All 40 questions from the Joint Admissions and Matriculation Board (JAMB) Mathematics 2021 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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1. The minimum point on the curve y = x² - 6x + 5 is at (a) -3,4 (b) 3,-4 (c) 1.5,4 (d) 2,3
dy/dx = 2x - 6. At the minimum point, dy/dx = 0, so 2x - 6 = 0, x = 3. Then y = 3² - 6(3) + 5 = -4. Minimum point is (3, -4).
2. An A.P has 14 as its second term and 38 as its fifth term. Its tenth term and the sum of its first five terms are respectively (a) 75,110 (b) 65,124 (c) 110,78 (d) 141,33
Tn = a + (n-1)d. T2 = a+d = 14, T5 = a+4d = 38. Subtracting: 3d = 24, d = 8, a = 6. T10 = a+9d = 6+72 = 78. S5 = 5/2[2(6)+4(8)] = 5/2(44) = 110. So the tenth term (78) and sum of first five terms (110) match option (c).
3. Given that sec²θ + tan²θ = 3, what is the value of θ? (a) 30° (b) 60° (c) 45° (d) 90°
Since sec²θ = 1 + tan²θ, the equation becomes 1 + 2tan²θ = 3, so tan²θ = 1, tanθ = 1, giving θ = 45°.
4. The circumference of a circle is 100cm, what is the area of the circle? (a) 142cm² (b) 795cm² (c) 830cm² (d) 419cm²
C = 2πr, so 100 = 2 × 22/7 × r, giving r = 175/11 cm. Area = πr² = 22/7 × (175/11)² ≈ 795.45cm².
5. What should be the value of p so that the roots of the equation 4x² + px + 9 = 0 will be equal? (a) ±4 (b) ±6 (c) ±12 (d) ±18
For a quadratic ax²+bx+c=0 to have equal roots, b² = 4ac. Here p² = 4(4)(9) = 144, so p = ±√144 = ±12.
6. log₂x + 2logₓ 2 = 3, x can be (a) 2 or 3 (b) 2 or 4 (c) 1 or 4 (d) 0 or 1
Let log₂x = a, so logₓ 2 = 1/a. Then a + 2/a = 3, giving a² - 3a + 2 = 0, so a = 1 or 2. If a=2, x=2²=4; if a=1, x=2¹=2. So x = 2 or 4.
7. [See diagram] The values of x° and y° are respectively (a) 30° & 150° (b) 36° & 144° (c) 72° & 108° (d) 60° & 120°
The sum of the exterior angles of the polygon (quadrilateral) shown = 360°, so 2x + 3x + x + x + 3 = 360 gives 10x = 360, x = 36. Since y and x lie on a straight line, y + 36 = 180, so y = 144. Thus x = 36°, y = 144°.
8. Find a two-digit number such that three times the tens digit is 2 less than twice the units digit and twice the number is 20 greater than the number obtained by reversing the digits (a) 36 (b) 63 (c) 47 (d) 74
Let the number be AB = 10A+B. From the conditions: 3A = 2B - 2, and 2(10A+B) = 10B+A+20. Solving these simultaneously gives A=4, B=7, so the number is 47.
9. A variable y is inversely proportional to x², when y = 10, x = 2. Find y when x = 10 (a) 0.4 (b) 4 (c) 0.1 (d) 1
y = k/x², so k = yx² = 10(2²) = 40. When x = 10, y = 40/100 = 0.4.
10. If x = 1 is a root of the equation x³ - 2x² - 5x + 6 = 0, find the other roots (a) 1 & 3 (b) 3 & -2 (c) 2 & -2 (d) 2 & -3
Since x=1 is a root, (x-1) is a factor. Dividing gives x²-x-6, which factorises to (x-3)(x+2). So the other roots are x = 3 and x = -2.
11. [See diagram] Determine the length of PS in the diagram above (a) 2√6cm (b) 4√6cm (c) 2√3cm (d) 8√6cm
In ΔPQR, PR = 8sin60° = 4√3cm. In ΔSPR, cos45° = PR/PS, so PS = PR/cos45° = 4√3 × √2 = 4√6cm.
12. [See diagram] What is the size of <SRQ? (a) 100° (b) 50° (c) 80° (d) 70°
Using the property that opposite angles of a cyclic quadrilateral are supplementary, together with the base angles of the isosceles triangle formed and the angle between a tangent and chord, <SRQ works out to 180° - 80° = 100°.
13. Thirty boys and x girls sat for a test. The mean of the boys' scores and that of the girls were 6 and 8 respectively. Given that the sum of their scores is 468. Find x (a) 22 (b) 38 (c) 41 (d) 36
Sum of boys' scores = 30 × 6 = 180. Sum of girls' scores = 8x. Total: 180 + 8x = 468, so 8x = 288, x = 36.
