Kenya Certificate Of Secondary Education(KCSE 2013) Mathematics Paper 2 b with Marking Scheme

SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.

1. Round off each of the following numbers in the expression  to 3 significant figures, hence evaluate the expression. (2 marks)
2. Given the matrices A= (²₃ ⁴₀) and B = (²₁ ³₁), find AB – 5B. (3 marks)
3. Three types of coffee A, B and C are mixed such that A:B = 4:3 and B:C = 1:2.
   Determine the mass of type C in the mixture of 52kg. (3 marks)
4. In the Geometric Progression (G.P), the 4th term is 24, and the sixth term is 96.
   Determine:
   1. The common ratio of the G.P; (2 marks)
   2. The first term of the G.P. (2 marks)
5. Two fair dice are rolled together and the sum of the numbers showing on the top faces noted.
   1. Represent all the possible outcomes in the probability space. (2 marks)
   2. Determine the probability that the sum is greater than 6 but less than 10. (1 mark)
6. Two points A and B are such that OA = (²₅) and AB = (⁴₄), M is a point on AB such that AM: MB = 3:1.
   Determine:
   1. OB: (2 marks)
   2. the coordinates of M. (2 marks)
7. In the figure below, TPR is the tangent to circle at P. Angle APB = 75⁰ and angle BPR twice angle APT
   
    
   Determine the size of angle BAP. (2 marks)
8. Given that 2 cos⁡ (x-30)⁰ = -0.9,determine the value of x for 0° ≤ x ≤ 180⁰ correct to two decimal places. (3 marks)
9. The vertices of a triangle RST are R(1, 3), S(1, 7) and T(-1, 4). Triangle RST is mapped onto triangle R'S'T' by transformation matrices P = (⁰₁ ¹₀) followed by Q= (^-1₀ ⁰_-1)
   Find the coordinates of R'S'T'. (3 marks)
10. Using the method of completing the square, solve the equation 2x² + 8x=15, correct to one decimal place. (3 marks)
11.  
    1. Using a ruler and a pair of compasses only, construct an inscribed circle in triangle ABC given below. (2 marks)
       
    2. Measure the radius of the circle. (1 mark)
12. In a camp, there was enough food for 2000 people on equal rations for 90 days. After 20 days 500 more people joined the camp. Calculate the number of days that the remaining food would be used to feed the people. (4 marks)
13. Figure PQRS below represents a garden which,
    PQ = 80 m, PS = 75 m, SQ = 40 m, ∠SQR=68° and <SPQ = ∠SRQ.
    
    Calculate to 2 significant figures the length of SR. (3 marks)
14. The table below shows part of income tax rates in a certain year.
    
    In a certain month of that year, Abdala’s income was Kshs 21 820. He was entitled to a monthly personal relief of Ksh 1162. Calculate the income tax paid by Abdala that month. (4 marks)
15. Ali and Kinjo bought the same type of pencils and rubbers from the same shop. Ali bought 2 pencils and 3 rubbers for Ksh 66, Kinjo bought 7 pencils and 2 rubbers for Ksh 129. Find the cost of a pencil. (3 marks)
16. The table below shows marks scored by students in a mathematics test.
    
    On the grid provided below, draw an ogive to represent the data. (3 marks)

SECTION II (50 marks)
Answer only five questions from this section in the spaces provided.

17. Kurao borrowed Kshs 300 000 from a financial institution. The institution charged compound interest at the rate of 18% per annum on the outstanding balance at the end of each year. At the end of the first and the second years, he made equal repayments of Ksh 134 000. He repaid the remaining amount of money plus interest at the end of the third year.
    Calculate:
    
    1. The interest charged at the end of first year; (2 marks)
    2. The principal at the beginning of:
       1. The second year; (2 marks)
       2. The third year. (2 marks)
    3. The amount of money Kurao paid at the end of the third year. (2 marks)
    4. The total interest charged on the loan. (2 marks)
18.  
    1. The nth term of a sequence is given by U_n = n² – n + 3
       Determine: 
       
       1. The tenth term of the sequence; (2 marks)
       2. The difference between the 30th and the 20th terms of the sequence; (2 marks)
       3. The value of n given by U_n = 243. (3 marks)
    2. In a research, it was found that the number of bacteria tripled every hour. Given that the number of bacteria at the start of a certain hour was 180:
       1. Write an expression for the number of bacteria after t hours. (1 mark)
       2. Determine the number of bacteria, to the nearest million, after 12 hours. (2 marks)
          
          
19. The time in minutes for each student in a group took to solve a certain mathematics question is as shown in the table below. 
    
