BUNYORE- MARANDA (BUMA II) JOINT EXAMINATIONS Kenya Certificate of Secondary Education

MATHEMATICS PAPER 121/2, 2013 MARKING SCHEME

1.SECTION 1 (50 MKS)

V0.002621 x 49.4Number S.form logarithms0.002621 2.621 xlO’3 3.4185

49.4 4.94x10' 1.69373.696 3.696 xl0° 0.5677

2.5445lO-'x 10° 27225 2 + 0.5445

2 20.1872 0.1872 1.27225Answer = 0.1872

2 The length and width of a rectangular shape measured to the nearest millimeter are 6.8 cm and 5.3cm respectively. Find to three significant figures the percentage error in the area of the rectangular shape.Actual area = Lx W = 6.8cm x 5.3cm = 36.04cm2 Maximum area = LxW

(6.85 x 5.35) cm2 = 36.6475cm2

Minimum area =Lx W = 6.75 x 5.25cm = 35.4375cm2

Absolute erroi- Maximum - minimum Area = 36.6475cm2 -35.4375cm2

2= 0.6050cm3

%error = Absolute error x 100 Actual area

= 0.605cm2 x 100 = 1.6787%36.04 cm2

= 1.68%(4s.f)

Va2 + x2iL * x

iL=_e x Va2 + x2

X Vi \2= e2

,£.+,X'

x2(e2) = y2(a2 + x2) x2e2 = y2a2 + y2x2 x2e2-y2x2 = y2a2 x2(e^-\2) - v2a2

e2-y2 e2y2

x2 = v2a2 2 2

e - yVx2 =

e-y* x = Vv;+a2

4. A quantity B is directly proportional to the square of C and inversely proportional to the square

root of D.Find the percentage change in B if C is increased by 25%.and D is decreased by 36%B =_Cf B = kcl

VD VDIncreased by 25%C, = 125%C D decreased by 36% New D = 64%

= 1.953IS -B x 100

B, = k( 1.25)

B\ 0 64 D

= kc2 fl.25)2

B= 0.953 IB x 100 =

BIncreased by 95.3 i%

VD x V0.64 B, = 1.953 IB

%ofB = B,-Bx 100 B

95.31%

=<2x-yQ)5 = 2x5 - 5x4y + 5xV

5. Write down the expansion of (2x-y/2)5 up to thefourth term.Hence evaluate(19.95)2 correct to 4s.f Co effients 1 5 10 10 5 5 1l(2x)5 + 5(2x)4 (-y/2)1 + 10(2x)3 (-v/2): + 10(2x)2(-y/3)3 +5 (2k)1 (-y/3)4 + l(2x)°(-y/3)5

5x4 (-y) + 5x3y2- 20x2(-y/9) +1 Ox <v4/81) + (-y5/243).........................................-20xV + j0xy4-v£

9 81 243 = (2x~y5 = (19.95)2

2= 2x - y = 2

2

61.+V2 + 1-V2 V5+V3 V5-V3

1+V2 + 1-V2 1+V2(V5 -V3) -t- 1-V2(V5 + V3)V5 +V3 V5-V3 V5 +V3) (V5-V3)

= lfV5-V3H V2fV5-V3') + HV5+V3W2fV.5+V3') V5(V5-V3) + V3(V5-V3)

= NumeratorV5-V3+V10 - V6 + V5+V3-V10-V6 2V5-2V6

Denominator 5-Vl5+Vl5-3 5-3 = 2 2V5 - 2V6 = l V 5 - lV6

2

7. A,B and C are points with position vectors3 -1 and 6

6 4 - 7 .42 0

AB = B -A Position vectors of D

A : B A+ ’/3 (B)'A % D-2A

OD= 2/3(3 6 4

8. The velocity v meters moved by a particle along a straight line after t seconds in motion is given bv v=7+8t2-2t3

© 2013 BUMA II JOINT EXAM IN A TIONS MA THEMA TICS P2 1

Y= 7-+8t'-2t3

Intergrate ds = 7+8t2-2t3 dt.

