BECE Mathematics past questions and answers

Reflection in the x-axis (the line y = 0) sends every point (x, y) to (x, −y). The x-coordinate stays; the y-coordinate flips sign.

Triangle PQR sits in Quadrant I, so every vertex has x > 0 and y > 0. After reflection: x stays positive, y becomes negative.

(positive x, negative y) → Quadrant IV.

Why the other options are wrong: A (Quadrant II) — Q II has negative x, positive y. Would come from reflecting in the y-axis, not the x-axis. C (Quadrant I) — the original quadrant; would only happen if no reflection were applied. D (Quadrant III) — Q III has negative x and negative y. Would come from a 180° rotation about the origin (or reflecting in BOTH axes).

The angles ∠BEF and ∠EFD are co-interior angles (also called "C-angles" or allied angles) on parallel lines AB and CD.

Co-interior angles sum to 180°: (3x + 30) + (2x + 20) = 180 5x + 50 = 180 5x = 130 x = 26

A perpendicular bisector cuts a line segment into two equal halves at a 90-° angle. The construction shown uses compass arcs from both endpoints to find points equidistant from A and B, then connects them to form the bisector.

The angle at M is 60°. The right angle is at P.

MP = 6 cm is the side adjacent to the 60° angle (it touches M and is one of the two sides forming that angle).

NP is the side opposite the 60° angle (the side that does not touch M).

tan 60° = opposite / adjacent
√3 = NP / 6
NP = 6√3 cm

The answer is D.

BECE tip: in a 30-60-90 triangle, the side opposite 60° is √3 times the side opposite 30°.

Two steps: find the equation of the boundary line, then choose the right inequality sign from the shading.

Multiply every term by the LCM of 2 and 3, which is 6: 3x + 2y = 6

Test the origin (0, 0) in 3x + 2y: 3(0) + 2(0) = 0. So we need an inequality that allows 0 to satisfy it.
0 ≤ 6 is true. 0 ≥ 6 is false. So the sign is ≤.

The inequality is 3x + 2y ≤ 6.

Why the other options are wrong:
A (2x + 3y ≤ 6) — has the coefficients of x and y swapped. The intercepts would be (3, 0) and (0, 2), not (2, 0) and (0, 3).
C (3x + 2y ≥ 6) — correct line but wrong direction; this is the region NOT containing the origin.
D (2x + 3y ≥ 6) — both line wrong and direction wrong.

When a circle is INSCRIBED in a square, the circle touches all four sides of the square. The circle's DIAMETER equals the square's SIDE.

Given the square has side 2 cm: Diameter of circle = 2 cm → radius r = 1 cm.

Shaded area = Area of square − Area of circle.

Area of square = side² = 2² = 4 cm².
Area of circle = π r² = (22/7) × 1² = 22/7 cm² ≈ 3.143 cm².

Shaded area = 4 − 22/7
= 28/7 − 22/7
= 6/7 cm²
≈ 0.857 cm² ≈ 0.86 cm² (to 2 d.p.).

Why the other options are wrong:
A (6.28 cm²) — equals 2π using π ≈ 3.14, which is the circumference of a circle of radius 1, not an area.
B (3.00 cm²) — looks like a rounded value of the circle area itself (3.14), not the shaded region.
C (12.56 cm²) — equals 4π, which is the area of a circle of radius 2 — wrong shape.

AB is parallel to CD and GE is a transversal cutting both lines (at G on AB, at F on CD).

Angles AGE and CFG sit on the SAME side of the transversal and BETWEEN the two parallel lines, so they are co-interior (same-side interior) angles. The key property:

Co-interior angles between parallel lines add up to 180°.

So k = 20 (matching option C: 20°).

Why the other options are wrong:
A (16°) — comes from 80÷5, ignoring the 180° relationship.
B (80°) — confuses k with the given angle.
D (160°) — uses sum = 180° but forgets to divide by 5.

The slope (gradient) of a line through two points is: m = (y₂ − y₁) / (x₂ − x₁) = rise / run.

Using the two points shown on the diagram, (0, 3) and (1, 1):
m = (1 − 3) / (1 − 0)
= −2 / 1
= −2.

Why the other options are wrong: A (1) — wrong sign and wrong magnitude; the line clearly falls left-to-right, so slope must be negative. B (2) — correct magnitude but wrong sign; +2 would mean the line rises. C (−1) — wrong magnitude; computed (1 − 3)/(2 − 0) = −1 by reading x₂ as 2 instead of 1.