JAMB Mathematics 2023 practice collection

C: 8!/(2! × 2!)

There are eight letter positions. S occurs twice and L occurs twice; the other letters occur once. Temporarily distinguish repeated letters: there are 8! arrangements. Each actual arrangement is counted 2! × 2! times. Divide: 40320/4 = 10080.

A: 2

Combine powers with base 16: 16^(0.16 + 0.04) = 16^0.2. Rewrite 16 as 2⁴, so 16^0.2 = 2^0.8. Multiply powers with base 2: 2^(0.8 + 0.2) = 2¹ = 2.

D: 15

Apply ★ inside the left brackets: (−4)² × 2 = 32. Apply ★ inside the right brackets: 7² × (−1) = −49. Apply ◇ to the two results: 2 × 32 + (−49) = 15.

B: −8/5

For a power xⁿ, increase the exponent by one and divide by that new exponent. An antiderivative is 2x² − (18/5)x^(5/3). Differentiating it returns 4x − 6x^(2/3). At 1 this is 2 − 18/5 = −8/5; at 0 it is zero. Subtract the endpoint values.

D: 15.45%

The decrease is 1230 − 1040 = 190. Use the original population as the denominator: 100 × 190/1230 = 15.447154…%. Round only the final percentage to two decimal places: 15.45%.

A: dodecagon

An interior angle and its adjacent exterior angle add to 180°. If the exterior angle is e, then 5e + e = 180°, so e = 30°. The exterior angles total 360°. Therefore the polygon has 360/30 = 12 sides, making it a dodecagon.

D: 0.299 cm/s

Since A=πr², the radius at this instant is √(2/π) cm. Differentiating with respect to time gives dA/dt=2πr(dr/dt). Therefore dr/dt=1.5/[2π√(2/π)]≈0.2992067 cm/s, which is 0.299 cm/s to three significant figures.

D: x = −(9c + 5dy)/(4y − 3)

Multiply by 4x + 5d to obtain 4xy + 5dy = 3x − 9c. Collect the terms containing x: 4xy − 3x = −9c − 5dy. Factor to obtain x(4y − 3) = −(9c + 5dy). Divide by 4y − 3: x = −(9c + 5dy)/(4y − 3). The original expression requires 4x + 5d ≠ 0, and this rearrangement requires 4y − 3 ≠ 0.

A: x ≤ −3

Expand both brackets: 3x − 3 ≤ 2x − 6. Subtract 2x from both sides, then add 3: x ≤ −3. The inequality direction stays the same because these steps do not multiply or divide by a negative number. At x = −3 both sides equal −12, so the boundary value is included.

A: 72°

The radii to P and R meet their tangents at right angles. In quadrilateral formed by the centre, P, S and R, the central angle subtending minor arc PR is 180°−36°=144°. The angle at Q on the major arc subtends that minor arc and is half its central angle: 72°.

C: 3 hours

The speed is 225 ÷ 4.5 = 50 km/h. The time for 150 km is distance divided by speed: 150 ÷ 50 = 3 hours.

D: 10101₂

1101001₂ is 64 + 32 + 8 + 1 = 105 in base ten. 101₂ is 5. Divide 105 by 5 to get 21, then express 21 as 16 + 4 + 1 = 10101₂.

A: a = 9, b = −5

Multiply top and bottom by 2 − √3. The denominator becomes 4 − 3 = 1; the numerator expands to 6 − 3√3 − 2√3 + 3 = 9 − 5√3. Thus a = 9 and b = −5.

C: −3

Their vertical separation is −5 − (−14) = 9. The distance formula gives 9² = (t + 3)² + 9², so (t + 3)² = 0 and t = −3.

C: 20 years

At 10 years the interest is 2P, where P is the principal. Reaching 5P requires interest of 4P, twice as much. Simple interest grows linearly with time, so the required time is 20 years.

A: 4

Factor the numerator as (x − 2)(x + 6) and the denominator as x(x − 2). For x other than 2, cancel x − 2. The limit is (2 + 6)/2 = 4.

A: 7.7 years

Use class midpoints 5.5, 7.5 and 9.5. Their weighted total is 29×5.5 + 40×7.5 + 38×9.5 = 820.5. Divide by 107 pupils to get about 7.668 years, or 7.7 to one decimal place.

C: 66

A female majority means either three women and two men, or four women and one man. Count C(4,3)C(6,2) + C(4,4)C(6,1) = 4×15 + 1×6 = 66.

C: −23

Expand along the first row: 2(1×1 − 2×(−3)) − (−1)(4×1 − 2×1) + 3(4×(−3) − 1×1). This is 14 + 2 − 39 = −23.

B: 323°

The displacement is northwest. Its angle west of north is arctan(6/8) ≈ 36.87°. Bearings increase clockwise from north, so the bearing is 360° − 36.87° ≈ 323°.

A: 7/8

The complement is no heads, meaning three tails. Its probability is (1/2)³ = 1/8. Hence the probability of at least one head is 1 − 1/8 = 7/8.

