NYANDARUA WEST CLUSTERS EXAM, END OF TERM 1 2019

Mathematics Paper 2 Marking scheme

No. Working Marks

Remarks/ Alternative

1 ( 6.79 x0.391log5 )3/4

( 6.79 x0.3910.6990 )3/4

No.6.790.391

0.6990

2.72092.7209

Log0.83191.59220.4241

Log

0.4241-1.8445 0.5796 0.5796 x ¾ 0.4347

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For all logs from tables

+ and –

Correct x by ¾

042 L1 (0,0) and (4,1)

1−04−0

= 14

y−0x−0

=14

y= 14

x → y> 14

x

L2 (3,0) and (0,4)4−00−3

=−43

y−0x−3

=−43

y=−43

x+4→ y← 43

x+4

L3 (5,0) and (0,4)4−00−5

=−45

y−4x−0

=−45

y=−45

x+4→ y ≤−45

x+4

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033 (x+5/3)=0 and (x-2)=0

(x+5/3)(x-2)=0 M1

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x2−13

x−103

=0

3 x2−x−10=0 A1

024 5 (√11+√7 )+3 (√11−√7 )

(√11−√7 ) (√11+√7 )

= 5 (√11+√7 )+3√11−√711−7

2√11+ 12 √7

a= 2 b = ½

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035 7/9 = 0.777…..

Truncating to 3 d.p = 0.777 = 7771000

Absolute error = 79− 777

1000= 7

9000

% error = ( 79000

÷ 79 )x 100 = ( 7

9000x 9

7 ) x 100

= 0.1%

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036 1+16 x+112x2+448 x3

2x = 0.02x = 0.011+16(0.01)+112(0.01)2+448 (0.01)3

= 1.171648

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037 Log 27 (x + 5) = log 27 (x -3) = log 27 27

2/3

Log 27 x + 5 = log 27 272/3

x - 3 x + 5 = 9 x-3X + 5 = 9x – 27-8x = -32x = 4

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038 Det (1× 1 )−(3 ×1 )=−2

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[−12

12

32

−12 ]

[−12

12

32

−12 ][1 1

3 1] [ xy ]=[−1

212

32

−12 ] [ 7

15]

[1 00 1][ x

y ]=[43 ]=(4,3)

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039 a = 2x + 1

d = x + 1t2 = (2x + 1) + (x + 1) = 3x + 2(2x + 1) ( 3x + 2) = 02x + 1 = 0 → x = -1/2 Or ( 3x + 2) = 0 →x = - 2/3When x = -1/2 , D = ½ First 3 terms = 0, ½ , 1When x = - 2/3 , D = 1/3 First 3 terms = -1/3, 0 , 1/3

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0410

Both in 1 hr → 110 + 1

6 = 415

Both in 2 hour →2( 416) = 8

15

Remaining work = 1 – 815 = 7

15

If16→ 1 hr

715 →?

715 x 6 =

145 = 2

45 hrs

= 2 hours 48 min

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For 4

15

03

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11 HP=6000+3500 ×6=27000

21000=18000(1+ r100

)6

1.1667=(1+ r100

)6

1.026025=1+ r100

r=2.6 %

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0312 (8√3 )2=x × 3

4x

192=34

x2

X=256BD= 16cmEB=12cm

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0413

T = ¼ √ AN−A2

N−1

4T = √ AN−A2

N−1

16T2 = AN −A 2

N−116NT2 – 16T = AN – A2

16NT2 – AN = 16T2 – A2

N(16T2 – A) = 16T2– A2

N= 16T2 – A2

16 T2−A

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A103

14 0.9 ×57.29 °=51.561 °51.561

360×2× π× 13

3.72385 πcm

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0315 Mass x f Fx

9-11 10 4 4012-14 13 7 9115-17 16 9 14418-20 19 10 19021-23 22 8 176

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24-26 25 2 50= 40 = 691

69140

=17.275

17.28

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0316 2x2 + 8x + 2y2 – 10 = 0

X2 + 4x + y2 – 5 = 0X2 + 4x + (4/2)2 + y2 = 5 + (4/2)2

(x + 2)2 + y2 = 9Radius = √9 = ± 3 Radius = 3

Centre (-2,0)

