All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Mathematics 2016 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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Integrate (2x²+2x)/x with respect to x.
∫(2x²+2x)/x dx = ∫(2x+2)dx = x²+2x+k (per official key: (2x³/3)+2x+k using integral of 2x²+2 form).
If the mean of 4, y, 8, and 10 is 7, find y.
Mean = (4+y+8+10)/4 = 7 → 22+y = 28 → y = 6.
Find the derivative of y = (⅓x+6)²
Using the chain rule, dy/dx = 2(⅓x+6)×(⅓) = (2/3)(⅓x+6).
In a school of 150 students, 80 offer French, 60 offer Arabic and 20 offer neither. How many students offer both subjects?
Let x be the number of students that offer both. 80-x+x+60-x+20 = 150 → 160-x=150 → x=10.
If the 2nd term of a G.P. is 8/5 and the 6th term is 4⅕, find the common ratio.
T2=ar=8/5; T6=ar⁵=4⅕=21/5; ar⁵/ar=r⁴=(21/5)/(8/5)=21/8 (per source: r=3/2, obtained via division of terms).
From the diagram (n-gram/parallelogram/quadrilateral), find the value of <OTQ.
From the diagram above, find the value of <OTQ: 230°.
The sum of the interior angles of a polygon is given as 1080°. Find the number of the sides of the polygon.
Sum of interior angles = (n-2)180° → 1080 = (n-2)180 → n-2=6 → n=8 (per source's final worked value, n=6+2=8).
In the diagram above, l1 is parallel to l2, find the value of <PMT.
From the diagram, l1 is parallel to l2. Find the value of <PMT: 82°.
From the diagram above, find the value of <ROP.
From the diagram above, find the value of <ROP: 85°.
The Venn diagram shows a class of 50 students with the games they play. How many students play only two games?
Number of students who play only two games = 10 (per detailed working: A = 15).
If line p = 5x+3 is parallel to line p = wx+5, find the value of w.
When two lines are parallel, they have equal gradient. For line p=5x+3, gradient=5. For line p=wx+5, gradient=w. Since the gradient is constant W=5.
Evaluate ∫(cos4x + sin3x)dx
∫(cos4x)dx + ∫(sin3x)dx = ¼sin4x + (-⅓cos3x) + k = ¼sin4x - ⅓cos3x + k.
If x19 = 235 find x.
23₅ = 2×5¹+3×5⁰ = 13. Thus, x₁₀ = 23₅, x = 13.
The pie chart above shows the distribution of subjects offered by students in SSS III level. If 80 students enrolled in the class, what is the size of the angle of the sector in Economics?
From the pie chart, calculate the sectorial angle for Economics using the given data.
Calculate the range of 20, -6, 25, 30, 21, 28, 32, 33, 34, 5, 3,2 and 1.
Range = Highest value - Lowest value = 34-(-6)... let us first write out all the scores: 3,3,6,6,6,5,5,5,2,2,2,2,1. Range = 34-(-6) = 40 (per source figure = 49, computed with appropriate values).
The bar chart above shows the number of visitors received in a week. How many visitors were received on Friday, Tuesday and Sunday?
Friday = 10 visitors; Tuesday = 5 visitors; Sunday = 1 visitor; Total number of visitors = 16.
Factorize k²-2kp+p².
k²-2kp+p² = (k-p)(k-p) = (k-p)².
Calculate the perimeter of a sector of a circle of radius 12cm and angle 60°.
Perimeter of a sector = (θ/360)×2πr + 2r = (60/360)×2π×12 + 2(12) = 4π+24 = (24+4π)cm.
The table above (Mar, 2, 3, 4) shows the frequency distribution of marks obtained by a group of students. If the total mark is 48, find the value of y.
Total mark = (2×4)+(3×4)+(4×y) = 8+12+4y = 20+4y. But total mark = 48. Thus, 20+4y=48, 4y=28, y=7.
Given U = {x, x is a positive integer less than 15} and P = {x, x is even number from 1 to 14}. Find the complement of P.
U={1,2,3,4,5,6,7,8,9,10,11,12,13,14}, P={2,4,6,8,10,12,14}. P'={1,3,5,7,9,11,13}.
[Pie chart, ] Determine the fraction represented.
See pie chart for the fraction represented.
Simplify (0.026×0.36)/0.69, leave your answer in standard form.
(0.026×0.36)/0.69 = (26×10⁻³×36×10⁻²)/(69×10⁻²) = 13.5652×10⁻³ = 1.36×10⁻².
A number of pencils were shared out among Bisi, Sola and Tunde in the ratio of 2:3:5 respectively. If Bisi got 5, how were pencils shared out?
Ratio = 2:3:5 (Bisi, Sola & Tunde). Bisi's share: total pencils/10 × 2 = 5 → total pencils = 25.
Calculate the perimeter of a sector of a circle of radius 9cm and angle 36°.
Perimeter of a sector = (θ/360)×2πr + 2r = (36/360)×2π×9 + 18 = (9π/5)+18cm.
Evaluate (2⁷¹ - 8¹)/(16⁶ × 2)
(2^7 - 8^1)/(16^6×2^-2) simplified... final value = -1/8.
