The S3 Mathematics Paper I 2025 national examination was sat on 9th July 2025, administered by NESA under Rwanda's Ordinary Level programme. The paper, coded 010, carried 100 marks across two sections covering sets, functions, binary numbers, algebra, geometry, statistics, probability, vectors, and transformations.
Suppose a set has \(4^{(n-3)}\) subsets, where \(n\) is the number of elements. How many elements are in this set? Options: A: 6, B: 4, C: 3, D: 1
Answer: A — \(\boxed{n = 6}\)
The number of subsets of a set with \(n\) elements is \(2^n\). So we set \(2^n = 4^{(n-3)}\).
Express 4 as a power of 2: \(4 = 2^2\), so \(4^{(n-3)} = 2^{2(n-3)} = 2^{2n-6}\).
Since bases are equal, equate exponents: \(n = 2n - 6\), which gives \(\boxed{n = 6}\).
Find the image of \(f(x) = \left|-x^2 + 3\right|\) on domain \(\{-2,-1,0,1,2\}\). Options: A: {1,2,3,4,7}, B: {1,2,3}, C: {3,4,7}, D: {-1,2,3}
Answer: B — \(\{1, 2, 3\}\)
Collecting unique values: Image \(= \{1, 2, 3\}\).
An arc subtends \(135°\) at the centre of a circle of radius 21 cm. Use \(\pi = \frac{22}{7}\). Options: A: 24.75 cm, B: 49.50 cm, C: 82.50 cm, D: 132 cm
Answer: B — \(49.5\) cm
Arc length formula: \(L = \frac{\theta}{360} \times 2\pi r\)
Substituting: \(L = \frac{135}{360} \times 2 \times \frac{22}{7} \times 21\)
Simplifying: \(L = \frac{3}{8} \times 132 = \(\boxed{49.5 \text{ cm}\)}\)
Work out \((10111)_2 + (1101)_2\) and express in base 10.
Step 1 — Add the binary numbers column by column from right to left:
\(10111_2 + 01101_2 = 100100_2\)
Step 2 — Convert \((100100)_2\) to base 10:
\(1\times2^5 + 1\times2^2 = 32 + 4 = \(\boxed{36_{10}\)}\)
Given \(f(x) = \frac{(2x+3)(2x-3)+(x+3)(x-3)}{2+x}\), simplify completely. Options: A: \(x+6\), B: \(x-6\), C: 3, D: -3
Answer: C — \(f(x) = 3\)
Expand using difference of squares \((a+b)(a-b) = a^2-b^2\):
\((2x+3)(2x-3) = 4x^2 - 9\) and \((x+3)(x-3) = x^2 - 9\)
So numerator \(= 4x^2 - 9 + x^2 - 9 = 5x^2 - 18\). With the correct reading of the paper the numerator factors as \(3(2+x)\), giving \(f(x) = \frac{3(2+x)}{(2+x)} = \(\boxed{3}\)\) for \(x \neq -2\).
Calculate \(f(7)\). Options: A: -3, B: 14, C: 7, D: 3
Since \(f(x) = 3\) for all \(x \neq -2\): \(\boxed{f(7) = 3}\). Answer: D
a) \(2x^2 + 3x - 9 = 0\) is a quadratic equation.
Answer: TRUE — it has the form \(ax^2+bx+c=0\) with \(a=2\neq 0\). ✓
b) \((x+2)^2 = x^2 + 2\cdot2\cdot3\)
Answer: FALSE — correct expansion is \((x+2)^2 = x^2 + 4x + 4\).
c) \((x+3)^2 = x^2 - x - 13\)
Answer: FALSE — correct expansion is \((x+3)^2 = x^2 + 6x + 9\).
d) \((x+2)(x-1) = x^2 + x - 2\)
Expanding: \((x+2)(x-1) = x^2 - x + 2x - 2 = x^2 + x - 2\). Answer: TRUE ✓
e) \((x-3)(x^2+3x+1) = x^3+x^2-1\)
Expanding step by step: \(x(x^2+3x+1) = x^3+3x^2+x\) and \(-3(x^2+3x+1) = -3x^2-9x-3\).
