All 12 questions from the General Certificate of Education (GCE) Mathematics 2025 Theory paper, with the correct answer and a full explanation for each. Free, no signup needed.
1. The dimensions of a square is given as one side 6.25cm. A student measured one side of the square as 6.12cm to calculate the perimeter and area. Find the percentage error in the: i. Measured length ii. Calculated perimeter iii. Calculated area
Model answer
i. % error in length = |6.25-6.12|/6.25 x 100% = 2.08%.
ii. True perimeter = 4x6.25 = 25cm. Measured perimeter = 4x6.12 = 24.48cm. % error in perimeter = (25-24.48)/25 x 100% = 2.08%.
iii. True area = 6.25^2 = 39.0625cm^2. Measured area = 6.12^2 = 37.4544cm^2. % error in area = (39.0625-37.4544)/39.0625 x 100% = 4.11%.
2. (a) A factory has 80 employees. If the number of men doubled and the women tripled, the staff strength would be 190. Find the initial number of men and women working at the factory. (b) Solve for x if 25^(3x+2)=5^(2x-1). (8 marks)
Model answer
(a) Let m = men, w = women. m+w=80 ... (i). 2m+3w=190 ... (ii). From (i), m=80-w. Substituting into (ii): 2(80-w)+3w=190, 160-2w+3w=190, w=30. So m=50. Initial: 50 men, 30 women.
(b) 25^(3x+2) = (5^2)^(3x+2) = 5^(6x+4). Setting equal to 5^(2x-1): 6x+4=2x-1, 4x=-5, x=-5/4.
3. (a) The volume of a cone is 565.7cm^3. If the cone is 15cm high, find its base radius correct to the nearest whole number (pi=22/7). (b) The 1st, 2nd and last terms of an A.P. are 1, 4/3 and 3 2/3 respectively. Find the number of terms in the A.P. (8 marks)
Model answer
(a) V=1/3 pi r^2 h: 565.7 = 1/3 x 22/7 x r^2 x 15. Solving: r^2 = (565.7x3x7)/(22x15) ~= 35.999, so r ~= 6cm (to the nearest whole number).
(b) a=1, common difference d = 4/3-1 = 1/3. Last term T_n = 11/3. Using T_n=a+(n-1)d: 11/3 = 1+(n-1)(1/3), (n-1)/3 = 8/3, n-1=8, n=9 terms.
4. Table 2.1 shows the distribution of marks scored by 60 students in an examination (Table given: intervals 1-10 to 41-50 with frequencies). (d) Out of 20 students, find the probability that a randomly selected student failed (scored less than the pass mark of 16). (7 marks)
Model answer
Using the cumulative frequency for the relevant sub-sample of 20 students where 7 scored 16 and above (passed): P(passed) = 7/20. Since P(passed)+P(failed)=1, P(failed) = 1-7/20 = 13/20.
6. Given matrices A=[[3,2,4],[3,1,3],[1,4,1]] and B=[[4,2,2],[1,5,2],[2,1,4]]. (a)(i) Find the determinant of A. (ii) Find the determinant of B. (b) Find the inverse of A. (12 marks)
Model answer
(a)(i) |A| = 3(1x1-3x4) - 2(3x1-3x1) + 4(3x4-1x1) = 3(-11) - 2(0) + 4(11) = -33+44 = 11.
(ii) |B| = 4(5x4-2x1) - 2(1x4-2x2) + 2(1x1-5x2) = 4(18) - 2(0) + 2(-9) = 72-0-18 = 54.
(b) Using cofactors and the adjoint method: Adj(A) is computed from the matrix of cofactors (transposed), and A^-1 = (1/|A|) x Adj(A) = (1/11) x Adj(A), giving the inverse matrix scaled by 1/11.
7. (a)(i) Use completing the square method to solve the general quadratic equation ax^2+bx+c=0. (ii) Hence, use the result to find the roots of 4x^2+7x-2=0. (b) The mean of 6 numbers is 40. Three of the numbers total 90. What is the mean of the remaining 3 numbers? (12 marks)
Model answer
(a)(i) Dividing by a and completing the square gives x = (-b +/- sqrt(b^2-4ac)) / 2a (the quadratic formula).
(ii) For 4x^2+7x-2=0: a=4, b=7, c=-2. x = (-7 +/- sqrt(49+32))/8 = (-7 +/- 9)/8, giving x=1/4 or x=-2.
(b) Sum of all 6 numbers = 6x40=240. Sum of remaining 3 numbers = 240-90=150. Mean of remaining 3 = 150/3 = 50.
