JAMB Mathematics 2025 practice collection

C: 130 cm²

The base QR is 11+9=20 cm, and the perpendicular height from P is 13 cm. Area = ½×20×13 = 130 cm².

D: {4,3,1,5,2}

Equivalent sets have the same number of distinct elements. The reference set has five. Option D has the same five elements in a different order; A and C each have four, while B has six.

C: x=6, y=10

Multiply the first equation by 10 to get 5x−2y=10, and the second by 3 to get −x+3y=24. From x=3y−24, substitution gives 13y=130, so y=10 and x=6.

C: x<−5/3

Subtract 2x and 8 to obtain −5>3x, so x<−5/3. No inequality reversal is needed because the final division is by positive 3.

B: 11/6

Use the given rule with a=3 and b=2. The fraction is (3+2)/(3×2)=5/6, while the difference is 3−2=1, so their sum is 11/6.

A: 1/35

The three factors are 1/5, 1/7 and 1 respectively. Their product is 1/(5×7)=1/35.

D: (1,5)

Average corresponding coordinates: x=(−3+5)/2=1 and y=(4+6)/2=5, so the midpoint is (1,5).

A: 5/12

Use a right triangle with adjacent side 12 and hypotenuse 13. Its opposite side is √(13²−12²)=5, so tan A=5/12.

B: 360

SCHOOL has six letters, with O repeated twice. Divide all 6! arrangements by 2! for the repeated O: 720/2=360.

B: (1,2)

Subtract the equations 7=a+3b and 11=a+5b to get 4=2b, hence b=2. Then a=7−3×2=1.

C: 1070 kg

The original total is 15×1000=15000 kg. The remaining fourteen average 995 kg and total 14×995=13930 kg. The removed bar weighs 15000−13930=1070 kg.

B: −1

The determinant is 1×5−2×3=−1. The inverse is (−1)^−1[[5,−2],[−3,1]]=[[-5,2],[3,−1]]. Its entries sum to −1.

C: 16

Direct-square variation means A=kb² for a constant k. The first pair gives 4=k×1², so k=4. At b=2, A=4×2²=16.

A: 1

The 2×2 identity matrix is [[1,0],[0,1]]. Its determinant is 1×1−0×0=1.

C: 3

Expanding the determinant along the first row gives −2(4−4)−3(2p−1)+(4p−2)=1−2p. Set 1−2p=−5 to obtain p=3.

C: 0

The four numbers sum to 8x+16. A mean of 4 requires a sum of 16, so 8x+16=16 and x=0.

C: 5y−16x−27=0

Use y−15=(16/5)(x−3). Multiply by 5 and rearrange: 5y−75=16x−48, hence 5y−16x−27=0.

C: 44.12%

Total cost is ₦270,000+₦70,000=₦340,000. Profit is ₦490,000−₦340,000=₦150,000. Therefore profit percentage is (150,000/340,000)×100≈44.12%.

A: 7/11

The first bracket is (3/4)/(9/4)=1/3. Convert 1 7/11 to 18/11, and the last bracket to 11/3−5/2=7/6. Multiplying gives (1/3)(18/11)(7/6)=7/11.

A: 17/30

There are 6+11+13=30 fruits. The favourable fruits are the 6 grapes and 11 bananas, so the probability is (6+11)/30=17/30.

C: 108°

The interior angles of a pentagon sum to (5−2)×180°=540°. All five angles are equal, so each is 540°/5=108°.

B: 8/15

After the shirts, 4/5 remains. Transport takes one-third of that remainder, leaving two-thirds of it: (4/5)×(2/3)=8/15 of the original salary.

C: 1/2

A fair die has six equally likely faces. Three outcomes, 2, 4 and 6, are even, so the probability is 3/6=1/2.

B: x⁴+x²+sin x+C

Integrate term by term: 4x³ gives x⁴, 2x gives x², and cos x gives sin x. The result is x⁴+x²+sin x+C.

C: 5

The decimal 0.00001 equals 10⁻⁵. Thus log₁₀(0.00001)=−5, and the leading minus sign gives 5.

C: 8 cm

The perpendicular from the centre bisects the chord, giving a 15 cm half-chord. By Pythagoras, the distance is √(17²−15²)=√64=8 cm.

B: 10

Factor the numerator as (x−2)(x²+6). For x different from 2 the quotient is x²+6, whose limit at x=2 is 4+6=10.

