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Tẹ & Di mu lati Gbe Yika
Pịa Ebe a ka Imechi

JAMB UTME - Mathematics - 1985

Sọ silẹ

Ajụjụ 1 Ripọtì

A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively?

5kg

16kg

19kg

6kg

Akọwa Nkọwa

$\frac{1 \sqrt{3}}{(\frac{1}{2})^{2}}$
= $\frac{4}{\sqrt{3}}$
= $\frac{\sqrt{3}}{\sqrt{3}}$
= $\frac{4 \sqrt{3}}{\sqrt{3}}$
m = 64kg, V = $\frac{4 π r^{3}}{3}$
= $\frac{4 π (4)^{3}}{3}$
= $\frac{256 π}{3}$ x 10-6m3
density(P) = $\frac{Mass}{Volume}$
= $\frac{64}{\frac{256 π}{3 \times 10^{- 6}}}$
= $\frac{64 \times 3 \times 10^{- 6}}{256}$
= $\frac{3}{4 \times 10^{- 6}}$
m = PV = $\frac{3}{4 π \times 10^{- 6}}$ x $\frac{4}{3}$ $π$ [32 - 22] x 10-6
$\frac{3}{4 \times 10^{- 6}}$ x $\frac{4}{3}$ x 5 x 10-6
= 5kg

Ajụjụ 2 Ripọtì

( $\sqrt{2} + 4 \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) is equal to

1

(

\sqrt{2} + 4 \sqrt{2}

)

(6

\sqrt{2}

8

Akọwa Nkọwa

( $\sqrt{2} + 4 \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ )
( $\sqrt{3} + \sqrt{2}$ + ( $\sqrt{3} + \sqrt{2}$ ) = 3 + ( $\sqrt{6} - \sqrt{6}$ ) - 2
= 1

Ajụjụ 3 Ripọtì

John gives one-third of his money to Janet who has ₦105.00. He then finds that his money is reduced to one-fourth of what Janet now has. Find how much money john has at first

₦45.00

₦48.00

₦52.00

₦60.00

₦53.00

Akọwa Nkọwa

Let x be John's money, Janet already had ₦105, $\frac{1}{3}$ of x was given to Janet
Janet now has $\frac{1}{3^{2}}$ x + 105 = $\frac{x + 315}{3}$
John's money left = $\frac{2}{3}$ x
= $\frac{\frac{1}{4} (x + 315)}{3}$
= $\frac{2}{3}$
24x = 3x + 945
∴ x = 45

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Ajụjụ 4 Ripọtì

The number of goals scored by a football team in 20 matches is shown below:

$\begin{array}{cc} No. of goals & 0 & 1 & 2 & 3 & 4 & 5 \\ No. of matches & 3 & 5 & 7 & 3 & 1 & 0 \end{array}$

What are the values of the mean and the mode respectively?

(1.75, 5)

(1.75, 2)

(1.75, 1)

(65, 74)

Akọwa Nkọwa

$\begin{array}{cc} x & f & f x \\ 0 & 3 & 0 \\ 1 & 5 & 5 \\ 2 & 7 & 14 \\ 3 & 4 & 12 \\ 4 & 1 & 4 \\ 5 & 0 & 0 \end{array}$
$\sum f$ = 20
$\sum f x$ = 35
Mean = $\frac{\sum f x}{\sum f}$
= $\frac{35}{20}$
= $\frac{7}{4}$
= 1.75
Mode = 2
= 1.75, 2

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Probability

Ajụjụ 5 Ripọtì

What is the probability that a number chosen at random from the intergers between 1 and 10 inclusive is either a prime or a multiple of 3?

\frac{7}{10}

\frac{3}{5}

\frac{4}{5}

\frac{3}{10}

Akọwa Nkọwa

Simple Space: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10)
Prime: (2, 3, 5, 7)
multiples of 3: (3, 6, 9)
Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6)
Probability = $\frac{6}{10}$
= $\frac{3}{5}$

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Ajụjụ 6 Ripọtì

List all integers satisfying the inequality -2 $\leq$ 2 x -6 < 4

2, 3, 4, 5

2, 3, 4

2, 5

3, 4, 5

4, 5

Akọwa Nkọwa

-2 $\leq$ 2x - 6 < 4 = 2x - 6 < 4

= 2x < 10

= x < 5

2x $\geq$ -2 + 6 $\geq$

= x $\geq$ 2

∴ 2 $\leq$ x < 5 [2, 3, 4]

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Inequalities

Ajụjụ 9 Ripọtì

If three numbers P, Q, R are in ratio 6 : 4 : 5, find the value of $\frac{3 p - q}{4 q + r}$

