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NATIONAL BUSINESS AND TECHNICAL EXAMINATIONS BOARD MAY/JUNE 006 NBC/NTC EXAMINATION MATHEMATICS 0.085 0.67 1(a) Evaluate, 3.36 leaving your answer in standard form. If the angles of a polygon are given as x o, (x+0) o, (x-10) o, (3x+10) o and (x-10) o, Find the value of x. (a) 0.085 x 0.67 3.36 = 0.0076095 3.36 = 0.00647 0.0065 in standard form we have it as.65 x10-3 sum of angles = 540 o x + x + 0 + x -10 + 3x +10 + x 10 = 540 o x = 66.5 o ALITER for 1 The corresponding exterior angles are (180 x) 0, (160 - x) 0, (190 - x) 0, (170 3x) 0, (190 - x) 0 Equating sum of the exterior angles to 360 o, we have x = 66.5 o (a) If the angles of a pentagon are x o, x o, x o, (x+ 40) o, (x +10) o. Find the value of the biggest angle. Given that 0 o < < 90 o and Sin = 5/13, Find without using tables or calculator Cot Sum of angles of a pentagon is 540 o x + x + x + x + 40 + x + 10 = 540 o Therefore x = 61.5 o

By Pythagoras theorem x = Cot θ = Cot 13 5 = 1 Cos 1 Sin Tan 1 5 144 5 19 5 5 Adjacent Opposite ALITER Cot + 1 = Cosec or equivalent Cosec = 169 5 Cot 19 = 5 5 3(a) The 1 st and 5 th terms of a G.P are 81 and 1 respectively. Determine the three terms between the two numbers. In the figure AB is the diameter of a semi-circle, AOBC. If /AB/ = 10cm and /AC/ = 8cm. Calculate the area of the shaded portion. (a) a = 81 and ar 5-1 = ar 4 = 1 Solving the equation, we have r 4 = 1 81 r = 1/3 Using ar n-1, we have between 81 and 1, n=, 3 and 4; 7, 9 and 3 b) Area of a circle = πr = x 5 = 78.55cm

7 Area of semicircle = 39.75cm Area of triangle = ½ bh Height h = 9.6cm ½ x5cm x 9.6cm = 4cm 4(a) t 1 Solve for t in the equation, 6 t 4 3 The dimensions of a room are 15m long, 8m wide and 10m high. Calculate the angle which the diagonal of the room makes with the floor. t 1 1 (a) 6 t, collecting like terms, we have t t 6 4 3 3 4 t 5 ALITER t 7 Solving and clearing fractions, 6 t 4 3 t 5 (a) the diagonal of the floor of the room = 17m; by Pythagoras theorem. Opposite 10 Tan = = = 0.588 Adjacent 17 tan = 0.533 and = 30.47 o

5(a) Find the sum of the first 100 even numbers In the figure AD//BC, BE is perpendicular to AD. If the area of ABCD is 40.5cm, find the length of BC. 5(a) Even numbers, a =, d =, n = 100 n S n a ( n 1) d 100 S n 4 (99) = 10,100 Total Area = ½ (4.5) (10.6 + BC) = 40.5 Solving for BC, we have BC = 7.4cm

r r 6(a) Given that p, q, express q+8p in terms of r in the simplest form. r 4 r (c) Solve the equation 3x +4x-7 = 0, leaving your answer to decimal places. Simplify: log6+ ½ log 81- log7 (a) Substituting the values of p, q and taking L.C.M. p+8q = r (r - ) + 8r (r + ) (r - ) r = r (3x+7) (x-1) = 0 3x+7=0 or x-1 = 0 either x = -.333 or x = 1 ALITER x 4 16 4(3)(7) (3) either x = -.33 or x = 1. (c) log 6 + ½ log 81 log 7; NB: log 81 log 81 = log 6 + log 81 log 7 = log 6 + log 9 log 7 = 6 9 log 7 = 54 log 7 = log 1 7(a) The difference between the reciprocal of a number and the reciprocal of the sum of 3 and the number is 3/40. What is the number?

In the figure given ABC is an isosceles triangle inscribed in a circle centre 0. If /AB/=AC/ = 13cm and /BC/ = 10cm. Calculate the radius of the circle to the nearest whole number. (a) Let the number be x. From the question we have: 1 1 = 3 x x+3 40 Simplifying, we obtain x + 3x 40 = 0 x = -8 or x = 5 r = x + 5. (i) AD = 13 5 = 1cm x = 1 r.. (ii) Solving for r: r = (1-r) + 5 r = 7.04 r = 7.0 cm

8(a) In an opinion survey, a group of people were interviewed about the daily newspaper read by them. 14 read the Tribune, 1 read the Champion and 118 read the Guardian. 66 read both the Tribune and the Champion. 48 read the Champion and the Guardian while 58 read the Tribune and the Guardian. 3 read all the three newspapers. (i) Illustrate the information using a Venn diagram. (ii) How many people were interviewed? (iii) How many people read the Guardian only? (i) (ii) x + 6 + 34 + 3 = 14 x = 50 y + 34 + 16 + 3 = 1 y =40 z + 6 + 3 + 16 = 118 z = 44 Total number of people that were interviewed = 50+40+44+6+34+16+3 = 4 (iii) z + 6 + 3 + 16 = 118 z = 44 (read Guardian only) 9(a) Complete the table of values for the relation y = + x - x for -3 x 4 x -3 - -1 0 1 3 4 y -4 (c) Using a scale of cm to 1 unit on the x-axis and units along the y-axis, draw the graph of y = + x - x with your values in 9(a) From your graph

