1993 - JAMB Mathematics Past Questions and Answers - page 3

x \(\ast\) y = x^y

x \(\ast\) 2 = x²

x \(\ast\) 2 = x

∴ x² - x = 0

x(x - 1) = 0

x = 0 or 1

Initial salary = N540

increment = N36 (every 6 months)

Period of increment = 2 yrs and 6 months

amount(increment) = N36 x 5 = N180

The man's new salary = N540 = N180

= N720.00

A rectangular polygon has each interior angle to be 150^o

let the polygon has n-sides

therefore, Total interior angle 150 x n = 150n

hence 150n = (2n - 4)90

150n = 180n - 360

360 = (180 - 150)n

30n = 360

n = 12

Diameter = 8cm

∴ Radius = 4cm

Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\)

∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4

= \(\frac{22\frac{1}{2} \times 8\pi}{360}\)

= \(\frac{180}{360}\)

= \(\frac{\pi}{2}\)

2x + 3 \(\neq\) 2x - 3 for any value of x

∴ for the \(\bigtriangleup\) to be isosceles, either

2x - 3 = x + 3 or 2x + 3 = x + 3

solve the two equations we arrive at

x = 6 or x = 0

When x = 6, the sides are 9, 15, 9

When x = 0, the sides are 3, 4, -3 since lengths of a \(\bigtriangleup\)can never be negative then the value of x = 6

Surface area = 154cm² (area of sphere)

4\(\pi\)r² = 154

r\(\sqrt{\frac{154}{4\pi}}\)

= 3.50cm

Area of sector

\(\frac{\theta}{360}\) x \(\pi\)r², \(\theta\) = 60^o, r = 3m

= \(\frac{60}{360}\) x \(\frac{12}{7}\) x 3 x 3

\(\frac{1}{6}\) x \(\frac{22}{7}\) x 9

= \(\frac{33}{7}\)

= 4.7m²

The angle between 2 latitudes one in northern hemisphere and the other in southern hemisphere and the other in southern hemisphere is the sum of their latitudes.

∴ Total angle difference = (30 + 13) = 43^o

sin \(\theta\) = cos \(\theta\) 0 \(\leq\) \(\theta\) \(\leq\) 360^o

The acute angle where sin \(\theta\) = cos \(\theta\) = 45^o

But at the fourth Quadrant Cos \(\theta\) = +ve

at the 4th quadrant, value with respect to Q is

(360 - Q) where Q = acute angle

(360 - 45) = 315^o

The two solution are 45^o, 315^o