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2025-maths-f3-pp2-ms-end-t3-s2-teacher_co_ke

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  • UPLOADED BY Unknown
  • DATE 07 Dec 2025
  • SIZE 2.04 MB
  • DOWNLOADS 0
  • TAGS
    notes
About This Document

Document Type: This is a Exam Paper, designed for Testing knowledge and exam technique.

Context: Standard material from the 2025 academic period.

Key Content: Likely covers essential definitions, problem solving, and structured questions to test your proficiency.

Study Strategy: Attempt these questions under timed conditions to simulate a real exam environment, then check against your notes.

Recommendation: comprehensive resource for students aiming to achieve top grades in their final assessments.

869 words

Detailed Content Overview

5 min read Intermediate Level 869 words
Introduction

This notes resource titled "2025-maths-f3-pp2-ms-end-t3-s2-teacher_co_ke" provides comprehensive exam preparation materials designed to test and enhance your understanding. This resource is structured to facilitate effective learning and retention of important information.

Key Topics Covered
1 a) i) Taxable income 21200 + 12000 + 1100 + 2000 = 36,300 ii) Payee 8400 First 2
Learning Objectives
  • Master key concepts required for examination success
  • Practice answering exam-style questions effectively
  • Develop time management skills for timed assessments
  • Identify and address knowledge gaps in understanding
Detailed Summary

MATHEMATICS FORM 3 PAPER 2 END OF YEAR EXAM 2025 MARKING SCHEME Download this and other FREE materials from https://teacher. ke/notes 1 MAIN SCHEME Perimeter = 2(l + w) Absolute error for both length and width = 0. 5) = 282 Actual perimeter = 2(80 + 60) = 280 Percentage error = 2 3 282−280 280 x 100 = 0. 714 COMMENTS B1 For either max, actual or min perimeter M1 A1 03 Det () = (4 x -2) – (3 x 5) = -8 – 15 = -23 1 −2 − ( 23 −5 +2 = ( 23 5 B1 −3 ) 4 3 B1 23 −4) 23 23 1 −2 −3 4 3 )( ) (𝑥 ) −5 4 5 −2 𝑦 1 −2 −3 18 = − ( )( ) 23 −5 4 11 𝑥 +3 ( )= ( ) 𝑦 +2 x = 3, y = 2 = − 23 ( M1 A1 04 5(5x) – 15(5 ) + 10 = 0 Let 5x be A 5A2 – 15A + 10 = 0 (A – 2) (A – 1) = 0 A=2 or 1 x 5 =2 or 5x = 1 2 x log 2 x = log 5or x = 0 4 ≅ 0. 7 MARKS x + 3x – 30 = 90 4x = 120 x = 30 o tan 30o = 1 √3 M1 M1 A1 For ✓quad equation formed B1 For both Or subst in quadration formula 04 B1 o 30 2 √3 1 B1 B1 03 Download this and other FREE revision materials from https://teacher. ke/notes 5 WXZ = 180 – 60o = 120o ZWX = 180 – (50 + 120) = 10o YZW = 10o alternate angles to ZWX B1 B1 02 a) Cost of the mixture 120 × 6 + 200 × 5 11 720+1000 = 11 Download this and other FREE materials from https://teacher. 36 ≅ 156/- to the nearest shilling b) % profit =.

Study Tips & Recommendations
Time Management

Practice under timed conditions to improve speed and accuracy. Allocate specific time limits to each section.

Active Practice

Attempt all questions before checking answers. Review mistakes to understand where improvements are needed.

Mark Scheme Review

Study marking schemes carefully to understand how examiners award points and structure your answers accordingly.

Regular Review

Schedule periodic reviews to reinforce learning and combat forgetting. Use spaced repetition for optimal retention.

Content Preview

MATHEMATICS FORM 3 PAPER 2 END OF YEAR EXAM 2025 MARKING SCHEME Download this and other FREE materials from https://teacher.co.ke/notes 1 MAIN SCHEME Perimeter = 2(l + w) Absolute error for both length and width = 0.5 Max perimeter = 2(80.5 + 60.5) = 282 Actual perimeter = 2(80 + 60) = 280 Percentage error = 2 3 282−280 280 x 100 = 0.714 COMMENTS B1 For either max, actual or min perimeter M1 A1 03 Det () = (4 x -2) – (3 x 5) = -8 – 15 = -23 1 −2 − ( 23 −5 +2 = ( 23 5 B1 −3 ) 4 3 B1 23 −4) 23 23...

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