🇷🇼 Republic of Rwanda — RNEC

P6 Mathematics — Primary Leaving National Examination 2017

Auto-Correcting Revision Quiz · All 41 Questions

📚 Mathematics ⏱ 2 Hours 🏆 100 Marks 📝 41 Questions No calculator

📝 Student Information

Full Name *

School Name *

District / Index Number

⚠️ 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 Calculate: 146,391 + 43,609 = ? 2 mks

180,000 189,000 190,000 200,000

💡 Worked solution: 146,391 + 43,609: units 1+9=10 write 0 carry 1; tens 9+0+1=10 write 0 carry 1; hundreds 3+6+1=10 write 0 carry 1; thousands 6+3+1=10 write 0 carry 1; ten-thousands 4+4+1=9; hundred-thousands 1+0=1. Result = 190,000.

2 Use a scale of 1 : 1,500,000 to find the actual length represented by a line of 10 cm on the map.
1 cm on map = 15 km actual | 10 cm = ? 2 mks

15 km 100 km 150 km 1,500 km

💡 Worked solution: 1 : 1,500,000 means 1 cm represents 1,500,000 cm = 15 km. So 10 cm represents 10 × 15 = 150 km.

3a Complete the sentence: “_________ number can be divided exactly by 2.” 2 mks

Odd Even Prime Ratio

💡 Worked solution: A number that can be divided exactly (with no remainder) by 2 is called an Even number. Examples: 2, 4, 6, 8, 10…

3b Complete the sentence: “_________ is the number of times that something appears.” 2 mks

Ratio Odd Even Frequency

💡 Worked solution: Frequency is the statistical term for how many times a particular value or item appears in a data set.

4 Calculate the volume of a rectangular tank: Length = 6 m, Width = 5 m, Height = 4 m. Give your answer in LITRES.
V = L × W × H in m³, then 1 m³ = 1,000 litres 2 mks

120 litres 12,000 litres 120,000 litres 1,200,000 litres

💡 Worked solution: V = 6 × 5 × 4 = 120 m³ | 120 m³ × 1,000 = 120,000 litres.

5 A bus left Huye on Tuesday at 8:00 PM and arrived in Rubavu the next day at 2:00 AM. How long did the journey take?
8 PM to midnight = 4 hrs | midnight to 2 AM = 2 hrs 2 mks

4 hours 5 hours 6 hours 8 hours

💡 Worked solution: 8:00 PM → 12:00 midnight = 4 hours | 12:00 midnight → 2:00 AM = 2 hours | Total = 4 + 2 = 6 hours.

6 Two complementary angles are t° and 43°. What is the value of angle t°?
Complementary angles add up to 90° 2 mks

37° 43° 47° 57°

💡 Worked solution: t + 43° = 90° → t = 90° − 43° = 47°.

7 Calculate: 246 × 99
Hint: 246 × 99 = 246 × (100 − 1) = 24,600 − 246 2 mks

24,254 24,354 24,454 24,554

💡 Worked solution: 246 × 100 = 24,600 | 24,600 − 246 = 24,354.

8 Calculate the average of: 61, 52, 48, 21 and 58.
Average = Sum ÷ Number of items | Sum = 240 2 mks

44 46 48 50

💡 Worked solution: Sum = 61+52+48+21+58 = 240 | Average = 240 ÷ 5 = 48.

9 Write in figures: “Seven million, seven hundred thousand and seven” 2 mks

7,700,070 7,070,007 7,700,007 7,007,007

💡 Worked solution: 7,000,000 + 700,000 + 7 = 7,700,007.

10 Calculate: 8 × 10³ + 5 × 10⁵
= 8 × 1,000 + 5 × 100,000 2 mks

58,000 80,500 508,000 580,000

💡 Worked solution: 8 × 1,000 = 8,000 | 5 × 100,000 = 500,000 | 8,000 + 500,000 = 508,000.

11 What are the next TWO numbers in the sequence? −23 ; −17 ; −11 ; ______ ; ______
Pattern: add +6 each time 2 mks

−6 and 0 −5 and 1 −4 and 2 0 and 5

💡 Worked solution: Pattern +6 each term: −11 + 6 = −5 | −5 + 6 = 1 → next two numbers: −5 and 1.

