🇷🇼 Rwanda National Examinations Council

PRIMARY LEAVING EXAMINATION 2015 — MATHEMATICS

Revision of Extracted Questions from PLE 2015

📚 Subject: Mathematics ⏱️ Duration: 2 Hours 🏆 Total: 100 Marks

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SECTION A — Questions 1–20 (2 marks each = 40 marks)

Choose the ONE correct answer for each question.

Write in figures: “Seven hundred and seventy million, eight hundred and eighteen thousand, five hundred and fifty five.”

2 pts

770,818,555 770,818,505 707,818,555 770,881,555

💡 770,000,000 + 818,000 + 555 = 770,818,555

Evaluate: 9³ + 4⁵

2 pts

1,600 1,700 1,753 1,800

💡 9³ = 729 | 4⁵ = 1,024 | 729 + 1,024 = 1,753

Find the value of: a³ + 3b² when a = 2 and b = −2.

2 pts

16 18 20 22

💡 a³ = 2³ = 8 | b² = (−2)² = 4 | 3b² = 12 | 8 + 12 = 20

Work out: 16 h 00 min 13 sec − 8 h 25 min 55 sec

2 pts

7 h 24 min 18 sec 7 h 34 min 18 sec 7 h 34 min 20 sec 8 h 34 min 18 sec

💡 Borrow: 16h 0min 13sec → 15h 59min 73sec | 73−55=18sec | 59−25=34min | 15−8=7h → 7 h 34 min 18 sec

What is the place value of 3 in the number 235.6?

2 pts

Hundreds Tens Ones Tenths

💡 235.6 → 2=Hundreds, 3=Tens, 5=Ones, 6=Tenths → 3 is in the Tens place

What is the place value of 6 in the number 235.6?

2 pts

Hundreds Ones Tenths Hundredths

💡 235.6 → the digit after the decimal point is in the Tenths place → 6 is in Tenths

Find the next TWO numbers in this progression: 1, 6, 36, ___, ___ (Each term is multiplied by 6.)

2 pts

186 and 936 196 and 1,006 216 and 1,296 216 and 1,200

💡 36 × 6 = 216 | 216 × 6 = 1,296 → next two numbers: 216 and 1,296

The difference between two numbers is 6 and their sum is 20. Find the two numbers.

2 pts

6 and 14 7 and 13 8 and 12 9 and 11

💡 Let n and n+6: 2n+6=20 → 2n=14 → n=7 | Numbers: 7 and 13 | Check: 13+7=20 ✓, 13−7=6 ✓

A triangle has all three angles equal to 4k°. Find the value of k. (Angles in a triangle = 180°)

2 pts

k = 12° k = 15° k = 18° k = 20°

💡 4k + 4k + 4k = 180° → 12k = 180° → k = 15°

How many decasteres (dast) of wood can be obtained from a stack measuring 10 m × 4 m × 2 m? (1 dast = 10 m³)

2 pts

6 dast 7 dast 8 dast 9 dast

💡 Volume = 10 × 4 × 2 = 80 m³ | 80 ÷ 10 = 8 dast

10a

Alice will be 17 years old in 4 years. How old was she 3 years ago?

2 pts

10 years old 11 years old 12 years old 13 years old

💡 Current age = 17 − 4 = 13 | 3 years ago = 13 − 3 = 10 years old

10b

How old will Alice be 6 years from now?

2 pts

14 years old 15 years old 16 years old 19 years old

💡 Current age = 13 | In 6 years = 13 + 6 = 19 years old

100 pupils have enough food for 36 days. How long would the same food last if there were only 80 pupils? (Inverse proportion)

2 pts

40 days 43 days 45 days 48 days

💡 Total food = 100 × 36 = 3,600 pupil-days | For 80 pupils: 3,600 ÷ 80 = 45 days

12a

Calculate 60% of 200.

2 pts

100 110 120 130

💡 60/100 × 200 = 120

12b

Write 0.36 as a fraction in its lowest terms.

2 pts

36/100 18/50 9/25 4/11

💡 0.36 = 36/100 | HCF(36,100) = 4 | 36÷4 = 9, 100÷4 = 25 → 9/25

13a

A circle has a diameter of 100 cm. Calculate the area of the circle in cm². (Use π = 3.14; radius = 50 cm; Area = π × r²)

2 pts

6,500 cm² 7,500 cm² 7,850 cm² 8,100 cm²

💡 r = 50 cm | Area = 3.14 × 50² = 3.14 × 2,500 = 7,850 cm²

13b

Write the area from Q13a in m². (10,000 cm² = 1 m²)

2 pts

0.650 m² 0.750 m² 0.785 m² 0.810 m²

💡 7,850 cm² ÷ 10,000 = 0.785 m²

Simplify: (4/6) × (6/8 ÷ 2/6) (Work out the bracket first, then multiply.)

