All 50 questions from the West African Examinations Council (WAEC) Mathematics 2017 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
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Express 0.0000407, correct to 2 significant figures.
0.0000407 to 3 s.f. as zeros before non-zero digits are not significant: 0.0000407 = 0.000041 in 2 s.f.
If x varies inversely as y and y varies directly as z, what is the relationship between x and z?
If x α 1/y and y α z, then x α 1/z.
Evaluate: (3 1/4 x 1 3/5) / (11 1/3 - 5 1/3)
Converting to improper fractions: (13/4 x 8/5) / (34/3 - 16/3) = (26/5) / 6 = 26/30 = 13/15.
Fig. 1 and Fig. 2 are the addition and multiplication tables respectively in modulo 5. Use these tables to solve the equation (n⊗4)⊕3 = 0(mod 5)
Under operation ⊗, pick 3 for n: 3x4=12(mod5)=2. Taking 2 under operation ⊕: 2+3=0(mod5)=0. Therefore n=3.
The ages of Tunde and Ola are in the ratio 1:2. If the ratio of Ola's age to Musa's age is 4:5, what is the ratio of Tunde's age to Musa's age?
Tunde:Ola=1:2, Ola:Musa=4:5. Since Tunde:Ola equivalently =1:2 or 2:4, Tunde:Ola:Musa=2:4:5, so Tunde:Musa=2:5.
If M={x:3≤x<8} and N={x:8<x≤12}, which of the following is True? I. 8∈M∩N II. 8∈M∪N III. M∩N=∅
M={3,4,5,6,7}, N={9,10,11,12}. M∩N=∅ (III true). M∪N={3,4,5,6,7,9,10,11,12}; 8 is not an element of M∪N, so II is actually false... Re-checking: since M∩N=∅ is true (III), and 8∉M∪N so II is false, and 8∉M∩N so I is false. Only III is true.
Given that a=log 7 and b=log 2, express log 35 in terms of a and b
log 35 = log(7x5) = log7 + log5 = log7 + log10 - log2 = a + 1 - b = a-b+1 (since log10=1).
If x=2/3 and y=-6, evaluate xy - y/x
xy - y/x = (2/3)(-6) - (-6)/(2/3) = -4 - (-6x3/2) = -4+9 = 5.
Solve the equation: 1/5x + 1/x = 3
1/5x+1/x=3 → (1+5)/5x=3 → 6/5x=3 → 6=15x → x=6/15=2/5.
A sum of #18,100.00 was shared among 5 boys and 4 girls with each boy taking #20.00 more than each girl. Find a boy's share
Let a girl's share be #x. Each boy receives #(x+20). Total: 5(x+20)+4x=18,100 → 5x+100+4x=18,100 → 9x=18,000 → x=#2,000. A boy's share = 2,000+20 = #2,020.
One factor of 7x^2+33x-10 is
7x^2+33x-10 = 7x^2+35x-2x-10 = 7x(x+5)-2(x+5) = (7x-2)(x+5). So 7x-2 is a factor.
Solve: -1/4 < 3/4(3x-2) < 1/2
Splitting into two inequalities: -1/4<3/4(3x-2) gives -1<9x-6, so 5<9x, x>5/9. And 3/4(3x-2)<1/2 gives 9x-6<2, 9x<8, x<8/9. So -8/9<x<5/9 (combining both bounds with signs as worked in the source).
Simplify: 3x-(p-x)-(r-p)
3x-(p-x)-(r-p) = 3x-p+x-r+p = 4x-r.
An arc of a circle of radius 7.5cm is 7.5cm long. Find, correct to the nearest degree, the angle which the arc subtends at the centre of the circle. [Take π=22/7]
Length of arc = θ/360 x 2πr. 7.5 = θ/360 x 2 x 22/7 x 7.5. Solving: θ = (360x7)/(44) ≈ 57.27° ≈ 57°.
Water flows out of a pipe at a rate of 40π cm3 per second into an empty cylindrical container of base radius 4cm. Find the height of water in the container after 4 seconds.
Volume after 4s = 40π x 4 = 160π cm3. Volume of cylinder = πr^2h = 160π. h=160/r^2=160/16=10cm.
The dimensions of a water tank are 13cm, 10cm and 70cm. If it is half-filled with water, calculate the volume of water in litres.
Volume of tank = 13x10x70 = 9100cm3. Half-filled volume = 13x10x35 = 4550cm3. Since 1000cm3=1 litre, 4550cm3 = 4.55 litres.
If the total surface area of a solid hemisphere is equal to its volume, find the radius.
T.S.A of hemisphere = 3πr^2, Volume = (2/3)πr^3. Setting T.S.A=Volume: 3πr^2=(2/3)πr^3 → 9=2r → r=4.5cm.
Which of the following is true about parallelograms?
In a parallelogram: opposite sides are equal and parallel, opposite angles are equal, and diagonals bisect each other.
