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Initial salary = N540 increment = N36 (every 6 months) Period of increment = 2 years 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 150o 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, \(\theta\) = 60o, 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) = 43o

sin \(\theta\) = cos \(\theta\) 0 \(\leq\) \(\theta\) \(\leq\) 360o The acute angle where sin \(\theta\) = cos \(\theta\) = 45o But at the third Quadrant Cos \(\theta\) = -ve; sin \(\theta\) = -ve. at the 3rd quadrant, value with respect to Q is (180 + Q) where Q = acute angle (180 + 45) = 225^\circ The two solution are 45^\circ, 225^\circ

Let each of the unknown side be x. \(10^{2} = x^{2} + x^{2} - 2(x)(x) \cos 120\) \(100 = 2x^{2} - 2x^{2} \cos 120\) \(100 = 2x^{2} + x^{2} = 3x^{2}\) x = \(\sqrt{\frac{100}{3}}\) = \(\frac{10}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) x = \(\frac{10\sqrt{3}}{3}\)cm

Angle corresponding to 7 in a pie chart will be \(\frac{7 \times 360}{\text{sum of items}}\) = \(\frac{7 }{18}\) x 360 = 140o

Number of students scoring at least 50 marks = Number of students scoring 50 and above From the table 53, 70, 84, 59, 90, 60, 81, 73, 50, 37, 67, 68, 64, 52. Hence, 14 students scored at least 50 marks

Mode = a + \(\frac{(b - a)(F_m - F_b)}{2F_m - F_a - F_b}\) = \(L_1 + \frac{\Delta_1 x^\text{c}}{\Delta_1 + \Delta_2}\) = \(10 + \frac{(20 - 10)(27 - 10)}{2(27) - 10 - 19}\) = 10 + \(\frac{170}{25}\) = 10 + 6.8 = 16.8

Average age of 110 students = 16 ∴ Total age = 16 x 10 = 160 years Age of teachers = x, total number of people now = 11 mean age = 19 Total age of new group = 19 x 11 = 209 Age of teachers = x = (209 - 160) = 49 yrs

Median = L1 + (\(\frac{Ef}{fm}\)) - fo \(\frac{\sum f}{2}\) = \(\frac{20}{2}\) = 10, L1 = 10.5, fo = 6, fm = 5 Median = 10.5 + \(\frac{(10 - 6)}{5}\)5 = 10.5 + 4 = 14.5

Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors itself and 1 The prime numbers among the group are 23, 29. Probability of choosing a prime number = \(\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}\) = \(\frac{2}{11}\)