All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Mathematics 2015 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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The bar chart shows the allotment of time (in minutes) per week for selected subjects in a certain school. What is the total time allocated to the six subjects per week?
Sum of all bars: Biology 80 + English 160 + Maths 200 + Geography 80 + Chemistry 120 + Physics 80 = 720 minutes.
Find the gradient of the line joining the points P(5, 6) and Q(3, 3).
Gradient = (y₂−y₁)/(x₂−x₁) = (3−6)/(3−5) = −3/−2 = 3/2.
If gt² − k − w = 0, make g the subject of the formula.
gt² = k + w ⟹ g = (k+w)/t².
Determine the maximum value of y = 3x² − x³.
dy/dx = 6x−3x² = 0 ⟹ x=0 or x=2. At x=2, y = 3(4)−8 = 4 (maximum).
In the figure, find the value of x.
By the sine rule, x/sin60° = 10/sin90°, so x = 10×(√3/2) = 5√3cm.
If the end points of one side of square RSTU are R(-1,-1) and S(3,-1), find the area of the square.
Side length = distance RS = 4. Area = 4² = 16 square units.
Find the median of 5, 9, 1, 10, 3, 8, 9, 2, 4, 5, 5, 5, 7, 3 and 6.
Arranged in order the 8th of 15 values is 5, the median.
Factorise 2y² − 15xy + 18x².
2y²−15xy+18x² = 2y(y−6x)−3x(y−6x) = (2y−3x)(y−6x).
The mean of 2−t, 4+t, 3−2t, 2+t and t−1 is
Sum = 10 (the t terms cancel). Mean = 10/5 = 2.
The table represents the outcome of throwing a die 100 times (Numbers 1–6, Frequencies 18,22,20,16,10,14). What is the probability of obtaining at least a 4?
P(≥4) = (16+10+14)/100 = 40/100 = 2/5.
Simplify (√147 − √75)/√48.
√147=7√3, √75=5√3, √48=4√3. (7√3−5√3)/4√3 = 2√3/4√3 = 1/2.
If P = [[6,4],[5,7]], then |P| is
|P| = (6×7) − (4×5) = 42 − 20 = 22.
If y = 3sin(−4x), dy/dx is
dy/dx = 3×(−4)cos(−4x) = −12cos(−4x).
If the angle of a sector of a circle with radius 10.5cm is 120°, find the perimeter of the sector.
Perimeter = 2r + (θ/360)×2πr = 21 + (1/3)×2×(22/7)×10.5 = 21+22 = 43cm.
The length a person can jump is inversely proportional to his weight. If a 20kg person can jump 1.5m, find the constant of proportionality.
L = k/W ⟹ 1.5 = k/20 ⟹ k = 30.
A man donates 10% of his monthly net earnings to his church. If it amounts to ₦4500, what is his net monthly income?
10% of x = 4500 ⟹ x = 4500×10 = ₦45,000.
If y = x² − 3x + 4, find dy/dx at x = 5.
dy/dx = 2x−3. At x=5: 2(5)−3 = 7.
Evaluate (1.25×0.025)/0.05 correct to 1 decimal place.
(1.25×0.025)/0.05 = 0.03125/0.05 = 0.625 ≈ 0.6.
Solve the inequality (1/3)x + 1/4 > (1/2)x + 1/3.
Multiplying through by 12 and simplifying gives −2x > 1, so x < −1/2.
If P = [[5,3],[2,1]] and Q = [[4,2],[3,5]], find 2P + Q.
2P = [[10,6],[4,2]]; 2P+Q = [[14,8],[7,7]].
In how many ways can seven directors sit round a table?
Circular permutations of n items = (n−1)! = 6! = 720.
Evaluate lim(x→2) (x²−1)/(x+1).
(x²−1)/(x+1) = (x−1)(x+1)/(x+1) = x−1. As x→2, limit = 1.
The pie chart shows the sectoral allocation of fruits. Find the allocation for oranges.
Sector angles sum to 360°: 18x = 360°, x = 20°. Oranges = 4x = 80°.
Simplify 1/(2−√3) in the form a + b√3.
Multiply numerator and denominator by (2+√3): (2+√3)/(4−3) = 2+√3.
If y = x² + √x, find dy/dx.
y = x²+x^(1/2). dy/dx = 2x + (1/2)x^(−1/2).
In the figure, KL∥MN, LN bisects ∠KNM. If angle KLN is 54° and angle MKN is 35°, calculate the size of angle KMN.
