JAMB MATHEMATICS PAST QUESTIONS 2021. The questions below are the mathematics past questions for 2021. Remember, it is going to be a computer-based test (CBT) examination and timing really matters. Before we start solving these questions, let us look at some hints that you must know as you prepare to sit for the JAMB UTME examinations. This post contains questions 1 to 10, subsequent post will show the rest of the 40 questions.

  1. The examination is going to be a computer-based test (CBT) examination and timing is very important;
  2. You will be tested on forty (40) objectives questions to solve;
  3. You will be required to spend less than 30 minutes on mathematics question;
  4. So be prudent in time usage, I suggest you spend 40 seconds or less on one question;
  5. Do not rush, it is better to solve 30 questions correctly at that allotted time than answer all questions and not be sure of getting 30 correctly.
  6. Do your possible best, and pray that other things go well! Wish you the best in your examination!

QUESTION 1: Solve the following equation \frac{2}{(2r-1)}-\frac{5}{3}=\frac{1}{(r+2)}   JAMB 2021 MATHEMATICS

a. (-1, \frac{5}{2})   b. (1, -\frac{5}{2})  c. (\frac{5}{2},1)  d. (2,1)


Collect terms that are alike


Find the LCM of the LHS,


Simplify the left-hand side;

\frac{2r+4-2r+1}{(2r-1)(r+2)}=\frac{5}{3} , from the numerator clear 2r,


cross multiply, 5[(2r-1)(r+1)]=5\times 3

Divide through by 5, \frac{5[(2r-1)(r+2)]}{5}=\frac{5\times 3}{5}


Simplify further gives 2r^{2}+4r-r-2=3

2r^{2}+3r-2-3=0, this will give 2r^{2}+3r-5=0

Solve finally, using quadratic equation, 2r^{2}+3r-5=0

From the equation we have, 2 × 5 = 10 and 3. So, look for two values that when multiplied will to give -10 and when added gives +3. That number is -5 and +2.




2r-5=0 and r+1=0

r=\frac{5}{2}, and r=-1

r= (-1, \frac{5}{2}).

The second question is on permutation and combination. From this question, you will learn how to solve questions on combination.

QUESTION 2: In how many ways can 2 students be selected from a group of 5 students in a debating competition. JAMB 2021 MATHEMATICS

a. 25 ways b. 10 ways c. 15 ways d. 20 ways


This is a question on combination. Note Combination involves selection, and the order or arrangement is not important. For more learnings on Permutation and Combination, read the article “Permutation and Combination“.



\frac{5!}{2!3!}=\frac{5\times4\times 3\times 2\times 1}{2! 3\times 2\times 1}

=\frac{5\times 4}{2!}

=\frac{5\times 4}{2\times 1}=\frac{20}{2}

{\color{Green} =10ways (B)}

The third question is on the rate of change. This question is related to differentiation, but you will be required to find the change at a specific value.

QUESTION 3: What is the rate of change of the volume, V of a hemisphere with respect to its radius r, when r = 2. JAMB 2021 MATHEMATICS

a. 8π   b. 16π   c. 2π   d. 4π


This is a question on differentiation, since it involves the rate of change.

Volume of a Hemisphere, V=\frac{2}{3}\barwedge r^{3}

\frac{dv}{dr}\frac{2}{3}\barwedge r^{3}

= \frac{2}{3}\times 3\barwedge r^{3-1}

= \frac{2}{3}\times 3\barwedge r^{2}

=2\barwedge r^{2}

At r=2

The rate of change of the volume, V will be

=2\barwedge (2)^{2}

=2\barwedge \times 4

{\color{Green} =8\barwedge (A)}

Next, is a question on the maximum and minimum value of a limit. This question is a question that is frequently asked in JAMB UTME examination.

QUESTION 4: Determine the maximum value of y=3x^{2}+5x-3 JAMB 2021 MATHEMATICS

a. 6 b. 0 c. 2 d. No correct option


At maximum value, \frac{dy}{dx}=0 (equal to zero)

and \frac{d^{2}y}{dx^{2}}< 0 (less than 0)

So, given the equation, y=3x^{2}+5x-3



Note the result above do not meet the criteria for the equation at maximum value. So, the correct option will be d (No correct option)

The fifth question is on the mensuration of shapes. This question will teach you how to solve questions on trapezium.

QUESTION 5: A trapezium has two parallel sides of length 5cm and 9cm. If the area is 91cm^{2}, find the distance between the parallel sides. JAMB 2021 MATHEMATICS

a. 13cm b. 4cm c. 6cm d. 7cm


Given the following parameters, a=5cm, b=9cm and Area, A=91cm^{2} and the distance between the parallel sides is the height, h=?

