JAMB UTME - Mathematics - 1984

Question 1 Report

Two fair dice are rolled. What is the probability that both show up the same number of points.

1

36

7

36

1

2

1

3

1

6

Answer Details

A dice has 6 faces, 2 dice has 6 x 6 = 36 combined face
Prob. of both showing the same number of points
=

\frac{6}{36}

=

\frac{1}{6}

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Probability

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Question 2 Report

The cumulative frequency function of the data below is given below by the equation $y = cf(x)$. What is $cf(5)$?

Score($n$)	Frequency($f$)
3	30
4	32
5	30
6	35
8	20

30

32

55

62

92

Answer Details

If y = cf(x)
cf(5) = 30 + 32 + 30
= 92

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Probability

Linear Inequalities

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Question 3 Report

In the figure, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP.

4cm

16cm

9cm

14cm

Impossible, insufficient data

Answer Details

\frac{P Q}{6 + R T} = \frac{6}{12} = \frac{6}{P S}

(Similar triangles)

$\frac{6}{12} = \frac{8}{P S}$

PS = $\frac{12 \times 8}{6}$

= 16cm

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Question 4 Report

A variable point p(x, y) traces a graph in a two-dimensional plane. (0, 3) is one position of P. If x increases by 1 unit, y increases by 4 units. The equation of the graph is

-3 =

\frac{y + 4}{x + 1}

4y = -3 + x

\frac{y}{x}

=

\frac{- 3}{4}

4x = y + 3

Answer Details

P(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4

\frac{x}{1}

=

\frac{y}{4}

y = 4x
Hence the sign of the graph is y + 3 = 4x

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Algebraic Expressions

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Question 5 Report

$ \dfrac{0.0001432}{1940000} = k \times 10^n $ where $ 1 \le k < 10 $ and n is a whole number. The values K and n are

7.381 qnd -11

2.34 and 10

3.871 and 2

7.831 and -11

Answer Details

$\frac{0.0001432}{1940000}$ = k x 10n

where 1 $\leq$ k $\leq$ 10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10-11

k = 7.381, n = -11

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Number Bases

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Question 6 Report

Bola choose at random a number between 1 and 300. What is the probability that the number is divisible by 4?

1

4

6

4

7

4

Answer Details

Numbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c. To get the number of figures divisible by 4, We solve by method of A.P
Let x represent numbers divisible by 4, nth term = a + (n - 1)d
a = 4, d = 4
Last term = 4 + (n - 1)4
288 = 4 + 4n - n
=

\frac{288}{4}

= 72
rn(Note: 288 is the last Number divisible by 4 between 1 and 300)
Prob. of x =

\frac{72}{288}

=

\frac{1}{4}

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Probability

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Question 7 Report

Tunde and Shola can do a piece of work in 18 days. Tunde can do it alone in x days, whilst Shola takes 15 days longer to do it alone. Which of the following equations is satisfied by x?

x2 - 5x - 10 = 0

x2 - 20x + 360 = 0

x2 - 21x - 270 = 0

3x2 - 65x + 362 = 0

Solution Of Linear Equations

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Question 8 Report

If the price of oranges was raised by $ \frac{1}{2} $k per orange. The number of oranges a customer can buy for ?2.40 will be less by 16. What is the present price of an orange?

2

\frac{1}{2}

\frac{2}{15}

5

\frac{1}{3}

20

Answer Details

Let x represent the price of an orange and
y represent the number of oranges that can be bought
xy = 240k, y =

\frac{240}{x}

.....(i)
If the price of an oranges is raised by

\frac{1}{2}

k per orange, number that can be bought for ?240 is reduced by 16
Hence, y - 16 =

\frac{240}{x + \frac{1}{2}}

=

\frac{480}{2 x + 1}

=

\frac{480}{2 x + 1}

.....(ii)
subt. for y in eqn (ii)

\frac{240}{x}

- 16
=

\frac{480}{2 x + 1}

=

\frac{240 - 16 x}{x}

=

\frac{480}{2 x + 1}

= (240 - 16x)(2x + 1)
= 480x
= 480x + 240 - 32x2 - 16
480x = 224 - 32x2
x2 = 7
x =

\sqrt{7}

= 2.5
= 2

\frac{1}{2}

k

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Variation

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Question 9 Report

Find the area of the shaded portion of the semicircular figure.

