# Mathematics Past Questions

Mathematics studies measurement, relationships, and properties of quantities and sets, using numbers and symbols. Arithmetic, algebra, geometry, and calculus are popular topics in mathematics.

Study the following Mathematics past questions and answers for JAMB, WAEC, NECO and Post UTME. Prepare yourself with official past questions and answers for your upcoming examinations.

Sort Past Questions by Exam Body:

Correct answers are highlighted in green.

### Question 1

Differentiate $(2x+5)^2(x-4)$ with respect to x.

### Options

A) $4(2x+5)(x-4)$

B) $4(2x+5)(4x-3)$

C) $(2x+5)(2x-13)$

D) $(2x+5)(6x-11)$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

To differentiate $(2x+5)^2(x-4),$ you first need to know that it is a product function.
Using the product rule you have $\frac{dy}{dx}=\frac{udv}{dx}+\frac{vdu}{dx}$
Let $(2x+5)^2$ be $u$ and $(x-4)$ be $v.$

To find $du$ we use chain rule which is $\frac{du}{dx}=\frac{du}{dw} \times \frac{dw}{dx}$
We say let $(2x+5)$ be $w$
We then have a new function $u=w^2$
$\frac{du}{dw}=2w$
$\frac{dw}{dx}=2$
So $\frac{du}{dx}=2 \times 2w$ which equals $4w$ and $w$ was $(2x+5)$
$\frac{du}{dx}=4(2x+5)$
$\frac{dv}{dx}=1$

Substituting everything into the product rule we have:
$\frac{dy}{dx}=(2x+5)^2(1) + 4(2x+5)(x-4)$
$\frac{dy}{dx}=(4x^2 + 20x + 25) + (8x^2 - 12x - 80)$
$\frac{dy}{dx}=12x^2 + 8x - 55$
$\frac{dy}{dx}=(2x+5)(6x-11)$

### Question 2

Find the area bounded by the curves $y = 4 - x^2$ and $y = 2x + 1$

### Options

A) $20\frac{1}{3}$ sq. units

B) $20\frac{2}{3}$ sq. units

C) $10\frac{2}{3}$ sq. units

D) $10\frac{1}{3}$ sq. units

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

$y = 4 - x^2$ and $y = 2x + 1$
$\Rightarrow 4 - x^2 = 2x + 1$
$\Rightarrow x^2 + 2x - 3 = 0$
$(x+3)(x-1) = 0$
Thus, $x = -3$ or $x = 1.$

Integrating $x^2 + 2x - 3$ from $(-3$ to $1)$ with respect to $x$ will give $\frac{31}{3} = 10\frac{1}{3}$

### Question 3

Find the rate of change of the volume, $V$ of a sphere with respect to its radius, $r$ when $r = 1.$

### Options

A) $12\pi$

B) $4\pi$

C) $24\pi$

D) $8\pi$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

Volume of sphere, $V = \frac{4}{3} \times \pi r^3$
Rate of change of $V = \frac{dv}{dr}$
Thus if, $V = \frac{4}{3} \times \pi r^3,$
$\Rightarrow \frac{dv}{dr} = 4\pi^2$

At $r = 1,$ Rate $= 4 \times \pi \times 1 = 4\pi$

### Question 4

If $y = x \; \mathrm{sin} \; x,$ find $\frac{dy}{dx}$ when $x = \frac{\pi}{2}.$

### Options

A) $\frac{- \pi}{2}$

B) $-1$

C) $1$

D) $\frac{\pi}{2}$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

$y = x \; \mathrm{sin} \; x$
$\frac{dy}{dx} = 1 \times \mathrm{sin} \; x + x \; \mathrm{cos} \; x$
$= \mathrm{sin} \; x + x \; \mathrm{cos} \; x$
At $x = \frac{\pi}{2},$ $= \mathrm{sin} \; \frac{\pi}{2} + \frac{\pi}{2} \; \mathrm{cos} \frac{π}{2}$
$= 1 + \frac{\pi}{2} \times (0) = 1$

### Question 5

Find the dimensions of a rectangle of greatest area which has a fixed perimeter $p.$

### Options

A) square of sides $p$

B) square of sides $2p$

C) square of sides $\frac{p}{2}$

D) square of sides $\frac{p}{4}$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

Let the rectangle be a square of sides $\frac{p}{4}.$
So that perimeter of square $= 4p$
$4 \times \frac{p}{4} = p.$

### Question 6

Given the scores: $4, 7, 8, 11, 13, 8$ with corresponding frequencies: $3, 5, 2, 7, 2, 1$ respectively. Find the square of the mode.

### Options

A) $49$

B) $121$

C) $25$

D) $64$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

Mode = score with highest frequency $= 11.$
Square of $11 = 121$

### Question 7

Teams $P$ and $Q$ are involved in a game of football. What is the probability that the game ends in a draw?

### Options

A) $\frac{2}{3}$

B) $\frac{1}{2}$

C) $\frac{1}{3}$

D) $\frac{1}{4}$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

$P$ (games end in draw)
$\Rightarrow$ Team $P$ wins and $Q$ wins
$P$ ($P$ wins) $= \frac{1}{2}$
$P$ ($Q$ wins) $= \frac{1}{2}$
Therefore, $P$ (games ends in draw) $= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

### Question 8

Given the scores: $4, 7, 8, 11, 13, 8$ with corresponding frequencies: $3, 5, 2, 7, 2, 1$ respectively. The mean score is

### Options

A) $7.0$

B) $8.7$

C) $9.5$

D) $11.0$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

Mean $= \frac{\sum fx}{\sum f}$
Mean $= \frac{(4 \times 3) + (7 \times 5) + ... + (8 \times 1)}{20}$
Mean $= \frac{174}{20} = 8.7$

### Question 9

If $^6\mathrm{P}_r = 6,$ find the value of $^6\mathrm{P}_{r + 1}$

### Options

A) $30$

B) $33$

C) $35$

D) $15$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

$^6\mathrm{P}_r = 6$ Thus $r = 1$
N.B: $^6\mathrm{P}_1 = \frac{6!}{(6 - 1)!}$
$= \frac{6!}{5!}$ $= \frac{6 \times 5!}{5!} = 6$
$^6\mathrm{P}_{r + 1} = ^6\mathrm{P}_2$ $= \frac{6!}{(6 - 2)!} = \frac{6!}{4!}$
$= \frac{6 \times 5 \times 4!}{4!} = 30$

### Question 10

Given distribution of color beads: blue, black, yellow, white and brown with frequencies $1, 2, 3, 4,$ and $5$ respectively. Find the probability that a bead picked at random will be blue or white.

### Options

A) $\frac{7}{15}$

B) $\frac{2}{5}$

C) $\frac{1}{3}$

D) $\frac{1}{15}$

Exam Body: Joint Admissions and Matriculation Board (JAMB)

### Explanation:

Total number of beads $= 15.$
Number of white beads $= 4. \Rightarrow \mathrm{P}$(white) $= \frac{4}{15}.$
Number of blue beads $= 1. \Rightarrow \mathrm{P}$(blue) $= \frac{1}{15}.$
$\mathrm{P}$(white or blue) $= \mathrm{P}$(white) $+ \mathrm{P}$(blue) $= \frac{5}{15} = \frac{1}{3}$