KCSE 2012 Mathematics Paper 2 with Marking Scheme

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

1.  Evaluate   giving the answer to 4 significant figures. (2 marks)
2. Make n the subject of the equation. (3 marks)
   
3. An inlet tap can fill an empty tank in 6 hours . It takes 10 hours to fill the tank when the inlet tap and an outlet tap are both opened at the same time .Calculate the time the outlet tap takes to empty the full tank when the inlet tap is closed. (3 marks)
4. Given that P = 2i - 3j + k.Q = 3i - 4j - 3k and R =3P + 2Q . Find the magnitude of R to 2 significant figures. ( 3 marks)
5. Solve the equation Sin (2t + 10)⁰ = 0.5 for 0⁰ ≤ t ≤ 180⁰ (2 marks)
6. Construct a circle center x and radius 2.5 cm. Construct a tangent from a point p,6 cm from x to touch the circle at R . Measure the length PR. (4 marks)
7. Kago deposited Ksh 30000 in a financial institution that paid simple interest at the rate of 12% per annum . Nekesa deposited the same amount of money as Kago in another financial institution that paid compound interest .After 5 years ,they had equal amounts of money in the financial institutions.
   Determine the compound interest rate per annum, to 1 decimal place , for Nekesa’s deposit. (4 marks)
8. The masses in kilograms of 20 bags of maize were : 90,94,96,98,99,102,105, 91 ,102, 99,105, 94, 99, 90, 94, 99, 98, 96,102 and 105 .
   Using an assumed mean of 96kg, calculate the mean mass ,per bag of the maize. (3 marks)
9. Solve the equations (4 marks)
   x + y =17
   xy - 5x = 32
10. Simplify ,leaving the answer in the form a+b√c,where a,b and c are integers .( 2 marks)
11. The base and height of a right angled triangle were measured as 6.4cm and 3.5 cm respectively. Calculate the maximum absolute error in the area of the triangle. (3 marks)
12. 1. Expand (1+x)⁷ up to the 4th term. (1 mark)
    2. Use the expansion in part (a) above to find the appropriate value of (0.94)⁷. (2 marks)
13. The graph below shows the relationship between distance s metres and time t seconds in the interval 0 ≤ t ≤ 6.
    
    Use the graph to determine ;
    1. The average rate of change of distance between t = 3 seconds and t =6 seconds: (2 marks)
    2. The gradient at t =3 seconds. (2 marks)
14. In the figure below, the tangent ST meets chord VU produced at T .Chord SW passes through the center, O, of the circle and intersects chord VO at X .Line ST =12 cm and UT = 8 cm.
    
    
    1. Calculate the length of chord VU. (2 marks)
    2. If WX = 3 cm and VX:XU = 2:3 find SX. ( 2 marks)
15. Three quantities P,Q and R are such that P varies directly as Q and inversely as the square root of R .When P = 8,Q =10 and R =16. Determine the equation connecting P,Q and R .(3 marks)
16. In the figure below, VABCD is a right pyramid on a rectangular base. Point O is vertically below the vertex V, AB = 24cm,BC=10 cm and CV = 26cm
    
    Calculate the angle between the edge CV and the base ABCD. (3 marks)
    

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

17. Amaya was paid an initial salary of Ksh 180000 per annum with a fixed annual increment .Bundiwas paid an initial salary of ksh 150000 per annum with 10% increment compounded annually.
    
    1. Given that Amaya annual salary in the 11th year was ksh 288 000, determine:
       1. his annual increment .( 2 marks)
       2. The total amount of money Amaya earned during the 11 years. (2 marks)
    2. Determine Bundi’s monthly earnings correct to the nearest shilling during the eleventh year. (2 marks)
    3. Determine, correct to the nearest shilling:
       1. The total amount of money Bundi earned during the 11 years .( 2 marks)
       2. The difference between Bundi’s and Amaya average monthly earnings during the 11 years. (2 marks)
18. OABC is a parallelogram with vertices O (0,0),A(2,0),B(3,2) and C (1,2)
    O’A’B’C’ is the image of OABC under transformation matrix (^-2₀ ⁰_-2)
    
