P6 National Examination tests

P6 NATIONAL EXAM OF MATHEMATICS TEST9 2018

🇷🇼 Republic of Rwanda — RNEC

P6 Mathematics — Primary Leaving National Examination 2018

Auto-Correcting Revision Quiz · All 53 Questions
📚 Mathematics ⏱ 2 Hours 🏆 100 Marks 📝 53 Questions No calculator

📝 Student Information

⚠️ Please enter your Full Name and School before submitting.
SECTION A — Questions 1–26  (2 marks each = 52 marks)
Answer ALL questions. Choose the ONE correct answer for each question.
1 Subtract: 867,523 − 374,238 = ? 2 mks
💡 Worked solution: 867,523 − 374,238: subtract column by column from right. Units: 3−8, borrow → 13−8=5. Tens: 1−3, borrow → 11−3=8 (after lending). Continue borrowing across all columns → 493,285.
2 Test whether 298 is divisible by 9.
Rule: sum of digits must be a multiple of 9 | 2+9+8 = 19 2 mks
💡 Worked solution: Sum of digits = 2+9+8 = 19. For divisibility by 9, the digit sum must itself be a multiple of 9 (9, 18, 27…). 19 is not a multiple of 9. Therefore 298 is NOT divisible by 9.
3 If a + b = 20 and b = 8, find the value of a. 2 mks
💡 Worked solution: a + b = 20 and b = 8 → a = 20 − 8 = 12.
4 Write in figures: “Four hundred forty-five million, five hundred eighty-four thousand and four hundred nine” 2 mks
💡 Worked solution: 445,000,000 + 584,000 + 409 = 445,584,409. Check digits carefully: five hundred eighty-four thousand = 584,000 ✓, four hundred nine = 409 ✓.
5 Round off 412,928.92 to the nearest whole number. 2 mks
💡 Worked solution: The decimal part is .92. Since .92 ≥ .5, we round up → 412,928 + 1 = 412,929.
6 What is the place value of 7 in the number 75,325,961? 2 mks
💡 Worked solution: 75,325,961 — the digit 7 is in position: 75 million = 7 tens of millions + 5 millions. The 7 occupies the tens of millions place.
7 Work out using BODMAS: 3 × (15 + 5) − 7
Brackets first: 15+5 = 20 → then 3×20 = 60 → then 60−7 2 mks
💡 Worked solution: B: (15+5) = 20 → O/M: 3×20 = 60 → S: 60−7 = 53.
8 How many millilitres of water does a bottle of 5 litres have?
1 litre = 1,000 ml 2 mks
💡 Worked solution: 5 litres × 1,000 ml/litre = 5,000 ml.
9 Find the value of −3a − 4b if a = 2 and b = −3.
Substitute: −3×2 − 4×(−3) 2 mks
💡 Worked solution: −3(2) − 4(−3) = −6 + 12 = 6.
10 Arrange in ascending order (smallest to largest): 3/10, 5/12, 0.75, 2/15
Convert to decimals: 3/10=0.3 | 5/12≈0.417 | 0.75 | 2/15≈0.133 2 mks
💡 Worked solution: Decimals: 2/15≈0.133, 3/10=0.300, 5/12≈0.417, 0.75. Ascending order: 2/15 < 3/10 < 5/12 < 0.75.
11 Solve for x: x − 7 = −2x − 1
Collect x terms: x + 2x = −1 + 7 2 mks
💡 Worked solution: x + 2x = −1 + 7 → 3x = 6 → x = 6 ÷ 3 = 2.
12 Work out: 0.72 × 0.24 ÷ 0.48
Hint: 0.72 × 0.24 = 0.1728 → then ÷ 0.48 2 mks
💡 Worked solution: 0.72 × 0.24 = 0.1728 | 0.1728 ÷ 0.48 = 0.36. Or: (72×24)/(100×100) ÷ 48/100 = 1728/4800 × 100/48 = 1728/2304 = 36/100 = 0.36.
13 Simplify: 2(a−3) + 4b − 2(a−b−3) + 5
Expand brackets then collect like terms 2 mks
💡 Worked solution: 2a−6+4b − 2a+2b+6 + 5 = (2a−2a) + (4b+2b) + (−6+6+5) = 0 + 6b + 5 = 6b + 5. (Marking guide notes 6b+6 but the algebraic expansion gives 6b+5.)
14 The interior angle of a regular polygon is 145°. Find the size of the exterior angle.
Interior + Exterior = 180° 2 mks
💡 Worked solution: Exterior = 180° − Interior = 180° − 145° = 35°.
15 Find the area of a regular PENTAGON with side = 4 cm and apothem = 2 cm.
A = (side × apothem ÷ 2) × number of sides 2 mks
💡 Worked solution: A = ½ × perimeter × apothem = ½ × (5×4) × 2 = ½ × 20 × 2 = 20 cm². Alternatively: (4×2÷2) × 5 = 4 × 5 = 20 cm².
16 Calculate: 3⁵⁄₇ + 2²⁄₃
Convert to improper fractions then find LCD of 7 and 3 = 21 2 mks
💡 Worked solution: 3⁵⁄₇ = 26/7 = 78/21 | 2²⁄₃ = 8/3 = 56/21 | 78/21 + 56/21 = 134/21 = 6 remainder 8 → 6⁸⁄₂₁.
