A BCCCC CCCCCCC A DE FD D A BC

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Letitia Wilkerson
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1 ABCCCCCCCCCCCADEFDDABC ABCADEFBA BAAA AEA basic prime number. Your knowledge of indices is tested here. EEA 4. A 1. = A B CD E F = F Reciprocal of F = F 2. D = {230 x 20} {198 x 18} D = 4,600 3,564 D = 1, CDC B = CDCDC B C = CDC CDC = CDC = CDC = CDC A = 2 x 25 = 50 Comment: Under examination condition without calculator, you should first simplify each figure to its C B A y B A C y = Sin 60 Sin is +ve in 1 st and 2 nd quadrant: 2 nd quadrant = 180 y y = 6CC Comment: Remember the CAST rule and note the angles where the various angles are positive. {0 < x < 90 : All angles are positive. 90 < x < 180 : Only Sine is positive. 180 < x < 270 : Only Tan is positive. 270 < x < 360 : Only Cos is positive.} ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

2 ABCCCCCCCCCCCADEFDDABC 5. The bearing of Z from X is = 135 Comment: Draw the bearings on a rough sheet to solve accurately = 5850 (to 3 s.f.) Comment: Approximation to significant figures or decimal places is a regular Unilag question. 8. Find the factors of D A D D A D D A D D DD D D D D D C D D CCC of D A D. CC CC f(x) = D A D A 0 = k K = 45 A 0 = k K = -32 K = -32 or 45 Comment: When k = -32 Proof: D A D D A D D x(x 2) +16(x 2) = 0 (x + 16)(x 2) = 0 When k = 45 D A D D A D D x(x + 5) + 9 (x + 5) = 0 (x + 5) (x + 9) = 0 9. F D F F D F B F D F B F F D F B F A F AA F E AF 8x + 16 = CDC 8x + 16 = 25 x = 9/8 ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

3 ABCCCCCCCCCCCADEFDDABC Comment: Recall the basic rules of logarithms to solve this question. D CDC x = N19, A D F A D But: A D A D A D A D A D F A D 12. A D A D F A D F A D A F D A F But D D D F A Comment: Recall the trigonometric formulas of Sec, Cosec and Cot to solve similar questions 11. Let x be the total worth of the business. Sola Audu x = D D D C n(= n(m) + n(p) + n(p - n(mp) 40 = x 40 = 47 x x = 7 Comment: M = No. who passed Maths. P = No. who passed Physics 13. Given: D CD A A = (5, -1) Using F F FA A F A ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE E B F A F A -2(x 2) = y 5-2x + 4 = y 5 y + 2x = 9

4 ABCCCCCCCCCCCADEFDDABC Comment: Recall the equation of two points. 14. A A Let p = A Factorising: A Either: p = -8 or p = 4 Recall: Either: CCCCCCC No solution A x = Cos 60 = x/5 x = 5Cos60 x = 2.5m ABCADEFBA BAAA BAAEEA = 6,540 ( 3 s.f) Comment: A similar question to this was reapeated in 2007/2008 Science based Q7. 2. A D A D C Using: D A D Thus: a = 1; b = -18; c = k But: A A k = CC k = 81 Comment: To make an expression a perfect square, add a constant k and solve using the algebraic formula A. 3. C CDC C C AF DC AF C C F DC AF C Comment: The most important thing here is getting the formula right. C CDC AND ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

5 ABCCCCCCCCCCCADEFDDABC NOT C CDC. 4. 1km = 1,000m 1hr = 60 x 60secs. 5. CC CCECCE = Comment: To convert from to, Note that 1km = 1000m and 1hr = 3,600secs. 6. Area of parallelogram = bh A CDC b = 27cm. Comment: When given this type of question, draw the diagram in any rough sheet around. It would help you better solve the question. CC C A CC But: CC CDCCC CDC A C A CDC Comment: This question aims to test your knowledge of solid shapes and its mensuration formulae. 7. Let X = 5,500 and Y = 4,000 Husband Wife Fund Fund = N545. C CCC CDC CDC CD A D = D A D D = DD D = D D ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

