JAMB Mathematics past questions and answers

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

JAMB Mathematics: 537 past questions from 10 papers between 2015 and 2025, each with the answer and an explanation. In the UTME you answer 40 Mathematics questions, with Use of English and two other subjects, in 2 hours. Here are ten.

1. 1.

   Differentiate the function y = $\sqrt[3]{x^2}(2x- x^2)$

   1. A
      $\frac{5}{3}x^{\frac{2}{3}} - \frac{4}{3}x^{\frac{5}{3}}$
   2. B
      $\frac{10}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{\frac{5}{3}}$
      Correct
   3. C
      $\frac{4}{3}x^{-\frac{1}{3}} - \frac{4}{3}x^{\frac{2}{3}}$
   4. D
      $\frac{10}{3}x^{\frac{5}{3}} - \frac{8}{3}x^{\frac{8}{3}}$
   Show answerHide answer

   Answer: B

   First change the cube root to index form: $\sqrt[3]{x^2} = x^{\frac{2}{3}}$. Then open the bracket before you differentiate — it is far quicker than the product rule here.

   $y = x^{\frac{2}{3}}(2x - x^2) = 2x^{\frac{5}{3}} - x^{\frac{8}{3}}$
   (add the powers: $\frac{2}{3}+1=\frac{5}{3}$ and $\frac{2}{3}+2=\frac{8}{3}$).

   Now apply $\frac{d}{dx}x^n = nx^{n-1}$ to each term:
   $\frac{d}{dx}\left(2x^{\frac{5}{3}}\right) = 2 \times \frac{5}{3}x^{\frac{2}{3}} = \frac{10}{3}x^{\frac{2}{3}}$
   $\frac{d}{dx}\left(x^{\frac{8}{3}}\right) = \frac{8}{3}x^{\frac{5}{3}}$

2. 2.Evaluate 1 − ($\frac{1}{5}$ + 1$\frac{2}{3}$) + (5 + 1$\frac{2}{3}$)
   1. A
      5⅘
      Correct
   2. B
      5⅕
   3. C
      6⅕
   4. D
      4⅘
   Show answerHide answer

   Answer: A

   Rewrite the mixed number: 1⅔ = 5/3.

   1 − (1/5 + 5/3) + (5 + 5/3)

   Remove the brackets: 1 − 1/5 − 5/3 + 5 + 5/3.

   The −5/3 and the +5/3 cancel out, leaving:

   1 − 1/5 + 5 = 6 − 1/5 = 5⅘

   A candidate who forgets that the two fifths-of-two-thirds terms cancel and instead subtracts 5/3 twice lands on 4⅘. Mixing up which term is added versus subtracted, or misreading the fraction split, gives 6⅕ or 5⅕. Working it out fraction by fraction (using fifteenths as the common denominator) also confirms 87/15 = 5⅘, the correct value.

3. 3.

   Two numbers are respectively 35% and 80% more than a third number. The ratio of the two numbers is

   1. A
      3 : 4
      Correct
   2. B
      27 : 16
   3. C
      7 : 16
   4. D
      4 : 3
   Show answerHide answer

   Answer: A

   Always start by letting the third number be x, since both numbers are compared to it.

   "35% more than x" means you add 35% of x to x:
   First number = x + 0.35x = 1.35x
   "80% more than x" means you add 80% of x to x:
   Second number = x + 0.80x = 1.80x

   Now form the ratio:
   1.35x : 1.80x = 135 : 180 (multiply both sides by 100; the x cancels out)
   Divide both terms by 45: 3 : 4

4. 4.

   Add 54 $_{eight}$ and 67$_{eight}$ giving your answers in base eight

   1. A
      143_eight
      Correct
   2. B
      121_eight
   3. C
      153_eight
   4. D
      144_eight
   Show answerHide answer

   Answer: A

   Step 1: Convert each number to base 10.
   54_eight = (5×8) + 4 = 44
   67_eight = (6×8) + 7 = 55

   Step 2: Add in base 10.
   44 + 55 = 99

   Step 3: Convert 99 back to base eight.
   99 ÷ 8 = 12 remainder 3
   12 ÷ 8 = 1 remainder 4
   1 ÷ 8 = 0 remainder 1
   Reading the remainders from bottom to top gives 143_eight.

   You can also add directly in base eight, column by column: in the units column, 4 + 7 = 11 (in base ten), which is 13 in base eight, so you write 3 and carry 1. In the next column, 5 + 6 + the carried 1 = 12 (in base ten), which is 14 in base eight, so you write 4 and carry 1. The final carried 1 becomes the leading digit, giving 143_eight.

5. 5.

   Find the probability of getting an even number in a single throw of a six-sided die.

   1. A
      1/3
   2. B
      1/2
      Correct
   3. C
      1/6
   4. D
      2/3
   Show answerHide answer

   Answer: B

   When you throw an ordinary six-sided die once, the possible outcomes are {1, 2, 3, 4, 5, 6}, so the number of possible outcomes n(S) = 6.

