WAEC Mathematics 2012 Objective Past Questions

All 50 questions from the West African Examinations Council (WAEC) Mathematics 2012 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

Mathematics 2012 Objective — Question 1

Express 302.10495 correct to five significant figures.

• A. 302.10Correct
• B. 302.11
• C. 302.105
• D. 302.1049

Explanation

302.10495 correct to 5 significant figures = 302.10.

Mathematics 2012 Objective — Question 2

Simplify: (3√5×4√6)/(2√2×3√3)

• A. √2
• B. √5
• C. 2√2
• D. 2√5Correct

Explanation

(3√5×4√6)/(2√2×3√3) = 12√30/6√6 = 2√(30/6) = 2√5.

Mathematics 2012 Objective — Question 3

In 1995, the enrollments of two schools X and Y were 1,050 and 1,190 respectively. Find the ratio of the enrollments of X to Y.

• A. 50:11
• B. 15:17Correct
• C. 13:55
• D. 12:11

Explanation

Ratio 1050:1190; divide through by 70 = 15:17.

Mathematics 2012 Objective — Question 4

Convert 35(10) to a number in base 2.

• A. 1011
• B. 10011
• C. 100011Correct
• D. 11001

Explanation

35 in base 10 = 100011 in base 2.

Mathematics 2012 Objective — Question 5

The nth term of a sequence is Tn = 5 + (n−1)². Evaluate T4 − T6.

• A. 30
• B. 16
• C. -16Correct
• D. -30

Explanation

T4 = 5+(4−1)² = 5+9 = 14. T6 = 5+(6−1)² = 5+25 = 30. T4−T6 = 14−30 = −16.

Mathematics 2012 Objective — Question 6

Mr. Manu travelled from Accra to Pamfokrom, a distance of 720km in 8 hours. What will his speed be in m/s?

• A. 25m/sCorrect
• B. 150m/s
• C. 250m/s
• D. 500m/s

Explanation

Distance in m = 720×1000 = 720,000m. Time in s = 8×3600 = 28,800s. Speed = distance/time = 720,000/28,800 = 25m/s.

Mathematics 2012 Objective — Question 7

If ₦2,500.00 amounted to ₦3,500.00 in 4 years at simple interest, find the rate at which the interest was charged.

• A. 5%
• B. 7½%
• C. 8%
• D. 10%Correct

Explanation

I = PRT/100. I=1000, P=2500, T=4. 1000 = (2500×R×4)/100 = 100R. R = 10%.

Mathematics 2012 Objective — Question 8

Solve for x in the equation: 1/x + 2/3x = 1/3

• A. 5Correct
• B. 4
• C. 3
• D. 1

Explanation

1/x + 2/(3x) = 1/3. Combine: (3+2)/(3x) = 1/3, so 5/(3x)=1/3, 15=3x, x=5.

Mathematics 2012 Objective — Question 9

Simplify: (54k²−6)/(3k+1)

• A. 6(1−3k²)
• B. 6(3k²−1)
• C. 6(3k−1)Correct
• D. 6(1−3k)

Explanation

54k²−6 = 6(9k²−1) = 6(3k−1)(3k+1). Dividing by (3k+1): result = 6(3k−1).

Mathematics 2012 Objective — Question 10

Represent the inequality −7<4x+9≤13 on a number line.

• A. Option A
• B. Option BCorrect
• C. Option C
• D. Option D

Explanation

−7−9<4x≤13−9 ⇒ −16<4x≤4 ⇒ −4<x≤1. This range (open at -4, closed at 1) is represented by option B.

Mathematics 2012 Objective — Question 11

Make p the subject of the relationship: q = 3p/r + s/2

• A. p=(2q−rs)/6
• B. p=2qr−sr−3
• C. p=(2qr−s)/6
• D. p=(2qr−rs)/6Correct

Explanation

q = 3p/r + s/2. Multiply through by 2r: 2qr = 6p + rs. So 6p = 2qr−rs, p = (2qr−rs)/6.

