2024-kcse-maths-p1-ms-teacher_co_ke_4
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DATE 07 Dec 2025
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About This Document
Document Type: This is a Study Notes, designed for Reviewing core curriculum material.
Context: Standard material from the 2024 academic period.
Key Content: Likely covers essential definitions, problem solving, and theoretical concepts necessary for mastery of the subject.
Study Strategy: Summarize these notes into flashcards or mind maps to aid active recall and long-term retention.
Recommendation: comprehensive resource for students aiming to deepen their understanding of General Studies.
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This notes resource titled "2024-kcse-maths-p1-ms-teacher_co_ke_4" contains valuable educational content for academic study and reference. This resource is structured to facilitate effective learning and retention of important information.
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Detailed Summary
Minimum length = LCM of 15 and 24 LCM = 23 31 51 = 120 cm 2 x ) − 32 ( 4x2 − 9 = 2 x 2 + x − 6 2 x 2 + 4 x − 3x − 6 ( 2 x + 3)( 2 x − 3) = 2x ( x + 2) − 3( x + 2) 2 2. 5 20 7 7 = 693 + 1320 ( 2 x − 3) ( x + 2 ) 1 2x − 3 x+2 1 Given y = − x + 2, Q ( 0, 2 ). −3 1 m2 = − − = −1 =3 1 3 y = 3x + c At Q, 2 = 3 ( 0 ) + c hence c = 2. Equation of L 2 is y = 3 x + 2. 360 = 60 6 Area of hexagon POQ = 1 6 32 sin 60 = 23. 25 7 44 2 = 40 cm Area = 52 x = 53 2 x = 3 x= 22 c r c = l 7 2 2 = 4. 5 + 10 10 = 34 x y The bank buys 1 S. 5) x −1 = 7 x − 1 = 14 hence x = 15 2 y+7 = 8. 25 3 =1 3 Area = h ( y1 + y2 + y3 ) h= The bank sells Tsh. 3x 30 3 x = 30 tan 50 x y 2 x + 3 y = 1. 672m Page 2 of 6 3 9 SECTION II (50 marks) (a) 17. Farmer Abdul Chebet x 2x Sold Remaining 32 − x 56 − 2x k Selling price 1. 05k 32 − x 3 = 56 − 2 x 5 160 − 5 x = 168 − 6 x x =8 Abdul sold 8 goats Chebet sold 2 8 = 16 goats 19. 4k = 97600 k = 4000 Chebet sold a goat at Ksh. 4 000 (b) Abdul sold a goat at 1. 4 200 (b) Mark T; 5 cm (on paper) from R along SQ. 6 200 = 720 km 720 4 =1 h 400 5 1hr 48 min Time taken to reach T = Abdul's earnings : Chebet's earnings ( 4200 8 ) : ( 4000 16 ) The aircraft will be at T at: 33600 : 64000 0800 + 1hr 48 min = 0948h//9.
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SECTION I (50 marks) 0.039 − 0.003 0.036 1000 = 0.09 0.09 1000 4 1. 36 4 = = 9 1 10 10 = 0.4 15 = 31 51 , 24 = 23 31 6. Minimum length = LCM of 15 and 24 LCM = 23 31 51 = 120 cm 2 x ) − 32 ( 4x2 − 9 = 2 x 2 + x − 6 2 x 2 + 4 x − 3x − 6 ( 2 x + 3)( 2 x − 3) = 2x ( x + 2) − 3( x + 2) 2 2. ( 2 x − 3) ( 2 x − 3) 1 = 7. 22 22 SA = 2 10.52 + 2 10.5 20 7 7 = 693 + 1320 ( 2 x − 3) ( x + 2 ) 1 2x − 3 x+2 1 Given y = − x + 2, Q ( 0, 2 ) . 3 = = 2013 cm 2 −1...
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