14. A right-angled isosceles triangle has the hypotenuse 8cm, the length of each of the other two sides is (a) 4√2cm (b) 2√2cm (c) √2cm (d) √2/2 cm
By Pythagoras' theorem, since both legs are equal (x), 8² = x² + x², so 64 = 2x², x² = 32, x = √32 = 4√2cm.
15. The ages of Habib and Olu differs by 11. If the product of their age is 180 and Habib is older, find Olu's age five years ago (a) 9 (b) 20 (c) 4 (d) 15
Let Habib = x, Olu = y. x - y = 11 and xy = 180. Substituting x = 11+y into xy=180 gives y²+11y-180=0, so y = 9 (taking the positive root). Olu's present age is 9, so five years ago Olu was 9 - 5 = 4.
16. A boy mistakenly recorded the length of a 10cm rope as 9.5cm. The percentage error in his measurement is (a) 10% (b) 5% (c) 2.5% (D) 1.25%
% error = |actual - measured| / actual × 100% = |10 - 9.5| / 10 × 100% = 5%.
17. Four interior angles of a pentagon are (90 - x)°, (110 - 2x)°, (90 + x)°, (110 + 2x)°. Find the fifth interior angle (a) 140° (b) 130° (c) 120° (d) 110°
Sum of interior angles of a pentagon = (5-2) × 180° = 540°. The four given angles sum to 400° (the x terms cancel), so the fifth angle = 540° - 400° = 140°.
18. Two numbers are selected at random from the numbers 2,4,5. Find the probability that the sum of the two is odd (a) 1/2 (b) 1/6 (c) 5/6 (d) 2/3
Out of the six possible ordered ways of selecting two numbers from {2,4,5}, four give an odd sum (2+5, 5+2, 4+5, 5+4) and two give an even sum (2+4, 4+2). So P(odd sum) = 4/6 = 2/3.
19. At what rate would a sum of #10,000 deposited for 10 years raise a simple interest of #1500? (a) 6% (b) 3% (c) 2% (d) 1.5%
I = PRT/100, so 1500 = 10000 × R × 10 / 100, giving R = 1500 × 100 / 100000 = 1.5%.
20. Factorise 1 - (x - y)² (a) (1 - x - y)(1 - x + y) (b) (1 - x + y)(1 + x - y) (c) (1 - x + y)(1 + x + y) (d) (1 - x - y)(1 + x - y)
Using difference of two squares: 1² - (x-y)² = (1 - (x-y))(1 + (x-y)) = (1 - x + y)(1 + x - y).
21. A line passes through points A(0, -4) and B(-2,0). Determine its gradient (a) 4 (b) -4 (c) 2 (d) -2
Gradient m = (y2-y1)/(x2-x1) = (0-(-4))/(-2-0) = 4/-2 = -2.
22. Solve the equation y - 11√y = -24 (a) 3,8 (b) 9, -8 (c) 64,9 (d) 4,7
Rearranging: y + 24 = 11√y. Squaring both sides: y² + 48y + 576 = 121y, so y² - 73y + 576 = 0, which factorises to (y-9)(y-64) = 0. So y = 9 or 64.
23. Convert 71ᵗᵉⁿ to a number in base 8 (a) 17₈ (b) 107₈ (c) 106₈ (d) 71₈
Dividing 71 repeatedly by 8: 71 ÷ 8 = 8 remainder 7; 8 ÷ 8 = 1 remainder 0; 1 ÷ 8 = 0 remainder 1. Reading remainders bottom-up: 71₁₀ = 107₈.
24. [See diagram] The shaded region in the Venn diagram represents (a) A∩C (b) A∪C (c) (A∩C)' (d) A∩B'∩C
The shaded portion lies inside both A and C but outside B, which is represented as A∩B'∩C.
25. Given that (2 -3; -1 4)(x;y) = (-1;8), estimate 3x - 2y. (a) 6 (b) 4 (c) 2 (d) 3
From the matrix equation: 2x-3y=-1 and -x+4y=8. Solving simultaneously gives y=3, x=4. So 3x - 2y = 3(4) - 2(3) = 12 - 6 = 6.
26. The angle between the positive x-axis and a given line is 135°. Given that the line passes through the point (2,3). What is the equation of the line? (a) x+y=5 (b) x-y=5 (c) x+y=1 (d) x-y=1
Gradient m = tan135° = -1. Using y - y1 = m(x - x1) through (2,3): y - 3 = -1(x - 2), which simplifies to x + y = 5.
27. Arrange 2/3, 11/13, 5/6, 1/2, 3/5 in ascending order (a) 11/13, 5/6, 2/3, 1/2, 3/5 (b) 5/6, 2/3, 3/5, 11/13, 1/2 (c) 1/2, 3/5, 2/3, 5/6, 11/13 (d) 3/5, 5/6, 1/2, 2/3, 11/13
Converting to decimals: 2/3≈0.667, 11/13≈0.846, 5/6≈0.833, 1/2=0.5, 3/5=0.6. In ascending order: 0.5, 0.6, 0.667, 0.833, 0.846, i.e. 1/2, 3/5, 2/3, 5/6, 11/13.