Mathematics 2012 Objective — Question 12

If x+y=2y−x+1=5, find the value of x.

• A. 3
• B. 2Correct
• C. 1
• D. -1

Explanation

From x+y=5 (i) and 2y−x+1=5 i.e. 2y−x=4 (ii). Multiply (i) by 2: 2x+2y=10 (iii). Subtract (ii) from (iii): 3x=6, x=2.

Mathematics 2012 Objective — Question 13

The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation.

• A. 12n−6=0
• B. 3n−12=0
• C. 2n−35=0
• D. 5n−33=0Correct

Explanation

12 + n/3 = 2n+1. Multiply by 3: 36+n = 6n+3. 33 = 5n. So 5n−33=0.

Mathematics 2012 Objective — Question 14

Solve the inequality: −m/2 − 5/4 ≤ 5m/12 − 7/6

• A. m≥5/4
• B. m≤5/4
• C. m≥−1/11Correct
• D. m≤−1/11

Explanation

Multiplying through by 12 and collecting like terms: −11m≤1, so m≥−1/11 (inequality reverses when dividing by −11).

Mathematics 2012 Objective — Question 15

The curved surface area of a cylindrical tin is 704cm². If the radius of its base is 8cm, find the height. [Take π=22/7]

• A. 14cmCorrect
• B. 9cm
• C. 8cm
• D. 7cm

Explanation

Curved surface area = 2πrh. 704 = 2×(22/7)×8×h. h = 14cm.

Mathematics 2012 Objective — Question 16

The lengths of the minor and major arcs of a circle are 54cm and 126cm respectively. Calculate the angle of the major sector.

• A. 306°
• B. 252°Correct
• C. 246°
• D. 234°

Explanation

Total arc corresponds to 360°. Angle of major sector = (126/(54+126))×360° = (126/180)×360° = 252°.

Mathematics 2012 Objective — Question 17

A sector of a circle which subtends 172° at the centre of the circle has a perimeter of 600cm. Find, correct to the nearest cm, the radius of the circle. [Take π=22/7]

• A. 120cm
• B. 116cm
• C. 107cmCorrect
• D. 100cm

Explanation

Perimeter of sector = arc length + 2r. Arc length = (172/360)×2πr. 600 = (172/360)×2×(22/7)×r + 2r, solving gives r ≈ 107cm.

Mathematics 2012 Objective — Question 18

In the diagram above, |QR|=10m, |SR|=8m, ∠QPS=30°, ∠QRP=90° and |PS|=x. Find x.

• A. 1.32m
• B. 6.32m
• C. 9.32mCorrect
• D. 17.32m

Explanation

tan30° = 10/(8+x). 8+x = 10/tan30° = 17.32. x = 17.32−8 = 9.32m.

Mathematics 2012 Objective — Question 19

In ΔXYZ, |XY|=8cm, |YZ|=10cm and |XZ|=6cm. Which of these relations is true?

• A. |XY|+|YZ|=|XZ|
• B. |XY|−|YZ|=|XZ|
• C. |XZ|²=|YZ|²−|XY|²Correct
• D. |YZ|²=|XZ|²−|XY|²

Explanation

Since 6²+8²=36+64=100=10², the triangle is right-angled at X, so |XZ|² = |YZ|²−|XY|² (Pythagoras: hypotenuse² = sum of squares of the other two, rearranged).

Mathematics 2012 Objective — Question 20

In the diagram above, O is the centre of the circle PQRS and ∠PSR=86°. If ∠POR=x°, find x°.

• A. 274°
• B. 172°Correct
• C. 129°
• D. 86°

Explanation

The angle at the centre is twice the angle at the circumference subtended by the same arc: x° = 2×86° = 172°.

Mathematics 2012 Objective — Question 21

The diagram above is a circle centre O. If ∠SPR=2m° and ∠SQR=n°, express m° in terms of n°.

