P6 Mathematics – Primary Leaving National Examination Test 8 2017
P6 Mathematics – Primary Leaving National Examination 2017
Republic of Rwanda | Total Marks: 100 | Duration: 2 Hours | Answer ALL 35 questions. Section A (Q1–26): 2 marks each | Section B (Q27–35): 3–7 marks each
Student Information
Section A — Questions 1–26 (2 marks each)
1Calculate: 146,391 + 43,609 = ?2 mks
💡 Worked solution: Add the units column: 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.
2Use 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 in reality. Multiply 10 × 15.2 mks
💡 Worked solution: Scale 1 : 1,500,000 means 1 cm represents 1,500,000 cm = 15 km. For 10 cm: 10 × 15 = 150 km.
3aComplete the sentence: “_________ number can be divided exactly by 2.”2 mks
💡 Worked solution: An Even number is any integer divisible exactly by 2 with no remainder (e.g. 2, 4, 6, 8…). Odd numbers leave a remainder of 1 when divided by 2.
3bComplete the sentence: “_________ is the number of times that something appears.”2 mks
💡 Worked solution:Frequency is the statistical term for how many times a particular value or event appears in a data set. It is used in tally charts and frequency tables.
4Calculate the volume of a rectangular tank: Length = 6 m, Width = 5 m, Height = 4 m. Give your answer in litres. V = L × W × H, then 1 m³ = 1,000 litres.2 mks
💡 Worked solution: V = 6 × 5 × 4 = 120 m³. Convert: 120 × 1,000 = 120,000 litres.
5A 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 → midnight = 4 hrs; midnight → 2 AM = 2 hrs.2 mks
💡 Worked solution: From 8:00 PM to midnight = 4 hours. From midnight to 2:00 AM = 2 hours. Total = 4 + 2 = 6 hours.
6Two complementary angles are t° and 43°. What is the value of angle t°? Complementary angles add up to 90°.2 mks
💡 Worked solution: Complementary angles sum to 90°. So t° = 90° − 43° = 47°.
20aSet A = {3, 7, 9, 11, 15, 17, 27, 37} and Set B = {3, 11, 27}. List the members of A ∩ B (intersection).2 mks
💡 Worked solution: A ∩ B lists elements in BOTH sets. Elements of B: 3, 11, 27 — all appear in A. So A ∩ B = {3, 11, 27}.
20bUsing the same sets: Set A = {3, 7, 9, 11, 15, 17, 27, 37} and Set B = {3, 11, 27}. Describe the relationship between Set A and Set B.2 mks
💡 Worked solution: Every element of B (3, 11, 27) is also in A, but A has more elements. Therefore Set B is a subset of Set A (B ⊂ A).
21A child sold a hen at 4,200 Frw and made a loss of 16%. How much did he/she buy it for? Selling price = 84% of cost price. Cost = SP ÷ 0.84.2 mks
💡 Worked solution: A 16% loss means the hen was sold for 84% of its cost. Cost price = (4,200 ÷ 84) × 100 = 50 × 100 = 5,000 Frw.
22Write in words: 75.272 mks
💡 Worked solution: 75.27 = 75 whole + 27 hundredths (the second decimal place is hundredths). Written: Seventy-five and twenty-seven hundredths.
23Find the Lowest Common Multiple (LCM) of 624 and 208. 208 = 2⁴ × 13; 624 = 2⁴ × 3 × 13. LCM takes highest powers of all prime factors.2 mks
💡 Worked solution: 208 = 2⁴ × 13. 624 = 2⁴ × 3 × 13. LCM = 2⁴ × 3 × 13 = 16 × 39 = 624. (Note: 624 is already a multiple of 208 since 624 = 208 × 3.)
24Find the area of a square garden whose perimeter is 164 m. Step 1: Side = 164 ÷ 4. Step 2: Area = side².2 mks
💡 Worked solution: Side = 164 ÷ 4 = 41 m. Area = 41 × 41 = 1,681 m².
25Work out: 6 − 2.1742 mks
💡 Worked solution: 6.000 − 2.174: borrow across zeros. 6.000 − 2.174 = 3.826.
26Calculate: (¹²⁄₃₆) × (⁶⁄₉) + ²⁵⁄₅₀ Simplify each fraction first, then multiply, then add.2 mks
💡 Worked solution: ¹²⁄₃₆ = ⅓; ⁶⁄₉ = ⅔; ²⁵⁄₅₀ = ½. So ⅓ × ⅔ + ½ = ²⁄₉ + ½ = ⁴⁄₁₈ + ⁹⁄₁₈ = ¹³⁄₁₈. The marking guide simplifies as: ½ × 1 + ½ = 1. (Accept 1 per official marking scheme.)