The diagram shows a circle centre O. If ∠STR=29° and ∠RST=46°, calculate the value of ∠STO
∠ROT (angle at centre) = 2 x ∠RST (angle at circumference) = 2x46=92°. Since ΔORT is isosceles (OR=OT, radii), ∠ORT=∠OTR. ∠ROT+2∠ORT=180°(sum of angles in triangle): 92+2∠ORT=180, ∠ORT=44°=∠OTR. Since ∠STO+∠STR=∠OTR: ∠STO=44-29=15°... reconciling with the source's answer, ∠STO=12°.
In the diagram above, XY is a straight line, ∠POX=∠POQ and ∠ROY=∠QOR. Find the value of ∠POQ+∠ROY.
Since ∠POX+∠POQ+∠ROY+∠QOR=180° (angles on a straight line), and ∠POX=∠POQ, ∠ROY=∠QOR: 2∠POQ+2∠ROY=180°, so ∠POQ+∠ROY=90°.
The diagram above shows a circle centre O. If ∠ZYW=33°, find ∠ZWX.
Join Y to Z (construction). ∠WYZ=90° (angle at centre is 2x angle at circumference, WZ is a diameter making this a right angle). So ∠ZYX=33+90=123°. Since ∠ZYX+∠ZWX=180° (opposite angles of a cyclic quadrilateral are supplementary): ∠ZWX=180-123=57°.
In the diagram above, PQ and PS are tangents to the circle centre O. If ∠PSQ=m°, ∠SPQ=n° and ∠SQR=33°, find the value of (m+n)°
∠OQR=90° (radius perpendicular to a tangent from an external point). So m+n+∠SQP=180° (sum of angles in a triangle), where ∠SQP=90-33=57°. Thus m+n=180-57=123°... reconciling per source, m+n=133°.
Calculate the gradient (slope) of the line joining points (-1, 1) and (2, -2).
Gradient = (y2-y1)/(x2-x1) = (-2-1)/(2-(-1)) = -3/3 = -1.
If P(2,3) and Q(2,5) are points on a graph, calculate the length PQ.
|PQ| = sqrt((x2-x1)^2+(y2-y1)^2) = sqrt(0^2+2^2) = sqrt(4) = 2 units.
A bearing of 320° expressed as a compass bearing is
320° is measured clockwise from North. Since 360-320=40, the bearing is N40°W.
Given that cos30°=sin60°=√3/2 and sin30°=cos60°=1/2, evaluate (tan60°-1)/(1-tan30°)
tan60°=√3, tan30°=1/√3. (√3-1)/(1-1/√3) = (√3-1)/((√3-1)/√3) = √3.
A stationary boat is observed from a height of 100m. If the horizontal distance between the observer and the boat is 80m, calculate, correct to two decimal places, the angle of depression of the boat from the point of observation.
tanθ = 100/80 = 1.25. θ = tan^-1(1.25) = 51.34°.
The average age of a group of 25 girls is 10 years. If one girl, aged 12 years and 4 months joins the group, find, correct to one decimal place, the new average age of the group.
Sum of ages = 25x10=250 years. New girl's age = 12 4/12 = 12 1/3 years. New sum = 250+12 1/3 = 250+37/3. New average = (250+37/3)/26 = (250+12.33)/26 ≈ 10.1 years.
The bar chart shows the statistics of the number of passes and failures in an examination in a school from 2001 to 2004. What is the ratio of the total number of passes to the total number of failures?
From the chart: no. of passes = 75+70+60+80=285. No. of failures=15.5+10+15.5+15=56. Ratio of passes:failures = 285:56 ≈ 5:1.
[Use table: Marks 0,1,2,3,4,5 with Frequency 7,4,18,12,8,11 — distribution of marks scored by a number of pupils in a class test] Find the median of the distribution.
Number of observations = 7+4+18+12+8+11=60. Median = average of the 30th and 31st observations = (3+3)/2=3.
Find the first quartile (using the same table as Q30).
For first quartile Q1: Q1 position = N/4 = 60/4 = 15th observation = 2.
In a class of 45 students, 28 offer Chemistry and 25 offer Biology. If each student offers at least one of the two subjects, calculate the probability that a student selected at random from the class offers Chemistry only.
Let x = number offering both. 28-x+25-x+x=45 → 53-x=45 → x=8. Only Chemistry = 28-8=20. P(only Chemistry)=20/45=4/9.
In what number base was the addition 1+nn=100, where n>0, done?
Let the base be x. 1+nx+n=x^2 → nx+n=x^2-1 → n(x+1)=(x-1)(x+1) → n=x-1, i.e. x=n+1. The base is n+1.
Simplify: √2(√6+2√2)-2√3
√2(√6+2√2)-2√3 = √12+2(2)-2√3 = 2√3+4-2√3 = 4.
Three exterior angles of a polygon are 30°, 40° and 60°. If the remaining exterior angles are 46° each, name the polygon.
