JAMB Mathematics 2022 past questions

Checked by the Skolra question team · Last updated 15 October 2026

Ten questions from the 2022 JAMB Mathematics paper, with answers and explanations. We hold 39 questions from this paper, and the Skolra app has all of them.

JAMB Mathematics 2022 questions and answers

1. 1.

   A rectangular pyramid has an area 24cm$^2$ and height 7.5cm. Find its volume?

   1. A
      45cm³
   2. B
      60cm³
      Correct
   3. C
      90cm³
   4. D
      180cm³
   Show answerHide answer

   Answer: B

   The volume of any pyramid is given by V = ⅓ × base area × height.

   Here, base area = 24cm² and height = 7.5cm.

   V = ⅓ × 24 × 7.5 = ⅓ × 180 = 60cm³

   A candidate who forgets the ⅓ factor and simply multiplies base area by height gets 180cm³, which is the volume of a prism with the same base and height, not a pyramid. Using ½ instead of ⅓ (a formula mix-up with the area of a triangle) gives 90cm³. Dividing 180 by the wrong number altogether gives 45cm³. The correct approach is always to take one-third of the product of base area and height.

2. 2.

   Factorize 4a$^2$ - 9b$^2$

   1. A
      (2a + 3b)²
   2. B
      (2a - 3b)²
   3. C
      (2a - 3b)(2a + 3b)
      Correct
   4. D
      (4a - 9b)(a + b)
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   Answer: C

   4a² - 9b² is a difference of two squares, since 4a² = (2a)² and 9b² = (3b)².

   Using the identity x² - y² = (x - y)(x + y) with x = 2a and y = 3b, we get:

   4a² - 9b² = (2a - 3b)(2a + 3b)

   Check by expanding: (2a - 3b)(2a + 3b) = 4a² + 6ab - 6ab - 9b² = 4a² - 9b². This confirms it.

3. 3.

   In how many ways can the letter of ZOOLOGY be arranged?

   1. A
      1680
   2. B
      840
      Correct
   3. C
      5040
   4. D
      2520
   Show answerHide answer

   Answer: B

   ZOOLOGY has 7 letters in total: Z, O, O, L, O, G, Y.

   Out of these, the letter O appears 3 times, while Z, L, G and Y each appear only once.

   When arranging letters that include repeated letters, we use:
   Number of arrangements = n! ÷ (repeats!)

   Here, n = 7 (total letters) and the O's repeat 3 times, so:
   7! ÷ 3! = 5040 ÷ 6 = 840.

   A candidate who forgets to divide by the repeated letters at all gets 5040, treating all 7 letters as if they were different. Dividing by only 2! (mistaking the repeat count) gives 2520. Dividing by an extra factor beyond what is needed gives 1680. The correct working accounts for exactly the three repeated O's, giving 840 arrangements.

4. 4.

   Evaluate $^{n^2 + 1}$C$_{n+5}$ if n = 3

   1. A
      36
   2. B
      120
   3. C
      90
   4. D
      45
      Correct
   Show answerHide answer

   Answer: D

   Substitute n = 3 into the upper and lower numbers first.

   Upper number: n² + 1 = 3² + 1 = 9 + 1 = 10.
   Lower number: n + 5 = 3 + 5 = 8.

   So we need ¹₀C₈ = 10! ÷ (8! × 2!).

   Using the combination shortcut, ¹₀C₈ = ¹₀C₂ (since choosing 8 out of 10 to include is the same as choosing 2 out of 10 to leave out) = (10 × 9) ÷ (2 × 1) = 90 ÷ 2 = 45. That gives the correct value.

   Where the wrong answers come from:
   90 is what you get if you multiply 10 × 9 and forget to divide by 2! (forgetting to divide out the repeated arrangements).
   36 comes from using n² instead of n² + 1, i.e. treating the upper number as 9 and computing ⁹C₂ = 36.
   120 comes from miscalculating the reduced combination as ¹₀C₃ instead of ¹₀C₂.

5. 5.

   Solve for k in the equation $\frac{1}{8}^{k+2}$ = 1

   1. A
      -2
      Correct
   2. B
      2
   3. C
      0
   4. D
      -1
   Show answerHide answer

   Answer: A

   Rewrite the base as a power of 8. Since $\frac{1}{8} = 8^{-1}$, the equation becomes:

   $8^{-1(k+2)} = 1$

   Rewrite the right side as a power of 8 too. Any nonzero number to the power 0 equals 1, so $1 = 8^0$.

   This gives $-(k+2) = 0$, so $k + 2 = 0$, and therefore k = -2.

   Checking: $\left(\frac{1}{8}\right)^{-2+2} = \left(\frac{1}{8}\right)^0 = 1$, which confirms the value.

6. 6.

   find the limit of y = $\frac{x^3 + 6x - 7}{x-1}$ as x tends to 1

   1. A
      10
   2. B
      7
   3. C
      8
   4. D
      9
      Correct
   Show answerHide answer

   Answer: D

   Substituting x = 1 directly gives $\frac{0}{0}$, an indeterminate form, so the numerator must be factorised first.