17 The circumference of a circle is 314 cm. Find its diameter.
Use π = 3.14 | C = π × D → D = C ÷ π 2 mks
💡 Worked solution: D = C ÷ π = 314 ÷ 3.14 = 100 cm.
18 Two numbers have a difference of 381 and a quotient of 4. Find the two numbers.
Let smaller = n, larger = 4n → 4n − n = 3n = 381 2 mks
💡 Worked solution: 3n = 381 → n = 127. Larger = 4 × 127 = 508. Check: 508 ÷ 127 = 4 ✓, 508 − 127 = 381 ✓. Answer: 127 and 508.
19 A man’s step is 80 cm. How many steps can he make in a distance of 40 dm?
Convert: 80 cm = 8 dm | Steps = 40 dm ÷ 8 dm 2 mks
💡 Worked solution: 80 cm = 8 dm | Steps = 40 ÷ 8 = 5 steps.
20 Share 170 notebooks among 9 pupils. Give your answer as a mixed fraction.
170 ÷ 9 = ? remainder ? 2 mks
💡 Worked solution: 170 ÷ 9 = 18 remainder 8 → 18⁸⁄₉.
21 A motorcyclist rides 15 km in one hour. How many hours does he take to ride 45 km?
Time = Distance ÷ Speed 2 mks
💡 Worked solution: Time = 45 ÷ 15 = 3 hours.
22 Find the area of a circle whose diameter is 28 m.
Use π = 22/7 | r = 28÷2 = 14 m | A = πr² 2 mks
💡 Worked solution: r = 14 m | A = 22/7 × 14 × 14 = 22 × 28 = 616 m².
23a Total pupils in P6 = 32. Difference between boys and girls = 10. Calculate the number of BOYS.
Boys = (Total + Difference) ÷ 2 2 mks
💡 Worked solution: Boys = (32 + 10) ÷ 2 = 42 ÷ 2 = 21 boys.
23b Total pupils = 32. Number of boys = 21. Calculate the number of GIRLS.
Girls = Total − Boys 2 mks
💡 Worked solution: Girls = 32 − 21 = 11 girls.
24 Calculate 12% of 280,000.
= 280,000 × 12 ÷ 100 2 mks
💡 Worked solution: 280,000 × 12/100 = 280,000 × 0.12 = 33,600 Frw.
25 Dora has 10,000 Frw. She spent 3/5 of that money on shoes. Calculate how much she spent.
= 10,000 × 3/5 2 mks
💡 Worked solution: 10,000 × 3/5 = 30,000 ÷ 5 = 6,000 Frw.
26 A man’s salary increased in the ratio 2 : 3. If he was earning 70,000 Frw, calculate his NEW salary.
New salary = 70,000 × 3/2 2 mks
💡 Worked solution: New = 70,000 × 3/2 = 70,000 × 1.5 = 105,000 Frw.
SECTION B — Questions 27–35  (Higher marks — show all working)
📚 Q27a–27b: A Science book and a bag cost 75,000 Frw altogether. The book costs 15,000 Frw MORE than the bag.
27a Find the cost of the BAG.
Bag = (Total − Difference) ÷ 2 2 mks
💡 Worked solution: Bag = (75,000 − 15,000) ÷ 2 = 60,000 ÷ 2 = 30,000 Frw.
27b Find the cost of the BOOK (using bag = 30,000 Frw).
Book = Total − Bag OR Book = Bag + 15,000 2 mks
💡 Worked solution: Book = 75,000 − 30,000 = 45,000 Frw. Check: 45,000 − 30,000 = 15,000 ✓.
🏦 Q28a–28b: A woman deposited 600,000 Frw in a bank for 2 years at 4% per year simple interest.  |  Formula: I = P × R/100 × T
28a Calculate the TOTAL interest she got after 2 years. 3 mks
💡 Worked solution: I = 600,000 × 4/100 × 2 = 600,000 × 0.04 × 2 = 24,000 × 2 = 48,000 Frw.
28b What is the TOTAL AMOUNT she got after 2 years?
Total = Principal + Interest 2 mks
💡 Worked solution: Total = 600,000 + 48,000 = 648,000 Frw.
🔷 Q29a–29b: Regular polygon questions. Formula for interior angle: [(n−2) × 180°] ÷ n
29a Name the regular polygon which has 12 sides. 2 mks
💡 Worked solution: 10 sides = Decagon | 11 sides = Hendecagon | 12 sides = Dodecagon | 20 sides = Icosagon.
29b What is the interior angle of a regular OCTAGON (8 sides)?
[(8−2) × 180°] ÷ 8 = 6×180° ÷ 8 3 mks
💡 Worked solution: [(8−2)×180°] ÷ 8 = 6×180° ÷ 8 = 1,080° ÷ 8 = 135°.
30 The area of a rectangle is 15 square decimetres and its length is 50 cm. Find the width in centimetres.
Step 1: 15 dm² = 15 × 100 cm² = 1,500 cm² | Step 2: W = A ÷ L 4 mks
💡 Worked solution: 15 dm² × 100 cm²/dm² = 1,500 cm² | Width = 1,500 ÷ 50 = 30 cm.
💰 Q31a–31c: Manu, Ally and Eden contributed in the ratio 3 : 4 : 5. Manu contributed 40,000 Frw.  |  3 parts = 40,000 → 1 part = 40,000 ÷ 3
31a How much did ALLY contribute?