    1. 1. Find the coordinates of O’A’B’C’. (2 marks)
       2. On the grid provided below draw OABC and O’A’B’C’. (2 marks)
    2. 1. Find O’’A’’B’’C’’, the image of O’A’B’C’ under the transformation matrix
          (¹₀ ⁰_-2)  (2 marks)
       2. On the same grid draw O’’A’’B’’C’’ (1 mark)
    3. Find the single matrix that maps O’’A’’B’’C’’ onto OABC (3 marks)
19. In the triangle OPQ below, OP = p, OQ = q Point M lies on OP such that OM:MP = 2:3 and point N lies on OQ such that ON :NQ = 5:1 .line PN intersects MQ at X.
    
    
    1. Express in terms of p and q:
       1. PN: (1 mark)
       2. QM: (1 mark)
    2. Given that PX=kPN and QX =rQM ,where k and r are scalars:
       1. Write two different expressions for OX in terms of p,q k and r: (2 marks)
       2. Find the values of k and r : (4 marks)
       3. Determine the ratio in which x divides line MQ. (2 marks)
20. In June of a certain year, an employee’s basic salary was ksh 17000. The employee was also paid a house allowance of ksh 6000 , a commuter allowance of ksh 2500 and a medical allowance of ksh 1800. In July of that year the employee’s basic salary was raised by 2%.
    
    1. Calculate the employees;
       1. Basic salary for July : (2 mark)
       2. Total taxable income in July of that year. (2 marks)
    2. In that year ,the income Tax Rates were as shown in the table below:
       
       Given that the monthly personal Relief was ksh 1056, calculate the net tax paid by the employee, (6 marks)
21. 1. On the same diagram construct:
       1. Triangle ABC such that AB =9 cm,AC=7 cm and angle CAB=60⁰ : (2 marks)
       2. The locus of a point P such that P is equidistant from A and B: (1 mark)
       3. The locus of point Q such that CQ ≤ 3.5 cm (1 mark)
    2. On the diagram in part (a)
       1. Shade the region R, containing all the points enclosed by the locus of P and the locus Of Q such that AP≥BP. (2 marks)
       2. Find the area of the region shaded in part (b)(i) above (4 marks)
22. A tourist took 1 h 20 minutes to travel by an aircraft from town T(3°S,35°E) to town U (9°N,35°E).
    (Take the radius of the earth to be 6370km and π= ²²/₇)
    
    1. Find the average speed of the aircraft. (3 marks)
    2. After staying at town U for 30 minutes ,the tourist took a second aircraft to town V(9°N,5°E) The average speed of the second aircraft was 90% that of the first aircraft. Determine the time to the nearest minute ,the aircraft took to travel from U to V. ( 3 marks)
    3. When the journey started at town T, Local time was 0700h. Find the local time at V when the tourist arrived. (4 marks)
23. A box contains 3 brown, 9 pink and 15 white clothes pegs. The pegs are identical except for the colour.
    
    1. Find the probability of picking:
       1. A brown peg: (1 mark)
       2. A pink or white peg:(2 marks)
    2. Two pegs are picked at random, one at a time, without replacement .Find the probability that
       1. A white peg and a brown peg are picked: (3 marks)
       2. Both pegs are of the same colour : (4 marks)
24. The acceleration of a body moving along a straight line is (4-t) m/s² and its velocity is v m/s after t seconds
    1. 1. if the initial velocity of the body is 3m/s express the velocity v in terms of t (3 marks)
       2. Find the velocity of body after 2 seconds. (2 marks)
    2. Calculate:
       1. Time taken to attain maximum velocity. (2 marks)
       2. The distance covered by the body to attain the maximum velocity. (3 marks)

MARKING SCHEME

1.   5log4 − 4log5
   ¹/₅log4 +¼log5
   = 3.010299957 − 2.795880017
      0.1204119980 + 0.174742501
   =0.726466785
   ≃ 0.7265  (4s.f)
2.  
   