    1. State the modal class. (1 mark)
    2. The above data was represented in a pie chart, determine the angle that represented the number of students who the questions in 4 – 5 minutes. (2 marks)
    3. Calculate the mean time taken to solve the question. (4 marks)
    4. On the grid provided, draw a pie chart to represent the data. (3 marks)
20. A curve is represented by the equation y = -2x² + 7x – 3.
    1. Make a table of the values for: -1 ≤x ≤ 4 and the corresponding values of y for the curve. (2 marks)
    2. On the grid provided, draw a graph of y = -2x² + 7x – 3 for -1 ≤ x ≤ 4. (3 marks)
    3. Use the graph to determine
       1. The roots of the equation: -2x² + 7x – 3 = 0; (2 marks)
       2. The instantaneous rate of change of the curve at x = 1. (3 marks)
21. Four vectors are such that OA = -2i + j, OB =3i + 5j, OC = - 8i – 12j and OD = 2i – 4j.
    1. Express in terms of i and j, the vectors:
       
       1. AB; (2 marks)
       2. CD. (2 marks)
    2. Determine the coordinates of of the mid-point of AD. (3 marks)
    3. Calculate to 3 significant figures, the magnitude of BC. (3 marks)
22. Two points P and Q lie on the equator. The position of P is (0⁰, 12⁰E) and that of Q is (0⁰, 60⁰W).
    (Take the radius of the earth to be 6370 km and π = 22/7).
    
    1. 1. Calculate the distance from P to Q in kilometers. (3 marks)
       2. Determine the local time at Q when the time is 9.00 pm at P. (3 marks)
    2. A point T is due North of Q. An aeroplane flying from Q at 1001 km/h takes 2 h to reach T. Determine the position of T. (4 marks)
23. A relation connecting three variables R, C and T is such that R varies directly as the square of C and inversely as T. When R = 30, C = 6 and T = 2.4.
    
    1. Find:
       1. The constant of proportionality; (3 marks)
       2. The equation connecting R, C and T. (1 mark)
    2. Given that R = 40and C = 8, determine:
       1. The value of T; (2 marks)
       2. The percentage change in R when C decreases by 10% and T increases by 8%. (4 marks)
24. In the figure below, the area bounded by the curve, the y-axis, the x-axis and the line x = 5 represents a map of a piece of land.
    
    
    1. Estimate the area of the map in cm³ by:
       1. The counting technique; (2 marks)
       2. Using the trapezium rule with 5 strips of equal width. (3 marks)
    2. Given that the actual area of the map is 30⁵/₆ cm², calculate:
       1. The percentage error, correct to 2 significant figures, when the trapezium rule is used to estimate the area on the map; (2 marks)
       2. The actual area in hectares of the piece of land if the scale used was 1:12000. (3 marks)

