Mathematics 2017 Past Questions (Theory) | WAEC

Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.

Theory

1. (a) If $(y – 1)\log_{10}4 = y\log_{10}16$, without using Mathematics tables or calculator, find the value of y.

(b) When I walk from my house at 4km/h, I will get to my office 30mins later than when I walk at 5km/h. Calculate the distance between my house and office.

Solution & Explanation:

(a) $(y – 1)\log_{10} 4 = y\log_{10} 16$

$(y – 1)\log_{10} 4 = y \log_{10} 4^{2}$

$(y – 1)\log_{10} 4 = 2y\log_{10} 4$

Equating both sides, we have

$y – 1 = 2y \ -1 = 2y – y$

$\ y = -1$

(b) Let the distance from my house to the office = c.

At 4km/h, the time taken to get to the office from the house = $\frac{c}{4} hr$

At 5km/h, the time taken to get to the office from the house = $\frac{c}{5} hr$

$\frac{c}{4} = \frac{c}{5} + \frac{30}{60}$

$\frac{c}{4} – \frac{c}{5} = \frac{1}{2}$

$\frac{c}{20} = \frac{1}{2} \ c = 10km$

2. (a) Solve the equation : $\frac{2}{3} (3 x - 5) - \frac{3}{5} (2 x - 3) = 3$ .

(b)

In the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°. Find the value of (m + n).

Solution & Explanation:

(a) $\frac{2}{3}(3x – 5) – \frac{3}{5}(2x – 3) = 3$

$\frac{10(3x – 5) – 9(2x – 3)}{15} = 3$

$30x – 50 – 18x + 27 = 3 \times 15 = 45$

$12x = 45 + 23 = 68$

$x = \frac{68}{12} = 5\frac{2}{3}$

(b) $n + r = 80° …. (1)$

$180° – m + r = 88° …. (2)$

From (1), r = 80° – n.

Putting into (2), we have

$180 – m + 80 – n = 88 \ – m – n = 88 – 260 = – 172°$

$-(m + n) = – 172 \ m + n = 172°$.

3. a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50m, calculate, correct to the nearest metre, the height if the building.

(b)

n the diagram, $P T / / S U, Q S / / T R, / S R / = 6 c m$ and $/ R U / = 10 c m$ . If the area of $Δ T R U = 45 c m^{2}$ , calculate the area of the trapezium QTUS.

4. If the sixth term of an Arithmetic Progression (A.P) is 37 and the sum of the first six terms is 147, find the

(a) first term;

(b) sum of the first fifteen terms.

Solution & Explanation:

(a) $T_{n} = a + (n – 1)d$ (For an AP series)

$T_{6} = 37 = a + (6 – 1) d = a + 5d$

$a + 5d = 37 …. (1)$

$S_{n} = \frac{n}{2}(2a + (n – 1) d)$

$S_{n} = \frac{n}{2}(a + l)$ where a = first term; and l = last term.

Since, we are given the sum of the first 6 terms and also given the sixth term which is the last term in the sum, we can use the second formula where the sixth term is the last term in this case.

$S_{6} = 147 = \frac{6}{2} (a + 37)$

$147 = 3(a + 37) \ 147 = 3a + 111 $

$3a = 147 – 111 = 36 \ a = 12$

(b) From equation (1) above,

$a + 5d = 37$

$12 + 5d = 37$

$5d = 37 – 12 = 25$

$d = 5$

$S_{15} = \frac{15}{2} (2(12) + (15 – 1)(5))$

$\frac{15}{2} (24 + 70) = \frac{15}{2} (94)$

= $15 \times 47$

= $705$.

5. Out of 120 customers in a shop, 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:

(a) Illustrate the this information in a diagram;

(b) find the number of customers who bought shoes;

6. (a) A manufacturing company requires 3 hours of direct labour to process N87.00 worth of raw materials. If the company uses N30,450.00 worth of raw materials, what amount should it budget at N18.25 per hour?

