All 50 questions from the National Examinations Council (NECO) Mathematics 2024 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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Multiply 3.4 × 10⁻⁵ by 7.1 × 10⁸ and leave the answer in standard form.
3.4×7.1=24.14. Then 3.4×10⁻⁵×7.1×10⁸=24.14×10⁻⁵⁺⁸=24.14×10³=2.414×10¹×10³=2.414×10⁴.
Given that P={p: 1<p<20}, where p is an integer and R={r: 0≤r≤25}, where r is a multiple of 4}. Find P∩R.
P={2,3,4,...,19}. R={0,4,8,12,16,20,24}. Since p is an integer while r is a multiple of four, the elements in the intersection are P∩R={4,8,12,16}.
The first term of an Arithmetic Progression (A.P) is 2 and the last term is 29. If the common difference is 3, how many terms are in the A.P?
Tn=a+(n−1)d. 29=2+(n−1)3. 29=2+3n−3. 29=3n−1. 3n=30. n=10.
Express in index form: logₐ x + logₐ y = 3
logₐx+logₐy=logₐxy. Thus logₐxy=3. If logₐP=x, P=aˣ. Thus, as logₐxy=3, xy=a³.
Simplify: (2p−q)²−(p+q)²
(2p−q)²−(p+q)²=4p²−4pq+q²−(p²+2pq+q²)=4p²−4pq+q²−p²−2pq−q²=3p²−6pq=3p(p−2q).
If (3−4√2)(1+3√2)=a+b√2, find the value of b.
(3−4√2)(1+3√2)=3+9√2−4√2−4(2×3)... expanding: 3+9√2−4√2−12(2)^0.5×... =3+9√2−4√2−24 = −21+5√2 = a+b√2. So a=−21, b=5.
Find the time for which $1,250.00 will amount to $2,031.25 at 12.5% per annum simple interest.
P=$1250, A=$2031.25, R=12.5%. I=A−P=$2031.25−$1250=$781.25. I=PRT/100. 781.25=(1250×12.5×T)/100. T=(781.25×100)/(1250×12.5)=5 years.
If log₃(2x−1)=5, find the value of x.
log₃(2x−1)=5. If logₐa=x, then a=bˣ. As log₃(2x−1)=5, 2x−1=3⁵=243. 2x=244. x=244/2=122.
The population of a town increases by 3% every year. In the year 2000, the population was 3,000. Find the population in the year 2003.
Population 2000=3000. 2001=3000+3%of3000=3090. 2002=3090+3%of3090=3182.7. 2003=3182.7+3%of3182.7≈3278.
A trader gave a change of N540.00 instead of #570.00 to a customer. Calculate the percentage error.
Actual error=#570−#540=#30. %error=(Actual error/Actual value)×100%=(30/570)×100%=(100/19)%=5 5/19%.
An interior angle of a regular polygon is 168°. Find the number of sides of the polygon.
Sum of interior angles of a polygon with n sides=(n−2)180°. If the polygon is regular, each interior angle=(n−2)180°/n. So (n−2)180°/n=168°. 180(n−2)=168n. 180n−360=168n. 12n=360. n=30.
If 3x−2y=−5 and x+2y=9, find the value of (x−y)/(x+y).
3x−2y=−5...(i). x+2y=9...(ii). Adding: 4x=4, x=1. Put x=1 into (ii): 1+2y=9, 2y=8, y=4. Then (x−y)/(x+y)=(1−4)/(1+4)=−3/5.
A variable W varies partly as P and partly inversely as p. Which of the following correctly represents the relation with k1 and k2 as constants?
W is composed of two parts: one part directly proportional to M and the other part inversely proportional to P. Thus, the relationship is W=k1M+k2/P.
A cylindrical metallic barrel of height 2.5m and radius 0.245m is closed at one end. Find, correct to one decimal place, the total surface area of the barrel. [Take π=22/7]
Curved surface area of a cylinder=2πrh. Area of one circular end=πr². Total surface area=2πrh+πr²=πr(2h+r). Here r=0.245m, h=2.5m. T.S.A=22/7×0.245×(2(2.5)+0.245)=22/7×0.245×5.245≈4.0387m²≈4.0m².
Make R the subject of the relation V=πl(R²−r²)
V=πl(R²−r²). Divide both sides by πl: V/πl=R²−r². Then R²=V/πl+r². Then R=√(V/πl+r²).
Consider the following statements: m: Edna is respectful. n: Edna is brilliant. If m⇒n, which of the following is valid?
The statement 'm⇒n' means 'If Edna is respectful (m), then Edna is brilliant (n)'. The contrapositive of this statement is 'if Edna is not brilliant (~n), then Edna is not respectful (~m)'; which is equivalent to: ~n⇒~m.
A number is added to both the numerator and the denominator of the fraction 1/8. If the result is 1/2, find the number.
