JAMB Mathematics 2023 past questions

Checked by the Skolra question team · Last updated 23 October 2026

Ten questions from the 2023 JAMB Mathematics paper, with answers and explanations. We hold 78 questions from this paper, and the Skolra app has all of them.

JAMB Mathematics 2023 questions and answers

1. 1.

   Find the area and perimeter of a square whose length of diagonals is 20$\sqrt2$ cm.

   1. A
      Area = 200 cm², Perimeter = 40$\sqrt2$ cm
   2. B
      Area = 400 cm², Perimeter = 40 cm
   3. C
      Area = 800 cm², Perimeter = 80$\sqrt2$ cm
   4. D
      Area = 400 cm², Perimeter = 80 cm
      Correct
   Show answerHide answer

   Answer: D

   In any square, the diagonal cuts the square into two right-angled triangles whose two shorter sides are the sides of the square. So by Pythagoras, if the side is s, then d² = s² + s² = 2s², giving d = s√2.

   Here the diagonal is 20√2 cm, so:
   s√2 = 20√2
   s = 20 cm.

   Now use the side:
   Area = s² = 20 × 20 = 400 cm²
   Perimeter = 4s = 4 × 20 = 80 cm
   That is why an area of 400 cm² with a perimeter of 80 cm is the correct pair.

2. 2.

   Evaluate the following limit: $lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}$

   1. A
      8
   2. B
      4
      Correct
   3. C
      1/4
   4. D
      2
   Show answerHide answer

   Answer: B

   First test the point. Putting x = 2 directly gives numerator 4 + 8 − 12 = 0 and denominator 4 − 4 = 0, that is 0/0. That is an indeterminate form, not the answer — it is simply a signal that (x − 2) is a factor of both the top and the bottom, so factorise and cancel.

   Numerator: x² + 4x − 12. Look for two numbers with product −12 and sum +4; they are +6 and −2.
   So x² + 4x − 12 = (x + 6)(x − 2).
   Denominator: x² − 2x = x(x − 2).

   Now cancel the common factor (x − 2), which is allowed because x is only approaching 2, it is not equal to 2:
   (x + 6)(x − 2) / [x(x − 2)] = (x + 6)/x.

   Then substitute x = 2 into the simplified expression:
   (2 + 6)/2 = 8/2 = 4.

3. 3.

   It takes 12 men 8 hours a day to finish a piece of work in 4 days. In how many days will it take 4 men working 16 hours a day to complete the same piece of work?

   1. A
      12 days
   2. B
      6 days
      Correct
   3. C
      24 days
   4. D
      2 days
   Show answerHide answer

   Answer: B

   Whenever a question mixes men, hours per day and days, the safest tool is the man-hour. Find the total amount of work first, then share it out under the new conditions.

   Step 1 – Total work done:
   Work = men × hours per day × days
   Work = 12 × 8 × 4 = 384 man-hours

   Step 2 – Work done per day under the new arrangement:
   4 men × 16 hours = 64 man-hours per day

   Step 3 – Number of days:
   Days = 384 ÷ 64 = 6 days

   You can also reason with ratios. Fewer men means more days, so multiply by 12/4; more hours each day means fewer days, so multiply by 8/16. That gives 4 × 12/4 × 8/16 = 6 days, the same answer.

4. 4.

   At simple interest, a man made a deposit of some money in the bank. The amount in his bank account after 10 years is three times the money deposited. If the interest rate stays the same, after how many years will the amount be five times the money deposited?

   1. A
      25 years
   2. B
      16 2/3 years
   3. C
      20 years
      Correct
   4. D
      15 years
   Show answerHide answer

   Answer: C

   Let the money deposited (the principal) be P and the rate be R% per annum. For simple interest, Amount = Principal + Interest, where Interest = PRT/100.

   Step 1: Find the rate.
   After 10 years the amount is 3P, so the interest earned in those 10 years is 3P − P = 2P (not 3P — the amount already contains the principal).
   2P = (P × R × 10)/100 ⇒ 2 = 10R/100 ⇒ R = 20% per annum.

   Step 2: Find the time for the amount to be 5P.
   Interest required = 5P − P = 4P.
   4P = (P × 20 × T)/100 ⇒ 4 = T/5 ⇒ T = 20 years.

   Notice the neat check: the interest grows by 2P every 10 years, so it takes another 10 years after the first 10 for the interest to reach 4P — that is 20 years in total, which is why 20 years is the correct choice.

5. 5.

   A rectangular plot of land has sides with lengths of 38 m and 52 m corrected to the nearest m. Find the range of the possible values of the area of the rectangle

   1. A
      1931.25 m² ≤ A < 2021.25 m²
      Correct
   2. B
      1887.00 m² ≤ A < 2067.00 m²
   3. C
      1968.75 m² ≤ A < 1982.75 m²
   4. D
      1975.50 m² ≤ A < 1976.50 m²
   Show answerHide answer

   Answer: A

   When a measurement is corrected to the nearest metre, the true value can be anything within half a metre of the stated figure. So the actual sides lie in these intervals:

   37.5 m ≤ L < 38.5 m
   51.5 m ≤ B < 52.5 m

   Area A = L × B. Since both sides are positive quantities, the area is smallest when both sides take their smallest values, and largest when both take their largest values.