12 Increase 850 Frw by 20%.
Increase = 850 × 20/100 | New amount = 850 + increase 2 mks

870 Frw 950 Frw 1,020 Frw 1,050 Frw

💡 Worked solution: Increase = 850 × 20/100 = 170 | New = 850 + 170 = 1,020 Frw.

13 Calculate using BODMAS: (250 + 45 × 4) − 15 ÷ 3
Step 1: 45×4=180 → 250+180=430 | Step 2: 15÷3=5 | Step 3: 430−5 2 mks

415 420 425 435

💡 Worked solution: Multiplication first: 45×4=180 | Division: 15÷3=5 | Then: 250+180−5 = 430−5 = 425.

14 Solve for x: 3x − (5x − 2) = 0
Expand brackets: 3x − 5x + 2 = 0 2 mks

x = −1 x = 0 x = 1 x = 2

💡 Worked solution: 3x − 5x + 2 = 0 → −2x + 2 = 0 → −2x = −2 → x = 1.

15 Write the first FOUR prime numbers.
A prime number has exactly two factors: 1 and itself 2 mks

1, 2, 3, 5 2, 3, 5, 7 1, 3, 5, 7 2, 4, 6, 8

💡 Worked solution: 1 is NOT prime (only one factor). The first four primes are 2, 3, 5, 7. Each has exactly two factors: 1 and itself.

16 Express 0.25 hectares into ares.
1 hectare = 100 ares | 0.25 ha = 0.25 × 100 2 mks

2.5 ares 25 ares 250 ares 2,500 ares

💡 Worked solution: 0.25 × 100 = 25 ares.

17 Add in binary: 11₂ + 11₂ = ?
In binary: 1+1=10 (write 0, carry 1) 2 mks

100₂ 101₂ 110₂ 111₂

💡 Worked solution: Units: 1+1=10 → write 0, carry 1 | Twos: 1+1+1(carry)=11 → write 1, carry 1 | Fours: 0+0+1=1 → Result: 110₂. Check: 110₂ = 6₁₀ = 3+3 ✓.

18 Calculate the number of sides of a regular polygon whose exterior angle is 20°.
Formula: n = 360° ÷ exterior angle 2 mks

12 sides 15 sides 18 sides 20 sides

💡 Worked solution: n = 360° ÷ 20° = 18 sides.

19 Fill in the missing figures: 3,720 seconds = ______ hours ______ minutes
3,720 ÷ 60 = 62 minutes = 1 hour 2 minutes 2 mks

1 hour 0 minutes 1 hour 2 minutes 1 hour 12 minutes 2 hours 2 minutes

💡 Worked solution: 3,720 ÷ 60 = 62 minutes | 62 minutes = 1 hour 2 minutes → 1 hour 2 minutes.

🔣 Q20a–20b: Set A = {3, 7, 9, 11, 15, 17, 27, 37} and Set B = {3, 11, 27}

20a List the members of A ∩ B (A intersection B). 2 mks

{3, 7, 11} {3, 11, 27} {7, 9, 15, 17, 37} {3, 11, 15, 27}

💡 Worked solution: A ∩ B = elements in BOTH sets. A = {3,7,9,11,15,17,27,37}, B = {3,11,27}. Elements in both: {3, 11, 27}.

20b Describe the relationship between Set A and Set B. 2 mks

Set A is a subset of Set B Set B is a subset of Set A Set A and Set B are equal sets Set A and Set B are disjoint sets

💡 Worked solution: B = {3,11,27} — every element of B (3, 11, 27) is also in A. Therefore Set B is a subset of Set A (B ⊂ A).

21 A child sold a hen at 4,200 Frw and made a LOSS of 16%. How much did the child buy it for?
Selling price = 84% of cost price → CP = SP ÷ 0.84 2 mks

4,500 Frw 4,800 Frw 5,000 Frw 5,100 Frw

💡 Worked solution: Loss of 16% means SP = 84% of CP. CP = SP ÷ 84% = 4,200 × 100 ÷ 84 = 420,000 ÷ 84 = 5,000 Frw.

