Question Number MARKING SCHEME PAPER IIB MAY 2010 SESSION Answer Mark Notes 1(a) 630 = 63 10 = 3 21 5 2 = 2 3 3 5 7 Factorisation process [at least two steps or 2 factors] Accept also in index form and in any order (b) 27 = 3 3 3 36 = 2 2 3 3 99 = 3 3 11 LCM = 2 2 3 3 3 11 = 1188 For two correct steps 1188 2(i) (iii),,,, Fraction nearest to is 0.8 0.2 = 0.6 Comparing size of fractions and/or changing all fractions to decimal form or 0.6 [or any equivalent value] Subtracting smallest from largest fraction 0.6 Using decimal form or LCM method to obtain boundaries In decimal form: and Examples: Any value written as a fraction or decimal LCM method: and Page 1 of 9
Examples: 3(i) Multiplying both sides by 100 = 4% Substitution in formula obtained in (i) or substitution in original formula 4% 4(i) 4km in 1 hour 1km in ¼ hour 2.6km in ¼ hour 2.6 = (13 20) 60 minutes = 39 minutes Direct proportion method up to 1 4 2.6 [or equivalent in minutes] 60 to convert from hours to minutes 39 minutes 8.15am 39min = 7.15am + 60min 39min = 7.15am + 21min = 7.36am ft Correct subtraction of minutes from minutes 7.36am [ft for incorrect part (i)] Total: 5 marks Page 2 of 9
5(i) Side of square = 15 5 = 3cm x = 7 3 = 21cm For 15 5 21cm Perimeter = 15 + 15 + 21 + 21 = 72cm Concept of Perimeter [Adding outer sides of figure] 72cm 6 Substituting given information to form an equation in either r or h Multiplying to remove denominator subject r = 3.8551 So h = 2 3.8551 = 7.7102 = 7.7cm 7.7cm 7(i) x + 2 x + 4 x + 2 x + 4 Adding x to the sum of both answers in part (i) to form equation Collecting like terms [up to 3x = 408 6 or 3x 408 = 0] 134 Total: 5 marks Page 3 of 9
8(i) P(vowel) = 6 in numerator 13 in denominator P(A or I) = 4 in numerator 13 in denominator 9 x = 40 [corresponding angles] EFC = 40 o [alternate angles] BFC = 40 15 = 25 o In BFC: y = 180 (90 + 25) [angles in sum = 180] y = 65 40 Any equivalent reason For angle BFC or EBF For y = 180 (90 + BFC) or y = 90 EBF 65 Award if all reasons for obtaining y are given 10(i) 5, 6, 7, 7, 10, 15, 24, 40, 45, 81 Median = (10 + 15) 2 = 12.5 Ascending or descending order and selecting 10 and 15 or 5.5 th term or between 5 th and 6 th term 12.5 or equivalent Mean = sum of ages total number of clients = (5 + 6 + +81) 10 = 240 10 = 24 For knowing how to use formula for determining mean 24 (iii) Mode = 7 7 Page 4 of 9
(iv) Mode is not a good average since sample is too small [or spread of data is too large] Award for any reasonable explanation 11(a)(i) 2 0 and 2 1 [or 1 and 2] Both correct and in this order 2 3 + 2 2 + 2 1 + 2 0 + 2 1 Conversion to fractions/decimals and addition of all five terms 3.875 or equivalent (b) 2 in numerator in numerator in numerator or in denominator 12 Longest distance = 90 + 70 + (90 40) = 210m By Pythagoras : AB 2 = 40 2 + 70 2 AB 2 = 6500 AB = 80.623m Shortest distance = 80.623m Difference = 210 80.623 210m Pythagoras theorem and substitution 80.623m Subtracting shortest from longest distance 130m Page 5 of 9
= 129.377 = 130m Total: 5 marks 13(i) Since D is mid-point of AC, then ABD and ACD are congruent by SSS or RHS AB = BC [given] AD = CD [D bisects AC] BD common side [or BD = BD] Or BDA = BDC = 90 o [property of isosceles ] For correct proof [that is; 3 statements, even without reasons] For at least one reason For stating SSS, SAS or RHS in conclusion AC 2 = 6.7cm In ABD: sinabd = 6.7 8.2 ABD = 54.793 o Hence ABC = 2 ABD = 109.586 = 110 o Using ½ AC Up to sinabd = AD AB Multiplying ABD by 2 110 o Total: 7 marks 14(i) a = 106 [corresponding angles] b = 106 [ext. angle = int. opp. angle of cyclic quad.] c = 180 106 = 74 106 Reason 106 Reason 74 Page 6 of 9
[opp. angles of cyclic quad. are supp.] Reason PR is not a diameter since PQR is not a right angle Or similar Total: 7 marks 15 0.876 = 1 1 = 1 0.876 270 = 1 0.876 270 = 308.219 = 308 Dividing by 0.876 Multiplying by 270 308.2 or higher accuracy 308 16 (9 3) (4 2) (2 1) = 27 8 2 =17 points Total: 3 marks Multiplying any one score with corresponding points + and assigned correctly 17 17 10.4cm 250 000 = 2 600 000cm = 26 000m = 26km For 10.4 250 000 100 to convert to metres and 1000 to convert to kilometers 26km Page 7 of 9
Total: 3 marks 18 Area large circle = 22 12 12 7 = 452.571 Area of small circle = 22 5 5 7 = 78.571 Shaded area = ¾ (452.571 78.571) = 280.5cm 2 Obtaining area of one of the circles Area of largest area of smallest circle Obtaining ¼ or ¾ of difference in areas 280.5cm 2 19(i) y Substituting any 2 values of x in equation to obtain the corresponding values of y Coordinates of one point on line Coordinates of another point on line Plotting the obtained two points correctly Accuracy of line drawn x ft ft x = 1.6 [accept values from 1.5 to 1.7, both values inclusive] y = 1.8 [accept values from 1.9 to 1.7, both values inclusive] [ft only if coordinates of point of intersection are read off correctly from graph] Page 8 of 9
Total: 7 marks 20(i) Area of trapezium = ½ (9.2 + 14.8) h = ½ 24h = 12h cm 2 Formula for area of trapezium Substitution of all values 12h cm 2 12h 21.8 = 1831.2 h = 1831.2 21.8 = 7cm For formula and substitution h subject 7cm Page 9 of 9