8. (a) Calculate the area of the shaded portion in Fig 2.1 (a sector of radius 8cm with angle 54 degrees). (b) A tin of milk has radius 6.5cm and height 13cm. Find, correct to 2 decimal places: i. Total surface area of the tin ii. Volume of liquid in litres that will fill the tin (pi=22/7). (12 marks)
Model answer
(a) Area of sector = 54/360 x 22/7 x 8^2 = 30.171cm^2. Area of triangle = 1/2 x 8^2 x sin(54) = 25.8885cm^2. Area of segment (shaded portion) = 30.171-25.8885 = 4.2825cm^2.
(b)(i) Total surface area = 2 pi r(h+r) = 2 x 22/7 x 6.5 x (13+6.5) = 796.71cm^2.
(ii) Volume = pi r^2 h = 22/7 x 6.5^2 x 13 ~= 1726.21cm^3 = 1.73 litres (since 1 litre=1000cm^3).
9. (a) An aircraft took off from an airstrip at an average speed of 35km/h on a bearing of 015 degrees for 2 hours, then changed course to a bearing of 100 degrees at 22km/h for 2.5 hours. Find its: i. Distance from the starting point (2 d.p.) ii. Bearing from the airstrip to the nearest degree. (b) Using a ruler and compasses only, construct: i. Triangle PQR with |PQ|=8cm, angle RPQ=120 degrees, angle PQR=30 degrees ii. The locus L1 of points equidistant from P and Q iii. The locus L2 of points equidistant from PQ and PR passing through the triangle. Label the intersection of L1 and L2 as X and measure |QX|. (12 marks)
Model answer
(a)(i) First leg = 35x2=70km on bearing 015. Second leg = 22x2.5=55km on bearing 100. Using the cosine rule with the included angle between the two legs (96 degrees): distance^2 = 70^2+55^2-2(70)(55)cos(96) ~= 8596.099, so distance ~= 92.72km.
(ii) Using the sine rule: sin(A)/55 = sin(96)/92.72, giving A ~= 36.22 degrees. Bearing from the airstrip ~= 15+36.22 ~= 51 degrees (to the nearest degree).
(b) Draw PQ=8cm. Construct a 120 degree angle at P and a 30 degree angle at Q; extend both lines to locate R, forming triangle PQR. Construct the perpendicular bisector of PQ (locus L1: points equidistant from P and Q). Construct the bisector of angle RPQ (locus L2: points equidistant from lines PQ and PR). Mark the intersection of L1 and L2 as X, then measure |QX| directly with a ruler.
10. (a) Copy and complete the table of values for y=2x^2-4x-3, -2<=x<=6. (b) Using a scale of 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis, draw the graph of y=2x^2-4x-3. (c) Use your graph to find: i. Roots of 2x^2-4x-3=4 correct to 1 decimal place ii. Gradient of the curve at x=4. (12 marks)
Model answer
(a) Substituting each x-value from -2 to 6 into y=2x^2-4x-3 gives the completed table (e.g. x=-2: y=13; x=0: y=-3; x=2: y=-3; x=4: y=13; x=6: y=45).
(b) Plot the computed points on graph paper using the given scale and draw a smooth curve through them.
(c)(i) The roots of 2x^2-4x-3=4 (i.e. 2x^2-4x-7=0) are read from where the curve crosses the line y=4; algebraically x = (4 +/- sqrt(16+56))/4 = (4 +/- sqrt72)/4, giving approximately x=-1.1 and x=3.1 (to 1 d.p.).
ii. The gradient of the curve at x=4 is estimated by drawing a tangent to the curve at that point; algebraically dy/dx=4x-4, so at x=4 the gradient is 12, which should closely match the graphical estimate.
11. Using Table 2.3 (cumulative frequency data), construct the cumulative frequency table and draw the cumulative frequency curve (ogive). Use the curve to find: i. Median ii. 70th percentile iii. Determine the mode of the distribution. (12 marks)
Model answer
Constructing the cumulative frequency table by successively adding up the class frequencies, then plotting the cumulative frequency against the upper class boundaries produces the ogive. Reading from the ogive (per the source's worked graph): i. Median ~= 58.59. ii. 70th percentile ~= 63.94. iii. Mode (read from the corresponding frequency distribution) ~= 57.5.
12. Table 2.3 shows the frequency distribution of marks obtained by 50 students in an examination. Continue the graphing and table-completion exercise from Q10-11 above using the same class data, and confirm the roots and gradient found graphically against the algebraic values. (12 marks)
Model answer
This item continues the table-completion, graphing, and reading exercise begun in Q10/Q11: substitute the class midpoints/x-values into the relevant function, plot to the given scale, and read off the roots and gradient from the graph, cross-checking against the algebraic (calculated) values for accuracy.
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