3

2

2

3

2

3

18

Akọwa Nkọwa

P : Q : r = 6 : 4 : 5
5 = 6 + 4 + 5
= 15
P = $\frac{6}{15}$ , q = $\frac{4}{15}$ , r = $\frac{5}{15}$ = $\frac{1}{3}$
To find $\frac{3 p - q}{4 q + r}$
3p - q = 3 x $\frac{6}{15}$ - $\frac{4}{15}$
$\frac{18}{15}$ - $\frac{4}{15}$ = $\frac{14}{15}$
∴ 4q + r = 4 x $\frac{4}{15}$ + $\frac{5}{15}$
$\frac{16}{15}$ = $\frac{16}{15}$ + $\frac{5}{15}$
= $\frac{21}{15}$
$\frac{14}{15}$ x $\frac{15}{21}$ = $\frac{14}{21}$
= $\frac{2}{3}$

Ajụjụ 10 Ripọtì

Arrange the following numbers in ascending order of magnitude $\frac{6}{7}$ , $\frac{13}{15}$ , 0.8650

$\frac{6}{7}$ < 0.865 < $\frac{13}{15}$

$\frac{13}{15}$ < $\frac{6}{7}$ < 0.865

$\frac{6}{7}$ < $\frac{13}{15}$ < 0.865

0.865 < $\frac{6}{7}$ < $\frac{13}{15}$

Akọwa Nkọwa

$\frac{6}{7}$ , $\frac{13}{15}$ , 0.8650

In ascending order, we have 0.8571, 0.8650, 0.8666

i.e. $\frac{6}{7}$ < 0.8650 < $\frac{13}{15}$

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Ajụjụ 12 Ripọtì

Find correct to two decimals places 100 + $\frac{1}{100}$ + $\frac{3}{1000}$ + $\frac{27}{10000}$

100.02

1000.02

100.22

100.01

100.51

Akọwa Nkọwa

100 + $\frac{1}{100}$ + $\frac{3}{1000}$ + $\frac{27}{10000}$
$\frac{1000, 000 + 100 + 30 + 27}{10000}$ = $\frac{1, 000.157}{10000}$
= 100.02

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Number Bases

Ajụjụ 13 Ripọtì

If cos $θ$ = $\frac{\sqrt{3}}{2}$ and $θ$ is less than 90o. Calculate cos $\frac{90 - θ}{s i n^{2} θ}$

\frac{4}{\sqrt{3}}

\sqrt{\frac{3}{2}}

\sqrt{\frac{76}{2}}

4

\sqrt{\frac{76}{2}}

Akọwa Nkọwa

cos $\frac{90 - θ}{s i n^{2} θ}$
= $\frac{\tan θ}{s i n^{2} θ}$
Sin2 $θ$ = $\frac{1}{4}$
$\frac{c o s (90 - θ)}{s i n^{2} θ}$
= $\frac{1}{\sqrt{3}}$
= $\frac{4}{\sqrt{3}}$

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

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Ajụjụ 14 Ripọtì

In the figure, GHIJKLMN is a cube of side a. Find the length of HN.

\sqrt{3 a}

3a

3a2

a

\sqrt{2}

a

\sqrt{3}

Akọwa Nkọwa

HJ2 = a2 + a2 = 2a2

HJ = $\sqrt{2 a^{2}} = a \sqrt{2}$

HN2 = a2 + (a $\sqrt{2}$ )2 = a2 + 2a2 = 3a2

HN = $\sqrt{3 a^{2}}$

= a $\sqrt{3}$ cm

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Gụọ akwụkwọ nkuzi na Mensuration (JAMB)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Polynomials

Mensuration

Inequalities

Ajụjụ 15 Ripọtì

A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?

\frac{2}{3}

\frac{2}{15}

\frac{1}{2}

\frac{1}{3}

Akọwa Nkọwa

P(R1) = $\frac{6}{10}$
= $\frac{2}{3}$
P(R1 n R11) = P(both red)
$\frac{3}{5}$ x $\frac{5}{9}$
= $\frac{1}{3}$

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Probability (JAMB)

Probability

Probability

Ajụjụ 16 Ripọtì

Solve the simultaneous equations 2x - 3y + 10 = 10x - 6y = 5

x = 2

\frac{1}{2}

, y = 3

\frac{1}{2}

x =

\frac{1}{2}

, y =

\frac{3}{2}

x = 2

\frac{1}{4}

, y = 3

\frac{1}{2}

x = 2

\frac{1}{3}

, y = 3

\frac{1}{2}

x = 2

\frac{1}{3}

, y = 2

\frac{1}{2}

Akọwa Nkọwa

2x - 3y = -10; 10x - 6y = -5
2x - 3y = -10 x 2
10x - 6y = 5
4x - 6y = -20 .......(i)
10x - 6y = 5.......(ii)
eqn(ii) - eqn(1)
6x = 15
x = $\frac{15}{6}$
= $\frac{5}{2}$
x = 2 $\frac{1}{2}$
Sub. for x in equ.(ii) 10( $\frac{5}{2}$ ) - 6y = 5
y = 3 $\frac{1}{2}$

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Ajụjụ 17 Ripọtì

Two points X and Y both on latitude 60oS have longitude 147oE and 153oW respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude (Take 2 $π$ R = 4 x 104km, where R is the radius of the earth)

