JAMB / UTME
MATHEMATICS
MASTER GUIDE
150 Original UTME-Style Questions + Detailed Step-by-Step Solutions
150 QUESTIONS 5 SYLLABUS SECTIONS
DETAILED SOLUTIONS FORMULAS + SHORTCUTS
NUMBER & NUMERATION | ALGEBRA | GEOMETRY & TRIGONOMETRY | CALCULUS |
STATISTICS
REVISION EDITION
ORIGINAL PRACTICE RESOURCE
, HOW TO USE THIS BOOK
This book follows the five broad sections of the UTME Mathematics syllabus: Number and Numeration, Algebra,
Geometry/Trigonometry, Calculus, and Statistics. The syllabus is designed to test computational skills, logical reasoning,
interpretation of graphs/diagrams/data, and application of mathematics to everyday problems.
Important: The 150 questions in this edition are original practice questions written in a UTME style. They are not claimed to
be official JAMB past questions.
CONTENTS
SECTION I - NUMBER AND NUMERATION: 30 questions
SECTION II - ALGEBRA: 30 questions
SECTION III - GEOMETRY AND TRIGONOMETRY: 30 questions
SECTION IV - CALCULUS: 30 questions
SECTION V - STATISTICS AND PROBABILITY: 30 questions
Final Answer Key + Formula Sheet + 7-Day Revision Plan
JAMB / UTME Mathematics Master Guide - Original Practice Edition Page 2
, SECTION I: NUMBER AND NUMERATION
30 focused questions with worked solutions
1. Convert 101101 base 2 to base 10.
A. 43 B. 45 C. 49 D. 47
Answer: B
Solution: 101101 base 2 = 1(32) + 0(16) + 1(8) + 1(4) + 0(2) + 1 = 45.
2. Convert 57 base 10 to base 2.
A. 110111 base 2 B. 101111 base 2 C. 111010 base 2 D. 111001 base 2
Answer: D
Solution: 57 = 32 + 16 + 8 + 1, so the binary digits are 111001 base 2.
3. Convert 234 base 5 to base 10.
A. 69 B. 79 C. 71 D. 64
Answer: A
Solution: 234 base 5 = 2(25) + 3(5) + 4 = 50 + 15 + 4 = 69.
4. Evaluate 1011 base 2 + 1101 base 2.
A. 10110 base 2 B. 11100 base 2 C. 11000 base 2 D. 10001 base 2
Answer: C
Solution: 1011 base 2 = 11 and 1101 base 2 = 13. Their sum is 24, which is 11000 base 2.
5. Evaluate 132 base 4 - 23 base 4.
A. 103 base 4 B. 105 base 4 C. 113 base 4 D. 101 base 4
Answer: A
Solution: 132 base 4 = 1(16) + 3(4) + 2 = 30. 23 base 4 = 2(4) + 3 = 11. Difference = 30 - 11 = 19. 19 = 1(16) + 0(4) + 3, so the
result is 103 base 4.
6. Convert 0.101 base 2 to base 10.
A. 0.5 B. 0.875 C. 0.625 D. 0.75
Answer: C
Solution: 0.101 base 2 = 1/2 + 0/4 + 1/8 = 4/8 + 1/8 = 5/8 = 0.625.
7. If x base 5 = 23 base 10, find x.
A. 34 base 5 B. 43 base 5 C. 41 base 5 D. 44 base 5
Answer: B
Solution: 23 = 4(5) + 3, so x = 43 base 5.
8. Find the value of 3A base 16 in base 10.
A. 48 B. 52 C. 62 D. 58
Answer: D
Solution: 3A base 16 = 3(16) + 10 = 48 + 10 = 58.
9. Calculate 3/4 + 5/8.
A. 10/8 B. 13/8 C. 9/8 D. 11/8
Answer: D
Solution: 3/4 = 6/8. Therefore 6/8 + 5/8 = 11/8 = 1 3/8.