D: 3y = 2x + 5

The given line has slope −3/2, so a perpendicular line has slope 2/3. Using y − 3 = (2/3)(x − 2) gives 3y = 2x + 5.

B: (10/3)x^(2/3) − (8/3)x^(5/3)

Expand first: y = 2x^(5/3) − x^(8/3). Apply d(xⁿ)/dx = nx^(n−1) term by term to obtain (10/3)x^(2/3) − (8/3)x^(5/3).

A: 2.72

The primes are 2, 3, 5, 7 and 11; their mean is 5.6. Their absolute deviations from 5.6 total 3.6 + 2.6 + 0.6 + 1.4 + 5.4 = 13.6. Divide by five to get 2.72.

D: 64/3

The curves meet when 2x² + 10 = 4x + 16, giving x = −1 and x = 3. The straight line is higher between these points. Integrate (4x + 16) − (2x² + 10) = −2x² + 4x + 6 from −1 to 3 to obtain 64/3.

C: 5

402ᵧ has value 4y² + 2. Set 4y² + 2 = 102 to obtain y² = 25. A numeral base is positive and must exceed its largest digit, 4, so y = 5.

B: 17

Both logarithms on the left require y>8. Combine them to obtain log₁₀(y²−64)=log₁₀(225), so y²=289. The algebraic roots are ±17, but −17 is outside the logarithm domain. Hence y=17.

A: (2x − y)(2x + y)(4x² + y²)

Write 16x⁴ − y⁴ as (4x²)² − (y²)² = (4x² − y²)(4x² + y²). Factor the first bracket again: 4x² − y² = (2x − y)(2x + y).

A: {1,3,5,8}

A − B contains the members only in A: 1, 3 and 5. B − A contains the member only in B: 8. Taking their union gives {1,3,5,8}.

C: 16/33

There are four white balls because their count must be half the eight red balls. With twelve balls in total, red then white has probability (8/12)(4/11)=8/33. White then red has the same probability. Adding the two mutually exclusive orders gives 16/33.

A: 7.725 years

There are 107 pupils, so the halfway position is 53.5 and lies in the 7–8 group. Using continuous class boundaries 6.5 to 8.5, estimate the median as 6.5 + ((53.5 − 29)/40)×2 = 7.725 years.

B: [[0,1],[1/2,1/2]]

Let A = [[a,b],[c,d]]. The product on the left is [[2b,a−b],[2d,c−d]]. Matching entries with [[2,−1],[1,0]] gives b=1, a=0, d=1/2 and c=1/2.

A: 16

The three bands starting at 25% or higher contain 5, 3 and 8 students. Add those frequencies: 5+3+8=16. The first band is below 25% and is not included.

C: −30

The remainder theorem says to substitute x = −1. The result is −2(−1)³ + 6(−1)² + 17(−1) − 21 = 2 + 6 − 17 − 21 = −30.

A: a*b = b*a

Commutativity means swapping the two operands leaves the result unchanged: a*b = b*a. The second identity describes associativity, and the third is a distributive form.

B: −2 ≤ x ≤ 3

Move 2 to the left: x² − x − 6 ≤ 0. Factor as (x − 3)(x + 2) ≤ 0. The upward-opening quadratic is nonpositive between its roots, including both endpoints.

A: 5n − 2

The sequence has first term 3 and common difference 5. Its nth term is 3 + (n−1)5 = 5n − 2; at n=1 it returns 3.

D: The two angle bisectors

Every point on either angle bisector of two intersecting lines has equal perpendicular distance from both lines. Together, the two bisectors give the full locus.

B: 3:4

If the reference is 100 units, the quantities are 135 and 180. Their ratio is 135:180, which reduces by 45 to 3:4.

D: 32 m

The horizontal gap is 24/tan60° = 8√3 m. The second roof rises (8√3)tan30° = 8 m above the first roof. Its height is 24 + 8 = 32 m.

C: 36.2 cm

The sine rule gives AB/sin56°=BC/sin82°. Hence AB=43.2 sin56°/sin82°≈36.1664 cm, which is 36.2 cm to three significant figures.

B: 10% loss

A ₦230 selling price at 15% profit means cost = 230/1.15 = ₦200. Selling for ₦180 loses ₦20, which is 20/200 × 100 = 10% of cost.

B: 30.6 cm

The perpendicular from the centre bisects the chord, leaving a half-chord of 12 cm. Half the central angle is arcsin(12/13), so the full angle in radians is 2arcsin(12/13). Arc length is radius times angle: 13×2arcsin(12/13)≈30.5761 cm, or 30.6 cm.

A: 1931.25 m² ≤ A < 2021.25 m²

The true sides lie in [37.5,38.5) and [51.5,52.5). All lengths are positive, so multiply the lower bounds for 1931.25 m² and upper bounds for 2021.25 m². The upper endpoint is excluded.

C: 54

After brand A, 8/9 of the stock remains. Brand C is 3/8 of that remainder, so C is (3/8)(8/9) = 1/3 of all stock. Total stock is 243; B is (5/8)(8/9)×243 = 135, or 54 more than 81.

