JAMB UTME - Mathematics - 1984

Tambaya 1 Rahoto

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

Bayanin Amsa

Total cost of production = ₦120.00
Labour Cost =

\frac{70}{120}

x

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

= 120o

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 2 Rahoto

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

Duba Amsa

Tambaya 3 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 4 Rahoto

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

60o

80o

120o

160o

Bayanin Amsa

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

Karanta bayanin darasi akan Angles (WAEC)

Angles

Duba Amsa

Tambaya 5 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Logarithms (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Fractions, Decimals And Approximations

Number Bases

Logarithms

Quadratic Equations

Duba Amsa

Tambaya 6 Rahoto

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

Duba Amsa

Tambaya 7 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 8 Rahoto

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

5

3

8

2

11

Solution Of Linear Equations

Algebraic Expressions

Duba Amsa

Tambaya 9 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 10 Rahoto

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

Duba Amsa

Tambaya 11 Rahoto

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

1

6

1

27

4

3

Bayanin Amsa

\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}

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Fractions, Decimals, Approximations And Percentage (JAMB)

Algebraic Expressions

Fractions, Decimals, Approximations And Percentage

Duba Amsa

Tambaya 12 Rahoto

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

Bayanin Amsa

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}

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Algebraic Fractions (WAEC)

Solution Of Linear Equations

Algebraic Fractions

Duba Amsa

Tambaya 13 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Variation (WAEC)

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Variation

Solution Of Linear Equations

Duba Amsa

Tambaya 14 Rahoto

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)

Bayanin Amsa

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

Duba Amsa

Tambaya 15 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Linear Inequalities

Duba Amsa

Tambaya 16 Rahoto

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

Duba Amsa

Tambaya 17 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Modular Arithmetic

Coordinate Geometry

Duba Amsa

Tambaya 18 Rahoto

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

Bayanin Amsa

\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}

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Duba Amsa

Tambaya 19 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Ratio, Proportions And Rates (WAEC)

Modular Arithmetic

Ratio, Proportions And Rates

Duba Amsa

Tambaya 20 Rahoto

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

29

^{\circ}

31

^{\circ}

31

^{\circ}

20'

31

^{\circ}

18'

Bayanin Amsa

\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'

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Triangles And Polygons

Duba Amsa

Tambaya 21 Rahoto

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)

Bayanin Amsa

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}

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Linear Inequalities

Duba Amsa

Tambaya 22 Rahoto

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

Bayanin Amsa

by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY2 = 62 + 32

XY2 = 36 + 9 = 45

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

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Triangles And Polygons

Duba Amsa

Tambaya 23 Rahoto

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

Bayanin Amsa

6 + 10 + 14 + 16 + 26 = 72

\frac{6}{72}

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

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Karanta bayanin darasi akan Lengths And Perimeters (WAEC)

Modular Arithmetic

Sets

Surds (radicals)

Lengths And Perimeters

Duba Amsa

Tambaya 24 Rahoto

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

Bayanin Amsa

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}

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 25 Rahoto

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

2

4

6

7

8

Bayanin Amsa

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

Karanta bayanin darasi akan Bearings (WAEC)

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Bearings

Circles

Surds (radicals)

Quadratic Equations

Duba Amsa

Tambaya 26 Rahoto

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)

Duba Amsa

Tambaya 27 Rahoto

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}

Bayanin Amsa

< 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)

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 28 Rahoto

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

Bayanin Amsa

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

Duba Amsa

Tambaya 29 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Statistics (WAEC)

Karanta bayanin darasi akan Simple Operations On Algebraic Expressions (WAEC)

Statistics

Simple Operations On Algebraic Expressions

Duba Amsa

Tambaya 30 Rahoto

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})

Bayanin Amsa

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 $△$

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

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

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

Karanta bayanin darasi akan Number Bases (JAMB)

Karanta bayanin darasi akan Areas (WAEC)

Number Bases

Areas

Duba Amsa

Tambaya 31 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Karanta bayanin darasi akan Statistics (WAEC)

Probability

Linear Inequalities

Statistics

Duba Amsa

Tambaya 32 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Number Bases (JAMB)

Number Bases

Duba Amsa

Tambaya 33 Rahoto

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

Bayanin Amsa

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}

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Logarithms (WAEC)

Modular Arithmetic

Logarithms

Duba Amsa

Tambaya 34 Rahoto

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

Bayanin Amsa

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

Duba Amsa

Tambaya 35 Rahoto

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

₦200

₦196

₦180

₦205

₦150

Bayanin Amsa

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

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Percentages (WAEC)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Percentages

Modular Arithmetic

Duba Amsa

Tambaya 36 Rahoto

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

Bayanin Amsa

\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}

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Sine, Cosine And Tangent Of An Angle

Duba Amsa

Tambaya 37 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Algebraic Expressions

Coordinate Geometry

Duba Amsa

Tambaya 38 Rahoto

What is the volume of this regular three dimensional figure?

160cm2

48cm2

120cm2

40cm2

Bayanin Amsa

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

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

= 6cm2

V = 6 x 8

= 48cm2

Karanta bayanin darasi akan Volumes (WAEC)

Volumes

Duba Amsa

Tambaya 39 Rahoto

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

Bayanin Amsa

\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

Duba Amsa

Tambaya 40 Rahoto

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

4cm

16cm

9cm

14cm

Impossible, insufficient data

Bayanin Amsa

\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

Duba Amsa

Tambaya 41 Rahoto

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}

Bayanin Amsa

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

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Triangles And Polygons

Duba Amsa

Tambaya 42 Rahoto

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

Bayanin Amsa

\frac{r}{h}

= tan yo
h =

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

= r cot yo

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Circles (WAEC)

Introductory Calculus

Circles

Duba Amsa

Tambaya 43 Rahoto

$ \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

Bayanin Amsa

$\frac{0.0001432}{1940000}$ = 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

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 44 Rahoto

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

Bayanin Amsa

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}

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 45 Rahoto

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

Bayanin Amsa

< 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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

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






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Danna nan don rufewa

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