∠LNM = 54° (alternate to ∠KLN); LN bisects ∠KNM so ∠KNL=∠LNM=54°, giving ∠KNM=108°. In triangle KMN: ∠KMN = 180°−108°−35° = 37°.
The sixth term of an A.P is 3 times the second term. If the first term is 2, find the common difference.
T₆=a+5d, T₂=a+d. a+5d=3(a+d) ⟹ 2d=2a ⟹ d=a=2.
Convert 27₁₀ to another number in base three.
27 = 3³, so 27₁₀ = 1000₃.
In a right angled triangle, if tanθ = 3/4. What is cosθ − sinθ?
Opposite=3, Adjacent=4, Hypotenuse=5. cosθ−sinθ = 4/5−3/5 = 1/5.
The third term of a G.P is 4 while the sixth term is 32. Find its common ratio.
ar²=4, ar⁵=32. Dividing gives r³=8, r=2.
Convert 101,100,011 (base 2, grouped) to base eight.
101₂=5, 100₂=4, 011₂=3, giving 543₈.
The table shows Values 1,2,3,4,5 with corresponding frequencies. Find the mode of the distribution.
The mode is the value occurring with the highest frequency, which is 4.
A matrix P has an inverse P⁻¹ = [[1,−3],[0,1]]. Find P.
|P⁻¹| = 1. P = adj(P⁻¹)/|P⁻¹| = [[1,3],[0,1]].
If P = {1,2,3,4,5} and P∪Q = {1,2,3,4,5,6,7}, list the elements in Q.
Since P∪Q−P=Q and the extra elements are 6 and 7, Q={6,7}.
In the figure, PQ∥RS and ∠PTR = 37°. What is the value of x?
Using angle relationships between the parallel lines PQ and RS with the transversal TS, x = 180°−37°−16° ... the exterior angle at S equals 127° by the co-interior/exterior angle relationship for this figure.
The locus of points that is equidistant from a fixed point is a
By definition, the locus of points equidistant from a fixed point is a circle centred at that point.
In how many ways can the letters in the word ELATION be arranged?
ELATION has 7 distinct letters, so the number of arrangements is 7!.
Factorize ax − by − ay + bx.
ax−by−ay+bx = x(a+b) − y(a+b) = (a+b)(x−y).
Solve the inequality 3(x+4) < 2(x+3).
3x+12<2x+6 ⟹ x < −6.
Evaluate ∫sin2x dx.
∫sin2x dx = −(1/2)cos2x + k.
The distribution of scores in a class test are 2, 8, 6, 5, 8, 6, 6, 5 and 6. Find the product of the modal and median score.
Arranged: 2,5,5,6,6,6,6,8,8. Mode=6, Median=6. Product = 36.
A bag contains 10 black balls and 15 white balls. If a ball is picked at random without replacement, what is the probability of picking a white ball?
P(white) = 15/(15+10) = 15/25 = 3/5.
In the diagram, O is the centre of the circle. If ∠TUR is 50°, find the value of ∠TOR.
The angle at the centre is twice the angle at the circumference: ∠TOR = 2×50° = 100°.
Find the value of the determinant |[0,3,2],[1,7,8],[0,5,4]|.
Expanding along the first column: −1×(3×4−2×5) = −1×2 = −2.
Find the midpoint of S(−5,4) and T(−3,−2).
Midpoint = ((−5−3)/2, (4−2)/2) = (−4,1).
P, Q and R are subsets of the universal set U. Which Venn diagram shows the relationship (P∩Q)∪R?
(P∩Q)∪R includes all of R plus the common elements of P and Q, as shown fully shaded in diagram C.
Simplify (0.00045×2.50)/(0.00009×0.5).
Expressing in standard form and simplifying gives 25.
U = even numbers between 0 and 30, P = multiples of 6 between 0 and 30, Q = multiples of 4 between 0 and 30. Find (P−Q)ᶜ.
P={6,12,18,24}, Q={4,8,12,16,20,24,28}. P−Q={6,18}; its complement in U is {2,10,14,22,26} (among the listed options).
Y varies directly as w². When y=8, w=2, find y when w=3.
k = y/w² = 8/4 = 2. When w=3, y = 2×9 = 18.
If some strips of sticks of lengths 7cm, 9cm and 27cm were cut independently from a single stem without a remainder, find the shortest possible length of that stem.
The shortest stem length is the LCM of 7, 9 and 27, which is 189cm.
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