1991 - JAMB Mathematics Past Questions and Answers - page 5

41
From a point Z, 60 m north of X, a man walks 60√3m eastwards to another point Y. Find the bearing of Y from X.
A
0.30o
B
045o
C
060o
D
090o
correct option: c

tan (\theta) = (\frac{60\sqrt{3}}{60}) = (\sqrt{3})

(\theta) = tan (\frac{1}{3}) = 60o

Bearing of y from x = 060o

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42
Find the total area of the surface of a solid cylinder whose base radius is 4cm and height is 5cm
A
56\(\pi\) cm2
B
72\(\pi\) cm2
C
96\(\pi\) cm2
D
192\(\pi\) cm2
correct option: b
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43
3% of a family's income is spent on electricity, 58% on food, 20% on transport, 11% on education and 7% on extended family. The angles subtended at the centre of the pie chart under education and food are respectively
A
76.8o and 25.2o
B
10.8o and 224.6o
C
112.4o and 72.0o
D
39.6o and 212.4o
correct option: d
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44
Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result
\(\begin{array}{c|c} \text{No. defective per box} & 4 & 5 & 6 & 7 & 8 & 9 \ \hline \text{No. of boxes} & 2 & 7 & 17 & 10 & 8 & 6\end{array}\)

The mean and the median of the distribution are respectively
A
6.7, 6
B
7.6, 5
C
5.7, 87
D
34, 6
correct option: a
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45
Fifty boxes each of 50 bolts were inspected for the number which were defective. The following was the result
\(\begin{array}{c|c} \text{No. defective per box} & 4 & 5 & 6 & 7 & 8 & 9 \ \hline \text{No. of boxes} & 2 & 7 & 17 & 10 & 8 & 6\end{array}\)
Find the percentage of boxes containing at least 5 defective bolts each
A
96
B
94
C
92
D
90
correct option: a

% of boxes containing at least 5 defective bolts each

= (\frac{7 + 17 + 10 + 8 + 6}{50})

= (\frac{48}{50}) x (\frac{100}{1})

= 96%

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46
A crate of soft drinks contains 10 bottles of Coca-cola, 8 of Fanta and 6 of sprite. If one bottle is selected at random, what is the probability that it is NOT a Cocacola bottle?
A
\(\frac{5}{12}\)
B
\(\frac{1}{3}\)
C
\(\frac{3}{4}\)
D
\(\frac{7}{24}\)
correct option: d

Coca-cola = 10 bottles, Fanta = 8 bottles

Sprite = 6 bottles, Total = 24

P(cola-cola) = (\frac{10}{24})

P(not coca-cola) = 1 - (\frac{10}{24})

(\frac{24 - 10}{24}) = (\frac{14}{24})

= (\frac{7}{24})

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47
In the figure above, Find the value of x
A
130o
B
110o
C
100o
D
90o
correct option: a
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48
PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PM = 5cm, PN = 12cm and PQ = 4.8cm, calculate the respective lengths of PR and PT in centimeters
A
7.3, 5.9
B
7.7, 12.5
C
12.5, 7.7
D
5.9, 7.3
correct option: c

(\frac{PQ}{PN} = \frac{PM}{PR} = \frac{QM}{NR})

(\frac{4.8}{12} = \frac{5}{PR})

PR = (\frac{5 \times 12}{4.8} = \frac{50}{4})

= 12.5

(\frac{PQ}{PN} = \frac{PM}{PT} = \frac{TM}{NT})

(\frac{PT}{12} = \frac{5}{PR})

PT2 = 60

PT = (\sqrt{60})

= 7.746

= 7.7

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