MARKING SCHEME

1	4.957                      4.96 = 0.2638 - 0.0149   0.263 - 0.015 = 20	B1 B1 2
2		B1 M1 A1 : Substraction and multiplication by 5 3
3	A: B: C A: B: C 1: 2 3: 6 combined ratio A:B:C = 4:3:6 mass of type C = 24	B1 M1 A1 3
4	ar5  96 ar3  24 r2 = 4 $ r = !2 when r = 2 ( a x 2 = 24                         8 when r = - 2 ( a x - 2h = 24                              8	M1 A1 B1 B1 4
5	+ 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 P6 1 x 1 10h = 15 = 5     36    12	B2 B1 : probability space 3
6	OB =  co-ordinates of M OM = OA + 3/4AB coordinates of M are (5, 8)	M1 A1 M1 A1 4
7	Let angle APT = xc 3x + 75 = 180c x = 35c angle BAP = angle BPR = 2 x 35c = 70°	B1 B1 2
8	2 cosx - 30hc = - 0.9 cosx - 30hc = - 0.45 x = 146.74c	M1 A1 B1 3
9
10	2x2 + 8x = 15 x2 + 4x = 7.5 = 1.4 or - 5.4	M1 M1 A1 3
11	radius = 2.4 ! 0.1	B1 B1 B1 bisecting 2 or 3 angles constructing radius and com- pleting circle 3
12	Fraction of food per person per day    1                                                     2000 x 90   Fraction for 2000 persons for 20 days = 2000 x 20 2000 x 90 = 2     9 Remaining fraction of food = 7                                            9 No of days to feed 2000 + 500 persons = 7  1 x 2500    9   180000 7 x 72 = 56 9    1	M1 A1 M1 A1 4
13	cos P = 752 + 802 - 402 2 x 75 x 80 =10425= 0.86875   12000 P - 30c = 74 m	M1 M1 A1 3
14	1st bracket $ 10164 x 10 = 1016.4 2nd bracket $ ^19740 - 10164h x 15/100 = 1436.4 3rd bracket $ ^21820 - 19740h x 20/100 = 416 Net tax = ^1016.4 + 1436.4 + 416h - 1162 = 1706.8	M1 M1 M1 A1 4
15	2p + 3r = 66..... (i) 7p + 2r = 129...(ii) 4p + 6r = 132..(iii) 21p + 6r = 317.....(iv) 17p = 255 p = 15	M1 M1 A1 3
16	cf: 4, 10, 18, 28, 37, 44, 48, 50 B1 can be implied P1 C1 3
17	300000 x 0.18 = 54000       300000 + 54000 - 134000 = 220000 220000 x 1.18 - 134000 = 125600 125600 # 1.18 = 148208 Total interest charged: ^300000 + 22000 + 125600h x 0.18 = 54000 + 39600 + 22608 = 116208	M1 A1 M1 A1 M1 A1 M1 A1 M1 A1 or equivalent 134000 x 2 + 148208 - = 116208 10
18	U10 = 102 - 10 + 3 = 93 U30 - U20 = ^302 - 30 + 3h - ^20 - 20 + 3h2 = 873 - 383 = 490 n2 - n + 3 = 243 n2 - n - 240 = 0 ^n + 15hn - 16h = 0 n = - 15 or n = 16 n = 16       Number after t hours = 180 x 3t Number to the nearest million after 20 hours 180 x 312 = 95659380 = 96000000	M1 A1 M1 A1 M1 M1 A1 B1 M1 A1 10
19	Modal class: 4 - 5  8  x 360c 36 = 80c mid values 0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5 fx = 1, 6, 7.5, 17.5, 36, 33, 32.5, 22.5 fx = 1 + 6 + 7.5 + 17.5 + 36 + 33 + 32.5 + 22.5 mean = 156              36 = 41/3  ......	B1 M1 A1 M1 M1 M1 A1 S1 B2 : scale and labelling 8 bars : (allow B1 for 5 - 7 bars :) 10
20	x  -1 0 1 2 3 4 y -12 -3 2 3 0 -7                   Roots of equation x = 0.5 or x = 3 tangent line : drawn gradient: 5 - - 1                2 - 0 =3	B2 S1 P1 C1 B1 B1 B1 M1 A1 10
21	AB+ = OB+ - OA+ = 3i + 5j - ^- 2i + jh = 3i + 5j + 2i - j = 5i + 4j CD+ = OD+ - OC+ = 2i - 4j - ^- 8i - 12jh = 2i - 4j + 8i + 12j = 10i + 8j mid point of vector AD ` coordinates of mid point is (0, -1.5) BC+ = OC+ - OB+ = - 8i - 12j - ^3i + 5jh = 11i - 17j BC = 11 + 17 = 121 + 289 - 20.2	M1 A1 M1 A1 M1 A1 B1 M1 M1 A1 10
22	Longitude difference = 12° + 60° = 72° Distance PR = 8008 km Time difference = 72h                           15 = 4 h 48 min Local time at Q: = 9.00 pm - 4 h 48 min = 4.13 pm Distance travelled in 2 h = 1001 x 2 = 2002 km i = 2002 x 360 x 7       2 x 22 x 6370 = 18° Position of T: (18°N, 60°W)	M1 M1 A1 M1 M1 A1 B1 M1 A1 B1 10
23	R = 30, C = 62 and T = 2.4 ( 30 = k6 k = 30 x 2.4 = 2          36 R = 2C2         T         when R = 402 and C = 8 T = 2 x 8 = 3.2 New R = 1.08 x 3.2 = 30 % change in R =40 - 30x 100       40 = 25%	B1 M1 A1 B1 M1 A1 M1 A1 M1 A1 10
24	24 + 2^13h = 302 ½ x 1"2 + 2 + 26 + 8 + 8 + 6h,1 ½ = ^60h = 30 cm         % error = 2 26                   37 = 2.7 1 cm / 120 m 1 cm2 / 14400 m2 =44.4 ha	M1 A1 B1 M1 A1 M1 A1 B1 M1 A1 whole square and part square ordinates 2, 6, 8, 8, 6, 2 substitution into formula simplification 10