Mathematics 2018 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

A used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of the year. If the car was bought on the 1st of March, 2011, calculate, correct to the nearest hundred naira, the value of the car on the 28th of February, 2015.

Solution & Explanation:

$Beginning of 1st year (March 2011 – February 2012) - N 900, 000$

Beginning of 1st year (March 2011 – February 2012)−N900,000

$Beginning of 1st year (March 2011 – February 2012) - N 900, 000$

$D e p r e c i a t i o n$

Price at the end of 1st year = N(900,000 – 270,000)

= N630,000.

$Beginning of 2nd year (March 2012 – February 2013) - N 630, 000$

Beginning of 2nd year (March 2012 – February 2013)−N630,000

$Beginning of 2nd year (March 2012 – February 2013) - N 630, 000$

$D e p r e c i a t i on$

Price at the end of 2nd year = N(630,000 – 138,600)

= N491,400.

$Beginning of 3rd year (March 2013 – February 2014) - N 491, 400$

Beginning of 3rd year (March 2013 – February 2014)−N491,400

$Beginning of 3rd year (March 2013 – February 2014) - N 491, 400$

$D e p r e c i a t i o n$

Price at the end of the 3rd year = N(491,400 – 108,108)

= N383,292.

$Beginning of 4th year (March 2014 – February 2015) - N 383, 292$

Beginning of 4th year (March 2014 – February 2015)−N383,292

$Beginning of 4th year (March 2014 – February 2015) - N 383, 292$

$D e p r e c i a t i o n$

Price at the end of the 4th year = N(383,292 – 84,324.24)

= N298,967.76 ≈

N299,000 (to the nearest hundred naira)

2. (a) The graph of

y=2px²−p²x−14

passes through the point (3, 10). Find the values of p.

(b) Two lines,

3 y - 2 x = 21

and $4 y + 5 x = 5$

intersect at the point Q. Find the coordinates of Q.

Solution & Explanation:

(a) y=2px²−p²x−14

At point (3, 10), y = 10 when x = 3.

$⟹ 10 = 2 p (3²) - p² (3) - 14$

10=18p−3p²−14

$- 10 = 3 p² - 18 p + 14 ⟹ 3 p^{²} - 18 p + 24 = 0$

$3 p^{2} - 12 p - 6 p + 24 = 0$

$(p - 4) (3 p - 6) = 0 ⟹ p = 2 or 4$

“>(p−4)(3p−6)=0⟹p = 2 or 4

$(p - 4) (3 p - 6) = 0 ⟹ p = 2 or 4$

(b)

$3 y - 2 x = 21 . . . (1)$

“>3y−2x=21...(1)

$3 y - 2 x = 21 . . . (1)$

$4 y + 5 x = 5 . . . . (2)$

Using elimination method, multiply (1) by 4 and (2) by 3.

(1)×4:12y−8x=84...(3)

$\times 3 : 12 y + 15 x = 15 . . . (4)$

Subtracting (3) – (4), we have:

$- 23 x = 69 ⟹ x = - 3$

Putting x = -3 in (1), we have

$3 y - 2 (- 3) = 3 y + 6 = 21$

“>3y−2(−3)=3y+6=21

$3 y - 2 (- 3) = 3 y + 6 = 21$

Hence, the coordinates of Q are (-3, 5).

3. (a) The diagonals of a rhombus are 10.2 cm and 9.3 cm long. Calculate, correct to one decimal place, the perimeter of the rhombus.

(b) Given that $\sin x = \frac{3}{5}, 0 ° < x < 90 °$ , find the value of $5 \cos x - 4 \tan x$

Solution & Explanation:

Let a side of the rhombus be n.

Then $n^{2} = {4.65}^{2} + {5.1}^{2}$

“>n²=4.65²+5.12

$n² = 4.65² + 5.1²$

= $21.6225 + 26.01$

“>21.6225+26.01

$21.6225 + 26.01$

$n^{2} = 47.6325$

“>n²=47.6325

$n =\sqrt 47.6325$

$n = \sqrt{47.6325} = 6.902 c m$

“>n=√47.6325 =6.902cm

Hence, the perimeter of the rhombus = 4n

= $4 \times 6.902$

“>4×6.902

$4 \times 6.902$

= $27.608 c m ≊ 27.6 c m$

“>27.608cm≊27.6cm

$27.608 c m ≊ 27.6 c m$

(to 1 d.p)

(b) Right- angled triangle ABC

$\sin x = \frac{3}{5}$

Thus, $m^{2} = 5^{2} - 3^{2} = 25 - 9 = 16$

$m = 4$ .