• A. m=n/2Correct
• B. m=2n
• C. m=n−2
• D. m=n+2

Explanation

∠SPR and ∠SQR are angles subtended by the same arc SR at the circumference, so they are equal: 2m°=n°, giving m=n/2.

Mathematics 2012 Objective — Question 22

In the diagram above, MQ//RS, ∠TUV=70° and ∠RLV=30°. Find the value of x°.

• A. 150°Correct
• B. 110°
• C. 100°
• D. 95°

Explanation

Using the properties of parallel lines and the given angles (alternate/corresponding angles and angles on a straight line), x° is found to be 150°.

Mathematics 2012 Objective — Question 23

In the diagram, MN, PQ and RS are three intersecting straight lines, which of the following statement(s) are true? I. t°=y° II. x°+y°+z°+m°=180° III. x°+m°+n°=180° IV. x°+n°=m°+z°

• A. I and IV onlyCorrect
• B. II only
• C. III only
• D. IV only

Explanation

Using vertically opposite angles (giving t°=y°, statement I) and angle relationships around the point of intersection, statement IV also holds; the other statements do not hold generally. So I and IV only are true.

Mathematics 2012 Objective — Question 24

If cos(x+40)°=0.0872, what is the value of x?

• A. 85°
• B. 75°
• C. 65°
• D. 45°Correct

Explanation

cos⁻¹0.0872 = 85°. So x+40=85, x=45°.

Mathematics 2012 Objective — Question 25

A kite flies on a taut string of length 50m inclined at an angle of 54° to the horizontal ground. The height of the kite above the ground is

• A. 50tan36°
• B. 50sin54°Correct
• C. 50tan54°
• D. 50sin36°

Explanation

Height = string length × sin(angle of inclination) = 50 sin54°.

Mathematics 2012 Objective — Question 26

The positions of three ships P, Q and R at sea are illustrated in the diagram. The arrows indicate the North direction. The bearing of Q from P is 050° and angle PQR=72°. Calculate the bearing of R from Q.

• A. 130°
• B. 158°Correct
• C. 222°
• D. 252°

Explanation

Bearing of R from Q = 180°−22° = 158° (using the angle relationships formed by the bearing of Q from P and the given angle PQR).

Mathematics 2012 Objective — Question 27

Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21. Calculate the standard deviation of the scores.

• A. √10
• B. 4
• C. 5Correct
• D. √30

Explanation

Standard deviation = √(Σ(x−x̄)²/n) = √[((15−21)²+(21−21)²+(17−21)²+(26−21)²+(18−21)²+(29−21)²)/6] = √[(36+0+16+25+9+64)/6] = √(150/6) = √25 = 5.

Mathematics 2012 Objective — Question 28

A bag contains 4 red and 6 black balls of the same size. The balls are shuffled briskly and two balls are drawn one after the other without replacement. Find the probability of picking balls of different colours.

• A. 8/15Correct
• B. 13/25
• C. 11/15
• D. 13/15

Explanation

P(different colours) = P(R then B) + P(B then R) = (4/10×6/9)+(6/10×4/9) = 24/90+24/90 = 48/90 = 8/15.

Mathematics 2012 Objective — Question 29

Use this bar chart, showing the frequency distribution of marks scored by students in a class test, to answer questions 29 to 31. How many students are in the class?

• A. 10
• B. 24
• C. 25Correct
• D. 30

Explanation

Total number of students = sum of frequencies from the bar chart = 25.

Mathematics 2012 Objective — Question 30

(Using the same bar chart as Q29) Calculate the mean of the distribution.

• A. 6.0
• B. 3.0
• C. 2.4Correct
• D. 1.8

Explanation

Mean = Σfx/Σf, computed from the bar chart's frequency distribution = 2.4.

Mathematics 2012 Objective — Question 31

(Using the same bar chart as Q29) What is the median of the distribution?

• A. 2Correct
• B. 4
• C. 6
• D. 8

Explanation

The median, found from the cumulative frequency position of the distribution shown in the bar chart, is 2.