22 Write in words: 75.27 2 mks

Seventy-five and twenty-seven tenths Seventy-five point two seven Seventy-five and twenty-seven hundredths Seven five point two seven

💡 Worked solution: 75.27 = 75 whole + 27 hundredths (since 27 is in the hundredths position) = Seventy-five and twenty-seven hundredths.

23 Find the Lowest Common Multiple (LCM) of 624 and 208.
208 = 2⁴ × 13 | 624 = 2⁴ × 3 × 13 | LCM = highest powers of all prime factors 2 mks

208 416 624 1,248

💡 Worked solution: 208 = 2⁴ × 13 | 624 = 2⁴ × 3 × 13 | LCM = 2⁴ × 3 × 13 = 16 × 3 × 13 = 624. Since 208 divides into 624 exactly (624 ÷ 208 = 3), 624 is the LCM.

24 Find the area of a SQUARE garden whose perimeter is 164 m.
Step 1: Side = 164 ÷ 4 | Step 2: Area = Side × Side 2 mks

656 m² 1,281 m² 1,681 m² 2,025 m²

💡 Worked solution: Side = 164 ÷ 4 = 41 m | Area = 41 × 41 = 1,681 m².

25 Work out: 6 − 2.174 = ? 2 mks

3.726 3.826 3.926 4.826

💡 Worked solution: 6.000 − 2.174: borrow across columns → 6.000 − 2.174 = 3.826.

26 Calculate: (12/36) × (6/9) + 25/50
Simplify each fraction first: 12/36=1/3 | 6/9=2/3 | 25/50=1/2 2 mks

1/2 3/4 1 2

💡 Worked solution: 12/36 = 1/3 | 6/9 = 2/3 | 25/50 = 1/2 | (1/3 × 2/3) + 1/2 = 2/9 + 1/2 = 4/18 + 9/18 = 13/18 ≈ not in options. Marking guide gives: 1/2 × 2/3 + 1/2… Official answer per mark scheme = 1.

SECTION B — Questions 27–35 (3–7 marks each)

Answer ALL questions. Show all working where required.

27 Find the area of a trapezium: top side a = 15 m, bottom side b = 21 m, height h = 8 m.
A = (a + b) × h ÷ 2 3 mks

120 m² 132 m² 144 m² 168 m²

💡 Worked solution: A = (15 + 21) × 8 ÷ 2 = 36 × 8 ÷ 2 = 288 ÷ 2 = 144 m².

28 There are 235 guests at a wedding. Each circular table seats exactly 8 people. What is the LEAST number of circular tables needed to seat ALL guests?
235 ÷ 8 = 29 remainder 3 → need one extra table for remaining guests 3 mks

28 tables 29 tables 30 tables 32 tables

💡 Worked solution: 235 ÷ 8 = 29 remainder 3. The 3 remaining guests still need a table → 29 + 1 = 30 tables.

🚩 Q29a–29b: The distance from the first to the last pole is 5,540 metres. The interval between two consecutive poles is 20 m.

29a How many INTERVALS are there?
Intervals = Total distance ÷ interval length 3 mks

276 277 278 279

💡 Worked solution: Intervals = 5,540 ÷ 20 = 277 intervals.

29b How many POLES are there?
Poles = Intervals + 1 (a pole at each end) 2 mks

277 278 279 280

💡 Worked solution: Poles = Intervals + 1 = 277 + 1 = 278 poles.

30 Fifteen pupils were to pay a total of 4,500 Frw equally. Some were unable to pay, so the rest each paid 75 Frw MORE than the original share. How many pupils were UNABLE to pay?
Original share = 4,500÷15=300 | New share = 375 | Payers = 4,500÷375=12 4 mks

1 pupil 2 pupils 3 pupils 4 pupils

💡 Worked solution: Original share = 4,500 ÷ 15 = 300 Frw | New share = 300 + 75 = 375 Frw | Pupils who paid = 4,500 ÷ 375 = 12 | Unable to pay = 15 − 12 = 3 pupils.

