WASSCE Elective Mathematics past questions and answers

Probability from a frequency table is the relative frequency: the number of times the outcome occurred divided by the total number of trials.

Two of the frequencies are written in terms of y, so find y first. The die was tossed 150 times, so the six frequencies add up to 150:

12 + 18 + y + 30 + 2y + 45 = 150

105 + 3y = 150

3y = 45, so y = 15.

The frequency of face 5 is 2y = 2 × 15 = 30.

P(obtaining a 5) = 30/150 = 1/5, option B.

Why the other options are wrong: 1/10 (option C) is 15/150, the frequency of face 3 (y) used in place of face 5 (2y). 3/10 (option A) is 45/150, the frequency of face 6. 1/6 (option D) is the theoretical probability of any face on a fair die with no data; this question asks for the experimental probability from the table, which is 1/5.

Slips to avoid: finding 2y = 30 and stopping, or dividing by 6 (the number of faces) instead of 150 (the number of tosses).

A check: with y = 15 the six frequencies are 12, 18, 15, 30, 30 and 45, which add up to 150.

The mode of a frequency distribution is the value that occurs most often. In this table the values are the faces 1 to 6, and the frequency row tells you how many times each face came up. So the mode is the face with the largest frequency, not the largest frequency itself.

Two of the frequencies are written in terms of y, so find y first. The die was tossed 150 times, so the six frequencies add up to 150:

12 + 18 + y + 30 + 2y + 45 = 150

105 + 3y = 150

3y = 45, so y = 15.

The frequencies are now: face 1 has 12, face 2 has 18, face 3 has 15, face 4 has 30, face 5 has 30 and face 6 has 45.

The largest frequency is 45, and it belongs to face 6. So the mode is 6, option B.

Why the other options are wrong: face 5 (option A) has frequency 2y = 30, which only ties with face 4 (option D) at 30; neither is the highest. Face 3 (option C) has frequency y = 15, one of the smallest. The face whose frequency was the unknown is not automatically the mode.

Slips to avoid: answering 45, which is the frequency of the mode and not the mode itself; or choosing face 5 because 2y looks large, without working out that 2y = 30 is still less than 45.

A check: with y = 15 the six frequencies are 12 + 18 + 15 + 30 + 30 + 45 = 150, as the question states.

The mean is the total of all the values divided by how many values there are.

Add the six speeds: 62 + 64 + 68 + 70 + 72 + 69 = 405

Divide by the number of readings:
mean = 405 / 6 = 67.5

So the mean speed is 67.5 m s⁻¹.

A quick check: every reading lies between 62 and 72, so the mean must lie between 62 and 72 as well. Any answer outside that range can be rejected straight away.

Why the other options are wrong. 68.5 comes from a slip in the addition, giving a total of 411 instead of 405. 69.5 comes from a total of 417. 70.5 comes from a total of 423. It also sits almost at the largest single reading, which the mean of six values spread from 62 to 72 cannot be.

Estimate the area by the trapezium rule with two strips of width 1.

Reading the graph: f(1) ≈ 2, f(2) ≈ 5, f(3) ≈ 4.5.

Trapezium estimate: ½ × 1 × (f(1) + f(2)) + ½ × 1 × (f(2) + f(3)) = ½(2 + 5) + ½(5 + 4.5) = 3.5 + 4.75 ≈ 8 sq units.

Check: (1,1) maps to (−1,−1). [0 −1; −1 0][1;1] = [−1;−1] ✓. (3,1) maps to (−1,−3). [0 −1; −1 0][3;1] = [−1;−3] ✓. This is reflection in the line y = −x.

2cos(x) = 1 ⇒ cos(x) = 1/2. In [0°, 360°]: x = 60° and x = 300°. The horizontal line y = 1 cuts the cosine curve at exactly 2 points.

The points lie close to a straight line with a positive gradient (as x increases, y increases). This indicates a strong positive correlation.

M = [0 −1; 1 0] is a 90° anticlockwise rotation. A(1,0) → (0,1) B(3,0) → (0,3) C(1,2) → (−2,1) All image points have non-positive x and positive y — the image lies in Q2.