(i) Find the co-ordinates of the point at which y is greatest. (ii) Estimate the roots of the equation + x - x = 0 (d) Using the same scale and axes, draw the graph of y = 1- x (e) Use your graphs to solve the equation 1+x - x = 0 (a) x -3 - -1 0 1 3 4 y -10-4 0 0-4 -10

(c) (i) The co-ordinates = (0.5,.3) + 0.1 (ii) The roots x = -1 or x = + 0.1 (d) The graph of y = 1- x Table of y = 1-x x -3 - -1 0 1 3 4 y 4 3 1 0-1 - -3 (e) From the graph, the roots of 1+ x - x = 0 are x = -0.4 + 0.1, and x =.4 + 0.1 10(a) With a pair of compasses and ruler only, construct parallelogram ABCD such that /AB/ = 5cm, /BC/ = 6.4 and ABC = 60 o (i) Bisect < BCD and let the bisector meet AD at M. (ii) Draw a circle centre 0 with CM as diameter. (iii) Measure /OD/ and <BCM. Given two points X (lat.50 o N, long 85 o E) and Y (lat. 50 o N, long 5 o E). Find the distance between them along the common parallel of latitude. Take R to be 6400km. π to be 3.14 and correct your answer to decimals. (a) The distance between the two points, difference in longitude = 60 o = 60 o x π x 6400 x Cos 50 o 360 = 4305.90km

11(a) The cost of sewing a set of suit is partly constant and partly varies inversely with the number of days it takes to sew the suit. If the suit takes 3 days to sew, the cost is 3,000. If it takes 5 days, the cost is,500. Find the cost if the tailor takes 4 days to sew a particular set of suit. Calculate the perimeter of the major segment of a circle radius 17.5cm determined at the centre of the circle. Take π = /7 We can represent the question with the following equations: 3a + b = 900 5a + b = 1,500 a = 1750 and b = 3750 The law of the variation, C = 1750 + For 4days, D = 4, C = 1750+937.5 =N 687.50 3750 D Perimeter of a segment of a circle = length of arc + length of chord 35 length of the semi circular arc = = 55cm 7 length of chord = 17.5cm + 17.5cm = 35cm perimeter = 55cm + 35cm = 90cm. 1(a) Mention two uses of averages List two merits and two demerits of the (i) mean, and (ii) median, as measures of central tendency. (c) Calculate (i) range, and (ii) standard deviation of the set of data: 4,5,8,3,9,10,15,1,8,16 1(a) Two () uses of average Listing of two () merits and two () demerits of mean and median.

(c) (i) range = 16-3=13 (ii) S. D = x x 174 x x 174 S.D. = 17. 4 N 10 S.D. = 17.4 4. 17 13(a) A man saves a sum of money in a bank which pays simple interest at the rate of 4% per annum. If after 3 years the money amounts to 0,000.00. Find the sum invested. Three boys did a business together and shared the profit they made. The first got 0% of the share and got 37½ % of the remainder. If the third boy received 450.00, how much was the profit they shared? (a) I = PRT = = 3P 100 5 Amount = A = P + 3p 5 sum invested = P + 3p = 0,000 5 P = 17857.14 remainder after the first boy s share = 100% - 0% = 80% The second boy s share = 75 % 4 5 80 100 4 5 = 300 = 30 30 = 3 10 100 10 Proportion left for the third boy = ½. If the third boy received 450, therefore the total profit shared would be 450 + 450 = 900.00

14(a) A man s total salary for 4 months in a particular year was 14,001.00. Then there was salary review in which monthly salaries were doubled. Calculate his total salary in that year. Ojo, Obi and Bala are three partners who own a business venture. It was agreed that the profit will be shared such that Ojo received /5. Obi receives /3 and Bala receives ¼ of the total profit made. If at the end of the year Ojo receives 3,00 less than Obi, what is Bala s share of the total profit? 14(a) Monthly pay before salary review = 14001 4 = 3,500.5 monthly salary after review = 3,500.5 x = 7000.50 8 months salary after review = 7000.50 x 8 = 56004.00 Total salary for the year = 56004 + 14001 = 70,005.00 Let x represent the total profit. Therefore we have the equation x + 300 = x 5 3 = x x = -300 5 3 L.C.M approach - 4x = - 300 15 x = 1,000 then Bala s share will be ¼ of 1,000 = 3,000.00 15(a) A man bought 10,000 1.00 shares of FELGRA PLC at 1.50 per share through a broker. If the brokerage commission rate is 5.4% plus 300 for any transaction of 5,000 or more, calculate the: (i) broker s commission and (ii) amount the man received if a dividend of 15% was declared. Afman is to pay back a loan of 48,000 monthly for 3 years. If the amount is borrowed at the rate of 1½ % per annum, what will be his monthly installment payment? (a) (i) Market value of the shares = 1.50 x 10,000 = 15,000 Broker s commission = N15,000 100 5. 4 + N300

= 810 + 300 = 1110.00 (ii) Amount the man received if a dividend of 15% was declared. Nominal value of the share = 10,000 Dividend received 1500.00 I = PTR 100 Interest on the loan = 48,000 x 3 x 1½ = 18,000 100 total money owed = 48,000 + 18,000 = 66,000 the monthly installment will be 66,000 36 months = 1833.33