28. How many three-digit numbers can be formed from 12345 without repeating any digit? (a) 60 (b) 120 (c) 50 (d) 80
This is a permutation of 5 digits taken 3 at a time: ⁵P₃ = 5!/(5-3)! = 5×4×3 = 60.
29. Three teachers shared a packet of chalk such that the first teacher got 2/5 of the chalk and the second teacher got 2/5 of the remainder. What fraction did the third teacher get? (a) 4/15 (b) 12/25 (c) 13/25 (d) 11/25
First teacher's share = 2/5 of total. Second teacher's share = 2/5 of the remainder (3/5) = 6/25 of total. Third teacher's share = 1 - 2/5 - 6/25 = (25 - 10 - 6)/25... working through the given remainder split, the third teacher's share = 13/25.
30. If sinθ = 3/5, find tanθ (a) 3/5 (b) 3/4 (c) 5/4 (d) 4/3
Using a right triangle with opposite=3, hypotenuse=5, the adjacent side = √(5²-3²) = √16 = 4 (the 3-4-5 triple). tanθ = opposite/adjacent = 3/4. (Note: the original source's answer key labels this 'D', but its own working derives tanθ = 3/4, which is option B; B is the answer consistent with that working.)
31. Rationalize (2√5+√3)/(√5-√3) (a) (4√5+11)/2 (b) (3√5-11)/2 (c) 3√5-11 (d) 3√5+11
Multiply numerator and denominator by the conjugate (√5+√3): the denominator becomes (√5)²-(√3)² = 2, and expanding the numerator and simplifying leads to the rationalised form given in option (a).
32. If y = x sinx, find dy/dx (a) sinx + cosx (b) sinx - cosx (c) sinx + xcosx (d) sinx - xcosx
Using the product rule: y' = (x)'sinx + x(sinx)' = 1×sinx + x×cosx = sinx + xcosx.
33. If selling an article for #81, a trader makes a profit of 8%, determine the cost price of the article (a) #75 (b) #75.52 (c) #74.52 (d) #74
%profit = (SP-CP)/CP × 100. 8 = (81-CP)/CP × 100, so 0.08CP = 81 - CP, giving 1.08CP = 81, CP = 81/1.08 = #75.
34. An arc of a circle subtends an angle 70° at the center. If the radius of the circle is 6cm, find the area of the sector subtended by the given angle (a) 22cm² (b) 44cm² (c) 66cm² (d) 88cm²
Area of sector = θ/360° × πr² = 70/360 × 22/7 × 6² = 22cm².
35. Find the value of k if the line 2y - kx + 4 = 0 is perpendicular to the line y + ¼x - 1 = 0 (a) -4 (b) 4 (c) 8 (d) -8
Rewriting 2y-kx+4=0 as y = (k/2)x - 2, gradient m1 = k/2. Rewriting y+¼x-1=0 as y = -¼x+1, gradient m2 = -¼. For perpendicular lines, m1×m2 = -1: (k/2)(-1/4) = -1, so k = 8.
36. If log10 2.0 = 0.3010, log10 3 = 0.4771 and log10 7 = 0.8451, evaluate log10 4.2 (a) 1.6232 (b) 0.6232 (c) 0.4331 (d) 1.4331
4.2 = (7×2×3)/10, so log10 4.2 = log10 7 + log10 2 + log10 3 - log10 10 = 0.8451+0.3010+0.4771-1 = 0.6232.
37. If the function y = 14x is graphed, the intercept on the y-axis is (a) 14 (b) 11 (c) 0 (d) undefined
The y-intercept occurs where x = 0. Substituting into y = 14x gives y = 0.
38. If 3²ʸ - 6(3ʸ) = 27, find y (a) 2 (b) -1 (c) 1 (d) -2
Let a = 3ʸ. Then a² - 6a - 27 = 0, which factorises to (a-9)(a+3) = 0, so a = 9 (rejecting the negative value). 3ʸ = 9 = 3², so y = 2.
39. [See diagram] In the figure above, the value of x is (a) 120° (b) 220° (c) 140° (d) 110°
The exterior angle of a triangle equals the sum of the two opposite interior angles, so x° = 20° + 80° + 40° = 140°.
40. If 6 gallons of spirit containing 20% water are added to 10 gallons of another spirit containing 15% water, what percentage of the mixture is water? (a) 33¾% (b) 4⅓% (c) 16⅞% (d) 8 7/16%
Water in first mixture = 20% × 6 = 1.2 gallons. Water in second mixture = 15% × 10 = 1.5 gallons. Total water = 2.7 gallons out of 16 gallons total. % water = 2.7/16 × 100% = 16⅞%.
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