7_ + 8r - 2tJ + c0+1 2+1 3+1 7t +. 8f' -It4 + c

3 4S is 0 when t = O S= 7t +_8t3 -_2t4

4 4 When t = 3S =7(3) 4ji(3)3-

Distance S :

-2i3)4

452.5 metres.

9.The sixth term of an arithmetic progression is 5 'A

and twelfth term is 8 'A .Find the difference in the sums of the 40 terms and SO terms of the progression.

1 term = a + 5d a+5d = 5 'A

12th term = a+] 1 d = 8 'A a+11 d=8.5 a+5d= 5.5 _6d = 3_= d = !/,

a+

= 1= d =5 6 "z5 = 5.5

a + 2.5 = 5.5 a =3

•2.5

Sn40 = n (2a +(n-l)d)2

Sn40 = 40 (2x3 + 14 (40-1)2

40 (25.5)2

=510Sn50 50(2x3 + ’A (50-1)

2= 762.5

Difference in sum of terms = 762.5 - 510 =252.50

10 Slvia sold her plot of land at sh 850,000 and invested the money in a financial institution which pays compound interest at 15% p.a compoandedsemi-annuaily.If she withdrew the amount after 2 years,hpw much did she receive?CI = P( i+r/100)n P=850,000 R= 15/100 T = 2yrs Rate = 'A (1 5/100)

Semi annually = j_5 100

»1 year = 2 halves 2 years = ?

2years has 4 halves Cl = 850,000 (1 +.15}

200 .

= 850,000 (1.075)4

Amount received = l,135,148.77Ksh

31.Height(cm) 10-

1920-29

30-39

40-49

50-59

60-69

No.ofseedlings

18 32 38 52 40 20

Use the table to calculate the semi-interquatile range.

Class f c.f L.c.b10-19 18 18 9.520-29 32 50 19.530-3? 38 88 29.540-49 52 140 39.550-59 40 180 49.560-69 20 200 59.5Upper Quartile 3(200) = 150

449.5 + (150-140) x 10 =52

40Lower quartile = %(n)

= V* x 200 =50 L.O = 19.5 +(50-18) x 10

32= 29.5

Semi I.R = U.O - L.O2

= 52-29.5 =11.252

12.Find the shortest distance between the places A(48°N,35°) and B(48°N,145°W) lying on the earths

. surface in nautical miles 2 inksA(48°N,35°) B(48°N,145°W)Difference in longitude 35+145 = 180°

Distance = 0x2n Cos360

Distance = 180 x 2x 22x 6371km360 7

= 20,023.1427km 20,023.1427 km Cos 48°

= 13,398.0976 km ■Inm = 1,853km

? = 13,398.0976km = 13.398.0976 x Inm

1.853km = 7,230.4898nm

13.P(A and Sam os a) or (Pb and Samosa)'A x 5 + 'A x 3

10 9 5.+J3 =90 + 30 = 120 10 18 180 180 = Probability of picking Samosa =J2

314.M = 2AB = EC’1 AB = A x BAB = 2 1 x 1 3 =5 -3

-3 4 3 1 9 10 2AB = 2[ 5 -3 =

9 102AB = 10 -6 = M = 2AB

18 20 M = 10 -6

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18 20

15. Find the values of A and nGradient = Av

Ax= log(3-l) log 2 = -2

Log(1-2) log-1 y= mx + c y = -2x + c at y intercepts ~0y = _2x + 0y =-2x A = -2 n =116. Let the length AB be x AC + BC = (CD)2

(x + 7cm) x 7cm = gem2 7xcm + 49cm2 8/cm2 - 49cm2 7xcm - 32 cm2 7cm 7cm

X = 4.5714cm AC = AB + BC 4.5714cm + 7cm = 11.5714cm

SECTION II 50MKSHeight 7- 10- 13- 16- 19- 22- 25-

9 12 15 18 21 24 27Frequency 5 9 18 28 20 14 6a) Draw a cumulative frequency graph to represent the data aboveHeight f c.f u.c.l7-9 5 5 9.510-12 9 14 12.513-15 18 32 15.516-15 28 60 18.519-21 20 80 21.522-24 14 94 24.525-27 6 100

Cm

Use the graph in a) to estimatei) The meridian height of the plant shoots incmcorrect to Idecimal place.