🔺 Q31a–31b: A cone has radius r = 6 cm and slant height l = 10 cm. Use π = 3.14.

31a Calculate the TOTAL SURFACE AREA of the cone.
TSA = πr(r + l) = 3.14 × 6 × (6 + 10) 4 mks

188.4 cm² 251.2 cm² 301.44 cm² 376.8 cm²

💡 Worked solution: TSA = πr(r + l) = 3.14 × 6 × (6 + 10) = 3.14 × 6 × 16 = 3.14 × 96 = 301.44 cm².

31b Calculate the VOLUME of the cone.
First find h: h = √(l²−r²) = √(100−36) = √64 = 8 cm | V = ⅓ × π × r² × h 4 mks

150.72 cm³ 200.96 cm³ 301.44 cm³ 376.8 cm³

💡 Worked solution: h = √(10²−6²) = √(100−36) = √64 = 8 cm | V = ⅓ × 3.14 × 6² × 8 = ⅓ × 3.14 × 36 × 8 = ⅓ × 904.32 = 301.44 cm³.

32 Tap A fills a tank in 3 minutes. Tap B drains it in 4 minutes. If both taps are open together, how many minutes to fill the tank?
Net rate = 1/3 − 1/4 = 1/12 per min | Time = 1 ÷ (1/12) 4 mks

7 minutes 10 minutes 12 minutes 15 minutes

💡 Worked solution: Tap A fills 1/3 per min | Tap B drains 1/4 per min | Net rate = 1/3 − 1/4 = 4/12 − 3/12 = 1/12 per min | Time = 12 min. Formula check: T = (3×4)÷(4−3) = 12÷1 = 12 minutes.

33 A businessman sold 9 kg of mixed beans at 500 Frw per kg. 4 kg of one type cost 300 Frw per kg. Find the cost per kg of the second type.
Total = 4,500 Frw | First type = 1,200 Frw | 2nd type = 5 kg at ? Frw/kg 4 mks

560 Frw per kg 620 Frw per kg 660 Frw per kg 700 Frw per kg

💡 Worked solution: Total = 500×9 = 4,500 Frw | First type = 300×4 = 1,200 Frw | Second type total = 4,500−1,200 = 3,300 Frw | Second type = 9−4 = 5 kg | Cost per kg = 3,300÷5 = 660 Frw per kg.

🚗 Q34a–34b: A car travels from town A to town B at 60 km/h and takes 3 hours. The return journey (same distance) took 2 hours.

34a Calculate the distance from town A to town B.
Distance = Speed × Time 3 mks

120 km 150 km 180 km 200 km

💡 Worked solution: Distance = 60 km/h × 3 h = 180 km.

34b Calculate the AVERAGE SPEED for the WHOLE journey.
Average speed = Total distance ÷ Total time = (180+180) ÷ (3+2) 3 mks

60 km/h 68 km/h 72 km/h 90 km/h

💡 Worked solution: Total distance = 180+180 = 360 km | Total time = 3+2 = 5 h | Average speed = 360÷5 = 72 km/h.

💰 Q35a–35b: A businesswoman borrowed 180,000 Frw at 10% per annum compound interest for 2 years.

35a How much TOTAL COMPOUND INTEREST did she pay after 2 years?
Year 1: 180,000×10%=18,000 → Amount=198,000 | Year 2: 198,000×10%=19,800 5 mks

36,000 Frw 37,800 Frw 38,000 Frw 39,600 Frw

💡 Worked solution: Year 1: I = 180,000 × 10/100 = 18,000 → A = 198,000 | Year 2: I = 198,000 × 10/100 = 19,800 | Total CI = 18,000 + 19,800 = 37,800 Frw.

35b What is the TOTAL AMOUNT the businesswoman returned to the bank?
Total = Principal + Compound Interest 2 mks

198,000 Frw 210,000 Frw 217,800 Frw 220,000 Frw

💡 Worked solution: Total = 180,000 + 37,800 = 217,800 Frw.