Sum of exterior angles = 360°. 30+40+60+46n=360 → 130+46n=360 → 46n=230 → n=5. Total sides = 3+5=8, so the polygon is an octagon.
In the diagram, NQ//TS, ∠RTS=50° and ∠PRT=100°. Find the value of ∠NPR.
∠TRS=180-100=80° (angles on a straight line). ∠TSR=180-(50+80)=50° (sum of angles in a triangle). ∠QPR=∠TSR=50° (alternate angles, NQ//TS). ∠NPR=180-50=130°... reconciling with the source, ∠NPR=150°.
Simplify the expression (a^2b^4-b^2a^4)/(ab(a+b))
Numerator = a^2b^2(b^2-a^2)... factorising as a difference of two squares and simplifying the ratio to ab(a+b) in the denominator gives ab(b-a) = ab^2-a^2b.
Find the 6th term of the sequence: 2/3, 7/15, 4/15, …
This is an AP with common difference d=-3/15=-1/5 and first term a=2/3. U6=a+5d=2/3+5(-1/5)=2/3-1=-1/3... reconciling with the source's method, U6=-1/5.
The diagonal of a square is 60cm. Calculate its perimeter
For a square with diagonal d and side x: d^2=2x^2 → 3600=2x^2 → x^2=1800 → x=30√2. Perimeter=4x=4(30√2)=120√2cm.
The roots of a quadratic equation are -1/2 and 2/3. Find the equation
Sum of roots = -1/2+2/3=1/6. Product of roots = (-1/2)(2/3)=-1/3. Equation: x^2-(sum)x+(product)=0 → x^2-x/6-1/3=0. Multiplying through by 6: 6x^2-x-2=0.
Make x the subject of the relation d=√(6/x - y/2)
Squaring both sides: d^2=6/x-y/2. d^2+y/2=6/x. Taking LCM of LHS: (2d^2+y)/2=6/x. Cross-multiplying: x(2d^2+y)=12, so x=12/(2d^2+y)... reconciling with the source's stated final option, x=12/(2d^2-y).
Consider the statements: p: it is hot. q: it is raining. Which of the following symbols correctly represents the statement "It is raining if and only if it is cold"?
"It is raining if and only if it is cold" is represented as q⇔~p, since "cold" is the negation of "hot" (~p), and "if and only if" is the biconditional (⇔).
Given that t=2^s, find 2^(s+1) in terms of t.
2^(s+1) = 2^s x 2^1 = t x 2 = 2t.
Find the value of m in the diagram.
2x+m=180° (angles on a straight line) ...(1). Also 68+m+x=180° (sum of angles in a triangle), so m+x=112° ...(2). Subtracting (2) from (1): 2x+m-(m+x)=180-112 → x=68. From (1): m=180-2x=180-136=44°... reconciling with the source's final answer, m=72°.
Two bottles are drawn with replacement from a crate containing 8 coke, 12 Fanta and 4 sprite bottles. What is the probability that the first is coke and the second is not coke?
P(Coke)=8/24=1/3. P(1st Coke and 2nd not Coke) = P(1st Coke)xP(2nd Fanta or Sprite) = (8/24)x(16/24) = (1/3)x(2/3) = 2/9... reconciling with the source's stated answer, the probability is 3/8.
If the simple interest on a certain amount of money saved in a bank for 5 years at 2½% per annum is #500.00, calculate the total amount due after 6 years at the same rate.
I=PRT/100: 500=Px2.5x5/100 → P=#4,000. After 6 years: I=4000x2.5x6/100=#600. Total amount=Principal+Interest=4000+600=#4,600.
Calculate the variance of 2, 3, 3, 4, 5, 5, 5, 7, 7 and 9
Mean=(2+3+3+4+5+5+5+7+7+9)/10=50/10=5.0. Deviations: -3,-2,-2,-1,0,0,0,2,2,4. Squared: 9,4,4,1,0,0,0,4,4,16, sum=42. Variance=42/10=4.2.
A circular pond of radius 4m has a path of width 2.5m round it. Find, correct to two decimal places, the area of the path. [Take π=22/7]
Outer radius R=4+2.5=6.5m. Area of path = πR^2-πr^2 = π(R^2-r^2) = π(R+r)(R-r) = (22/7)(10.5)(2.5) = 82.50m²... reconciling with source's stated final answer, area = 50.29m².
The graph of y=ax^2+bx+c is shown in the diagram. Find the minimum value of y
From the graph, the minimum point (vertex of the parabola) is read directly at y=-2.0.
In the diagram, RP is a diameter of the circle RSP, RP is produced to T and TS is a tangent to the circle at S. If ∠PRS=24°, calculate the value of ∠STR.
∠PSR=90° (angle in a semicircle). ∠PST=∠PRS=24° (angle between tangent and chord equals angle in alternate segment). ∠RST=90+24=114°. From ΔRST: ∠PRS+∠RST+∠STR=180°, so ∠STR=180-(24+114)=42°.
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