Area of a trapezium, A=\frac{1}{2}(a+b)h


91cm^{2}=\frac{14}{2}cm\times h

91cm^{2}=7cm\times h


{\color{Green} h=13cm(A)}

The next question will teach you how to solve questions on coordinate geometry.  

QUESTION 6: Find the value of p, if the line which passes through (-1, -P) and (-2, 2) is parallel to the line 2y+18x-17=0? JAMB 2021 MATHEMATICS

a. -\frac{2}{7} b. -\frac{7}{4} c. -\frac{4}{3} d.3


For parallel lines, the gradient, M_{1}=M_{2} and for perpendicular lines, M_{1}\times M_{2}=-1. With this question, the two lines are parallel.

Line 1 passes through two points, (-1,-p) and (-2,2)

Line 2 has the equation, 2y+8x-17=0

Firstly, let’s find the gradient of the second line, to do that we will have to re-arrange the equation,




From the equation, y=mx+c, the gradient, m_{2}=-4

So, using the formula for a gradient of a line given two points, we can find the value of -p.

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}  (x_{1}, y_{1}) (x_{2}, y_{2})

So, for (-1,-p)(-2,2),

m=\frac{2-(-p)}{-2-(-1)}, recall m_{1}=m_{2}




Cross multiply,

2+p=-4\times -1



{\color{Green} p=2 (D)}

The next question is on the comparison between similar shapes. On this question, you need to know more on the principles of comparing shapes.

QUESTION 7: The ratio of the length of two similar rectangle blocks is 2:3. If the volume of the larger block is 351cm^{2}, then the volume of the other block is? JAMB 2021 MATHEMATICS

a. 234.00cm^{3} b. 526.50cm^{3} c. 166.00cm^{3} d. 683cm^{3}


Given two rectangles, a bigger and smaller shape.

Small rectangle, Length, L_{s}= 2cm and the Volume, V_{s} = ?

Big rectangle, Length, L_{b}=3cm and the Volume, V_{b}=351cm^{3}

Comparing the two similar rectangles, \frac{L_{s}}{V_{s}}=\frac{L_{b}}{V_{b}}


Cross multiply,

V_{s}\times 3=2\times351cm^{3}

V_{s}=\frac{2\times 351}{3}


{\color{Green} V_{s}=234cm^{3} (A)}

QUESTION 8: find the derivative of the function y=2x^{2}(2x-1) at the point were x=-1? JAMB 2021 MATHEMATICS

a. 18 b. 16 c. -4 d. -6


The derivative of the function, y=2x^{2}(2x-1) at x=-1

Firstly, simplify the expression, by opening the bracket,


Recall, if y=ax^{n} where, a is a constant, 

\frac{dy}{dx}=nax^{n-1} this is a general equation for finding the derivatives of an equation.

\frac{dy}{dx}=4\times 3x^{3-1}-2\times 2x^{2-1}


At x=-1

= 12(-1)^{2}-4(-1)^{}

= 12+4

{\color{Green} =16 (B))}

The next question is on significant figures, many consider this kind of questions easy, but if you are not careful you might get the wrong answer. To solve this kind of questions accurately, watch this YouTube video “Significant Figure, all you need to Know“.

QUESTION 9: Correct 241.34 (3\times 10^{-3})^{2} to 4 significant figures.  JAMB 2021 MATHEMATICS

a. 0.0014 b. 0.001448 c. 0.0022 d. 0.002172


Note: For significant figure, the zeros after the decimal point and before any real number is not counted. But on the other hand, for decimal figure, the zeros after the decimal point and before any real number is counted.

So, simplify using a calculator, since you will be given a calculator.

241.34(3\times 10^{-3})^{2}

241.34\times (0.003)^{2} gives 241.34\times 0.000009


In 4 significant figures the value will give

{\color{Green}= 0.002172 (D)}

Let’s consider the final of the set of 10 questions which are the first of 40 questions for the 2021 JAMB UTME examination. Expect the next post which will feature the next ten (10) questions. 

QUESTION 10: Find the mean derivation of 1, 2, 3, and 4.  JAMB 2021 MATHEMATICS

a. 1.0 b. 1.5 c. 2.0 d. 2.5


First calculate the mean of the value. 1, 2, 3, and 4

Mean =\frac{\sum fx}{N}

Mean =\frac{(1\times 1)+(1\times 2)+(1\times 3)+(1\times 4)}{4}

Mean  =\frac{1+2+3+4}{4}=\frac{10}{4}=2.5

Mean deviation = \frac{\sum /x-\bar{x/}}{N} 

Mean Deviation =\frac{/1-2.5/+/2-2.5/+/3-2.5/+/4-2.5/}{4}

Remember, we are looking for the absolute values, for example, -1 will give /-1/ in an absolute. The result will be 1.

Mean deviation = \frac{1.5+0.5+0.5+1.5}{4}=\frac{4}{4}=1

∴ Mean deviation {\color{Green} = 1(A)}

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