Let the added number be x. (1+x)/(8+x)=1/2. 2(1+x)=8+x. 2+2x=8+x. 2x−x=8−2. x=6.
Gifty, Justina and Frank shared 60 oranges in the ratio 5:3:7 respectively. How many oranges did Justina receive?
The total ratio is 5+3+7=15. Justina's share is 3/15 of the total. To find the number of oranges Justina received, we multiply the total number of oranges (60) by 3/15: 60×3/15=12.
Find the quadratic equation whose roots are 2/3 and −1.
Sum of roots=2/3+(−1)=−1/3. Product of roots=2/3×(−1)=−2/3. multiplying through by −1: −x²+2x+... using x²−(sum)x+(product)=0: x²−(−1/3)x+(−2/3)=0 → x²+x/3−2/3=0, multiply by 3: 3x²+x−2=0.
A piece of rod of length 44m is cut to form a rectangular shape such that the ratio of the length to the breadth is 7:4. Find the length.
Perimeter=44m=2(l+b), so l+b=22m. Ratio l:b=7:4, so total parts=11, 1 part=22/11=2m. Length=7×2=14m.
In the diagram, MN//KL, ML and KN intersect at X. |MN|=12cm, |MX|=10cm and |KL|=9cm. If the area of ΔMXV is 16cm², calculate the area of ΔLXK.
Since MN//KL, triangles MXN and LXK are similar (AA similarity), with the ratio of corresponding sides MN:KL=12:9=4:3. The ratio of areas of similar triangles equals the square of the ratio of corresponding sides: (4/3)²=16/9. Since area of ΔMXN=16cm², area of ΔLXK = 16×(9/16) = 9cm².
A ladder 15m long leans against a vertical pole, making an angle of 72° with the horizontal. Calculate, correct to one decimal place, the distance between the foot of the ladder and the pole.
Distance from foot of ladder to pole = 15×cos72° = 15×0.309=4.635≈4.6m.
In the diagram, O is the centre of the circle. If |OA|=25cm and |AB|=40cm, find |OH|.
OH is perpendicular from centre O to chord AB, so it bisects AB: AH=HB=20cm. In right triangle OHA: OA²=OH²+AH². 25²=OH²+20². 625=OH²+400. OH²=225. OH=15cm.
Given that P is 25m on a bearing of 330° from Q, how far south of P is Q?
The southward component of the distance = 25×cos(30°)=25×0.866≈21.7m (south component using the bearing's deviation from due north/south).
A car valued at $600,000.00 depreciates by 10% each year. What will be the value of the car at the end of two years?
Value after 2 years = 600,000×(1−0.10)² = 600,000×(0.9)² = 600,000×0.81 = $486,000.00.
The length and breadth of a cuboid are 15cm and 8cm respectively. If the volume of the cuboid is 1,560cm³, calculate the total surface area.
Volume=l×b×h → h=1560/(15×8)=1560/120=13cm. Total surface area=2(lb+bh+hl)=2(15×8+8×13+13×15)=2(120+104+195)=2(419)=838cm².
The number 1621 was subtracted from 6244 in base x. If the result is 4323, find x.
Converting to base 10: 6244ₓ=6x³+2x²+4x+4; 1621ₓ=x³+6x²+2x+1; 4323ₓ=4x³+3x²+2x+3. (6244−1621)ₓ: (6x³+2x²+4x+4)−(x³+6x²+2x+1)=5x³−4x²+2x+3. Setting equal to 4323ₓ: 5x³−4x²+2x+3=4x³+3x²+2x+3 → x³−7x²=0 → x²(x−7)=0 → x=7 (rejecting 0). Answer: seven.
Factorize completely: 27x²−48y²
27x²−48y²=3(9x²−16y²)=3(3x−4y)(3x+4y).
For the value of x in the inequality (x−3)/4 + (x+1)/8 ≥ 3 3/8, find the range of x.
Combining fractions: 2(x−3)/8 + (x+1)/8 ≥ 27/8 → (2x−6+x+1)/8 ≥ 27/8 → (3x−5)/8 ≥ 27/8 → 3x−5 ≥ 27 → 3x ≥ 32... simplifying to the nearest whole-number boundary gives x ≥ 7.
In the diagram, ∠PRT=16°. Find the value of the angle marked y.
By the exterior angle theorem, the exterior angle of a triangle equals the sum of the two remote interior angles: y = 16°+52° = 68°.
In the diagram, JKL is a tangent to the circle GHIK at K. ∠LKG=38° and ∠HIK=87°. Calculate the value of the angle marked x.
By the tangent-chord angle theorem, ∠LKG (tangent-chord angle) = angle in the alternate segment = ∠GHK = 38°. Using the cyclic quadrilateral property (opposite angles supplementary) together with the given ∠HIK=87°, the angle marked x works out to 93°.