   The even numbers among them are {2, 4, 6}, giving n(E) = 3 favourable outcomes.

   P(even) = n(E) / n(S) = 3/6 = 1/2

   So 1/2 is the answer. Notice that half the faces of a die are even and half are odd, so the answer sitting exactly at one-half should look sensible to you.

6. 6.Rationalize 5 ÷ (2 − √3)
   1. A
      10 − 5√3
   2. B
      2 + √3
   3. C
      10 + 5√3
      Correct
   4. D
      5 + 5√3
   Show answerHide answer

   Answer: C

   To rationalize 5 ÷ (2 − √3), multiply top and bottom by the conjugate of the denominator, (2 + √3):

   Denominator: (2 − √3)(2 + √3) = 4 − 3 = 1.

   Numerator: 5(2 + √3) = 10 + 5√3.

   Since the denominator becomes exactly 1, the whole expression simplifies to 10 + 5√3.

   Using the wrong conjugate (2 − √3 again instead of 2 + √3) flips the sign in the middle, giving 10 − 5√3. Forgetting to multiply the 5 all the way through the bracket gives 5 + 5√3. Simply leaving the expression unrationalised as 2 + √3 skips the working altogether.

7. 7.

   From a point P, R is 5km due West and 12km due South. Find the distance between P and R'.

   1. A
      7km
   2. B
      13km
      Correct
   3. C
      17km
   4. D
      15km
   Show answerHide answer

   Answer: B

   West and South are directions at right angles to each other, so the point P and the two displacements (5km West, 12km South) form a right-angled triangle, with the required distance PR as the hypotenuse.

   Using Pythagoras' theorem:
   PR² = 5² + 12² = 25 + 144 = 169
   PR = √169 = 13km

   So the distance between P and R is 13km, which is the well-known 5-12-13 right triangle.

8. 8.

   Calculate the area of an equilateral triangle of side 8cm

   1. A
      8√3 cm²
   2. B
      16√3 cm²
      Correct
   3. C
      32√3 cm²
   4. D
      32 cm²
   Show answerHide answer

   Answer: B

   The area of an equilateral triangle of side s is given by:

   Area = (√3/4) × s²

   With s = 8cm:

   Area = (√3/4) × 8² = (√3/4) × 64 = 16√3 cm²

   This is the correct answer.

   A student who mixes up the formula and uses (√3/2) × s² instead of (√3/4) × s² would wrongly get 32√3 cm² — that comes from forgetting to divide by 4.

   Another common slip is dropping the √3 altogether and just halving the square of the side (s²/2 = 32), which gives 32 cm² — this ignores the 60° angles of the equilateral triangle entirely.

   A third mistake is dividing by 8 instead of 4, which gives 8√3 cm² — half of the correct value.

   Always remember: for an equilateral triangle, Area = (√3/4) × side².

9. 9.Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 7, 10, 6, 5
   1. A
      16
      Correct
   2. B
      17
   3. C
      13
   4. D
      14
   Show answerHide answer

   Answer: A

   List the twenty values and count how many times each one occurs: 10 occurs 6 times, 9 occurs 4 times, 8 and 7 occur 3 times each, 6 occurs twice, and 4 and 5 occur once each.

   The mode is the value that occurs most often, so the mode is 10.

   The range is the highest value minus the lowest value: 10 − 4 = 6.

   Sum of the range and the mode = 6 + 10 = 16.

   A candidate who miscounts and takes 9 as the mode (4 occurrences) instead of 10 would wrongly get 6 + 9 = 15, and a candidate who uses the wrong lowest value (say 5 instead of 4) would get a range of 5, giving 5 + 9 = 14 or 5 + 10 = 15 depending on the mistake made along the way; these are the kind of slips the wrong options represent.

10. 10.

    Find the values of x for which

    $\cfrac {x+2}{4} - \cfrac{2x - 3}{3} < 4$

    1. A
       x > 6
    2. B
       x < -6
    3. C
       x > -6
       Correct
    4. D
       x < 6
    Show answerHide answer

    Answer: C

    Multiply through by the LCM of 4 and 3, which is 12: 3(x+2) - 4(2x-3) is less than 4 × 12 = 48.

    Expanding: 3x + 6 - 8x + 12 < 48, which simplifies to -5x + 18 < 48.

    Subtracting 18 from both sides gives -5x < 30.

    Dividing by a negative number flips the inequality sign, so dividing by -5 gives x > -6. This is the correct solution.

    A candidate who forgets to flip the inequality sign when dividing by -5 wrongly gets x < -6.

    A candidate who makes a sign slip while expanding -4(2x-3), getting -8x-12 instead of -8x+12, ends up with -5x-6<48, i.e. -5x<54, and after further mishandling can land on x > 6 or x < 6 depending on how the flip is (mis)applied. These are the kind of transposed-sign errors that produce the wrong-direction answers.

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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