\frac{r^{2}}{4} (4 π - 3 \sqrt{3})

\frac{r^{2}}{4} (2 π - 3 \sqrt{3})

\frac{1}{2} r^{2} π

\frac{1}{8} r^{2} \sqrt{3}

\frac{r^{2}}{4} (4 π - 3 \sqrt{3})

Answer Details

Asector =

\frac{60}{360} \times π r^{2}

= $\frac{1}{6} π r^{2}$

A $△$ = $\frac{1}{2} r^{2} \sin 60^{o}$

$\frac{1}{2} r^{2} \times \frac{\sqrt{3}}{2} = \frac{r^{2} \sqrt{3}}{4}$

A $shaded portion$ = Asector -
A $△$

= ( $\frac{1}{6} π r^{2} - \frac{r^{2} \sqrt{3}}{4})^{3}$

= $\frac{π r^{2}}{2} - \frac{3 r^{2} \sqrt{3}}{4}$

= $\frac{r^{2}}{4} (2 π - 3 \sqrt{3})$

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Number Bases

Areas

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Question 10 Report

In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS.

60

^{\circ}

90

^{\circ}

120

^{\circ}

30

^{\circ}

80

^{\circ}

Answer Details

From the diagram, PQRSTW is a regular hexagon.

Hexagon is a six sided polygon.

Sum of interior angles of polygon = (2n - 4)90 $^{\circ}$ = [2 x 6 - 4] x 90 = 8 x 90 = 720 $^{\circ}$

each angle = $\frac{720^{o}}{6} = 120^{o}$ and TVS = $\frac{120}{2} = 60^{o}$

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Triangles And Polygons

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Question 11 Report

If $ \sin \theta = \frac{x}{y} $ and $0^o < 90^o$, then find $ \frac{1}{\tan \theta} $.

x

√y²−x²

y

x

√y²+ x²

y²−x²

x

Answer Details

\frac{1}{t a n θ}

=

\frac{c o s θ}{s i n θ}

sin

θ

=

\frac{x}{y}

cos

θ

=

\frac{\sqrt{y^{2} - x^{2}}}{y}

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Sine, Cosine And Tangent Of An Angle

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Question 12 Report

In a racing competition, Musa covered a distance 5x km in the first hour and (x + 10)km in the next hour. He was second to Nzozi who covered a total distance of 118km in the two hours. Which of the following inequalities is correct?

-1 < x < x < 0

-3 < x < 3

0

\leq

x < 18

15 < x < 18

Answer Details

Total distance covered by Musa in 2 hrs
= x + 10 + 5x
= 6x + 10
Ngozi = 118 km
If they are equal, 6x + 10 = 188
but 6x + 10 < 118

6x < 108

= x < 18

0 < x < 18 = 0 $\leq$ x < 18

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Linear Inequalities

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Question 13 Report

Find the x co-ordinates of the points of intersection of the two equations in the graph.

1, 1

0, 4

0, 0

4, 0

4, 9

Answer Details

If y = 2x + 1 and y = x2 - 2x + 1

then x2 - 2x + 1 = 2x + 1

x2 - 4x = 0

x(x - 4) = 0

x = 0 or 4

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Modular Arithmetic

Coordinate Geometry

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Question 14 Report

The larger value of y for which (y - 1)2 = 4y - 7 is

2

4

6

7

8

Answer Details

(y - 1)2 = 4y - 7
y2 - 2xy + 1 = 4y - 7
y2 - 6y + 8 = 0
(y - 4)(y - 2)
y = 4 or 2 = 4

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Question 15 Report

Simplify $ \frac{\left(\frac{2}{3}-\frac{1}{5}\right)-\frac{1}{3}\text{ of }\frac{2}{5}}{3-\frac{1}{1\frac{1}{2}}} $

1

7

1

3

Answer Details

$\frac{2}{3} - \frac{1}{5}$ = $\frac{10 - 3}{15}$
= $\frac{7}{15}$
$\frac{1}{3}$ Of $\frac{2}{5}$ = $\frac{1}{3}$ x $\frac{2}{5}$
= $\frac{2}{5}$
( $\frac{2}{3} - \frac{1}{5}$ ) - $\frac{1}{3}$ of $\frac{2}{5}$
= $\frac{7}{15} - \frac{2}{15}$ = $\frac{1}{3}$
3 - $\frac{1}{1 \frac{1}{2}}$ = 3 - $\frac{2}{3}$
= $\frac{7}{3}$
$\frac{\frac{2}{3} - \frac{1}{5} of \frac{2}{15}}{3 - \frac{1}{1 \frac{1}{2}}}$
= $\frac{\frac{1}{3}}{\frac{7}{3}}$
= $\frac{1}{3}$ x $\frac{3}{7}$
= $\frac{1}{7}$

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Logarithms

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Question 16 Report

A man invested a total of ₦50000 in two companies. If these companies pay dividends of 6% and 8% respectively, how much did he invest at 8% if the total yield is ₦3700?