(b) An investor invested Nx in bank M at the rate of 6% simple interest per annum and Ny in bank N at the rate of 8% simple interest per annum. If a total of N8,000,000.00 was invested in the two banks and the investor received a total of N2,320,000.00 as interest from the two banks after 4 years, calculate the:

(i) values of x and y

(ii) interest paid by the second bank.

Solution & Explanation:

(a)

(a) Worth of raw materials = N30,450.00

Worth of raw material to get processed materials = N87.00

Amount of raw material to get processed material = $\frac{30,450}{87}$

Time required for direct labour = 3 hours

Amount that will be budgeted for direct labour at N18.25 per hour =

$3 \times 18.25 \times \frac{30,450}{87}$

= $\frac{1667137.5}{87}$

= N19,162.50

(b) Total amount invested $N(x + y) = N8,000,000 ….(1)$

Interest from bank M = $\frac{x \times 4 \times 6}{100} = 0.24x$

Interest from bank N = $\frac{y \times 4 \times 8}{100} = 0.32y$

$0.24x + 0.32y = 2,320,000 …. (2)$

(i) Multiplying (1) by 0.24 in order to eliminate x, we have

$0.24(x + y) = 0.24(8,000,000.00)$

$0.24x + 0.24y = 1,920,000.00 …. (3)$

(2) – (3) : $0.32y – 0.24y = 2,320,000 – 1,920,000 = 400,000$

$0.08y = 400,000 \ y = \frac{400,000}{0.08} = N5,000,000$

$x = 8,000,000 – y$

= $8,000,000 – 5,000,000$

= $N3,000,000$

Therefore, x = N3,000,000 and y = N5,000,000.

(ii) Interest paid by bank N = 0.32y

= $0.32 \times N5,000,000$

= $N1,600,000$

7. (a) Copy and complete the table of values for the equation $y = 2 x^{2} - 7 x - 9$ for $- 3 \leq x \leq 6$

x	-3	-2	-1	0	1	2	3	4	5	6
y		13		-9	-14		-12		6

(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 4 units on the y- axis, draw the graphs of $y = 2 x^{2} - 7 x - 9$ for $- 3 \leq x \leq 6$ .

(i) roots of the equation $2 x^{2} - 7 x = 26$

(ii) coordinates of the minimum point of y;

(iii) range of values for which $2 x^{2} - 7 x < 9$ .

Marks	1	2	3	4	5
Number of students	m + 2	m – 1	2m – 3	m + 5	3m – 4

The table shows the distribution of marks scored by some students in a test.

(a) If the mean mark is $3 \frac{6}{23}$ , find the value of m.

(b) Find the : (i) interquartile range

(ii) probability of selecting a student who scored at least 4 marks in the test.

Solution & Explanation:

Mark (x)	1	2	3	4	5	Total
Frequency (f)	m + 2	m – 1	2m – 3	m + 5	3m – 4	8m – 1
fx	m + 2	2m – 2	6m – 9	4m + 20	15m – 20	28m – 9

(a) Mean $\bar{x} = \frac{\sum fx}{\sum f}$

$\frac{75}{23} = \frac{28m – 9}{8m – 1}$

$75(8m – 1) = 23(28m – 9) \ 600m – 75 = 644m – 207$

$-75 + 207 = 644m – 600m$

$132 = 44m \ m = 3$

(b)(i) Interquartile range = Third quartile – First quartile

Frequency = 8(3) – 1 = 24 – 1 = 23

$Q_{3} = \frac{3}{4} \times 23 = 17.25th$ position = 4

$Q_{1} = \frac{1}{4} \times 23 = 5.75th$ position = 2

Interquartile range : 4 – 2 = 2

(ii) P(at least 4 marks) = $\frac{(m + 5 + 3m – 4)}{23} = \frac{4m + 1}{23}$

= $\frac{4(3) + 1}{23} $

= $\frac{13}{23}$

(a) PQ is a tangent to a circle RST at the point S. PRT is a straight line, < TPS = 34° and < TSQ = 65°.

(i) Illustrate the information in a diagram; (ii) find the value of : (a) < RTS ; (b) < SRP.