Median 17.5ii) The quartile deviationUpper Quatile= % <n) % x 100= 75th position on c.f Upper quartile= 19.9

'/.x 100 = 25 Quatile deviation = U.O - L.O

c.f= 19.5 - 13.8

100 =0.057iii) The percentage of maize seedlings whose height was between 18.5cm and 23.5In 18.5cm = cf = 60 23.5cm, cf 87% of maize seedling = 87-60 x 100

100= 27 x 100

100 = 27%

18. Calculatea) Otieno’s monthly tax paid in kshs.Taxable pay = Basic salary + Allowances

= 23,600 + 12,500 + 2,800 + 830 = 39,730ksh x 12 = 23,838 kp.a

b) Otieno’s monthly tax paid in kshs.Is1 5808 x 2 = 11,616 monthlytax

5472 x 3 = 16,416 Tax less relief +other deductions .

5472x4 =21,888 86.964 =- 7,247-(1,162 + 450)

5472x5 =27,360 12 =5,635 + 10,000 1614x 6 =9,684 = 15,635

c) Otieno’s monthly net income from his earmings. Monthly Net income= 39,730 - 15,635 = 24,095ksh

d) If he received a 10% increase in his basic salary ina certain month,Calculate his new net tax month.Basic salary = 23,600 = 100%

= 110%

110x23600100

N.B.S -= 25,960Taxable income = 25.960 + (12.500 + 2.800 + 830)

20Taxable income= 25,254

X 0 30 60 90 120 150 180 210 240 270 300 330 360SinX

0.00 0.50 0.87 1.10 0.87 0.5 0 -0.5 -0.87 -1.00 -0.87 -0.5

. .Median = n+1 100+i 503t position on c.f

2 2

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5808 x2 = 11,616 5472x3 = 16,416

'5472x4 = 21.888 '5472x5 =27,360 3,030x6= 18,180

New tax for that month = 95.460 =7,955

1219, a)b)Draw the graph of the functions y=sin x andy=2sin(x+30) on the pair of axes lising the scale lcm represents 300 on the x-axis and 1 cm represents 0.5 units on the y-axss.

c) From your graph, find the roots of the equation 2sin(x+30) = sin x

X=126°, 306°

2sin(x+30)°-Sin =0d) State the range of values of x wbtcjj satsfy the

v(CF)2 = VsOOcm2 CF « 22.3607cm (4d.p)

c)The angle between the plane BCTS and the base BCDES-midpoint of FE SE= ’/a (FE)

Vz x 10 = 5 cm L.S.F = FS = FP

AC AD8 x 5 cm = FP x 8

10 8 FP= 40cm x 4cm

10cm SOHCAHTOA Tan = Opp 4cm =_I

Hyp 20cm 5 0 = tarf‘ 1

5= 11.31°

d) The angle between the planes FBC and BCBE SOHCAHTOATan = opp Tan 0 = 8cm

Adj 20cm© = Tan’1 8

20 '

0 = 21.80°

21. The product of the first three terms of an increasing geometric progression is 64.1f the first term is a and the common ratio is r.a)Express r in terms of a1st term = a 2nd term =ar3 rd term = ar2 ax ar x ar2 = 64