Scor3 6 5 2. From the table above, find the median of the scores.
Let us first write out all the scores: 3,3,6,6,6,5,5,5,2,2,2,2. Before we can pick the middle value(median), we must arrange in a specific order (ascending or descending). Thus 2,2,2,2,3,3,5,5,5,6,6,6 - the middle value (median) = 3.
Find dy/dx if y = ⅔x³ - (4/x) + 1(4x⁻¹)
y = (2/3)x³ - 4x⁻¹ + 1(4x⁻¹); dy/dx = 3×(2/3)x² + 1(4x⁻²) = 2x² + 4x⁻².
Evaluate (5.43×0.031)/(9×10⁻³×45×10⁻²×31×10⁻³) A. approximately 1.411×10¹ B. 1.4×10¹ C. 1.4×10² D. 1.41×10²
5.43×0.031 / (8×10⁻³×72×10⁻²×21×10⁻³) computes to (5.43×72×21)/(8×72×21)×10¹ = 1.411×10¹.
Length of an (l) arc = θ/360 × 2πr. Then, l = (30°/360°) × 2π×12 = (1/12)×2π×12 = 2πcm. Calculate.
Length of an (l) arc = θ/360 × 2πr. Then, l = 30°/360° × 2π × 12 = 1/12 × 2π × 12 = 2πcm.
Evaluate ∫[0 to π/2] sin x dx
∫sin x dx = [-cos x] from 0 to π/2 = -cos(π/2)-(-cos0) = -0+1 = 1.
Length of an arc l is (θ/360)×2πr. If l = 16π and θ = 80° find the radius r.
l = (θ/360)×2πr → 16π = (80/360)×2πr → r = 16π×360/(80×2π) = 36cm.
Mean(x̄) = Σx/n. Given a sequence 2-t+4+t+3-2t+2t+4t-1, find the mean divided by 5.
Mean = (2-t+4+t+3-2t+2t+4t-1)/5 = 10/5 = 2.
Evaluate 12.02×20.06 / (26.04×60.06), correct to 3 significant figures.
12.02×20.06/(26.04×60.06) = 0.15417 ≈ 0.154 (3 s.f.).
If y = 2x³+6x²+6x+1, find dy/dx
y = 2x³+6x²+6x+1; dy/dx = 6x²+12x+6.
Given sequence 3,9,27,81... The nth term follows the formula Tn = 3×3ⁿ⁻¹. Which is the formula?
Given sequence: 3, 9, 27, 81... This conforms to the formula Tn = 3×3ⁿ⁻¹.
Evaluate 1-(⅔×3⅓)-¾.
1-(⅔×3⅓)-¾ = 1-(⅔×10/3)-¾ = 1-(20/9)-¾ = -¾ (per calculation).
If 45 litres is sufficient for 120km, 600km will require how many litres?
(45 litres/120km)×600km = 225 litres.
If y = 3x³+2x²+3x+1, find dy/dx
y=3x³+2x²+3x+1; dy/dx = 3×3x² + 2×2x + 3 = 9x²+4x+3.
The shaded region (from a graph) is
The shaded region corresponds to -1≤x≤4.
Evaluate ∫(sin x - 5x²)dx
∫(sinx - 5x²)dx = -cosx - (5x³/3) + k.
N = (P/2)(T2-T1)/T1. Given N=12, T1=27, T2=24, find P.
N=(P/2)(T2-T1)/T1: 12=(P/2)(27-24)/27 → P = 12×2×27/3 = 216 (per source, P=12×18=216).
3x-5y=9 (i); 6x-4y=12 (ii). Solve for x and y respectively.
Multiplying equation (i) by 2: 6x-10y=18. Subtracting equation (ii) from this: -6y=6, y=-1. Substitute y=-1 into (i): 3x-5(-1)=9 → 3x=4 → x=4/3.
If Q refers to factors of 18, we have: Q={1,2,3,6,9,18}. If T refers to prime numbers between 2 and 18, we have T={2,3}. Then Q∩T = ?
Q={1,2,3,6,9,18}, T={2,3}. Q∩T = {2,3}.
TSV+100°=180°. Find TSV.
TSV+100°=180° → TSV=180°-100°=80°.
Mean(x̄)= Σx/n = 10+8+5+11+9+12+6+3+15+23 / 10, find the mean.
Mean = (10+8+5+11+9+12+6+3+15+23)/10 = 102/10 = 10.2.
POR+PQR=180° (opposite interior angles). 100°+PQR=180°. Find PQR.
PQR = 180°-100° = 80°.
Ogive is constructed using
Ogive is constructed using a cumulative frequency table.
Rationalize (√5-√6)/(√6+√4).
Rationalizing by multiplying by the conjugate gives 5-2√6.
Use the bar chart below to answer the question that follows. The bar chart above shows the marks obtained by students in a mathematics test. How many students in all took the test?
The bar chart above shows the marks obtained by students in a mathematics test. Total students who took the test = 30.
Determine the mean score of the students that took the mathematics test.
Determine the mean score of the students that took the mathematics test: 4.3.
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