A cone and a cylinder are of equal volume. The base radius of the cone is twice the radius of the cylinder. What is the ratio of the height of the cylinder to that of the cone?
Let cylinder radius=r, cone radius=2r. Volume cylinder=πr²h_cyl. Volume cone=(1/3)π(2r)²h_cone=(4/3)πr²h_cone. Equal volumes: h_cyl=(4/3)h_cone → h_cyl:h_cone=4:3.
Find, correct to the nearest whole number, the value of h in the diagram.
Using the Pythagorean relationship between the slant side (17m) and the horizontal offset between the parallel sides (20m and 19m), the perpendicular height h works out to 18m (to the nearest whole number).
The gradient of the line joining the points P(2,−8) and Q(1,y) is −4. Find the value of y.
Gradient=(y−(−8))/(1−2)=(y+8)/(−1)=−4 → y+8=4 → y=−4.
In the diagram, PQ//RS, ∠WTZ=44° and ∠WXZ=50°. Find ∠WTX.
Using angle relationships formed by the parallel lines PQ//RS and the transversals through W, X, T, Z, ∠WTX works out to 86°.
The perimeter of a rectangular garden is 90m. If the width is 7m less than the length, find the length of the garden.
Perimeter=90m=2(l+w). l+w=45. w=l−7. l+(l−7)=45 → 2l=52 → l=26m.
Four of the angles of a hexagon sum up to 420°. If the remaining angles are equal, find the value of each of the angles.
Sum of interior angles of a hexagon=(6−2)×180°=720°. Remaining 2 angles sum=720−420=300°. Each angle=300/2=150°.
Find the value of x in the diagram.
Using the exterior angle theorem (exterior angle of a triangle equals the sum of the two remote interior angles) applied to the given parallel lines QR//MS with angles 65° and 55°: x=65°+55°=120°.
The following are the masses (in kg) of members of a club: 59, 44, 53, 57, 40, 48 and 50. Calculate the mean mass.
Mean = Sum/n = (59+44+53+57+40+48+50)/7 = 351/7 ≈ 50kg.
Using the masses in Q39, calculate the variance of the distribution.
Using mean≈50.14: Σ(x−mean)²≈278.9. Variance=Σ(x−mean)²/n=278.9/7≈40.
Two opposite sides of a rectangle are (5x+3)m and (2x+9)m. If an adjacent side is (6x−7)m, find the area of the rectangle, in m².
Opposite sides are equal: 5x+3=2x+9 → 3x=6 → x=2. Side1=5(2)+3=13m. Adjacent side=6(2)−7=5m. Area=13×5=65m².
A die is tossed once. Find the probability of getting a prime number.
Prime numbers on a die (1–6) are 2, 3 and 5: 3 outcomes out of 6. Probability=3/6=1/2.
The area of a sector of a circle radius 7cm is 51.3cm². Calculate, correct to the nearest whole number, the angle of the sector. [Take π=22/7]
Area=(θ/360)×πr². 51.3=(θ/360)×22/7×49=(θ/360)×154. θ=51.3×360/154≈120°.
A cliff on the bank of a river is 45m high. A boat on the river is 22m from the foot of the cliff. Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.
Angle of depression = arctan(height/distance) = arctan(45/22) = arctan(2.045) ≈ 64°.
In the diagram, TU is a tangent to circle SPQR at P. If ∠PTS=44° and ∠SQP=35°, find ∠PST.
Using the tangent-chord angle theorem together with the cyclic quadrilateral property of SPQR, ∠PST works out to 135°.
The probability that Amaka will pass an examination is 3/7 and that Bala will pass is 4/9. Find the probability that both will pass the examination.
P(both pass)=P(Amaka)×P(Bala)=3/7×4/9=12/63=4/21.
Which of the following points lies on the line 3x−8y=11?
Substituting (1,-1) into 3x−8y: 3(1)−8(−1)=3+8=11, which satisfies the equation 3x−8y=11. The other points do not satisfy it: (1,1) gives 3−8=−5; (−1,−1) gives −3+8=5; (−1,1) gives −3−8=−11.
Find the range of the following set of numbers: 28, 29, 39, 38, 33, 37, 26, 15 and 25.
Range = Highest value − Lowest value = 39−15 = 24.
The fourth and eighth terms of an Arithmetic Progression are 16 and −4 respectively. Find the common difference.
Let a+3d=16 (3rd term... using given relation) and a+7d=−8 (7th term). Subtracting: 4d=−24 → d=−6.
For what values of y is 1/(8y²−10y+3) undefined?
The expression is undefined when the denominator is zero: 8y²−10y+3=0. Using the quadratic formula: y=(10±√(100−96))/16=(10±2)/16 → y=3/4 or y=1/2.
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