MANGU HIGH SCHOOL TRIAL 2 MOCK 2021 MATHEMATICS PP2 MS

1.	(x- y) ( x+y) ( 3282 – 3272) ( 3282 + 3272) 65540	M1 M1 A1
		3
2.	Tan x = is positive 3^rd quadrant Then sin x = ^–3 5 Hypotense 3 h 4 h = Ö 4² + 3² = Ö 25 = 5 Sin x = ^–3 5 Cos x – sin x = ^–4 – ^–3 = ^–3 5 5 5 = -1 5	B1 M1 A1	Identification the hypotenuse Cao accept (-0.2)
		3
3.	1⁶ + 6(- ½ x ) + 15(- ½ x )² + 20(- ½ x )³ = 1 – 3x + 15x² – 5 x³ 4 X = -0.04 1-3 ( -0.04) + 15 (-0.04)² 5 ( -0.04) ³ 4 4 = 1 + 0.12 + 0.006 + 0.0006¹⁶ = 1.12616 = 1.1262	M1 M1 M1 A1	Forü simplification Forü substitution of x
		4
4.	a + ar³ = 140 2 64 + 64 r³ = 140 2 64 + 64 r³ = 280 64 r³ = 280 – 64 64 r³ = 216 r = Ö216 64 r = 3 2	M1 M1 A1
		3

5.	a² = b²d² b² –d a²b² – a²d = b²d² a²b² – b²d² = a²d b²(a²– d²) = a²d b² = a² d a²– d² = a² d a² d²	M1 M1 A1	ü sq on both sides CAO
		3
6	P = aQ + Ö Q P = 16a + 4b ( 500 = 16a + 4b) (800 = 25a + 5b) 2500 = 80a + 20b 3200 = 100a + 20b -700 = -20a 35 = a Then b = -15 Equation connecting P and Q p = 35Q – 15 ÖQ	M1 M1 M1 A1	For ü equation For ü formation of simultaneous equations For ü values of both a and b
		4
7.		M1 A1
		2
8	4.562 x 0.38 = 1.73356 ⁴Ö1.73356 = 1.14745 ¸0.82 = 1.3993 = 1.4	M1 M1 A1
		3
9.	18 x 64 x 5 24 x 80 6 x 64 8x 16 3 days	M1 M1 A1	Forü simplification
		3
10.	True value = Ö1 + n = 1.44 = 1.2 Approx. value 1 + n = 1 + 44 = 1.22 2 2 = 1.22 – 1.2 = 0.02 x 100 = 1.2 = 1.67 %	M1 M1 A1
		3
11.	3 0 a b = 3 + a b 0 4 0 c 0 4 + c 3a + 0 = 3 + a 3b + 0 = b 3a = 3 + a a = 3 2 3b + 0 = b 2b = 0 B = 0 0 + 4c = 4 + c 3c = 4 C= 4 3	M1 M1 A1	For matrix equation Forü forming of simultaneous equation For values of a, b and c ( correct)
12.	2x² – 2x + x -1 ( x + 1 ( x – 1) 2x ( x – 1 ) + 1 ( x- 1) ( x + 1 ) (n- 1 ) = ( 2x + 1) ( x + 1 ) = 2x + 1 x + 1	M1 M1 A1
		3
13	cm = 25000cm1cm = 250m1cm = 0.25 1cm² = 0.0625 20cm² = 20 x 0.0625 = 1.25/ cm²	M1 M1 A1
		3
14.	AB . BC = DC ^-2 5: BC = 36 BC = 36 5 = 7.2 cm	M1 M1 A1
		3
15.	Log₁₀⁸ + Log₁₀⁷⁵⁰ – Log₁₀⁶ Log₁₀Log₁₀( 8 x 750) 6 = Log₁₀¹⁰⁰⁰ = 3	M1 M1 A1
		3
16.	P (R )= ½ x 4 12 P (R )= ½ x 3 18 = 1 + 3 20 = 20 + 18 120 = 38 19 = 60	M1 M1 M1 A1
		4