3. Fractionfilled by inlet tap in 1hr = ¹/₆
   Fraction filled when two taps open in 1h = ¹/₁₀
   ∴ fraction emptied by outlet tap in
   1h = ¹/₆ − ¹/₁₀ = ¹/₁₅
   Time for outlet tap to empty tank = 15h
4. R = 6i − 9j + 3k +6i − 8j − 6k
   =12i − 7j − 3k
   /R/ = √(12² + 17² + 3²)
   =√442
   -21.02 =21   (2 s.f)
5. Sin(2t + 10)° = 0.5
   2t + 10=30°, 150°
   t=10°, 70°
6.  
   
   Drawing circle
   Fixing point P
   Bisecting XP and drawing tangent
   RP = 5.4 ± 0.1cm
7. Amount for Kago
   =30000 + 12 × 30000 × 5
                  100
   =48000
   Compound interest rate for Nekesa
   30000(1 + ^r/₁₀₀)⁵ =48000
   (1 + ^r/₁₀₀)⁵ = 48000 = 1.6
                         30000
   1 + ^r/₁₀₀ = √1.6
   r = 100(1.0985605543 − 1)
   =9.9%
8. Differences from assumed mean
   −6−2+0+2+3+6+9−5+6+3+9−2+3−6−2+3+2+0+6+9=38
   ∴ mean = 96 + ³⁸/₂₀
              =97.9
9. x+y=17.....(i)
   xy−5x=32.....(ii)
   from (i) y=17−x
   substituting y=17−x in (ii)
   x(17−x) −5x =32
   17x −x² −5x = 32
   17x − x² −5x = 32
   x² −12x + 32 = 0
   (x−4)(x−8) = 0
   x=4 or x=8
   substituting x=4 in (i) 4+y =17 ⇒ y =13
   substituting x=8 in (ii) 8+y =17 ⇒ y = 9
10.   √5   =   √5   ×  √5+2
    √5−2   √5−2       √5+2
    = 5+2√5
         5−4
    =5 + 2√5
11. minimum possible area
    =½(6.35 × 3.45)
    =10.95375cm²
    maximum possible area 
    =½×6.45×3.55
    =11.44875cm²
    maximum absolute error in area
    = 11.4478 − 10.95375
                    2
    =0.2475cm²
12.  
    1. (1+x)⁷ = 1⁷+7×1⁶×x+21×1⁵× x²+35×1⁴×x³+.....
                    = 1+7x + 21x²+35x³
    2. (0.94)⁷ =(1+(−0.06))³
       =1+7×(−0.06)+21×(−0.06)² +35 (−0.06)³
       =1 − 0.42 + 0.0756 − 0.00756
       =0.64804
13.  
     
    1. Average rate of change between t=3 and t=6
       20 − 2 
         6 −3
       =¹⁸/₃ = 6
    2. Gradient at t = 3 seconds
         6 − 0    = 6
       4.3−2.3      2
       = 3 ± 0.1
14.  
    1. let UV be x cm: VT × Ut = ST²
       (x+8)8 = 12²
       8x = 144 − 64
        8x = 80
       x =10cm
    2. VX = ²/₅ × 10 = 4cm
       XU = 10 − 4 = 6cm
       SX × XW = VX × XU
       SX × 3 = 4 × 6 
       SX = 8cm
15. P α   Q ⇒ P = kQ
          √R         √R
    8 = k × 10
            √16
    k = 3.2
    P = 3.2Q
           √R
16. OC = 
         =13
    ∠VCO = cos ^−1 13/₂₆
    =60°
17.  
    1.  
       1. 180000 +(11−1)x = 288000
                               10x = 108000
                                   x = 10800
       2. S_n = ¹¹/₂(180000 + 288000)
              =2574000
    2. 150000 × 1.1¹⁰
                12
       =32422
    3.  
       1. (150000 ×(1.1ⁿ − 1))
                   (1.1−1)
          =2779675
       2. Difference between monthly averages for the 11 years
          2779675 − 2574000
                   11×12
          =1558
18.  
    1.  
       
       co-ordintes of O'A'B'C'
       O'(0,0), A'(–4,0), B'(–6, –4), C'(–2,–4)
       
    2.  
       