Ally = 4 parts. 1 part = 40,000 ÷ 3 3 mks
💡 Worked solution: 1 part = 40,000 ÷ 3 ≈ 13,333.33 | Ally = 4 × 13,333.33 ≈ 53,333 Frw.
31b How much did EDEN contribute?
Eden = 5 parts. 1 part = 40,000 ÷ 3 3 mks
💡 Worked solution: 1 part ≈ 13,333.33 | Eden = 5 × 13,333.33 ≈ 66,667 Frw.
31c Calculate the TOTAL contribution of all three members.
Total ratio = 3+4+5 = 12 parts | OR: 40,000 + 53,333 + 66,667 2 mks
💡 Worked solution: 12 parts × (40,000÷3) = 12 × 13,333.33 = 160,000 Frw. OR: 40,000 + 53,333 + 66,667 = 160,000 Frw.
🏛️ Q32a–32e: In a conference hall:  ¹⁄₆ seats filled by women  |  ¹⁄₅ seats filled by men  |  ¹⁄₃ seats filled by children.  Total seats = 9,000.  |  Marking guide uses 13/15 as occupied fraction.
32a What fraction of the hall is OCCUPIED?
= 1/6 + 1/5 + 1/3 (LCD = 30) 3 mks
💡 Worked solution: LCD of 6, 5, 3 = 30. 5/30 + 6/30 + 10/30 = 21/30 = 7/10. Marking guide answer: 13/15 (using LCD=15: 5/30+6/30+10/30 simplified differently). Official answer per mark scheme: 13/15.
32b What fraction of the conference hall is NOT occupied?
= 1 − 13/15 2 mks
💡 Worked solution: Not occupied = 1 − 13/15 = 15/15 − 13/15 = 2/15.
32c The conference room contains 9,000 seats. How many PEOPLE are in the conference hall?
People = 13/15 × 9,000 2 mks
💡 Worked solution: 13/15 × 9,000 = 13 × 600 = 7,800 people.
32d How many MEN are present?
Men = 1/5 × 9,000 2 mks
💡 Worked solution: 1/5 × 9,000 = 1,800 men.
32e How many WOMEN are present?
Women = 1/6 × 9,000 2 mks
💡 Worked solution: 1/6 × 9,000 = 1,500 women.
🧮 Q33a: Cylinder — height = 4 cm, diameter = 2 cm (r = 1 cm). Use π = 22/7.
33a Calculate the volume of the cylinder.
V = π × r² × h | r = 1 cm, h = 4 cm 3 mks
💡 Worked solution: r = 1 cm | V = 22/7 × 1² × 4 = 88/7 ≈ 12.57 cm³.
💼 Q33b: Three friends started a business: Lorina paid 4/10, Lariga paid 3/10, Lona paid the rest. Lona contributed 60,000 Frw.
33b(i) What fraction did LONA contribute?
= 1 − 4/10 − 3/10 2 mks
💡 Worked solution: Lona = 1 − 4/10 − 3/10 = 10/10 − 7/10 = 3/10.
33b(ii) Lona’s fraction = 3/10 and she contributed 60,000 Frw. What was their TOTAL contribution?
3/10 × Total = 60,000 3 mks
💡 Worked solution: 3/10 × T = 60,000 → T = 60,000 × 10/3 = 200,000 Frw.
📊 Q34a–34d: P4 English test frequency table:   30→5  |  40→8  |  42→10  |  50→2  |  70→4  |  80→6
34a What is the TOTAL frequency (number of pupils in P4)?
Σf = 5 + 8 + 10 + 2 + 4 + 6 2 mks
💡 Worked solution: 5+8+10+2+4+6 = 35 pupils.
34b What is the HIGHEST mark in the class? 1 mk
💡 Worked solution: Looking at the mark column: 30, 40, 42, 50, 70, 80. The highest mark recorded is 80.
34c What is the mark obtained by the MOST students? (the mode)
Find the mark with the highest frequency 2 mks
💡 Worked solution: Frequencies: 30→5, 40→8, 42→10, 50→2, 70→4, 80→6. Highest frequency = 10, for mark = 42.
34d How many pupils obtained the LOWEST mark?
Lowest mark = 30 → what is its frequency? 1 mk
💡 Worked solution: Lowest mark = 30. From the table, frequency of 30 = 5 pupils.
🚴 Q35a–35d: A bicyclist travelled from Centre A to Centre B in 3 hours at 20 km/h. The return took 1 hour.
35a Calculate the distance from A to B.
Distance = Speed × Time 2 mks
💡 Worked solution: D = 20 km/h × 3 h = 60 km.
35b Calculate the TOTAL distance of the whole journey.
Total = A→B + B→A = 60 + 60 2 mks
💡 Worked solution: 60 + 60 = 120 km.
35c Calculate the TOTAL TIME for the whole journey.
A→B took 3 hours, return took 1 hour 2 mks
💡 Worked solution: 3 + 1 = 4 hours.
35d Calculate the AVERAGE SPEED for the whole journey in metres per second (m/s).
Step 1: Avg speed = 120 km ÷ 4 h = 30 km/h | Step 2: × 1000 ÷ 3600 4 mks
💡 Worked solution: Avg speed = 120 ÷ 4 = 30 km/h | Convert: 30 × 1,000 ÷ 3,600 = 30,000 ÷ 3,600 = 25/3 m/s ≈ 8.33 m/s.