   Least area = 37.5 × 51.5 = 1931.25 m²
   Greatest area = 38.5 × 52.5 = 2021.25 m²

   Hence the answer 1931.25 m² ≤ A < 2021.25 m² is correct. Note the signs: the lower bound is attainable (≤) because 37.5 and 51.5 round up to 38 and 52, while the upper bound is not attainable (<) because 38.5 and 52.5 would round to 39 and 53.

6. 6.

   Solve the following quadratic inequality: $x^2- x$ - 4 ≤ 2

   1. A
      −2 ≤ x ≤ 3
      Correct
   2. B
      −1 ≤ x ≤ 2
   3. C
      x ≤ −2 or x ≥ 3
   4. D
      −3 ≤ x ≤ 2
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   Answer: A

   Always collect everything on one side first so that the right-hand side becomes zero:

   x² − x − 4 ≤ 2
   x² − x − 4 − 2 ≤ 0
   x² − x − 6 ≤ 0

   Where the others go wrong:
   −3 ≤ x ≤ 2 comes from reading the factors (x − 3)(x + 2) and writing the roots with their signs unchanged. Remember: (x − 3) = 0 gives x = +3, and (x + 2) = 0 gives x = −2 — the sign always flips.
   −1 ≤ x ≤ 2 comes from adding the 2 instead of subtracting it, giving x² − x − 2 ≤ 0 and (x − 2)(x + 1) ≤ 0. The 2 is on the right, so it must be taken over as −2.
   x ≤ −2 or x ≥ 3 has the correct roots but the wrong region. That outside region is the answer to x² − x − 6 ≥ 0; for a "≤ 0" inequality with an upward parabola you take the inside portion.

7. 7.

   Factorize: $16x^4 - y^4$

   1. A
      (2x − y)(2x + y)(4x² + y²)
      Correct
   2. B
      (2x − y)(2x + y)(2x² + y²)
   3. C
      (4x² − y²)(4x² + y²)
   4. D
      (2x − y)²(2x + y)²
   Show answerHide answer

   Answer: A

   16x⁴ − y⁴ is a difference of two squares: it is (4x²)² − (y²)², so it factorizes first as (4x² − y²)(4x² + y²).

   Look closely at the first bracket, 4x² − y². This is itself a difference of two squares, since 4x² = (2x)². So it splits further into (2x − y)(2x + y).

   The second bracket, 4x² + y², is a sum of two squares, not a difference, so it cannot be broken down any further.

   Putting it together, the complete factorization is (2x − y)(2x + y)(4x² + y²) — that is the correct answer.

   Stopping at (4x² − y²)(4x² + y²) is a common mistake: the candidate spots the first difference of squares but forgets to check whether either bracket can still be factorized further.

8. 8.

   Find the value of y if $402_y = 102_{ten}$

   1. A
      6
   2. B
      4
   3. C
      3
   4. D
      5
      Correct
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   Answer: D

   Write the number out in place values. In base y, the digits of 402 carry the weights y², y and 1:
   402_y = 4(y²) + 0(y) + 2(1) = 4y² + 2

   Now equate it to the base ten value:
   4y² + 2 = 102
   4y² = 100
   y² = 25
   y = 5 (a base is positive, so we drop −5).

   Check: 402_five = 4(25) + 0 + 2 = 100 + 2 = 102_ten. Correct.

9. 9.

   Find the volume of a cone which has a base radius of 5 cm and slant height of 13 cm.

   1. A
      942.86 cm³
   2. B
      314.29 cm³
      Correct
   3. C
      340.48 cm³
   4. D
      209.52 cm³
   Show answerHide answer

   Answer: B

   In a cone, the slant height is not the vertical height. The radius, the vertical height and the slant height form a right-angled triangle with the slant height as the hypotenuse.

   Step 1 — Find the vertical height h:
   l² = r² + h²
   13² = 5² + h²
   h² = 169 − 25 = 144
   h = 12 cm

   Step 2 — Apply the volume formula:
   V = ⅓πr²h
   V = ⅓ × 22/7 × 5² × 12
   V = ⅓ × 22/7 × 300 = 2200/7
   V ≈ 314.29 cm³ (that is 100π cm³)

10. 10.

    A student pilot was required to fly to an airport and then return as part of his flight training. The average speed to the airport was 120 km/h, and the average speed returning was 150 km/h. If the total flight time was 3 hours, calculate the distance between the two airports.

    1. A
       202.5 km
    2. B
       180 km
    3. C
       200 km
       Correct
    4. D
       400 km
    Show answerHide answer

    Answer: C

    Let the distance between the two airports be d km. Remember that time = distance ÷ speed, and the same distance d is covered going and coming back.

    Time going = d/120 hours
    Time returning = d/150 hours
    Total flight time = 3 hours, so:

    d/120 + d/150 = 3
    Multiply every term by the LCM of 120 and 150, which is 600:
    5d + 4d = 1800
    9d = 1800
    d = 200 km

    Check it: 200 ÷ 120 = 1 hour 40 minutes going, and 200 ÷ 150 = 1 hour 20 minutes returning. Together that is exactly 3 hours, so 200 km is the distance between the two airports.

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