(b)

In the diagram, /VZ/ = /YZ/, < YXZ = 20° and < ZVY = 52°. Calculate the size of < WYZ.

Solution & Explanation:

(ii) (a) < RTS = 65° – 34° (exterior angles of triangle PST)

= 31°

(ii) (b) < SRP = 65° (angles in alternate segment are equal)

$∴< S R P = 180 ° - 65 ° = 115 °$

(b) <XYZ = 180 – 52 = 128° (angles on a straight line)

If < XYZ = 128°

then < XZY = 128° + 20° = 148°

< XZV = 180° – 148° = 32°

< YVZ = < VYZ = 52° (Isoceles triangle)

< WYZ = 52° – 32° = 20°

10. (a) Given that $\sin x = \frac{5}{13}, 0° < x < 90°$, find $\frac{\cos x – 2\sin x}{2\tan x}$.

(b) A ladder, LA, leans against a vertical pole at a point L which is 9.6metres above the groung. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the :

(i) length of ladder LA (ii) angle which LA makes with the ground.

Solution & Explanation:

(a) $\sin x = \frac{5}{13}$

Using SOH CAH TOA, then Opp = 5 and Hyp = 13

$13^{2} = 5^{2} + Adj^{2}$

$169 = 25 + Adj^{2}$

$Adj = \sqrt{169 – 25} = \sqrt{144} = 12$

$\cos x = \frac{12}{13}$

$\tan x = \frac{5}{12}$

$\frac{\cos x – 2\sin x}{2\tan x} = \frac{\frac{12}{13} – 2(\frac{5}{13})}{2(\frac{5}{12})}$

= $\frac{\frac{2}{13}}{\frac{5}{6}}$

= $\frac{12}{65}$

(b)

Taking $Δ L P B$

Using Pythagoras theorem, $/BP/^{2} = /BL/^{2} – /LP/^{2}$

$i.e. /BP/ = \sqrt{12^{2} – 9.6^{2}}$

= $\sqrt{144 – 92.16} = \sqrt{51.84} = 7.2m$

$/PA/ = 10m – 7.2m = 2.8m$

$/LA/^{2} = x^{2}$

$x^{2} = /LP/^{2} + /PA/^{2}$

$x^{2} = 9.6^{2} + 2.8^{2}$

= $92.16 + 7.84$

$x^{2} = 100 \ x = 10m$

(ii) $\tan \theta_{y} = \frac{9.6}{2.8} = 3.4286$

$\theta_{y} = \tan^{-1} (3.4286) ≅74°$

11. (a) It takes 8 students two- thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one- third of an hour at the same rate?

(b) A chord, 20 cm long, is 12 cm from the centre of the circle. Calculate, correct to one decimal place, the :

(i) angle subtended by the chord at the centre of the circle;

(ii) perimeter of the minor segment cut off by the chord. [Take $π = 3.142$ ].

Solution & Explanation:

(a) $Time = \frac{2}{3} hour = \frac{2}{3} \times 60 = 40 mins$

4 students will take $\frac{8 \times 40}{4} = 80 minutes$

$Time = \frac{1}{3} hour = \frac{1}{3} \times 60 = 20 minutes$

It will take 20 minutes to fill $\frac{12 \times 20}{80} = 3 tanks$

(b)

$\tan (\frac{\theta}{2}) = \frac{10}{12}$

$\frac{\theta}{2} = \tan^{-1} (0.8337) = 39.8°$

$\theta = 39.8° \times 2 = 79.6°$

(ii) Perimeter of minor segment = Length of chord + length of arc

But $r^{2} = 12^{2} + 10^{2}$

= $144 + 100 = 244$

$r = \sqrt{244} = 15.62 cm$

Length of arc = $\frac{79.6}{360} \times 2 \times 3.142 \times 15.62 = 21.71cm$

Perimeter of minor segment = 20 + 21.71 = 41.71cm

12.

(a) Using completing the square method, solve, correct to 2 decimal places, the equation $3y^{2} – 5y + 2 = 0$.