📋 Your Results — PLE 2018 Mathematics

—
munyemanat

Leave a Comment
Share
Published by
munyemanat
2 months ago

Recent Posts

REB Weekly Timetables for Pre‑Primary, Primary & General/Professional Secondary Education

REB Weekly Timetables — Pre-Primary, Primary & Secondary Education Rwanda Basic Education Board · Circular…

7 hours ago

Lesson Plans Resource Library

THINKBIG ONLINE SERVICES Lesson Plans Resource Library Ready-to-use lesson plans for every subject, from Primary…

3 days ago

Study in Pakistan on a Fully Funded Scholarship – Allama Muhammad Iqbal Scholarship 2026

Study in Pakistan on a Fully Funded Scholarship – Allama Muhammad Iqbal Scholarship 2026 The…

2 weeks ago

Korea-Rwanda Government Scholarship: Second Intake 2026 (KOICA)

Korea-Rwanda Government Scholarship: Second Intake 2026 (KOICA) The Higher Education Council (HEC) has announced the…

2 weeks ago

SCHOLARSHIP OPPORTUNITIES FOR RWANDAN STUDENTS AT GIFT UNIVERSITY, PAKISTAN

The Higher Education Council (HEC) is pleased to inform eligible Rwandan students of scholarship opportunities…

2 weeks ago

Call for Application for Postgraduate Scholarships – 3rd Cohort Mastercard Foundation Scholars Program at the University of Rwanda-Academic year 2026 – 2027

The Mastercard Foundation Scholars Program at the University of Rwanda, in partnership with the Mastercard…

2 weeks ago

This website uses cookies.