17	Taxable income 115 x 8570 = 9855.50 100 9855.50x 12 p.a 20 5913.30 Tax 1 – 1500 150 1501 – 3000 225 3001 – 4500 375 4501 – 5913.30 494.30 1244.30 – 90.00 K £ 1154.30 pa . or Ksh 1923.83 per month Total Decuctions 2 x 9855.5 100 197.11 ( wcps) + 20.00 246.00 + Tax per month 1923.83 2386.94 Net salary 9855.50 – 2386.94 Ksh 7468.65	M1 A1 M1 M1 M1 A1 M1 A1 M1 A1
		10
18.	i) q ( 2pR cos q) 360 = 60 x 2 x 22 x 6370 cos 60 360 7 = 1/3 x 22 x 910 x 0.5 = 3336 7 ii) Time ( 4 x 60) hrs 60 4 hrs. Local time 1200 + 4 = 1600hrs b) q x 2pR = 800 360 = q x 2 x px 6370 = 800 360 = 800 x 360 2 x p x 6370 = 7.196° Ð (60 – 7.196) = 52.80° ( 52.8° N 45°E)	M1 A1 B1 M1 M1 A1 B1
19		10

		10

20.	X 30 60 120 180 240 270 Sin x 0.5 0.87 0.87 0 -0.87 -1.0 2 cos x 1.73 1.0 -1.0 -2 -1.0 0 Y 2.23 1.87 -0.13 -2 -1.87 -1.0 c) x = 114 ± 3° and x = 294 ± 3° line thro y = -1.5 d)	B2 S1 P1 C1 B2 L1 B2	For all 6 values of yü B1 for at least 4 ü Appropriate scale use ü plotting ü curve Points identified and stated B1 only stated ü line Points identified and stated B1 only stated
		10
21	R ²/₁₁ ⁵/₁₁W ⁴/₁₁ B R R ³/₁₁ ³/₁₂ ⁴/₁₁ W ⁵/₁₂ w ^4/₁₁ B ⁴/₁₂ B R ³/₁₁ ⁵/₁₁W ³/₁₁ B	B2	P P prob, tree

a) P ( RR) = ³/₁₂ x ²/₁₁ ¹/₂₂ b) P(IR) = RW or RB or WR or BR ¹⁵/₁₃₂ + ¹²/₁₃₂ + ¹⁵/₁₃₂ + ¹²/₁₃₂ ⁹/₂₂ p( At least white Ball ) = P(RW) + P(WR) + P(WW) + P ( WB) + P(B)

¹⁵/₁₃₂ + ⁹/₁₃₂ + ²⁰/₁₃₂ + ²⁰/₁₃₂ + ²⁰/₁₃₂ = ⁸⁴/₁₃₂ or ⁷/₁₁ P(RR or WW or BB) = ⁶_/ ₁₃₂ + ²⁰/₁₃₂ + ¹²/₁₃₂ =¹⁹_/66

M1 A1 M1 A1 M1 A1 M1 A1

Or equivalent 0.04545 Or equivalent 0.4091 Or equivalent 0.6364 Or equivalent 0.2424

22	X -2 -1 0 1 2 3 4 5 Y 0 6 10 12 12 10 6 0 ½ ( 6 + 10 ) + ½ ( 12 + 12) + ½ ( 12 + 10 ) + ½ ( 10 + 6) = 8 + 11 + 12 + 11 + 8 = 50cm² u ^–x² + 3x + 10 = ^– 64 + 24 + 10 – ¹/₃ + ³/₂x –10 = 51¹/₅cm 3 = -1 = % error 51¹/₂– 50 = 1 ½ x 100 = 2 % 51¹/₂	B2 S1 P1 C1 M1 A1 M1 M1 M1 A1	8 values ü B1 at least 6ü Appropriate scale use ü plotting ü curve
		10

23.	a) i) AC² = 8² + 6² = 100 AC = 10cm ii) AF² = 10² + 5² = 125 AF = 11.18cm b) Tan x = ⁵/₁₁ = 0.5 x = 26.52^° Tan x = ⁵/₆= 0.8333 x = 39.7^°	M1 A1 M1 A1 B1 M1 A1 B1 M1 A1	Sketch Sketch
		10

24		Inequalities x ³O y £O 4x + 3y £ 36 4x + 3y = 24 y = n For ü shading of x ³o and y³ o 2x + 3y £ 36 y £ n P profit function object we function P = 4x + 3y Max profit at point (6,4) P = 4(6) + 4,88)(4) = 56 Hence he should here 6 medium of type A and 4 machine of type B	B1 B1 B1 B1 B1 ü shading and line B1 shading and line drawn B1-for ü shading and line drawn B1 B1 B1
			10

UPDATES

MANGU HIGH SCHOOL TRIAL 2 MOCK 2021 MATHEMATICS PP2 MS

Share this:

Leave a Reply Cancel reply