Mathematics 2012 Objective — Question 32

Which of these statements about y=8√m is correct?

• A. log y = log8 × log√m
• B. log y = 3log2 × ½logm
• C. log y = 3log2 − ½logm
• D. log y = 3log2 + ½logmCorrect

Explanation

y=8√m = 2³×m^(1/2). Taking log of both sides: log y = log2³ + logm^(1/2) = 3log2 + ½logm.

Mathematics 2012 Objective — Question 33

If x+0.4y=3 and y=½x, find the value of (x+y).

• A. 1¼
• B. 2½
• C. 3¾Correct
• D. 5

Explanation

Substituting y=½x into x+0.4y=3: x+0.4(½x)=3 ⇒ x+0.2x=3 ⇒ 1.2x=3 ⇒ x=2.5. Then y=½(2.5)=1.25. x+y = 2.5+1.25 = 3.75 = 3¾.

Mathematics 2012 Objective — Question 34

Express 3−[(x−y)/y] as a single fraction.

• A. 3xy/y
• B. (x−4y)/y
• C. (4y+x)/y
• D. (4y−x)/yCorrect

Explanation

3−(x−y)/y = (3y−(x−y))/y = (3y−x+y)/y = (4y−x)/y.

Mathematics 2012 Objective — Question 35

Find the coefficient of m in the expansion of (m/2−1½)(m+⅔).

• A. -1/6
• B. -1/2
• C. -1
• D. -1⅙Correct

Explanation

Expanding: (m/2)(m) + (m/2)(⅔) − 1½(m) − 1½(⅔). The coefficient of m comes from (m/2)(⅔) − 1½(m) = m/3 − 3m/2 = (2m−9m)/6 = −7m/6, giving coefficient −1⅙ (i.e. −7/6).

Mathematics 2012 Objective — Question 36

In the diagram above, MN//PO, ∠PMN=112°, ∠PNO=129°, ∠NOP=37° and ∠MPN=y°. Find the value of y°.

• A. 51°
• B. 54°Correct
• C. 56°
• D. 68°

Explanation

Using angle sum properties of the quadrilateral/parallel lines in the figure, y° = 54°.

Mathematics 2012 Objective — Question 37

If P={prime factors of 210} and Q={prime numbers less than 10}, find P∩Q.

• A. {1,2,3}
• B. {2,3,5}
• C. {1,3,5,7}
• D. {2,3,5,7}Correct

Explanation

210 = 2×3×5×7, so P={2,3,5,7}. Q={2,3,5,7} (primes less than 10). P∩Q = {2,3,5,7}.

Mathematics 2012 Objective — Question 38

Alfred spent 1/4 of his money on food, 1/3 on clothing and saved the rest. If he saved ₦72,000.00, how much did he spend on food?

• A. ₦43,200.00Correct
• B. ₦43,000.00
• C. ₦42,200.00
• D. ₦40,000.00

Explanation

Let total money = M. Savings = M − M/4 − M/3 = 5M/12 = 72,000. M = 72,000×12/5 = 172,800. Amount spent on food = M/4 = 172,800/4 = ₦43,200.

Mathematics 2012 Objective — Question 39

Solve (27/125)^(−1/3) × (4/9)^(1/2)

• A. 10/9Correct
• B. 9/10
• C. 2/5
• D. 12/125

Explanation

(27/125)^(−1/3) = (125/27)^(1/3) = 5/3. (4/9)^(1/2) = 2/3. Product = (5/3)×(2/3) = 10/9.

Mathematics 2012 Objective — Question 40

The sum of the interior angles of a regular polygon is 1800°. How many sides has the polygon?

• A. 16
• B. 12Correct
• C. 10
• D. 8

Explanation

Sum of interior angles = (n−2)×180° = 1800°. n−2 = 10. n = 12 sides.

Mathematics 2012 Objective — Question 41

The diagram is a circle with centre O. PRST are points on the circle. Find the value of ∠PRS.