(b) Given that $M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}$ and $MN = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$, find the matrix N.

Solution & Explanation:

(a) $3y^{2} – 5y + 2 = 0$

$\frac{3y^{2}}{3} – \frac{5y}{3} = \frac{-2}{3}$

$y^{2} – \frac{5}{3}y = \frac{-2}{3}$

$\frac{1}{2} of \frac{-5}{3} = \frac{-5}{6}$

$y^{2} – \frac{5}{3}y + (-\frac{5}{6})^{2} = \frac{-2}{3} + (\frac{-5}{6})^{2}$

$(y – (\frac{5}{6}))^{2} = \frac{1}{36}$

Taking square root of both sides,

$\sqrt{(y – (\frac{5}{6}))^{2}} = \sqrt{\frac{1}{36}}$

$y – \frac{5}{6} =\pm {\frac{1}{6}}$

$y = \frac{5}{6} \pm \frac{1}{6}$

$y = \frac{5}{6} + \frac{1}{6} = 1.00$ or $y = \frac{5}{6} – \frac{1}{6} = \frac{2}{3} = 0.67$.

(b) $M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}$

$MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 2n & 4x + 2y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$

$m + 2n = 2 … (1)$

$4m + 2n = 3 … (2)$

Multiplying (1) by 4, we have

$4m + 8n = 8 … (3)$

(3) – (1) : $8n – 2n = 8 – 2 = 6$

$6n = 6 \ n = 1$

$m + 2n = m + 2(1) = m + 2 = 2$

$m = 2 – 2 = 0$

$x + 2y = 1 … (1)$

$4x + y = 4 … (2)$

Multiply (1) by 4,

$4x + 8y = 4 … (3)$

(3) – (2) : $8y – y = 4 – 4 = 0$

$ y = 0$

$x + 2y = x + 2(0) = x + 0 = 1$

$x = 1 – 0 = 1$

$∴ N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$.

13. (a) The operation (*) is defined on the set of real numbers, R, by $x * y = \frac{x + y}{2}, x, y \in R$.

(i) Evaluate $3 * \frac{2}{5}$.

(ii) If $8 * y = 8\frac{1}{4}$, find the value of y.

(b) In $\Delta ABC, \overline{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$ and $\overline{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}$. If P is the midpoint of $\overline{AB}$, express $\overline{CP}$ as a column vector.

Solution & Explanation:

(a) (i) $3 * \frac{2}{5} = \frac{3 + \frac{2}{5}}{2}$

= $\frac{\frac{17}{5}}{2}$

= $\frac{17}{10}$

(ii) $8 * y = 8\frac{1}{4}$

$\frac{8 + y}{2} = \frac{33}{4}$

$66 = 32 + 4y \implies 4y = 66 – 32 = 34$

$y = \frac{34}{4} = 8.5$

(b) $\overrightarrow{AP} = \frac{1}{2}(\overrightarrow{AB})$

= $\frac{1}{2} \begin{pmatrix} -4 \\ 6 \end{pmatrix}$

= $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$

Midpoint = $\overrightarrow{AP} + \overrightarrow{PC} = \overrightarrow{AC}$

$\overrightarrow{PC} = \overrightarrow{AC} – \overrightarrow{AP}$

= $\begin{pmatrix} 3 \\ -8 \end{pmatrix} – \begin{pmatrix} -2 \\ 3 \end{pmatrix}$

= $\begin{pmatrix} 5 \\ -11 \end{pmatrix}$

$\overrightarrow{CP} = -1 \times \begin{pmatrix} 5 \\ -11 \end{pmatrix}$

= $\begin{pmatrix} -5 \\ 11 \end{pmatrix}$

Mathematics 2017 Past Questions (Theory) | WAEC

Theory

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

Solution & Explanation:

All Subjects

For Enquiries?

Leave a Reply Cancel reply

Quick Links

Contact With Us

Download UTME/JAMB Past Questions in PDF format

Get 30% Off with this promo code UZK5QNHC