
35cm2
B
65cm2
C
70cm2
D
140cm2
correct option: c

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

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33
TQ is tangent to circle XYTR, < YXT = 32o, RTQ = 40o. fIND < YTR
A
108o
B
121o
C
140o
D
148o
correct option: a

< 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

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34
In the diagram, QTR is a straight line and < PQT = 30o. find the sin of < PTR
A
\(\frac{8}{15}\)
B
\(\frac{2}{3}\)
C
\(\frac{3}{4}\)
D
\(\frac{15}{16}\)
correct option: c

(\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

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35
A chord of a circle radius \(\sqrt{3cm}\) subtends an angle of 60o on the circumference of he circle. Find the length of the chord
A
\(\frac{\sqrt{3}}{2}\)
B
\(\frac{3}{2}\)
C
3
D
\(\sqrt{3}\)cm
correct option: a
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36
The pie chart shows the monthly expenditure pf a public servant. The monthly expenditure on housing is twice that of school fees. How much does the worker spend on housing if his monthly income is N7200?
A
N1000
B
N2000
C
N3000
D
N4000
correct option: b

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

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37
The bar chart shows the distribution of marks scored by 60 pupils in a test in which the maximum score was 10. If the pass mark was 5, what percentage of the pupils failed the test?
A
59.4%
B
50.0%
C
41.7%
D
25.0%
correct option: c

(\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%

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38
If 10112 + x7 = 2510, solve for X.
A
207
B
14
C
20
D
24
correct option: a

10112 + 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

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39
Evaluate [\(\frac{1}{0.03}\) \(\div\) \(\frac{1}{0.024}\)]-1 correct to 2 decimal places
A
3.76
B
1.25
C
0.94
D
0.75
correct option: b

[(\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

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40
If b3 = a-2 and c\(\frac{1}{3}\) = a\(\frac{1}{2}\)b, express c in terms of a
A
a-\(\frac{1}{2}\)
B
a\(\frac{1}{3}\)
C
a\(\frac{3}{2}\)
D
a\(\frac{2}{3}\)
correct option: a

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})

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