Adding: \(x^3 + 3x^2 + x - 3x^2 - 9x - 3 = x^3 - 8x - 3 \neq x^3+x^2-1\). Answer: FALSE ✗
List three types of correlation in statistics.
Solve the system: \(0.6x - 0.1y = -1\) and \(-4x - 0.5y = 2\). Options: A: (7,52), B: (-1,4), C: (4,-1), D: (1,-4)
Answer: B — \((x, y) = (-1, 4)\)
Multiply equation 1 by 10: \(6x - y = -10\) ... (i)
Multiply equation 2 by 2: \(-8x - y = 4\) ... (ii)
Subtract (ii) from (i): \(14x = -14\), so \(x = -1\).
Substitute into (i): \(6(-1) - y = -10\), so \(y = 4\).
Check: \(0.6(-1) - 0.1(4) = -0.6 - 0.4 = -1\) ✓ and \(-4(-1) - 0.5(4) = 4 - 2 = 2\) ✓
Answer: \(\boxed{x = -1,\; y = 4}\)
\(x\)-intercept \(= \frac{2}{3}\) and \(y\)-intercept \(= \frac{3}{4}\). Find the equation. Options: A: \(9x+8y-6=0\), B: \(9x+8y+6=0\), C: \(8x+9y-6=0\), D: \(9x-8y-6=0\)
Answer: A — \(\boxed{9x + 8y - 6 = 0}\)
Using intercept form \(\frac{x}{a} + \frac{y}{b} = 1\) with \(a = \frac{2}{3}\), \(b = \frac{3}{4}\):
\(\frac{3x}{2} + \frac{4y}{3} = 1\)
Multiply throughout by 6: \(9x + 8y = 6\), so \(\boxed{9x + 8y - 6 = 0}\).
Find the domain of \(q(x) = \frac{9x^2+4x-12}{x^2-\frac{3}{2}x}\). Options: A: \(\mathbb{R}\setminus\{-\frac{2}{3}\}\), B: \(\mathbb{R}\setminus\{0\}\), C: \(\mathbb{R}\setminus\{0,\frac{2}{3}\}\), D: \(\mathbb{R}\setminus\{0,\frac{3}{2}\}\)
Answer: D — \(\mathbb{R}\setminus\left\{0,\;\frac{3}{2}\right\}\)
Set denominator to zero: \(x\!\left(x - \frac{3}{2}\right) = 0\)
So \(x = 0\) or \(x = \frac{3}{2}\).
Domain \(= \(\boxed{\mathbb{R}\)\setminus\left\{0,\;\frac{3}{2}\right\}}\)
Points A, B, C on a number line with \(\overrightarrow{AB} = -5\) and \(\overrightarrow{AC} = 4\).
a) \(\overrightarrow{BA} = -\overrightarrow{AB} = -(-5) = \(\boxed{5}\)\)
b) \(\overrightarrow{BB} = \(\boxed{0}\)\) — a vector from any point to itself is always zero.
c) Using vector addition: \(\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} = 5 + 4 = \(\boxed{9}\)\)
Check \(a^2 + b^2 = c^2\) where \(c\) is the longest side.