₦15000

₦29600

₦27800

₦21400

₦35000

Answer Details

Total yield = ₦3,700
Total amount invested = ₦50,000
Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest
then yield on x =

\frac{6}{100}

x and yield on y =

\frac{8}{100}

y
Hence,

\frac{6}{100}

x +

\frac{8}{100}

y
= 3,700.........(i)
x + y = 50,000........(ii)
6x + 8y = 370,000 x 1
x + y = 50,000 x 6
6x + 8y = 370,000.........(iii)
6x + 6y = 3000,000........(iv)
Eqn (iii) - Eqn (ii)
2y = 70,000
y =

\frac{70, 000}{2}

= 35,000
Money invested at 8% is ₦35,000

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Modular Arithmetic

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Question 17 Report

Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below:

Class boundary in cm	frequency
3.35 − 3.45	3
3.45 − 3.55	6
3.55 − 3.65	7
3.65 − 3.75	4

What is the mean diameter of the copper spheres?

3.40cm

3.58cm

3.56cm

3.62cm

Answer Details

\begin{array}{cc} x(mid point) & f & f x \\ 3.4 & 3 & 10.2 \\ 3.5 & 6 & 21.0 \\ 3.6 & 7 & 25.2 \\ 3.7 & 4 & 14.8 \end{array}

\sum f

= 20

\sum f x

= 71.2
mean =

\frac{\sum f x}{\sum f}

=

\frac{71.2}{20}

= 3.56

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Question 18 Report

The table below is drawn for a graph $y = x^3 - 3x + 1$.

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y = x^3 - 3x + 1$	$1$	$-1$	$3$	$1$	$-1$	$3$	$1$

From x = -2 to x = 1, the graph crosses the x-axis in the range(s)

-1 < x < 0 and 0 < x < 1

-2 < x < -1 and 0 < x < 1

-2 < x < -1 and -1 < x < 0

0 < x < 1

Answer Details

If the graph of y = x3 - 3x + 1 is plotted,the graph crosses the x-axis in the ranges -2 < x < -1 and 0 < x < 1

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Question 19 Report

Find the angle of the sectors representing each item in pie chart of the following data 6, 10, 14, 16, 26

15o, 25o, 35o, 40o, 65o

60o, 100o, 140o, 160o, 260o

6o, 10o, 14o, 16o, 26o

30o, 50o, 70o, 80o, 130o

none of the above

Answer Details

6 + 10 + 14 + 16 + 26 = 72

\frac{6}{72}

x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively

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Question 20 Report

The sides of a triangle are $ (x + 4)\text{cm} $, $ x\text{cm} $ and $ (x - 4)\text{cm} $, respectively If the cosine of the largest angle is $ \frac{1}{5} $, find the value of $ x $

24cm

20cm

28cm

7cm

88

7

Answer Details

< B is the largest since the side facing it is the largest, i.e. (x + 4)cm

Cosine B = $\frac{1}{5}$
= 0.2 given
b2 - a2 + c2 - 2a Cos B
Cos B = $\frac{a^{2} + c^{2} - b^{2}}{2 a c}$
$\frac{1}{5}$ = $\frac{x^{2} + ? (x - 4)^{2} - (x + 4)^{2}}{2 x (x - 4)}$
$\frac{1}{5}$ = $\frac{x (x - 16)}{2 x (x - 4)}$
$\frac{1}{5}$ = $\frac{x - 16}{2 x - 8}$
= 5(x - 16)
= 2x - 8
3x = 72
x = $\frac{72}{3}$
= 24

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Algebraic Expressions

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Question 21 Report

A right circular cone has a base radius r cm and a vertical angle 2yo. The height of the cone is

r tan yo cm

r sin yo cm

r cot yo cm

r cos yo cm

r cosec yo cm

Answer Details

\frac{r}{h}

= tan yo
h =

\frac{r}{t a n y^{o}}

= r cot yo

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Introductory Calculus

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Question 22 Report

P varies directly as the square of Q and inversely as R. If P = 36, when Q = 3 and R = 4, find P when Q = 5 and R = 2.

