Junior Secondary School Mathematics Past Questions And Answers
Mathematics Past Questions | JAMB, WAEC, NECO and Post UTME 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 past questions and answers for your upcoming examinations.
JAMB Syllabus for Mathematics
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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) \((2x+5)(6x-11)\)
The correct answer is D.
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:
\(\frac{dy}{dx}=(2x+5)^2(1) + 4(2x+5)(x-4)\)
\(\frac{dy}{dx}=(4x^2 + 20x + 25) + (8x^2 - 12x - 80)\)
\(\frac{dy}{dx}=12x^2 + 8x - 55\)
\(\frac{dy}{dx}=(2x+5)(6x-11)\)
Comments (3)
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
The correct answer is C.
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 \(\frac{32}{3} = 10\frac{2}{3}\)
Comments (6)
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) \(8\pi\)
The correct answer is B.
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 \(= 4 \times \pi \times 1 = 4\pi\)
Comments (1)
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) \(\frac{\pi}{2}\)
The correct answer is C.
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 \(x = \frac{\pi}{2},\) \(= \mathrm{sin} \; \frac{\pi}{2} + \frac{\pi}{2} \; \mathrm{cos} \frac{π}{2}\)
\(= 1 + \frac{\pi}{2} \times (0) = 1\)
Comments (3)
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 \(\frac{p}{4}\)
The correct answer is D.
Explanation:
Let the rectangle be a square of sides \(\frac{p}{4}.\)
So that perimeter of square \(= 4p\)
\(4 \times \frac{p}{4} = p.\)
Comments (3)
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) \(64\)
The correct answer is B.
Explanation:
The mode is the value that occurs most frequently in a given set of data.
Mode = score with highest frequency \(= 11.\)
Scores | 4 | 7 | 8 | 11 | 13 | 8 |
Frequency | 3 | 5 | 2 | 7 | 2 | 1 |
The highest frequency is 7 and 11 is the number with that frequency i.e. the mode.
Square of \(11 = 121\)
Comments (8)
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) \(\frac{1}{4}\)
The correct answer is C.
Explanation:
In a football game, one team either wins or there is a tie. This means there are three possible outcomes: a win, a loss or a draw. Therefore, the probability of a draw is \(\frac{1}{3}\).
Comments (7)
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) \(11.0\)
The correct answer is B.
Explanation:
Mean \(= \frac{\sum fx}{\sum f}\)
Mean \(= \frac{(4 \times 3) + (7 \times 5) + (8 \times 2) + (11 \times 7) + (13 \times 2) + (8 \times 1)}{20}\)
Mean \(= \frac{174}{20} = 8.7\)
Comments (4)
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) \(15\)
The correct answer is A.
Explanation:
\(^6\mathrm{P}_r = 6\) Thus \(r = 1\)
N.B: \(^6\mathrm{P}_1 = \frac{6!}{(6 - 1)!}\)
\(= \frac{6!}{5!}\) \(= \frac{6 \times 5!}{5!} = 6\)
\(^6\mathrm{P}_{r + 1} = ^6\mathrm{P}_2\) \(= \frac{6!}{(6 - 2)!} = \frac{6!}{4!}\)
\(= \frac{6 \times 5 \times 4!}{4!} = 30\)
Comments (2)
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) \(\frac{1}{15}\)
The correct answer is C.
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) \(= \frac{5}{15} = \frac{1}{3}\)
Comments (3)
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Junior Secondary School Mathematics Past Questions And Answers
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