$\cos x = \frac{4}{5} ⟹ 5 \cos x = 5 \times \frac{4}{5} = 4$

$\tan x = \frac{3}{4} ⟹ 4 \tan x = 4 \times \frac{3}{4} = 3$

$5 \cos x - 4 \tan x = 4 - 3 = 1$

(a)

In the diagram

(i) The value of x; (ii) <RSQ

(b) If 2N4seven=15Nnine,

find the value of N.

Solution & Explanation:

In the diagram,

(i)

$x ° + 90 ° = (3 x + 15) °$

(Sum of opposite interior angles)

Thus,

$90 ° - 15 ° = 3 x - x ⟹ 75 ° = 2 x$

$x = 37.5 °$

(ii)

< RSQ = 90° – 37.5° = 52.5°

(b)

$2 N 4_{s e v e n} = 15 N_{n i n e}$

$2 N 4_{s e v e n} = (2 \times 7^{2}) + (N \times 7^{1}) + (4 \times 7^{0})$

= $98 + 7 N + 4$

= $102 + 7 N$

$15 N_{n i n e} = (1 \times 9^{2}) + (5 \times 9^{1}) + (N \times 9^{0})$

= $81 + 45 + N$

= $126 + N$

$⟹ 102 + 7 N = 126 + N$

$7 N - N = 126 - 102 = 24$

$6 N = 24 ⟹ N = 4$

Therefore, $244_{s e v e n} = 154_{n i n e}$ .

5. (a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n – 3), (s + 6), (p – 2) and (q + 8).

(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?

m + n + s + p + q / 5

=12

⇒ m + n + s + p + q = 12 x 5 = 60

(m+4)+(n−3)+(s+6)+(p−2)+(q+8) / 5 = m+n+s+p+q+4−3+6−2+8 / 5

=60+ 13 /5 =73/5 = 14.6

(b) In the community, there are 500 people.

The 75th percentile age = 65 years

Number of people in the 75th percentile = $\frac{75 \times 500}{100} = 375$

The 25th percentile = 15 years

Number of people in the 25th percentile = $\frac{25 \times 500}{100} = 125$

Number of people between 15 years and 65 years = 375 – 125 = 250 people

6. In a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only – 28, Clutch and Steering – 14; Clutch, Steering and Brakes – 8; Clutch and Brakes – 20; Brakes and Steering only – 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.

(a) Draw a Venn Diagram to illustrate this information.

(b) How many cars had : (i) Faulty Brakes? (ii) Only one fault?

7. (a) Find the equation of the line passing through the points (2, 5) and (-4, -7).

(b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and the bearing of P and R is 300°. If |PQ| = 5 km and |PR| = 8 km,

(i) Illustrate the information in a diagram.

(ii) Calculate, correct to three significant figures, the:

(1) distance between Q and R

(2) bearing of R from Q.

Solution & Explanation:

(a) Using the two- point form,

Y-Y1/Y2-Y1 = X-X1/X2-X1

Y-5/-7-5 = X-2/-4-2

Y-5/-12 = X-2/-6

y−5=2(x−2)

y−5=2x−4⟹y−2x=−4+5=1

Equation:y=2x+1

(b)

In $Δ P Q R, < Q P R = 60 ° + 30 ° = 90 °$

“>ΔPQR,<QPR=60°+30°=90°

$Δ P Q R, < Q P R = 60 ° + 30 ° = 90 °$

. PQR is a right- angled triangle.

(ii) (1) In $Δ P Q R, | Q R |^{2} = 5^{2} + 8^{2}$

“>ΔPQR,|QR|²=5²+8²

$25 + 64 = 89$

“>25+64=

$25 + 64 = 89$

$| Q R | = \sqrt{89} = 9.434 k m ≊ 9.43 k m$

“>|QR|=√89=9.434km≊9.43km

(2) In the diagram above, the bearing of R from Q is the obtuse angle NQR.

But $\tan α = \frac{8/}{5} = 1.6$

$α = \tan^{- 1} (1.6) = 58 °$

Hence, angle NQR = 360° – (a + 60° + 90°)

= 360° – (58° + 60° + 90°)

= 360° – 208°

= 152°

8. (a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book at a profit of x%. If James paid

N (6 x + \frac{3}{4})

more for the book than Lamin paid, find the value of x.