72

100

90

200

125

Answer Details

P \propto \frac{Q^{2}}{R}

∴

P = \frac{K Q^{2}}{R}

36 = \frac{K Q^{2}}{4}

K = \frac{36 \times 4}{9}

K = 16

P = \frac{16 \times 5^{2}}{2} = 200

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Question 23 Report

Thirty boys and X girls sat for a test. The mean of the boys' scores and that of the girls were respectively 6 and 8. Find 8 if the total score was 468.

38

24

36

22

41

Answer Details

Le the number of girls = A; Total score of boys = 30

\times

6 = 180
Total score of girls = A

\times

8 = 8A

∴

180 + 8A = 468
8A = 288
A = 36

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Statistics

Simple Operations On Algebraic Expressions

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Question 24 Report

Factorize $6x^2 - 14x - 12$

2(x + 3)(3x - 2)

6(x - 2)(x + 1)

2(x - 3)(3x + 2)

6(x + 2)(x - 1)

(3x - 4)(2x + 3)

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Question 25 Report

Using $ \triangle $ XYZ in the figure, find XYZ.

29

^{\circ}

31

^{\circ}

31

^{\circ}

20'

31

^{\circ}

18'

Answer Details

\frac{\sin y}{3} = \frac{\sin 120^{o}}{5}

sin 120 $^{\circ}$ = sin 60 $^{\circ}$

5 sin y = 3 sin 60 $^{\circ}$

sin y = $\frac{3 \sin 60^{o}}{5}$

$\frac{3 \times 0.866}{5}$

= $\frac{2.598}{5}$

y = sin-1 0.5196 = 30 $^{\circ}$ 18'

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Triangles And Polygons

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Question 26 Report

If $pq + 1 = q^2$ and $t = \frac{1}{p} - \frac{1}{pq}$ express $t$ in terms of $q$

1

P-Q

1

q-1

1

q+1

1 + q

1

1-q

Answer Details

Pq + 1 = q2......(i)
t =

\frac{1}{p}

-

\frac{1}{p q}

.........(ii)
p =

\frac{q^{2} - 1}{q}

Sub for p in equation (ii)
t =

\frac{1}{q^{2} - \frac{1}{q}}

-

\frac{1}{\frac{q^{2} - 1}{q} \times q}

t =

\frac{q}{q^{2} - 1}

-

\frac{1}{q^{2} - 1}

t =

\frac{q - 1}{q^{2} - 1}

=

\frac{q - 1}{(q + 1) (q - 1)}

=

\frac{1}{q + 1}

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Solution Of Linear Equations

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Question 27 Report

What is the volume of this regular three dimensional figure?

160cm2

48cm2

120cm2

40cm2

Answer Details

Volume of the three dimensional figures = v = A x h

A = $\frac{1}{2}$ x 4 x 3

= 6cm2

V = 6 x 8

= 48cm2

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Volumes

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Question 28 Report

A cone is formed by bending a sector of a circle Saving an angle or 210°. Find the radius of the base of the cone. If the diameter of the circle is 12cm.

7.00cm

1.75cm

21cm

3.50cm

2.21cm

Loci

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Question 29 Report

Simplify $ \frac{3^n - 3^{n-1}}{3^3 \times 3^n - 27 \times 3^{n-1}} $

1

6

1

27

4

3

Answer Details

\frac{3^{n} - 3^{n - 1}}{3^{3} \times 3^{n} - 27 \times 3^{n - 1}}

=

\frac{3^{n} - 3^{n - 1}}{3^{3} (3^{n} - 3^{n - 1})}

=

\frac{3^{n} - 3^{n - 1}}{27 (3^{n} - 3^{n - 1})}

=

\frac{1}{27}

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Fractions, Decimals, Approximations And Percentage

