JAMB UTME - Mathematics - 1984

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 2 Ripọtì

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)

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Linear Inequalities

Lee Azịza

Ajụjụ 3 Ripọtì

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

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 4 Ripọtì

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

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Fractions (WAEC)

Solution Of Linear Equations

Algebraic Fractions

Lee Azịza

Ajụjụ 5 Ripọtì

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

₦200

₦196

₦180

₦205

₦150

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Percentages

Modular Arithmetic

Lee Azịza

Ajụjụ 6 Ripọtì

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

Lee Azịza

Ajụjụ 7 Ripọtì

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

Akọwa Nkọwa

\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}

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Lee Azịza

Ajụjụ 8 Ripọtì

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

29

^{\circ}

31

^{\circ}

31

^{\circ}

20'

31

^{\circ}

18'

Akọwa Nkọwa

\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'

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Triangles And Polygons

Lee Azịza

Ajụjụ 9 Ripọtì

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

Akọwa Nkọwa

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

Lee Azịza

Ajụjụ 10 Ripọtì

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

1

6

1

27

4

3

Akọwa Nkọwa

\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}

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Fractions, Decimals, Approximations And Percentage (JAMB)

Algebraic Expressions

Fractions, Decimals, Approximations And Percentage

Lee Azịza

Ajụjụ 11 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Variation (WAEC)

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Variation

Solution Of Linear Equations

Lee Azịza

Ajụjụ 12 Ripọtì

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

Akọwa Nkọwa

by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY2 = 62 + 32

XY2 = 36 + 9 = 45

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

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Triangles And Polygons

Lee Azịza

Ajụjụ 13 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Ratio, Proportions And Rates (WAEC)

Modular Arithmetic

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 14 Ripọtì

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

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Modular Arithmetic

Logarithms

Lee Azịza

Ajụjụ 15 Ripọtì

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

Lee Azịza

Ajụjụ 16 Ripọtì

$ \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

Akọwa Nkọwa

$\frac{0.0001432}{1940000}$ = 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

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 17 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 18 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Linear Inequalities

Lee Azịza

Ajụjụ 19 Ripọtì

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

4cm

16cm

9cm

14cm

Impossible, insufficient data

Akọwa Nkọwa

\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

Lee Azịza

Ajụjụ 20 Ripọtì

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)

Akọwa Nkọwa

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

Lee Azịza

Ajụjụ 21 Ripọtì

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

Akọwa Nkọwa

\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

Lee Azịza

Ajụjụ 22 Ripọtì

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)

Lee Azịza

Ajụjụ 23 Ripọtì

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

2

4

6

7

8

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Bearings (WAEC)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Bearings

Circles

Surds (radicals)

Quadratic Equations

Lee Azịza

Ajụjụ 24 Ripọtì

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}

Akọwa Nkọwa

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 = $\frac{120}{2} = 60^{o}$

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Triangles And Polygons

Lee Azịza

Ajụjụ 25 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Modular Arithmetic

Coordinate Geometry

Lee Azịza

Ajụjụ 26 Ripọtì

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

Lee Azịza

Ajụjụ 27 Ripọtì

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})

Akọwa Nkọwa

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 $△$

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

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

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

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Number Bases

Areas

Lee Azịza

Ajụjụ 28 Ripọtì

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

Akọwa Nkọwa

< 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

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 29 Ripọtì

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

Akọwa Nkọwa

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

Lee Azịza

Ajụjụ 30 Ripọtì

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

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 31 Ripọtì

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

Akọwa Nkọwa

6 + 10 + 14 + 16 + 26 = 72

\frac{6}{72}

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

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Lengths And Perimeters (WAEC)

Modular Arithmetic

Sets

Surds (radicals)

Lengths And Perimeters

Lee Azịza

Ajụjụ 32 Ripọtì

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

Akọwa Nkọwa

\frac{r}{h}

= tan yo
h =

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

= r cot yo

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Introductory Calculus

Circles

Lee Azịza

Ajụjụ 33 Ripọtì

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

5

3

8

2

11

Solution Of Linear Equations

Algebraic Expressions

Lee Azịza

Ajụjụ 34 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Number Bases

Lee Azịza

Ajụjụ 35 Ripọtì

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

Akọwa Nkọwa

\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}

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Sine, Cosine And Tangent Of An Angle

Lee Azịza

Ajụjụ 36 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Algebraic Expressions

Coordinate Geometry

Lee Azịza

Ajụjụ 37 Ripọtì

What is the volume of this regular three dimensional figure?

160cm2

48cm2

120cm2

40cm2

Akọwa Nkọwa

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

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

= 6cm2

V = 6 x 8

= 48cm2

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Volumes

Lee Azịza

Ajụjụ 38 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 39 Ripọtì

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

Akọwa Nkọwa

Total cost of production = ₦120.00
Labour Cost =

\frac{70}{120}

x

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

= 120o

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 40 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Gụọ akwụkwọ nkuzi na Simple Operations On Algebraic Expressions (WAEC)

Statistics

Simple Operations On Algebraic Expressions

Lee Azịza

Ajụjụ 41 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Probability

Linear Inequalities

Statistics

Lee Azịza

Ajụjụ 42 Ripọtì

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

Lee Azịza

Ajụjụ 43 Ripọtì

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}

Akọwa Nkọwa

< 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)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 44 Ripọtì

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

Akọwa Nkọwa

$\frac{2}{3} - \frac{1}{5}$ = $\frac{10 - 3}{15}$
= $\frac{7}{15}$
$\frac{1}{3}$ 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 $\frac{3}{7}$
= $\frac{1}{7}$