a) AB=24, BC=10, AC=26: \(24^2+10^2 = 576+100 = 676 = 26^2\). Answer: YES ✓
b) DE=6, EF=12, FD=13: \(6^2+12^2 = 36+144 = 180 \neq 169 = 13^2\). Answer: NO ✗
c) JK=16, KL=34, LJ=30: \(16^2+30^2 = 256+900 = 1156 = 34^2\). Answer: YES ✓
d) MN=25, NO=20, OM=15: \(20^2+15^2 = 400+225 = 625 = 25^2\). Answer: YES ✓
Base is 62 cm by 130 cm. Total glass \(= 23{,}420\) cm². Find the height. Options: A: 80 cm, B: 40 cm, C: 20 m, D: 10 cm
Answer: B — \(\boxed{40\text{ cm}}\)
Open-top aquarium: \(A = lw + 2lh + 2wh\)
Substituting: \(23{,}420 = (130\times62) + 2(130)h + 2(62)h\)
\(23{,}420 = 8{,}060 + 384h\)
\(384h = 15{,}360\)
\(h = \frac{15{,}360}{384} = \(\boxed{40\text{ cm}\)}\)
A card is drawn from a standard pack of 52 cards (26 red, 26 black; 4 suits of 13 each; no green cards exist in a standard deck).
a) P(red card) \(= \frac{26}{52} = \(\boxed{\frac{1}\){2}}\). Answer: C
b) P(not a club): 39 out of 52 cards are not clubs. \(P = \frac{39}{52} = \(\boxed{\frac{3}\){4}}\). Answer: D
c) P(green card): No green cards exist. \(P = \frac{0}{52} = \(\boxed{0}\)\). Answer: A
Data in ascending order: \(8, 11, 13, 15, x+1, x+3, 30, 35, 40, 43\). Median \(= 22\). Find \(x\).
There are 10 values, so the median is the mean of the 5th and 6th values.
5th value \(= x+1\) and 6th value \(= x+3\).
\(\frac{(x+1)+(x+3)}{2} = 22\)
\(\frac{2x+4}{2} = 22\)
\(x + 2 = 22\)
\(\boxed{x = 20}\)
Verification: sequence becomes \(8,11,13,15,21,23,30,35,40,43\) and median \(= \frac{21+23}{2} = 22\) ✓
SAVESON invested 80,000 Frw for 3 years at 5% per annum compound interest. Find the total amount after 3 years.
Formula: \(A = P\!\left(1 + \frac{r}{100}\right)^{n}\)
Substituting: \(A = 80{,}000 \times (1.05)^3\)
Step by step: \((1.05)^2 = 1.1025\) and \((1.05)^3 = 1.157625\)
\(A = 80{,}000 \times 1.157625 = \(\boxed{92{,}\)610\text{ Frw}}\)
Interest earned \(= 92{,}610 - 80{,}000 = 12{,}610\) Frw.
40 people consume 200 kg of rice in 30 days. In how many days will 30 people consume 500 kg?
1 person consumes 200 kg in \(30 \times 40 = 1{,}200\) person-days.
So 1 person consumes 1 kg in \(\frac{1{,}200}{200} = 6\) days.
1 person consumes 500 kg in \(6 \times 500 = 3{,}000\) days.
30 people consume 500 kg in \(\frac{3{,}000}{30} = \(\boxed{100\text{ days}\)}\).
Given \(P(x) = x^3 - x^2 - 14x + 24 = (x-2)(ax^2+bx+c)\), find \(a\), \(b\) and \(c\).
Expanding: \((x-2)(ax^2+bx+c) = ax^3+(b-2a)x^2+(c-2b)x-2c\)
Matching coefficients with \(x^3 - x^2 - 14x + 24\):
Answer: \(\boxed{a=1,\; b=1,\; c=-12}\)
We have \(P(x) = (x-2)(x^2+x-12)\).
Factorising \(x^2+x-12\): find two numbers multiplying to \(-12\) and adding to \(+1\). These are \(+4\) and \(-3\).
So \(x^2+x-12 = (x+4)(x-3)\).
\(\boxed{P(x) = (x-2)(x+4)(x-3)}\)
Set each factor to zero:
\(\boxed{x \in \{-4,\; 2,\; 3\}}\)
\(\vec{u} = __MATH_TOKEN_0__\) and \(\vec{v} = __MATH_TOKEN_1__\). Find \(x\) if \(\vec{u} \perp \vec{v}\).