    3.   
       
19.  
    1.  
       1. PN = ⁵/₆q−p
       2. QM = ²/₅p−q
    2.  
       1. OX = p+k(⁵/₆q−p)
          OX =q+r(²/₅p − q)
       2. p+k(⁵/₆q−p) = q+r(²/₅p − q)
          p(1−k) + ⁵/₆kq =q(1−r) + ²/₅rp
          1−k = ²/₅r and 1−r = ⁵/₆k
          1−r = ⁵/₆(1−²/₅r)
          1−r = ⁵/₆ − ¹/₃r
          ¹/₆ = ²/₃r ⇒ r = ¼
          k = 1−²/₅r ⇒ k = 1− ²/₅ × ¼ = ⁹/₁₀
       3. QX = ¼OM
          MX = ¾OM
          ∴ MX:XO = ¾: ¼ = 3:1
          
20.   
    1.  
       1. July basic salary
          =17000 × 1.02
          =17340
       2. Total taxable income
          =17340+6000+2500+1800
          =27640
    2. Gross tax
       1^st bracket: 9680 × 10% = 968
       2^ndbracket:(18800 − 9680) × 15% = 1368
       3^rd bracket: (27640 − 18800) × 20% = 1768
       Gross tax: 968 + 1368 + 1768
                     =4104
       Net tax : 4104 − 1056 = 3048
21.  
    
    
    1. locus of P
       locus of Q
    2.  
       1. shading region R
       2. area of shaded region
          area of minor sector P₁CP₂
          ^{= 146}/₃₆₀ × π × 3.5²
          ≃ 15.6cm²
          area of ∆P₁CP₂
          ½ × 3.5² sin 146°
          ≃3.4cm²
          ∴ shaded area
          15.6 − 3.4
          = 12.2cm² 
22.  
    1. distance from T to U
       =2 × 6370 × ²²/₇ × ¹²/₃₆₀
       
       =1001km/h
    2.  
       
       = 3.658104965h
       ≃ 3h 39min
    3. Arrival time at U
       0700 + 1h 20min
       =0820h
       Departure time at U 
       0820 + 30min
       =0850h
       Time difference between U and V 
       35 − 5 × 24 
          360
       =2h
       Arrival time at V (local time)
       0850h + 3h 39min − 2h
       =1029h
23.  
    1.  
       1. P(brown) = ³/₂₇
          
       2. p(pink or white)
          = ⁹/₂₇ + ¹⁵/₂₇
          ^{= 8}/₉
    2.  
       1. P(white and brown)
          = ¹⁵/₂₇ × ³/₂₆ + ³/₂₇ × ¹⁵/₂₆
          ^{= 5}/₇₈ + ⁵/₇₈ = ⁵/₃₉
       2. white, white + pink, pink + brown, brown
          = ¹⁵/₂₇ × ¹⁴/₂₆ + ⁹/₂₇ × ⁸/₂₆ + ³/₂₇ × ²/₂₆
          = ³⁵/₁₁₇ + ⁴/₃₉ + ¹/₁₁₇ = ¹⁶/₃₉
24.  
    1.  
       1. ^dv/_dt = 4 − t
          v = ∫(4 − t)dt
          =4t − ½t² + c
          when t =0, v=3m/s
          ∴ 3 = 4 × 0 − ½ × 0² + c
             3 = c
          ∴ V = 4t − ½t² + 3
       2. when t = 2 seconds
          V = 4 × 2 − ½ × 2² + 3
             = 8 − 2 + 3
             = 9m/s
    2.  
       1. At maximum velocity ^dv/_dt = 0
          i.e.  4 − t = 0
                       t = 4 seconds
       2.  
          
          =(2 × 16 − ¹/₆ × 64 + 12) − 0
          =32 − 10²/₃ + 12 = 33¹/₃