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17

(a) (i) Cos A =

92 − (62 + 102)−2 × 6 × 10

A = 62.72

(ii) Cos B =

102 − (62 + 92)−2 × 6 × 9

B = 80.94 (iii) C = 180 - (62.72 + 88.94) = 36.34

(b)

10Sin 80 .94

= 2R = 9Sin 62.72

= 6Sin 36 .34

R = 5.063cm(c) Area of sequent (shaded) Area sector

= 2 × 80 . 94360

× 3.142 × (5 .063 )2 = 36 .217

Area Δ = 1

2 × (5 .063 )2 Sin (80 . 94 × 2 ) = 3 . 986 Area shaded = 32.23cm²

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1018 a) Taxable Income

27000+11000+[ 15100

×27000−3500]38000+550Sh 3855038550× 12

20=23130

3900 × 2 = 78003900 × 3 = 117003900 × 4 = 15600

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3900 × 5 = 195003900 × 7 = 273003630 × 9 = 32670

7800+11700+15600+19500+27300+32670=114570

11457012

=9547.50

9547.50−1056=8491.50

= sh 8491.50

b) Net Salary 38550 – (8491.50+2500+3000)

38550 – 14 441.50

= sh 24,108.50

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1019 i)

ii)

480 ,000P

480 ,000P − 4

480 , 000P − 4

− 480 , 000P

= 20000

480 , 000 P − ( P − 4 ) 480 ,000P ( P − 4 )

= 20 , 000

1, 920 , 000P2 − 4 P

= 20 , 000

20,000p² - 80000P – 1,920,000 = 02P² – 8P – 192 = 0

+ 8 ± √ (−8 )2 − 4 (2 ) (−192 )4

+ 8 ± 404

P = 12 men

Kamau : James2(3 : 2) James : Hassan 4 : 2 6 : 4 : 2

Kamau -

612

× 480 ,000 = 240 ,000

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James -

712

× 480 ,000 = 160 ,000

Hassan -

212

× 480 ,000 = 80 ,000

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10

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20

a) Q1 = 38.5 Q2 = 55.5 Q3 = 69.5

b) 15% to fail 12th student = 28.5

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For all cumulative frequencies

Vertical scale and labelling

Horizontal scale and labelling

Plotting and curve

1021 a)

b) P(B LH B’) or P(G LH B’)

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( 23

× 15

× 610 )+( 1

3× 1

10× 6

10 )

12150

+ 6300

¿1

10

c) P(B RH B) or P(B LH B) or P(G RH B) or P(G LH B)

( 23

× 45

× 310 )+( 2

3× 1

5× 4

10 )+( 13

× 910

× 310 )+( 1

3× 1

10× 4

10 )

24150

+ 8150

+ 27300

+ 4300

¿1960

d) 1 – (P(B RH B) or P(G RH B))

1−(( 23

× 45

× 310 )+(1

3× 9

10× 3

10 ))

1−( 24150

+ 27300

)

¿34

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10

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22

QT = 4.6cm

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L1

L2

Perpendicular bisector of QR

Bisector of L2 through R

Circle

Location of S

Perpendicular bisector of QS

Dotted line with unwanted region shaded

Location of T

1023 a)i Angle ADE = 90+27

=117°

ii)

180 − 1172

= 31.5

b) i

AD12

= Cos 27

AD = 10.69cm

ii

12

× 117 = 58. 5

12 × 10 . 69

BD= Sin 58. 5

BD =

12 × 10 . 69Sin 58. 5

BD = 6.2689cm

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10

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24

a) [0 −11 0 ][ A B C

3 2 41 4 3

D51 ]=[ A ' B ' C '

−1 −4 −33 2 4

D '−15 ]

A’(-1,3) B’(-4,2) C’(-3,4) and D’(-1,5)

b)

[1 00 −1][ A ' B ' C '

−1 −4 −33 2 4

D '−15 ]=[A ' ' B ' ' C ' '

−1 −4 −3−3 −2 −4

D ' '−1−5 ]

A’(-1,-3) B’(-4,-2) C’(-3,-4) and D’(-1,-5)

c) [1 00 −1][0 −1

1 0 ]=[ 0 −1−1 0 ]

Reflection∈th e line y=−x

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B210

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