10. Calculate 7/9 - 2/3.
JAMB / UTME Mathematics Master Guide - Original Practice Edition Page 3
MATHEMATICS
MASTER GUIDE
150 Original UTME-Style Questions + Detailed Step-by-Step Solutions
150 QUESTIONS 5 SYLLABUS SECTIONS
DETAILED SOLUTIONS FORMULAS + SHORTCUTS
NUMBER & NUMERATION | ALGEBRA | GEOMETRY & TRIGONOMETRY | CALCULUS |
STATISTICS
REVISION EDITION
ORIGINAL PRACTICE RESOURCE
, HOW TO USE THIS BOOK
This book follows the five broad sections of the UTME Mathematics syllabus: Number and Numeration, Algebra,
Geometry/Trigonometry, Calculus, and Statistics. The syllabus is designed to test computational skills, logical reasoning,
interpretation of graphs/diagrams/data, and application of mathematics to everyday problems.
Important: The 150 questions in this edition are original practice questions written in a UTME style. They are not claimed to
be official JAMB past questions.
CONTENTS
SECTION I - NUMBER AND NUMERATION: 30 questions
SECTION II - ALGEBRA: 30 questions
SECTION III - GEOMETRY AND TRIGONOMETRY: 30 questions
SECTION IV - CALCULUS: 30 questions
SECTION V - STATISTICS AND PROBABILITY: 30 questions
Final Answer Key + Formula Sheet + 7-Day Revision Plan
JAMB / UTME Mathematics Master Guide - Original Practice Edition Page 2
, SECTION I: NUMBER AND NUMERATION
30 focused questions with worked solutions
1. Convert 101101 base 2 to base 10.
A. 43 B. 45 C. 49 D. 47
Answer: B
Solution: 101101 base 2 = 1(32) + 0(16) + 1(8) + 1(4) + 0(2) + 1 = 45.
2. Convert 57 base 10 to base 2.
A. 110111 base 2 B. 101111 base 2 C. 111010 base 2 D. 111001 base 2
Answer: D
Solution: 57 = 32 + 16 + 8 + 1, so the binary digits are 111001 base 2.
3. Convert 234 base 5 to base 10.
A. 69 B. 79 C. 71 D. 64
Answer: A
Solution: 234 base 5 = 2(25) + 3(5) + 4 = 50 + 15 + 4 = 69.
4. Evaluate 1011 base 2 + 1101 base 2.
A. 10110 base 2 B. 11100 base 2 C. 11000 base 2 D. 10001 base 2
Answer: C
Solution: 1011 base 2 = 11 and 1101 base 2 = 13. Their sum is 24, which is 11000 base 2.
5. Evaluate 132 base 4 - 23 base 4.
A. 103 base 4 B. 105 base 4 C. 113 base 4 D. 101 base 4
Answer: A
Solution: 132 base 4 = 1(16) + 3(4) + 2 = 30. 23 base 4 = 2(4) + 3 = 11. Difference = 30 - 11 = 19. 19 = 1(16) + 0(4) + 3, so the
result is 103 base 4.
6. Convert 0.101 base 2 to base 10.
A. 0.5 B. 0.875 C. 0.625 D. 0.75
Answer: C
Solution: 0.101 base 2 = 1/2 + 0/4 + 1/8 = 4/8 + 1/8 = 5/8 = 0.625.
7. If x base 5 = 23 base 10, find x.
A. 34 base 5 B. 43 base 5 C. 41 base 5 D. 44 base 5
Answer: B
Solution: 23 = 4(5) + 3, so x = 43 base 5.
8. Find the value of 3A base 16 in base 10.
A. 48 B. 52 C. 62 D. 58
Answer: D
Solution: 3A base 16 = 3(16) + 10 = 48 + 10 = 58.
9. Calculate 3/4 + 5/8.
A. 10/8 B. 13/8 C. 9/8 D. 11/8
Answer: D
Solution: 3/4 = 6/8. Therefore 6/8 + 5/8 = 11/8 = 1 3/8.
10. Calculate 7/9 - 2/3.
JAMB / UTME Mathematics Master Guide - Original Practice Edition Page 3