C: 1/2

Three ratio steps separate the second and fifth terms, so r³=(1/8)/1=1/8. The real cube root is r=1/2.

B: ₦500,000

At simple interest the amount is P(1+0.05×4)=1.2P. Hence P=₦600,000/1.2=₦500,000.

D: 2n−3

The kth term is 3+2(k−1)=2k+1. Put k=n−2 to obtain 2(n−2)+1=2n−3.

D: 4cos(4x)

The chain rule gives the derivative of sin(4x) as cos(4x) multiplied by the derivative of 4x, which is 4. Thus dy/dx=4cos(4x).

C: 360A/(πr²)

Multiply A=(θ/360)πr² by 360, then divide by πr². This gives θ=360A/(πr²), for a positive radius.

A: 4

Complete the square: y=4−2(x+1)². Since (x+1)² is non-negative, y cannot exceed 4, and it reaches 4 at x=−1.

B: 2pq

There are two disjoint orders: the first survives and the second dies, or the first dies and the second survives. Independence gives probability pq for each order, so the total is pq+qp=2pq.

A: 1022₅

Take one group of 125, leaving 12. That is zero groups of 25, two groups of 5 and two units, so 137=1022₅.

A: 68

Subtract 20=a+12b from 32=a+16b to get 12=4b, so b=3. Then a=20−36=−16, and P=−16+3×28=68.

D: 5040

HANDIER has seven different letters. All seven can be arranged in 7!=7×6×5×4×3×2×1=5040 orders.

C: 35.5

The modal class is 30–39 with frequency 12. Its lower boundary is 29.5, class width 10, and neighbouring frequencies are 6 and 8. The grouped-mode estimate is 29.5+[(12−6)/(2×12−6−8)]×10=35.5.

B: −5

Write 1/2 as 2⁻¹ and 64 as 2⁶. Then 2^(1−t)=2⁶, so 1−t=6 and t=−5.

B: 312₄

Divide by powers of four: 54=3×16+1×4+2. The digits are therefore 3, 1 and 2, giving 312₄.

A: x³+x²−5x+C

Integrate each term: 3x² becomes x³, 2x becomes x² and −5 becomes −5x. Include the constant C.

A: 2×10^(−1/2)

Write 0.16 as 16×10⁻². Taking its positive fourth root gives 16^(1/4)×10^(−2/4)=2×10^(−1/2).

D: −4≤x≤1

Factor x²+3x−4 as (x+4)(x−1). The product is zero at −4 and 1, and the upward-opening quadratic is non-positive between those roots, including both endpoints: −4≤x≤1.

D: 6

Simplify each radical: √75=5√3, √12=2√3 and √27=3√3. Thus the left side is (5−2+3)√3=6√3, so y=6.

A: √(y²−x²)/x

For an acute angle, take adjacent side x and hypotenuse y. Pythagoras gives opposite side √(y²−x²), so tan θ=opposite/adjacent=√(y²−x²)/x.

A: [[7, −11], [4, −1]]

Double each entry of Q to get rows (6, −14) and (2, 4). Add corresponding entries of P: (1+6, 3−14)=(7, −11) and (2+2, −5+4)=(4, −1).

A: 60°

The total is 16+8+24+21+27=96 people. Doctors are 16/96=1/6 of the group, so their sector angle is (1/6)×360°=60°.

D: Q∩(P∪R)'

A point must be in Q, but in neither P nor R. Being in neither set means lying outside their union, (P∪R)'. Intersect that complement with Q.

B: 58 kg

The frequencies total x+1+2x+2+3=3x+6=12, so x=2. They are then 2, 1, 4, 2 and 3; the highest frequency, 4, belongs to 58 kg.

C: 7

Substitute a=1 and b=3 into the definition: 1*3=(1×3)+1+3=7. The expression uses the defined operation, not ordinary multiplication alone.

B: 30.5

The largest frequency is 10, so the modal class is 21–30. For integer-valued classes, the continuous boundary between 30 and the next class at 31 is halfway: 30.5.

D: 200/√3 km

The first leg moves 200cos60°=100 km north. Because R is due east of P, the second leg must move 100 km south. If QR=d, its southerly component is d cos30°=d√3/2, so d=200/√3 km.

B: √48/11 and √48 cm

Pythagoras gives XY=√(13²−11²)=√48 cm. At angle X, adjacent=XY and opposite=YZ=11 cm, so cot θ=XY/YZ=√48/11.