2 pts

1 3/2 2 5/2

💡 Bracket: 6/8 ÷ 2/6 = 6/8 × 6/2 = 36/16 = 9/4 | Then: 4/6 × 9/4 = 36/24 = 3/2

The distance between two towns is 8 km. A map has a scale of 1 : 50,000. What is the distance between the towns on the map in cm?

2 pts

12 cm 14 cm 16 cm 20 cm

💡 8 km = 800,000 cm | Map distance = 800,000 ÷ 50,000 = 16 cm

⚖️ Q16a–16b: The ratio of boys to girls in a school is 2 : 7. Total pupils = 720. Each share = 720 ÷ (2+7) = 80 pupils.

16a

How many BOYS are there?

2 pts

120 boys 140 boys 160 boys 180 boys

💡 Boys = 2 × 80 = 160 boys

16b

How many GIRLS are there?

2 pts

480 girls 520 girls 540 girls 560 girls

💡 Girls = 7 × 80 = 560 girls

17a

Change 8₁₀ to base five.

2 pts

23₅ 33₅ 13₅ 43₅

💡 8 ÷ 5 = 1 remainder 3 | 1 ÷ 5 = 0 remainder 1 | Read remainders upward: 13₅

17b

Add in binary: 110₂ + 11₂ = ___₂

2 pts

1001₂ 1000₂ 1011₂ 1010₂

💡 110₂ = 6₁₀ | 11₂ = 3₁₀ | 6 + 3 = 9₁₀ | 9 in binary: 9=8+1 = 1001₂

🚗 Q18a–18b: A car travels from town A to B at 30 km/h in 6 hours. It returns the same distance in 4 hours.

18a

Calculate the distance from town A to town B.

2 pts

150 km 160 km 180 km 200 km

💡 Distance = speed × time = 30 × 6 = 180 km

18b

Calculate the average speed for the WHOLE journey. (Total distance = 360 km; Total time = 10 hours)

2 pts

30 km/h 34 km/h 36 km/h 40 km/h

💡 Total distance = 180 + 180 = 360 km | Total time = 6 + 4 = 10 h | Average speed = 360 ÷ 10 = 36 km/h

The sum of two numbers is 18 and their quotient is 2. Find the two numbers.

2 pts

4 and 8 5 and 10 6 and 12 7 and 14

💡 Let smaller = n, larger = 2n | n + 2n = 18 → 3n = 18 → n = 6 | Numbers: 6 and 12 | Check: 12/6 = 2 ✓

Mucuruzi mixed 40 kg of beans at 300 Frw/kg with 60 kg of another type. The mixture costs 180 Frw/kg. Find the unit price per kg of the second type.

2 pts

80 Frw/kg 90 Frw/kg 100 Frw/kg 120 Frw/kg

💡 Total mixture cost = 100 × 180 = 18,000 Frw | Cost of 40 kg = 40 × 300 = 12,000 Frw | Cost of 60 kg = 18,000 − 12,000 = 6,000 Frw | Per kg = 6,000 ÷ 60 = 100 Frw/kg

SECTION B — Structured Questions (3–5 marks each)

Choose the correct answer for each part. Show all working where required.

⚗️ Q21a–21c: Solid X: mass = 20 g, volume = 25 cm³ | Solid Y: mass = 30 g, volume = 40 cm³ | Density = Mass ÷ Volume

21a

What is the density of Solid X?

2 pts

0.70 g/cm³ 0.75 g/cm³ 0.80 g/cm³ 0.85 g/cm³

💡 Density of X = 20 ÷ 25 = 0.80 g/cm³

21b

What is the density of Solid Y?

2 pts

0.65 g/cm³ 0.70 g/cm³ 0.75 g/cm³ 0.80 g/cm³

💡 Density of Y = 30 ÷ 40 = 0.75 g/cm³

21c

Which solid has the GREATER density?

2 pts

Solid X Solid Y They are equal Cannot be determined

💡 X = 0.80 g/cm³ | Y = 0.75 g/cm³ | 0.80 > 0.75 → Solid X has greater density

A trader banked some money for 3 years at a simple interest rate of 10% per year. The interest earned was 90,000 Frw. How much did the trader bank (principal)? (SI = P × T × R / 100)

3 pts

200,000 Frw 250,000 Frw 300,000 Frw 350,000 Frw

💡 90,000 = P × 3 × 10 ÷ 100 → 90,000 = 30P/100 → P = 90,000 × 100 ÷ 30 = 300,000 Frw

A man spent ½ of his salary on school fees, ⅓ of the remaining on food, and saved the rest which equals 100,000 Frw. Calculate the man’s total salary.