A: 1162 cm²

The nine ratio parts make sides 40, 60 and 80 cm; the semiperimeter is 90 cm. Heron's formula gives √[90(90−40)(90−60)(90−80)] = √1,350,000 ≈ 1161.895, which rounds to 1162 cm².

D: ₦2,612.48

The amount is ₦15,700×1.08² = ₦18,312.48. Subtract the principal to isolate compound interest: ₦18,312.48 − ₦15,700 = ₦2,612.48.

C: 198

From a+2d=6 and a+4d=12, subtract to obtain d=3 and then a=0. The first twelve terms sum to 12/2 × [2a+11d] = 6×33 = 198.

A: 1.66×10⁻⁴

Write every term with the same power of ten: (1.654−0.000676+0.0023)×10⁻⁴=1.655624×10⁻⁴. The fourth significant digit is 5, so to three significant figures the coefficient rounds from 1.655624 to 1.66.

A: 2.17%

The absolute error is |18 − 18.4| = 0.4 mL. Relative to the accepted 18.4 mL, percentage error is 0.4/18.4 × 100 ≈ 2.17%.

B: 13

Their exterior angles are 360/(n−1) and 360/(n+2). Their difference is 1080/[(n−1)(n+2)] = 6, so (n−1)(n+2)=180. The positive solution of n²+n−182=0 is n=13.

A: 5/12

Prime sums from 2 to 12 are 2, 3, 5, 7 and 11. Their ordered-outcome counts are 1, 2, 4, 6 and 2, totalling 15. There are 36 equally likely outcomes, so the probability is 15/36 = 5/12.

B: 2568 cm³

Split the solid into a full lower beam and the extra upper block. The lower volume is 36×6×8=1728 cm³. The added block is 10×6×(22−8)=840 cm³. Their sum is 2568 cm³.

D: 400 cm², 80 cm

For a square, diagonal = side×√2. The side is therefore 20 cm. Area = 20² = 400 cm² and perimeter = 4×20 = 80 cm.

A: 25.23%

Total cost is ₦10,700. Gain is ₦13,400 − ₦10,700 = ₦2,700. Divide gain by total cost: 2700/10700 × 100 ≈ 25.23%.

C: 7/40

Compute the division first, then multiply: (3/4) ÷ (5/12) × (1/4) = (9/5)×(1/4) = 9/20. Now 5/8 − 9/20 = 25/40 − 18/40 = 7/40.

C: 13 cm

Let the shorter side be x and the longer side x+6. The area increase is (x+2)(x+8) − x(x+6) = 4x+16. Set this equal to 68 to obtain x=13 cm.

A: −1/3

Write the second term as ar=−2/3 and the sum as a/(1−r)=3/2, where |r|<1. Eliminate a to get 9r²−9r−4=0. The roots are 4/3 and −1/3; only −1/3 meets the convergence condition.

B: 20

A point on a line must satisfy its equation. Substitute x=−2 and y=k: 3k + 6(−2) = 48. Thus 3k−12=48, so 3k=60 and k=20.

D: 60

Let the number of child tickets be c, so adult tickets number 5c. Revenue is 250c + 520(5c) = 2850c. Solve 2850c = 171000 to get c=60.

A: x = 2 or 4

The domain is 0<x<6. Add log₂(x) to both sides, giving log₂[x(6−x)] = 3. Hence x(6−x)=8 and x²−6x+8=(x−2)(x−4)=0. Both 2 and 4 satisfy the domain.

D: 100π cm³

The radius, height and slant height form a right triangle, so the vertical height is √(13²−5²) = 12 cm. Volume is (1/3)πr²h = (1/3)π×25×12 = 100π cm³.

A: 10 cm and 15 cm

Area = (height/2)×sum of parallel sides, so their sum is 2×200/16 = 25 cm. Split 25 in the ratio 2:3 to get 10 cm and 15 cm.

B: 200 km

If the one-way distance is d, total time is d/120 + d/150 = 3. Since 1/120 + 1/150 = 3/200, solving gives d = 200 km.

D: −2 + 2√2 m

Let each equal leg have length s, so the hypotenuse is s√2. Then s(2+√2)=2. Multiply 2√2/(2+√2) by the conjugate to obtain the hypotenuse 2√2−2 m.

A: 62 km

The longitude change is 8°. The radius of the small circle at latitude 86° is 6370cos86°. Arc distance is (8/360)×2×3.142×6370cos86° ≈ 62.1 km, or 62 km.

D: 10 cm

The perpendicular from the centre bisects the chord. One half has length √(13²−12²) = √25 = 5 cm, so the full chord is 10 cm.

D: 80 VIP, 120 regular

Let v be VIP sales, so regular sales are 200−v. Revenue gives 1200v + 700(200−v) = 180000. Then 500v=40000, so v=80 and regular sales=120.

A: 6 days

The job needs 12×8×4 = 384 worker-hours. Four workers at 16 hours each supply 64 worker-hours per day. Divide 384 by 64 to get 6 days.