   Since $1^3 + 6(1) - 7 = 0$, (x - 1) is a factor of $x^3 + 6x - 7$. Dividing gives:

   x³ + 6x − 7 = (x − 1)(x² + x + 7)

   So for x ≠ 1, y simplifies to $x^2 + x + 7$, and the limit as x tends to 1 is:

   $1^2 + 1 + 7 = 9$

   The value 9 is correct. A candidate who drops the middle x term during the factorisation gets x² + 7, giving 8 instead. One who makes a sign slip in the quadratic factor (x² − x + 7) lands on 7. One who miscounts the constant term (using x² + x + 8) arrives at 10. Always factor out (x − 1) fully and evaluate the reduced expression, not the original 0/0 form.

7. 7.

   The locus of points equidistant from a fixed point.

   1. A
      Sphere
   2. B
      Parabola
   3. C
      Straight line
   4. D
      Circle
      Correct
   Show answerHide answer

   Answer: D

   The locus of points equidistant from a fixed point is a circle. By definition, a circle is the set of all points in a plane that are the same distance (the radius) from one fixed point (the centre). So every point on that locus traces out a circle around the fixed point.

   A straight line is the locus of points equidistant from two fixed points (it is the perpendicular bisector of the line joining them), not from a single point, so that does not fit here.

   A sphere would be the correct answer if the question were set in three-dimensional space (points equidistant from a fixed point in space), but for a plane locus, as is standard in the JAMB syllabus, the answer is a circle.

8. 8.

   Evaluate Log$_2$ 8√2

   1. A
      3
   2. B
      4
   3. C
      3.5
      Correct
   4. D
      4.5
   Show answerHide answer

   Answer: C

   Write 8√2 as a single power of 2: 8 = 2³ and √2 = 2^1/2, so 8√2 = 2³ × 2^1/2 = 2^{3 + 1/2} = 2^3.5.

   Log₂ 8√2 = Log₂ 2^3.5 = 3.5, since Log₂ 2^x = x.

   A candidate who evaluates only Log₂ 8 and stops gets 3, forgetting the √2 factor entirely. Rounding √2 up to 2 (as if 8√2 were 16) gives Log₂ 16 = 4, another common slip. Adding an extra whole unit instead of a half unit to the 3 from Log₂ 8 gives 4.5. The correct approach keeps √2 as 2^1/2, which adds exactly 0.5 to the index, giving 3.5.

9. 9.

   Simplify $\frac{1}{3-√2}$ in the form of p + q√2

   1. A
      3/7 − (1/7)√2
   2. B
      3/7 + (1/7)√2
      Correct
   3. C
      1/7 + (3/7)√2
   4. D
      3/11 + (2/11)√2
   Show answerHide answer

   Answer: B

   To simplify 1/(3-√2), multiply the top and bottom by the conjugate of the denominator, which is (3+√2). This removes the surd from the denominator.

   Numerator: 1 × (3+√2) = 3+√2.
   Denominator: (3-√2)(3+√2) = 3² − (√2)² = 9 − 2 = 7.

   So the expression becomes (3+√2)/7, which splits into 3/7 + (1/7)√2. That matches the form p + q√2 with p = 3/7 and q = 1/7, and that is the correct simplification.

   A candidate who multiplies by the wrong conjugate, (3-√2) instead of (3+√2), ends up with (3-√2)/7 = 3/7 − (1/7)√2, which has the wrong sign on the surd term.

   Someone who mixes up p and q gets 1/7 + (3/7)√2 by swapping the numerators.

10. 10.

    If A =$\begin{pmatrix} 2& 1\\ 2& 3 \\ 1 & 2\end{pmatrix}$ and B =$\begin{pmatrix} 3& 2\\ 4& 2\end{pmatrix}$. Find AB

    1. A
       $\begin{pmatrix} 6 & 4\\ 6 & 4\\ 3 & 2\end{pmatrix}$
    2. B
       $\begin{pmatrix} 10 & 6\\ 18 & 10\\ 11 & 6\end{pmatrix}$
       Correct
    3. C
       $\begin{pmatrix} 11 & 6\\ 18 & 10\\ 10 & 6\end{pmatrix}$
    4. D
       $\begin{pmatrix} 8 & 10\\ 12 & 14\\ 7 & 8\end{pmatrix}$
    Show answerHide answer

    Answer: B

    A is 3×2 and B is 2×2, so AB is defined and will be a 3×2 matrix. Each entry of AB is found by multiplying a row of A by a column of B and adding the products.

    Row 1 of A is (2, 1):
    (2×3 + 1×4, 2×2 + 1×2) = (10, 6)

    Row 2 of A is (2, 3):
    (2×3 + 3×4, 2×2 + 3×2) = (18, 10)

    Row 3 of A is (1, 2):
    (1×3 + 2×4, 1×2 + 2×2) = (11, 6)

    So the correct product is the matrix with rows (10, 6), (18, 10) and (11, 6).

    The matrix with rows (8, 10), (12, 14), (7, 8) comes from multiplying A by the transpose of B instead of B itself — using B's rows as if they were its columns.

    The matrix with rows (11, 6), (18, 10), (10, 6) simply has the first and third rows swapped, a copying slip when reading off the rows of A.

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

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