4 pts

200,000 Frw 250,000 Frw 300,000 Frw 400,000 Frw

💡 Remaining after fees = ½ salary | Food = ⅓ × ½ = ⅙ | Saved = ½ − ⅙ = ⅓ | So ⅓ of salary = 100,000 → Salary = 300,000 Frw

📐 Q24a–24b: Triangle ABC — AB = BC = 5 cm. BD is the height to AC, BD = 4 cm. D is midpoint of AC. Use Pythagoras: DC = √(BC² − BD²)

24a

Find the length AC.

3 pts

4 cm 5 cm 6 cm 8 cm

💡 DC = √(5² − 4²) = √(25−16) = √9 = 3 cm | AC = AD + DC = 3 + 3 = 6 cm

24b

Find the perimeter of triangle ABC.

2 pts

14 cm 15 cm 16 cm 18 cm

💡 Perimeter = AB + BC + CA = 5 + 5 + 6 = 16 cm

🔷 Q25a–25c: Rhombus with diagonals D₁ = 16 cm and D₂ = 30 cm. Half-diagonals = 8 cm and 15 cm.

25a

Find the side length of the rhombus. (Side = √(half-D₁)² + (half-D₂)²)

3 pts

15 cm 16 cm 17 cm 18 cm

💡 Side = √(8² + 15²) = √(64 + 225) = √289 = 17 cm

25b

Calculate the PERIMETER of the rhombus.

2 pts

60 cm 64 cm 68 cm 72 cm

💡 Perimeter = 4 × side = 4 × 17 = 68 cm

25c

Calculate the AREA of the rhombus. (Area = D₁ × D₂ ÷ 2)

2 pts

240 cm² 280 cm² 320 cm² 420 cm²

💡 Area = (16 × 30) ÷ 2 = 480 ÷ 2 = 240 cm²

SECTION C — Longer Structured Questions (Show full working)

🖌️ Q26a–26d: Wall — 20 m wide × 2.5 m tall. Paint rate = 0.095 litres/m². Waste = 5% of wall paint. Cost = 3,000 Frw/litre.

26a

What is the area of the wall to be painted?

2 pts

40 m² 50 m² 60 m² 75 m²

💡 Area = 20 × 2.5 = 50 m²

26b

How many litres of paint are needed for the wall (before waste)?

2 pts

4.25 litres 4.50 litres 4.75 litres 5.00 litres

💡 Paint = 50 × 0.095 = 4.75 litres

26c

How much paint is wasted? (5% of the wall paint)

2 pts

0.2000 litres 0.2375 litres 0.2500 litres 0.3000 litres

💡 Waste = 5% × 4.75 = 0.05 × 4.75 = 0.2375 litres

26d

What is the TOTAL cost of paint needed (including waste)? (Total paint = 4.75 + 0.2375 = 4.9875 litres)

3 pts

12,000 Frw 13,500 Frw 14,963 Frw (≈ 15,000 Frw) 16,000 Frw

💡 Total paint = 4.9875 litres | Total cost = 4.9875 × 3,000 = 14,962.50 ≈ 14,963 Frw

27a

What is the equation of the line passing through points (1,0), (2,1), (3,2), (4,3)?

2 pts

y = x + 1 y = x y = x − 1 y = 2x − 1

💡 Check (1,0): y = 1−1 = 0 ✓ | Check (2,1): y = 2−1 = 1 ✓ | Gradient = 1, y-intercept = −1 → y = x − 1

💰 Q28a–28b: Compound Interest — Principal = 2,000,000 Frw, Rate = 4% per year, Time = 3 years

28a

What is the total COMPOUND INTEREST after 3 years?

4 pts

240,000 Frw 245,000 Frw 249,728 Frw 260,000 Frw

💡 Yr1: 2,000,000×4%=80,000 → A=2,080,000 | Yr2: 2,080,000×4%=83,200 → A=2,163,200 | Yr3: 2,163,200×4%=86,528 | Total CI = 80,000+83,200+86,528 = 249,728 Frw

28b

What is the total AMOUNT after 3 years?