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Question 30 Report

Simplify $ (1 + \frac{\frac{x - 1}{1}}{\frac{1}{x + 1}})(x + 2) $

(x2 - 1)(x + 2)

x2(x + 2)

x2 + 34

\frac{3 x^{2} - 1}{(x - 1)}

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Algebraic Expressions

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Question 31 Report

In the figure, 0 is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is

67

\frac{1}{2}

^{\circ}

22

\frac{1}{2}

^{\circ}

33

\frac{1}{2}

^{\circ}

45

^{\circ}

90

^{\circ}

Answer Details

< R = 180

^{\circ}

- 45

^{\circ}

(sum of angles on a straight line)

< R = < P = 45 $^{\circ}$ (corresponding angles)

< PSO = < P = 45 $^{\circ}$ ( $△$ PSO is a right angle)

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Circles

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Question 32 Report

The quadratic equation whose roots are $1 - \sqrt{13}$ and $1 + \sqrt{13}$ is

x2 + (1 -

\sqrt{13}

)x + 1 +

\sqrt{13}

= 0

x2 - 2x - 12 = 0

x2 - 2x + 12 = 0

x2 + 12 + 2x2 = 0

Answer Details

1 -

\sqrt{13}

and 1 +

\sqrt{13}

Product of roots = (1 -

\sqrt{13}

) (1 +

\sqrt{13}

) = -12
x2 - (sum of roots) x + (product of roots) = 0
x2 - 2x + 12 = 0

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Quadratic Equations

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Question 33 Report

If $a = \frac{2x}{1-x}$ and $b = \frac{1+x}{1-x}$, then $a^2 - b^2$ in the simplest form is $\frac{3x+1}{1-x}$

\frac{3 x + 1}{1 - x}

3x²−1

(x−1)²

3x−2

x−1

3x−2

1−X

3x²−1

(x−1)

Answer Details

a2 - b2 = (

\frac{2 x}{1 - x}

)2 - (

\frac{1 + x}{1 - x}

)2
= (

\frac{2 x}{1 - x} + \frac{1 + x}{1 - x}

)(

\frac{2 x}{1 - x} - \frac{1 + x}{1 - X}

)
= (

\frac{3 x + 1}{1 - x}

)(

\frac{x - 1}{1 - x}

)
=

\frac{3 x + 1}{x - 1}

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Linear Inequalities

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Question 34 Report

Rationalize $ \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}} $

-2 \sqrt{35}\)

4\sqrt{7}\) - 6

\sqrt{5}

-√35

4\sqrt{7}\) - 8

\sqrt{5}

√35

Answer Details

\frac{5 \sqrt{7} - 7 \sqrt{5}}{\sqrt{7} - \sqrt{5}}

=

\frac{5 \sqrt{7} - 7 \sqrt{5}}{\sqrt{7} - \sqrt{5}}

x

\frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}}

=

\frac{(5 \times 7) + (5 \sqrt{7} \times 5) - (7 \times \sqrt{5} \times 7) (- 7 \times 5)}{(\sqrt{7})^{2}}

=

\frac{5 \sqrt{35} - 7 \sqrt{35}}{2}

=

\frac{- 2 \sqrt{35}}{2}

= -

\sqrt{35}

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Positive And Negative Integers, Rational Numbers

Surds (radicals)

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Question 35 Report

The cost of production of an article is made up as follows: Labour ₦70, Power ₦15, Materials ₦30, Miscellaneous ₦5. Find the angle of the sector representing Labour in a pie chart

210o

105o

106o

175o

90o

Answer Details

Total cost of production = ₦120.00
Labour Cost =

\frac{70}{120}

x

\frac{360^{o}}{1}

= 120o

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Algebraic Expressions

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Question 36 Report

Find, without using logarithm tables, the value of $ \frac{\log_{3} 27-\log_{\frac{1}{4}} 64}{\log_{3} \frac{1}{81}} $

−3

9

−3

2

6

11

43

18

Answer Details

log327 = 3log33
=3 log3

\frac{1}{81}

= -4 log33 = -4
let log

\frac{1}{4}

64 = (

\frac{1}{4}

)x
= 64
4-x = 43

\frac{l o g_{3} 27 - l o g_{\frac{1}{4}} 64}{l o g_{3} \frac{1}{81}}

=

\frac{3 - (- 3)}{- 4}

=

\frac{- 6}{4}

=

\frac{- 3}{2}

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Modular Arithmetic

Logarithms

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Question 37 Report

P sold his bicycle to Q at a profit of 10%. How much did the bicycle cost P?

