JAMB Mathematics 2024 practice collection

C: 30

Expand along the first row: 3(5+3) − 4(10−1) + 6(6+1) = 24 − 36 + 42 = 30.

D: ₦230,000

Simple interest is ₦200,000 × 0.05 × 3 = ₦30,000. Add the principal to get the total amount of ₦230,000.

C: 1/5

For inverse variation, xy is constant. The first pair gives xy=(1/2)×4=2, so with y=10, x=2/10=1/5.

D: 135°

The circumference is πd=(22/7)×28=88 cm. The arc is 33/88 of a full circle, so its central angle is (33/88)×360°=135°.

A: −1/2

Write both sides as powers of 2: 16^x=2^(4x), while 0.25=1/4=2^(−2). Hence 4x=−2 and x=−1/2.

B: 5

The original committee has 15 men and 5 women. Keeping 15 men, a 3:2 ratio needs 10 women, so 5 more women must join.

A: 39

The operation gives p*2=4p+2 and p*1=3p+1. Their difference is p+1, so p+1=40 and p=39.

D: Mean

The mean describes the centre of a data set. Standard deviation, range and variance describe its spread rather than its central tendency.

A: 9 cm

In right triangle OTQ, OQ is the 9 cm hypotenuse and TQ is opposite the 30° angle. Thus TQ=9 sin 30°=4.5 cm. The perpendicular from the centre bisects a chord, so PQ=2×TQ=9 cm.

A: x>1

Every shaded point has an x-coordinate greater than 1 because it lies to the right of x=1. The boundary is dashed, so points with x=1 are excluded. There is no separate restriction on y.

A: 120

Fix one person's seat to remove rotational duplicates, then arrange the remaining five people in 5!=120 ways.

A: 3x+4y<12, x≥0, y≥0

The shading lies below 3x+4y=12, so the expression is less than 12. The diagonal is dashed and therefore excluded. The solid coordinate axes are included, giving x≥0 and y≥0.

A: [[4,4,−1],[−1,7,−2]]

Double each entry of B to obtain [[2,8,−4],[−6,6,−2]], then add corresponding entries of A. The result is [[4,4,−1],[−1,7,−2]].

A: 150

Because |−1/2|<1, the infinite-sum formula applies: S=a/(1−r). Rearrange to get a=S(1−r)=100[1−(−1/2)]=150.

A: 16√3 cm²

Equal sides PQ and QR give equal opposite angles at R and P. Since angle P is 60°, the triangle is equilateral. Its area is (1/2)×8×8×sin 60°=16√3 cm².

A: 15√3 m

The fruit at the tree top is 25−10=15 m above the observer. With tan30°=15/d=1/√3, the horizontal distance is d=15√3 m.

C: ₦10.80

Each orange costs ₦24/5=₦4.80. Four cost ₦19.20 and sell for ₦30, giving profit ₦30−₦19.20=₦10.80.

A: 20(5x+1)^3

By the chain rule, differentiate the outer fourth power and multiply by the inner derivative 5: dy/dx=4(5x+1)^3×5=20(5x+1)^3.

B: 4−√7

Multiply numerator and denominator by 3−√7. The numerator becomes 8−2√7 and the denominator becomes 9−7=2, giving 4−√7.

C: 3+√10+√13

The horizontal base is 3. The other two lengths are √[(4−3)^2+(5−2)^2]=√10 and √[(6−4)^2+(2−5)^2]=√13. Add them.

C: −156250

The common ratio is −5. The seventh term is a₇=a₁r^6=−10(−5)^6=−10×15625=−156250.

B: 3

Count the occurrences: 3 appears three times, while 2 and 5 each appear twice and the other scores once. Thus the mode is 3.

C: 8

The midpoint x-coordinate is (x+10)/2=9. Hence x+10=18 and x=8; the y-coordinate (4+8)/2=6 also agrees.

A: 3

Convert to powers of 2: the left side is 2^[4(2+x)−2]=2^(6+4x), while the right side is 2^(6x). Therefore 6+4x=6x and x=3.

D: {1,2,9}

The elements only in P form {2}; the elements only in Q form {1,9}. Their union is {1,2,9}.

B: 23¾

The whole-number part is 16+4+2+1=23. The fractional part is 1/2+1/4=3/4, so the result is 23¾.

B: 2456₇

Convert to base ten: 20045₇=4835 and 14256₇=3912. The difference is 923, which is 2456₇.

D: 30 years

The total age is 4×40+8×25=360 years. Divide by all 12 teachers to get an average of 30 years.

C: 32

Each of five elements can either be included or excluded, giving 2⁵=32 possible subsets.

B: 17

The seven values must total 7×22=154. Their known values total 103, so 103+3x=154 and x=17.

C: 152

Variance is the square of the standard deviation: 12.34²=152.2756, which rounds to 152.

A: √2

The mean is 6. The squared deviations sum to 1+4+0+1+4=10; dividing by five gives variance 2 and standard deviation √2.

A: 350

Choose the girls in C(5,2)=10 ways and the boys in C(7,3)=35 ways. Multiply to get 350 committees.

A: 105

Choose the one woman in C(3,1)=3 ways and the remaining four members from seven men in C(7,4)=35 ways. Multiplying gives 3×35=105 committees.

B: 65/121

Both red has probability (7/11)² and both black (4/11)². Add them to get 49/121+16/121=65/121.

B: 4

Write U=k√V. The first pair gives k=24/3=8. Then 16=8√V, so √V=2 and V=4.

B: 6

Use B=k/∛C. Since ∛27=3 and B=2, multiply both sides of 2=k/3 by 3 to obtain k=6.

C: 5, −3, and −2

The polynomial factors as (x−5)(x+3)(x+2): multiplying gives x³−19x−30. Setting each factor to zero gives the three roots 5, −3 and −2.

B: x=36/y

Inverse variation has the form x=k/y, or xy=k. Substituting x=9 and y=4 gives k=36, so the required relation is x=36/y for nonzero y.

B: [[0,0],[0,0]]

Matrix multiplication gives P²=[[5,8],[8,13]], while 4P=[[4,8],[8,12]]. Their difference is I, and subtracting I leaves the zero matrix [[0,0],[0,0]].

D: [[4,8],[2,3],[4,1]]

Transposing A gives [[3,4],[2,2],[1,−1]]. Add the corresponding entries of B to obtain [[4,8],[2,3],[4,1]].

B: 119/72

For an acute angle, an 8–15–17 right triangle gives sin θ=8/17 and cos θ=15/17. The numerator is −7/17 and the denominator is 64/289−8/17=−72/289. Dividing gives 119/72.

C: 0 and 1

By the operation rule, x*2=x². Thus x²=x, so x(x−1)=0. The only real solutions are x=0 and x=1; both satisfy the equation.

C: 42 kg

Count the occurrences in the listed data: 42 appears three times, 30 twice, and every other value once. The most frequent weight is therefore 42 kg.

B: 24

Sum all seven disjoint regions: 144=(42−x)+34+(30−x)+(6+x)+4x+3x+2x=112+8x, so x=4. Both M and P includes their pair-only region and the three-way region: 4x+2x=24.

A: q=rt²/(p−r³)

Square both sides: t²=pq/r−r²q=q(p−r³)/r. Multiply by r and divide by p−r³ to get q=rt²/(p−r³), with the stated nonzero denominators.

C: a=6, b=24

Expand the square: a+8a+2√(8a²)=9a+4a√2 for positive a. Matching rational and irrational parts gives 9a=54, so a=6, and then b=4a=24.