(b) Find the range of values of x which satisfies the inequality

3 x - 2 < 10 + x < 2 + 5 x

Solution & Explanation:

(a) Seliing Price SP = $C . P + \frac{}{100} \times C . P$

= $300 + (\frac{x}{100} \times 300)$

S.P = N(300 + 3x)

Therefore, Bola bought it at N(300 + 3x).

James paid $N (6 x + \frac{3}{4})$ extra from what Lamin paid, therefore Bola’s S.P = $N (300 + 6 x + \frac{3}{4})$

= N(300.75 + 6x).

Profit for Bola = $N (300.75 + 6 x - (300 + 3 x)) = N (0.75 + 3 x)$

$\frac{x}{100} \times N (300 + 3 x) = N (0.75 + 3 x)$

$300 x + 3 x^{2} = 75 + 300 x$

$⟹ 3 x^{2} = 75$

$x^{2} = 25 ∴ x = 5$

(b) $3 x - 2 < 10 + x < 2 + 5 x$

$3 x - 2 < 10 + x ⟹ 3 x - x < 10 + 2$

$2 x < 12 ⟹ x < 6$

$10 + x < 2 + 5 x$

$x - 5 x < 2 - 10$

$- 4 x < - 8 ⟹ x > 2$

The range = $2 < x < 6$

In the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm and < SPR = 30°. Calculate, correct to the nearest whole number:

(a) |SR| ;

(b) area of TQRS.

10.

90°, Find, correct to three significant figures, |PR|.

(b) The length of two ladders, L and M are 10m and 12m respectively. They are placed against a wall such that each ladder makes angle with the horizontal ground. If the foot of L is 8m from the foot of the wall.

(i) Draw a diagram to illustrate this information; (ii) Calculate the height at which M touches the wall.

11. (a) Copy and complete the table of values for $y = 2 x^{2} + x - 10$ for $- 5 \leq x \leq 4$

x	-5	-4	-3	-2	-1	0	1	2	3	4
y			5		-9	-10		0

(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, Draw the graph of $y = 2 x^{2} + x - 10$ for $- 5 \leq x \leq 4$ .

(i) $2 x^{2} + x = 10$

(ii) $2 x^{2} + x - 10 = 2 x$

Solution & Explanation:

(a)

x	-5	-4	-3	-2	-1	0	1	2	3	4
y	35	18	5	-4	-9	-10	-7	0	11	26

(b)

(c)i

c)(i) $2 x² + x = 10 \equiv 2 x² + x - 10 = 0$

Thus, the solution of $2 x² + x - 10 = 0$

are the values of x at which the curve cuts the x- axis.

x = -2.5 or x = 2.

(ii) $2 x² + x - 10 = 2 x$

To solve the given equation, we first draw the graph of $y = 2 x$

: when x = -2, y = 2(-2) = -4; when x = 3, y = 2(3) = 6.

The values of x at which $y = 2 x² + x - 10$

and $y = 2 x$

intersect give the solution of $2 x² + x - 10 = 2 x$

These are x = -2 or x = 2.5.

12.

Marks	10	20	30	40	50	60	70	80	90
Frequency	1	1	x	5	y	1	4	3	1

The frequency distribution shows the marks distribution of a class of 30 students in an examination.

The mean mark of the distribution is 52.

(a) Find the values of x and y.

(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.

(d) Use the histogram to estimate the mode

Solution & Explanation:

(a)

Marks(x)	Frequency (f)	fx
10	1	10
20	1	20
30	x	30x
40	5	200
50	y	50y
60	1	60
70	4	280
80	3	240
90	1	90
Total	16 + x + y	900 + 30x + 50y

$\sum f = 16 + x + y = 30$

$⟹ x + y = 14 . . . (1)$

$\bar{x} = \frac{\sum f x}{\sum f}$

$52 = \frac{900 + 30 x + 50 y}{30}$

$1560 = 900 + 30 x + 50 y ⟹ 30 x + 50 y = 660$

$3 x + 5 y = 66 . . . . (2)$

Solving equation (1) and (2),

From (1), x = 14 – y

$3 (14 - y) + 5 y = 42 - 3 y + 5 y = 66$

$2 y = 24 ⟹ y = 12$

$x = 14 - y = 14 - 12 = 2$

(x, y) = (2, 12).

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Mathematics 2018 Past Questions (Theory) | WAEC

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