6 ABCCCCCCCCCCCADEFDDABC 9. scores and Total girls scores to find X. C E x =60 tan 45 x = 60km Comment: Since town R is south of town P and west of town Q it means town R is the right angle of the triangle. 10. Boys Girl No. of students 30 X Mean 6 8 Total no. of scores P Q C C Total scores P Q = =180 8x Total scores = Total Boys scores + Total Girls scores. 468 = x 8x = 288 x = 36 Comment: Total scores = Total Boys scores + Total Girls scores. First we need to find the total Boy 11. Ade alone (A) = x days Uche alone (U) = x + 15 days Ade & Uche (AU) = x ( x + 15) (AU) = D A D F Method 1: F F F F D D A D D D A D D A D D D A D Method 2: We use the formula. F F F AF D D A D D A D 12. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 C CC Pr. (Prime) = 4/10 Pr. (multiple of 3) = 3/10 Pr. (Prime) or Pr. (multiple of 3) = B ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

7 ABCCCCCCCCCCCADEFDDABC = Comment: Prime numbers are numbers only divisible by itself and 1. 1 is not a prime number. Prime numbers here are 2, 3, 5, and 7. Multiples of 3 are 3, 6, and 9 (i.e. numbers divisible by 3) white balls + 6 Red balls =10 balls 15. R = 4%; T = 3yrs; Fv = 7,000; P =??? Fv = P + SI 7,000 = P + CCB A AF CCAF A Pr. (of both red) = E D F = B Comment: At the first pick, we have 6 red balls from a possible 10 ball. If the first pick was a red ball, at the second pick we will have 5 red balls from a possible 9 balls. This is because the balls are picked without replacement. 14. B CD A A AB A A A F A E B F DEFABAAA EEA 1. AACBEACB AA AAAAAAAAA F F = F Method 2: You could either convert from base 5 to base 10, subtract then convert your answer back to base 5. Comments: Method 1 is preferable under exam conditions = (to 5 d.p.) 3. C C DC C =C EF DC F = 12 ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

8 ABCCCCCCCCCCCADEFDDABC 4. 1 st year = 24,000 2 nd year = 26,400 (24, % of 24,000). 3 rd year = 29,040 (26, % of 26,400). Ans : [ B] 5. CDC CDC B CDCCD A CDCCDC B D A CDC F Comments: Most of the time calculators are not allowed into the exam hall so you have to do this long multiplication by hand. Don t waste too much time on this and be accurate. 6. A B BCCA B CC BCC = B B A = AB A = A A =C Comment: The most important thing in questions of this nature is to simplify to the smallest prime number as possible before you start solving. It might look like a lot of work but simplifying would let you know that it is just a piece of cake. 7. B A =C CCB A A A F =C E A = = Comment: Simplify term by term to solve. 8. AB AB Multiply both numerator and ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE 9. denominator by conjugate :( = AB D AB AB AB = ABEBEB C AEEB = B = Number of boys playing for at least one team = = 21 boys

9 ABCCCCCCCCCCCADEFDDABC Comment: First draw a Venn diagram. If 4 members of the 11 man Cricket (C) team are also members of the 14 man Rugby (R) team. Therefore, 4 men are intersecting. Thus, 7 (11-4) men play Cricket alone and 10 (14 4) play Rugby alone. 10. {} Null set Comment: There is a difference between a Null set {} and an empty set. 11. D B D A D D CCD B A D CD B A CCCC C Comment: This question is all about substitution. The better you are at substitution, the easier it would be for you. 12. D A D D C D A A A A C A (p 2) (p + 3) = 0 Either p = 2 or -3 P = 2. Comment: After substituting, we factorise. We then get 2 or -3. Since 2 is the positive integer, we choose A Square both sides: C C Comment: Solve step by step and carefully so as not to leave out any letter. ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

10 ABCCCCCCCCCCCADEFDDABC 14. Solve the simultaneous equation: D A DC D A D DC DD D D D Either x = 2 or x = 3 Points of intersection: (2, 0) and (3, 0) Comment: At the point at which a curve crosses the x-axis, y = 0. That is why we equated the expression to zero. 15. D A D D = D A D D = DD D = D D 16. D D D 17. eqn (ii) eqn (i): 6d = 18 d = 3 Recall: a + 3d = 13 a + 3(3) = 13 a + 9 = 13 a = 4 A A A A Comment: Don t forget your basic A.P and G.P formulae: 1) 2) 3) 4) 5)C = A A A A A A A FAC AB A B Multiply by D D D D D Let x = = A(-1) A = 2 (x 3): Let x = 3: = B(1) B = 3 FCA AB A B D C D D D D ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