JVr3=3V64 . A3

r= N64

a3

r=_4a

inequality.2sin(x+30)° — Sin x <02Sin(x+30)° < Sin x Range of x = 180° <300°

20.Calculatea) The length ofCFCF2 = CD2 +DF2

CF2 = (200cm)2 + (10cm)2

b) Given the sum of the three terms is 14 i)Find the values of a and r and hence write down two possible sequences each upto the 5th terma+a(4/9) +a(4/a) a + 4+9 x 16

a2

a+16 + 4 = 1 4 9 1

9f a2+16+491 =

2 _ 14

1 4 x 9

a* -10a+16=0 a2-2a-8a+16=0 a(a-2)-8(a-2)=0 (a-8) (a-2) = 0 a-8=0; a=8 a-2 = 0 a=2

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2*»d ,

a‘+] 6+4a = 14a a=8 ; a= 2a2 -10a+16=0

r= 4 r= 4; r = '/2 a 8

1st sequence when r= '/2, a=8 =8, 4,2,1,0.25

2nd sequence when r = 2, and a = 2 =2, 4, 8, 16, 32'

ii) Find the product of the 30th terms of the two sequences30 tertn =a(r)29 When a =8, r= Vi

=8 (Vz)29

1 sequence When a-2, r= 2 30th term =a(r)29 2(2)®

Product of the term 30th term 8(*/2 )29 x 2(2)29 = 16

22. a) A point P (3,3) is mapped ontoP‘(3,9)And Q (2, 1) is mapped onto Q‘(2,5) by a shear with x-axis invariant.i) Determine the matrix

representing this shearShear factor = 5J. = 2

2Matrix = (12

0 1.

ii) Draw a triangle PQR and its image triangle P'Q’R under the shear given that (-3, 5) is the coordinate of R

b) T.M x OM = 1M1 0 3 2-3 = 3 2-32-1 9 5-1 -3 -1 -5

Triangle = P(3,-3) Q= (2,-1) R(-3,-5)

c) Determine a single matrix of transformation that maps PnQllR11 onto PQR

12 1 0 = 5 - 2 0 1 2 -1 2 - 1

Transformation matrix = 5 -2 2 -1

23. An aeroplane leaves airport A (69°N,30°W) for an airport D (31°S,90°E) through airport C (69°N,90°E)

a) Calculate the distance in kilometers from A and D via C

A(69°N,30°W) D(31 °S,90°E) C(69°N,90°E) Difference in longitude

90°+ 30° =120°

1 2 3 4 5 6 7 8 9 10 11 12 13t 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5 12.5V "3.25 5.25 923 i5.25 23.25 33.25 45.25 59.25 75.25 93.25 113.25 135.25

Distance^ % JzdtCosO 360

35120 x 2x22 x 6371 x Cos69°360 7

= 4,783.7684km = 4,783.7684km

b) Calculate the distance between D and C in kilometers.D (31 °S, 90°E) C (69°N,90°E)Difference in latitudes =31° + 69° =100°©_x 2kR 360= 100 X 2x 22 x 6371 km

360= 11,123.9683kmc)Time = total distance

speed= 4.783.7684km + 11.123.9683

850km/hr = 15.907.7367km

850km/hr = 18.71 hr/19hrs

d) Departure time at = 8.00 a.min 24 hr clock. 0800+1900hrs-2700hrs - 2400hrs = 03.00 a.mThe local time at C 3:00 a.m the following day.

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Area = (Width x sum of mid ordinates)2(3.25+5.25+9.25+15.5+23.25+33.25+45.25+59.25+75.25+93.25+113.25+135.25)= 2 x 611 = 1,222 square units.

b) Determine the exact displacement of the particle between t=l and t=13y= I2 _ 2t +4 S,3ft2-2t+4dt = G - 2t +4t +c

3 2 1 s=i! -t2+4t+c)n

3Q3}! - (13)2+(l3) +c-IOf-(l)2+4( 1 )+c

3 3= 6151+c- 1 [3Jl+ c]

3 3= 6151-31 = 6 1 2 Units

3 3- =612units

ii)Calculate the percentage error arising from the estimate in (a) above.

% error = 1222-612 =610

= 610 x 100 1222

= 49.9182%

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