Section B — Questions 27–35 (3–7 marks each)
27Find the area of a trapezium: top side = 15 m, bottom side = 21 m, height = 8 m. Formula: A = (a + b) × h ÷ 2.4 mks
💡 Worked solution: A = (15 + 21) × 8 ÷ 2 = 36 × 8 ÷ 2 = 288 ÷ 2 = 144 m².
28There are 235 guests at a wedding. Each circular table seats exactly 8 people. What is the least number of tables needed to seat ALL guests? Divide 235 ÷ 8 and round up to the next whole number.3 mks
💡 Worked solution: 235 ÷ 8 = 29 remainder 3. The 3 remaining guests still need a table. So minimum tables = 29 + 1 = 30 tables.
29aThe distance from the first to the last pole is 5,540 m. The interval between two consecutive poles is 20 m. How many intervals are there? Number of intervals = Total distance ÷ interval size.3 mks
29bUsing the same pole problem (5,540 m total, 20 m interval): How many poles are there? Number of poles = Number of intervals + 1.3 mks
💡 Worked solution: Poles = intervals + 1 = 277 + 1 = 278 poles. (There is always one more pole than the number of gaps between them.)
30Fifteen 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? Find original share, then new share, then count payers.4 mks
💡 Worked solution: ① Original share = 4,500 ÷ 15 = 300 Frw. ② New share = 300 + 75 = 375 Frw. ③ Pupils who paid = 4,500 ÷ 375 = 12. ④ Unable to pay = 15 − 12 = 3 pupils.
31aA cone has radius r = 6 cm and slant height l = 10 cm. Calculate the Total Surface Area. Use π = 3.14. TSA = πr(r + l).4 mks
31bSame cone: radius = 6 cm, slant height = 10 cm. Find the Volume. Use π = 3.14. First find height: h = √(l² − r²). Then V = ⅓ × π × r² × h.4 mks
💡 Worked solution: h = √(10² − 6²) = √(100 − 36) = √64 = 8 cm. V = ⅓ × 3.14 × 36 × 8 = ⅓ × 904.32 = 301.44 cm³.
32Tap A takes 3 minutes to fill a tank. Tap B takes 4 minutes to drain the tank. If both taps are open together, how many minutes to fill the tank? Net rate = fill rate − drain rate. Time = 1 ÷ net rate.5 mks
💡 Worked solution: Tap A fills ⅓ of the tank per minute. Tap B drains ¼ per minute. Net rate = ⅓ − ¼ = ⁴⁄₁₂ − ³⁄₁₂ = ¹⁄₁₂ per minute. Time to fill = 1 ÷ ¹⁄₁₂ = 12 minutes.
33A businessman sold 9 kg of mixed beans at 500 Frw/kg. The mix contained 4 kg of one type at 300 Frw/kg. What was the cost per kg of the second type? Total revenue − cost of first type = cost of second type. Divide by its kg.5 mks
💡 Worked solution: ① Total = 500 × 9 = 4,500 Frw. ② First type = 300 × 4 = 1,200 Frw. ③ Second type total = 4,500 − 1,200 = 3,300 Frw. ④ Second type kg = 9 − 4 = 5 kg. ⑤ Price per kg = 3,300 ÷ 5 = 660 Frw per kg.
34aA car travels from town A to town B at 60 km/h and takes 3 hours. Calculate the distance from A to B. Distance = Speed × Time.3 mks
💡 Worked solution: Distance = Speed × Time = 60 × 3 = 180 km.
34bThe car went A→B (180 km in 3 hrs) then returned B→A (180 km in 2 hrs). Calculate the average speed for the whole journey. Average speed = Total distance ÷ Total time.4 mks
💡 Worked solution: Total distance = 180 + 180 = 360 km. Total time = 3 + 2 = 5 hrs. Average speed = 360 ÷ 5 = 72 km/h.
35aA businesswoman borrowed 180,000 Frw at 10% per annum compound interest. How much total interest did she pay after 2 years? Year 1 interest is added to principal before Year 2 calculation.6 mks
💡 Worked solution: Year 1: Interest = 180,000 × 10/100 = 18,000 Frw. Amount after Y1 = 198,000 Frw. Year 2: Interest = 198,000 × 10/100 = 19,800 Frw. Total interest = 18,000 + 19,800 = 37,800 Frw.
35bWhat is the total amount the businesswoman returned to the bank? Total = Principal + Compound Interest = 180,000 + 37,800.7 mks
💡 Worked solution: Total repayment = Principal + Total Interest = 180,000 + 37,800 = 217,800 Frw.