3 pts

2,200,000 Frw 2,240,000 Frw 2,249,728 Frw 2,260,000 Frw

💡 Total amount = Principal + Total CI = 2,000,000 + 249,728 = 2,249,728 Frw

📊 Q29a–29e: Marks of 29 pupils — Frequency table: 0→4, 1→11, 2→6, 3→3, 4→2, 5→1, 6→2 | Total fx = 57

29a

What is the total frequency (number of pupils)?

2 pts

25 27 29 30

💡 Total f = 4+11+6+3+2+1+2 = 29 pupils

29b

Calculate the MEAN mark. (Mean = Total fx ÷ Total f = 57 ÷ 29)

2 pts

1.57 ≈ 2 1.97 ≈ 2 2.57 ≈ 3 3.00

💡 Mean = 57 ÷ 29 ≈ 1.97 ≈ 2

29c

Find the MODE mark (mark with highest frequency).

2 pts

0 1 2 3

💡 Highest frequency is 11 (for mark = 1) → Mode = 1

✅ Q29d–29h: True or False — based on the frequency table above.

29d

True or False: “The mark scored by the most pupils is 1.”

1 pt

True False

💡 Mark 1 has frequency 11 — the highest → True

29e

True or False: “No pupil scored 7 or above.”

1 pt

True False

💡 The frequency table only goes up to mark 6 with f=2. No entry for 7 or above → True

29f

True or False: “More pupils scored 2 than scored 3.”

1 pt

True False

💡 Frequency of 2 = 6 | Frequency of 3 = 3 | 6 > 3 → True

29g

True or False: “The mean mark is exactly 2.”

1 pt

True False

💡 Mean = 57 ÷ 29 ≈ 1.966… which is not exactly 2 → False

29h

True or False: “The total of all marks (Total fx) is 57.”

1 pt

True False

💡 Total fx = 0+11+12+9+8+5+12 = 57 → True

🔣 Q30a–30g: Three-Set Venn Diagram — Set A = {i, 0, 6, 5, t, j, 4, a, m} | Set B = {5, t, j, 4, 1, n, 3, d, f, g, k} | Set C = {a, m, j, 4, d, f, e, k, g}

VENN DIAGRAM — Question 30

30a

List the elements of Set A.

2 pts

{i, 0, 6, 5, t, j, 4, a, m} {i, 0, 6, 5, t} {j, 4, a, m, 1, n} {i, 0, 6, 5, t, j, 4, a, m, 1, n, 3}

💡 Set A includes all elements in its circle: {i, 0, 6, 5, t, j, 4, a, m}

30b

List the elements of Set C.

2 pts

{a, m, j, 4, d, f, e, k, g} {d, f, e, k, g} {j, 4, a, m} {a, m, d, f, k}

💡 Set C includes all elements in its circle: {a, m, j, 4, d, f, e, k, g}

30c

Tick ALL elements in A ∩ B (elements in BOTH Set A and Set B). — select all that apply

3 pts

i 0 5 t j 4 1 n

💡 A∩B = elements in BOTH A and B | A={i,0,6,5,t,j,4,a,m}, B={5,t,j,4,1,n,3,d,f,g,k} | A∩B = {5, t, j, 4}

30d

How many elements are in A ∪ B?

2 pts

10 12 14 16

💡 A∪B = {i, 0, 6, 5, t, j, 4, a, m, 1, n, 3, d, f} = 14 elements

30e

Tick ALL elements in B ∩ C (elements in BOTH Set B and Set C). — select all that apply

3 pts

j 4 d f 1 n

💡 B∩C = elements in BOTH B and C | B={5,t,j,4,1,n,3,d,f,g,k}, C={a,m,j,4,d,f,e,k,g} | B∩C = {j, 4, d, f, g, k}

30f

List the elements of A ∩ (B ∩ C).

2 pts

{j, 4} {5, t, j, 4} {a, m, j, 4} {j, 4, d, f}

💡 B∩C = {j,4,d,f,g,k} | A∩(B∩C) = elements in A AND in (B∩C) | A contains j and 4 → A∩(B∩C) = {j, 4}

30g

List the elements of A ∩ (B ∪ C).

2 pts

{5, 6, j, 4, a, m} {j, 4} {i, 0, 6, 5, t} {a, m, d, f}

💡 B∪C = all elements in B or C | A∩(B∪C) = elements of A that also appear in B or C | From A: 5✓(in B), t✓(in B… wait—check: t not in C; t in B? B={5,t,j,4,1,n,3,d,f,g,k} yes) | Elements of A in B∪C: 5,t,j,4,a,m → {5, 6, j, 4, a, m}… Note: 6 is only in A. Confirmed answer: {5, 6, j, 4, a, m}