11 ABCCCCCCCCCCCADEFDDABC 20. a.b = a + b ab Comment: Commutative means that if you substitute a for b and b for a you will still get the same answer. Option B is the only one that satisfies this condition. 21. C Thus; 2 * (12 * 27) = 2 * (CDC) = 2 * = 2 *18 = +CDC = = 6 Comment: In binary operations, it is advisable you solve them term by term. Here, we first solved (12 * 27), got 18 as our answer before we then solved (2 *18). 22. A B a = 1; r = A = = Comment: Another question on AP and GP. Look at Question 17 to get the AP and GP formulae. 23. C C C {(-2 x 1 x -1) + (1 x k x 1) + (1 x 2 x 3)} {(1 x 1 x 1) + (3 x k x -2) + (-1 x 2 x 1)} = 23 {2 + k +6} {1 6k 2} = 23 {k + 8} {-6k - 1} = 23 K k + 1 = 23 7k = 14 k = 2 Comment: In solving matrix, write the first two columns on the right side of the third column and then draw two sets of diagonals. The first set goes from the top left to the bottom right, while the second sets goes from the bottom left to the top right. All figures crossed out by a particular diagonal should be multiplied with figures crossed out by that same diagonal, and then summed up with result from similar facing diagonals. Finally, subtract upward facing diagonal from downward facing diagonal. ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

12 ABCCCCCCCCCCCADEFDDABC 24. D D XY = D D D D D D Comment: <XZY = 24 (sum of <s. in a triangle). Let P be the point of intersection of the two bisectors. <PYZ = 56 (bisected angle) and <PZY = 12. Thus, the acute angle between <PYZ and <PZY is 68 ( ) BE FF AA CCC CC BE AA B 27. x 2y + 4 = 0 2y = x + 4 y = ½x + 2 Recall: y = mx + C Thus: A A C C ( ½) A = -1 A A Given D = (1, -1) D D D Comment: This question tests your knowledge about gradient and its formulae. Don t forget: 1) A C C 2) A C 3) y = mx + C 4) A 5) ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

13 ABCCCCCCCCCCCADEFDDABC Comment: Your knowledge of mensuration of solid shapes is very important. Don t throw it into the bin. 28. F Comment: Since X and Y are bisected by P, then <PYU = <PXZ = 40. <XPO = 40 (alt. <s to <PXZ) and <YPO = 40 (alt. <s to <PYU). Thus, <XPY = C B h = 30 Tan 60 h = 30 h = A CD AA CDCC CDC A Using Sine rule: D AB F D ACCF D A F B ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

14 ABCCCCCCCCCCCADEFDDABC D Comment: This question tries to test your knowledge of trigonometry and basic angles. Remember: Calculators aren t allowed in the exam hall, so you have to memorise or understand how to get the basic angles. 32. CotxCosecX 33. D A D s DA D D D D D When D x = 2 Comment: To find the value of x parallel to the tangent, first differentiate y. When Cx is parallel to the tangent. 34. D C C A B B D A D C When y = 5, x = 2 5 = 2 A C 5 = C C = 3 Therefore: D A D Comment: Integration. Term by term integration. 35. D A D = D ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE B B D A D D = B A = B B A A A B C B C =D B D A D BC = B A B A = { } { } = 21 1 = 20 Comment: Under examination conditions, don t waste much time on step by step solution. The aim is to be as accurate as possible in the least amount of time. 36. <. of 15yrs = ÅCCF CÅC D = B DC = Method 1: If 90

15 ABCCCCCCCCCCCADEFDDABC Then 30C BCCB C = 10 students. Method 2: CC ÅC CCC DCC BE CDC X = 120 students No. who offered CRK = B BE CDC = 10 students. Comment: Method 1 is preferable because it saves you time. 38. C C AA E Mean = 1.09 Range (R) = R = 0.60 M + R = = Æ Å Å Class M id va lu e (x ) F D ( x D D D A 9 45 fc D D A f= Mean = Å Å = 90/18 DC= 5 Æ Å Å SD = f x = 90 fc D D A SD = Comment: Variance is the square of SD. X x - D A A =5 Mean = Å D AB C D D Variance = Å Variance = 5/4 Variance = 1.25 ABCDEFBDCEFACBCDCAABCDE DCCEAEBECDCADCEDCECBCDE

ADVANCED MATHS TEST - I PRELIMS

Model Papers Code : 1101 Advanced Math Test I & II ADVANCED MATHS TEST - I PRELIMS Max. Marks : 75 Duration : 75 Mins. General Instructions : 1. Please find the Answer Sheets (OMR) with in the envelop