₦200

₦196

₦180

₦205

₦150

Answer Details

Let the selling price(SP from P to Q be represented by x
i.e. SP = x
When SP = x at 10% profit
CP =

\frac{100}{100}

+ 10 of x =

\frac{100}{110}

of x
when Q sells to R, SP = ₦209 at loss of 5%
Q's cost price = Q's selling price
CP =

\frac{100}{95}

x 209
= 220.00
x = 220
=

\frac{2200}{11}

= 200
= ₦200.00

Read lesson note on Sine, Cosine And Tangent Of An Angle (WAEC)

Read lesson note on Percentages (WAEC)

Read lesson note on Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Percentages

Modular Arithmetic

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Question 38 Report

Find a factor which is common to all 3 binomial expressions $4a^2 - 9b^2,\; 8a^3 + 27b^3,\; (4a + 6b)^2$.

4a + 6b

4a - 6b

2a + 3b

2a - 3b

None

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Question 39 Report

In the diagram, find the size of the angle marked ao

60o

80o

120o

160o

Answer Details

2 x s = 280o(Angle at centre = 2 x < at circum)

S = $\frac{280^{o}}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

Read lesson note on Angles (WAEC)

Angles

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Question 40 Report

XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY.

5

\sqrt{3}

cm

3

\sqrt{5}

cm

3

\sqrt{3}

cm

5cm

6cm

Answer Details

by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY2 = 62 + 32

XY2 = 36 + 9 = 45

XY = $\sqrt{45} = 3 \sqrt{3}$

Read lesson note on Triangles And Polygons (WAEC)

Triangles And Polygons

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Question 41 Report

If (x - 2) and (x + 1) are factors of the expression x3 + px2 + qx + 1, what is the sum of p and q

9

-3

3

17

3

2

3

Answer Details

x3 + px2 + qx + 1 = (x - 1) Q(x) + R
x - 2 = 0, x = 2, R = 0,
4p + 2p = -9........(i)
x3 + px2 + qx + 1 = (x - 1)Q(x) + R
-1 + p - q + 1 = 0
p - q = 0.......(ii)
Solve the equation simultaneously
p =

\frac{- 3}{2}

q =

\frac{- 3}{2}

p + q =

\frac{3}{2}

-

\frac{3}{2}

=

\frac{- 6}{2}

= -3

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

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Question 42 Report

A straight lie y = mx meets the curve y = x2 - 12x + 40 in two distinct points. If one of them is (5, 5) find the other

(5, 6)

(8, 8)

(8, 5)

(7,7)

(7, 5)

Answer Details

When y = 5, y = x2 - 12x + 40, becomes
x2 - 12x + 40 = 5
x2 - 12x + 40 - 5 = 0
x2 + 12x + 35 = 0
x2 - 7x - 5x + 35 = 0
x(x - 7) - 5(x - 7) = 0
= (x - 5)(x - 7)
x = 5 or 7

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Question 43 Report

If 2x + 3y = 1 and x - y = 11, find (x + y).

5

3

8

2

11

Solution Of Linear Equations

Algebraic Expressions

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Question 44 Report

A trader in country where their currency 'MONI'(M) is in base five bought 1035 oranges at M145 each. If he sold the oranges at M245 each, what will be his gain?

1035

10305

1025

20025

30325

Answer Details

Total cost of 1035 oranges at ₦145 each
= 1035 x 145
= 20025
Total selling price at ₦245 each
= (103)5 x 245
= 30325
Hence his gain = 30325 - 20025
= 10305

Read lesson note on Number Bases (WAEC)

Read lesson note on Number Bases (JAMB)

Number Bases

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Question 45 Report

If f(x) = 2(x - 3)2 + 3(x - 3) + 4 and g(y) = 5 + y, find g [f(3)] and f[g(4)].

3 and 4

-3 and 4

-3 and -4

3 and -4

0 and 5

Answer Details

f(x0 = 2(x - 3)2 + 3(x - 3) + 4
= (2 + 3) + (x - 3) + 4
5(x - 3) + 4
5x - 15 + 4
= 5x - 11
f(3) = 5 x 3 - 11
= 4

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
\(y = x^3 - 3x + 1\)	\(1\)	\(-1\)	\(3\)	\(1\)	\(-1\)	\(3\)	\(1\)






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Score(\(n\))	Frequency(\(f\))
3	30
4	32
5	30
6	35
8	20