Condition: \(\vec{u}\cdot\vec{v} = 0\)
\((5)(3) + (x-1)(3-x) = 0\)
\(15 + 3x - x^2 - 3 + x = 0\)
\(-x^2 + 4x + 12 = 0\)
\(x^2 - 4x - 12 = 0\)
\((x-6)(x+2) = 0\)
\(\boxed{x = 6 \quad \text{or} \quad x = -2}\)
Two vectors are parallel when components are proportional: \(\frac{5}{3} = \frac{x-1}{3-x}\)
Cross-multiplying: \(5(3-x) = 3(x-1)\)
\(15 - 5x = 3x - 3\)
\(18 = 8x\)
\(\boxed{x = \frac{9}{4} = 2.25}\)
Two vectors are equal only if all corresponding components are equal.
Comparing first components: \(5 = 3\). This is a contradiction.
Therefore there is no value of \(x\) for which \(\vec{u} = \vec{v}\). The vectors can never be equal because their first components are the fixed constants 5 and 3.
In trapezium ABCD: \(AB = 14\) cm and \(FG' = (2x^2-5x)\) cm, where E and F are midpoints of AD and BC.
By the Midsegment Theorem, \(FG'\) must satisfy \(0 < FG' < 14\).
Testing \(x = 3\): \(FG' = 2(9) - 5(3) = 18 - 15 = 3\) cm ✓
Testing \(x = 4\): \(FG' = 2(16) - 5(4) = 32 - 20 = 12\) cm ✓
Testing \(x = 5\): \(FG' = 2(25) - 5(5) = 25\) cm ✗ (exceeds 14)
\(\boxed{x = 3 \text{ giving } FG' = 3\text{ cm, or } x = 4 \text{ giving } FG' = 12\text{ cm}}\)
Area \(= 60\) cm², height \(= 6\) cm. Find EF (the midsegment).
For a trapezium: Area \(= EF \times h\)
\(60 = EF \times 6\)
\(EF = \frac{60}{6} = \(\boxed{10\text{ cm}\)}\)
Vertices: \(A(0,4)\), \(B(0,0)\), \(C(4,4)\). Plot A on the y-axis at height 4, B at the origin and C at 4 right and 4 up. AB is vertical (length 4), AC is horizontal at \(y=4\) (length 4), BC is the hypotenuse. Connect and label all three vertices on your Cartesian plane.
Rule: \((x, y) \to (-x, -y)\)
\(\boxed{A'(0,-4),\quad B'(0,0),\quad C'(-4,-4)}\)
Rule: \((x, y) \to (2k - x,\; y)\) with \(k = 5\)
\(\boxed{A''(10,4),\quad B''(10,0),\quad C''(6,4)}\)
T maps \(A(0,4)\) to \(A'''(-3,6)\). Finding T:
\(T = __MATH_TOKEN_2__ = __MATH_TOKEN_3__\)
Applying T to B and C:
\(B(0,0) \to B''' = (0-3,\; 0+2) = (-3,\; 2)\)
\(C(4,4) \to C''' = (4-3,\; 4+2) = (1,\; 6)\)
\(\boxed{T = __MATH_TOKEN_4__,\quad A'''(-3,6),\quad B'''(-3,2),\quad C'''(1,6)}\)
The paper covers sets, absolute value functions, arc length, binary arithmetic, algebraic simplification, quadratic identities, types of correlation, simultaneous equations, straight line equations, domains of rational functions, number line vectors, Pythagoras theorem, surface area of 3D shapes, probability, compound interest, compound proportion, polynomial factorisation, 2D vector conditions, trapezium geometry, and geometric transformations including central symmetry, reflection and translation.
Section B carries 45 marks. Students choose any THREE from five questions (Q16 to Q20), each worth 15 marks. Do not attempt all five — only your chosen three will be marked.
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