• A. 144°Correct
• B. 72°
• C. 40°
• D. 36°

Explanation

Using the reflex angle at the centre (2×72°=144°, or equivalent) and the circle theorem relating angle at centre to angle at circumference, ∠PRS = 144°.

Mathematics 2012 Objective — Question 42

The diagram above is a circle of radius |OQ|=4cm. Line /TR/ is a tangent to the circle at R. If ∠TPO=120°, find /PQ/.

• A. 2.32cm
• B. 1.84cm
• C. 0.62cmCorrect
• D. 0.26cm

Explanation

Using sin60° = 4/|PO| (from the tangent-radius right angle and the given angle), |PO| is found, and |PQ| = |PO|−|OQ| ≈ 0.62cm.

Mathematics 2012 Objective — Question 43

If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?

• A. y=kx
• B. y=k/xCorrect
• C. y=k√x
• D. y=x+k

Explanation

An inverse relationship is one where y is proportional to 1/x, i.e. y=k/x.

Mathematics 2012 Objective — Question 44

In the diagram, /SR/=/QR/, ∠SRP=65° and ∠RPQ=48°. Find ∠PRQ.

• A. 65°
• B. 45°
• C. 25°
• D. 19°Correct

Explanation

Since /SR/=/QR/, triangle SRQ (or the relevant sub-triangle) is isosceles; using the given angles 65° and 48° and angle sum properties, ∠PRQ = 19°.

Mathematics 2012 Objective — Question 45

The graph is that of y=2x²−5x−3. Use it to answer questions 45 and 46. For what values of x will y be negative?

• A. -½≤x<3
• B. -½<x≤3
• C. -½<x<3Correct
• D. -½≤x≤3

Explanation

y<0 between the roots of 2x²−5x−3=0, which are x=−½ and x=3 (exclusive, since y=0 exactly at the roots), i.e. −½<x<3.

Mathematics 2012 Objective — Question 46

(Using the same graph as Q45) What is the gradient of y=2x²−5x−3 at the point x=4?

• A. 11Correct
• B. 9
• C. 19
• D. -9

Explanation

dy/dx = 4x−5. At x=4: dy/dx = 16−5 = 11.

Mathematics 2012 Objective — Question 47

The diagram above is a polygon. Find the largest of its interior angles.

• A. 30°
• B. 100°
• C. 120°
• D. 150°Correct

Explanation

Using the sum of exterior angles of a polygon (=360°) and the given angle expressions in the figure, the largest interior angle = 180°−30° = 150°.

Mathematics 2012 Objective — Question 48

The volume of a cuboid is 54cm³. If the length, width and height are in the ratio 2:1:1 respectively, find its total surface area.

• A. 108cm²
• B. 90cm²Correct
• C. 85cm²
• D. 75cm²

Explanation

Let L:B:H = 2:1:1, so L=2H, B=H. Volume = L×B×H = 2H×H×H = 2H³ = 54, H³=27, H=3. So L=6, B=3, H=3. Total surface area = 2LB+2LH+2BH = 2(6×3)+2(6×3)+2(3×3) = 36+36+18 = 90cm².

Mathematics 2012 Objective — Question 49

A side and a diagonal of a rhombus are 10cm and 12cm respectively. Find its area.

• A. 20cm²
• B. 24cm²
• C. 48cm²
• D. 96cm²Correct

Explanation

Half the given diagonal = 6cm. Using Pythagoras with side 10cm: the half of the other diagonal = √(10²−6²) = √64 = 8cm, so the other diagonal = 16cm. Area = ½×d1×d2 = ½×12×16 = 96cm².

Mathematics 2012 Objective — Question 50

Factorise completely: 32x²y−48x³y²

• A. 16x²y(2−3xy²)Correct
• B. 8xy(4x−6x²y²)
• C. 8x²y(4−6xy²)
• D. 16xy(2x−3xy²)

Explanation

The highest common factor of 32x²y and 48x³y² is 16x²y. 32x²y−48x³y² = 16x²y(2−3xy).