16667km

28850km

8333km

2233km

Akọwa Nkọwa

Length of an area = $\frac{θ}{360}$ × 2 $π$ r
Longitude difference = 147 + 153 = 300oN
distance between xy = $\frac{θ}{360}$ × 2 $π$ R cos60o
= $\frac{300}{360}$ × 4 × 104 × $\frac{1}{2}$
= 1.667 × 104km (1667 km)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Ajụjụ 18 Ripọtì

Make c the subject of formula v = 1 - $\frac{a}{5}$ (b + $\frac{3 c}{7}$ )

[

\frac{7}{3}

+

\frac{5}{a}

(v - 1)] + b

[

\frac{7 b}{3}

+

\frac{5}{a}

(v - 1)]

\frac{7}{3}

[b +

\frac{5}{a}

(v - 1)]

\frac{7}{3}

[b +

\frac{5 v - 1}{a}

]

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Ajụjụ 19 Ripọtì

At what real value of x do the curves whose equations are y = x3 + x and y = x2 + 1 intersect?

-2

2

-1

9

1

Akọwa Nkọwa

y = x3 + x and y = x2 + 1
$\begin{array}{cc} x & - 2 & - 1 & 0 & 1 & 2 \\ Y = x^{3} + x & - 10 & - 2 & 0 & 2 & 10 \\ y = x^{2} + 1 & 5 & 2 & 1 & 2 & 5 \end{array}$
The curves intersect at x = 1

Gụọ akwụkwọ nkuzi na Coordinate Geometry (JAMB)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Coordinate Geometry

Quadratic Equations

Ajụjụ 20 Ripọtì

In $△$ XYZ, XKZ = 90O, XK = 15cm, XY = 35cm and YK = 8cm. Find the area of $△$ XYZ

190 sq.cm

20 sq.cm

210 sq.cm

160sq.cm

320sq.cm

Akọwa Nkọwa

By Pythagoras, KZ2 = 252 - 52

KZ2 = (25 + 15)(25 - 15) = 400

KZ = $\sqrt{400}$ = 20

area of XYZ = $\frac{1}{2} \times 28 \times 15$

= 210 sq. cm

Ajụjụ 21 Ripọtì

PRSQ is a trapezium of area 14cm2 in which PQ||RS. If PQ = 4cm and SR = 3cm, Find the area of SQR in cm2

7.0

6.0

5.2

5.0

Akọwa Nkọwa

Area of trapezium = 14cm2
Area of trapezium = $\frac{1}{2}$ (a + b)h
14 = $\frac{1}{2}$ (4 + 3)h
14 = $\frac{7}{2}$ h
h = $\frac{14 \times 2}{7}$
= 4
Area of SQR = $\frac{1}{2}$ (3 x 4)
= $\frac{12}{2}$
= 6.0

Ajụjụ 22 Ripọtì

The ratio of the length of two similar rectangular blocks is 2 : 3. If the volume of the larger block is 351cm3, then the volume of the other block is

234.00 cm3

526.50 cm3

166.00 cm3

687cm3

Akọwa Nkọwa

Let x represent total vol. 2 : 3 = 2 + 3 = 5
$\frac{3}{5}$ x = 351
x = $\frac{351 \times 5}{3}$
= 585
Volume of smaller block = $\frac{2}{3}$ x 585
= 234.00

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Ajụjụ 23 Ripọtì

If ex = 1 + $\frac{x}{2}$ + $\frac{x^{2}}{1.2}$ + $\frac{x^{3}}{1.2.3}$ + .....Find $\frac{1}{e^{x}}$

1 -

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{3}}{2^{4} .3}

+ .........

1 -

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{4}}{2.4.3}

+ ..

1 +

\frac{x}{2}

+

\frac{x^{2}}{1.2}

+

\frac{x^{3}}{1.2.3}

+

\frac{x^{4}}{1.23.4}

+ .........

1 - x +

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{3}}{2^{4} .3}

+ .........

1 +

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{4}}{1.2.6}

+ ........

Akọwa Nkọwa

ex = 1 + $\frac{x}{2}$ + $\frac{x^{2}}{1.2}$ + $\frac{x^{3}}{1.2.3}$ + $\frac{x^{4}}{1.23.4}$

Ajụjụ 24 Ripọtì

If the quadratic function 3x2 - 7x + R is a perfect square, find R

\frac{49}{24}

\frac{49}{12}

\frac{49}{13}

\frac{49}{3}

\frac{49}{36}

Akọwa Nkọwa

3x2 - 7x + R. Computing the square, we have
x2 - $\frac{7}{3}$ = - $\frac{R}{3}$
( $\frac{x}{1} - \frac{7}{6}$ )2 = - $\frac{R}{3}$ + $\frac{49}{36}$
$\frac{- R}{3}$ + $\frac{49}{36}$ = 0
R = $\frac{49}{36}$ x $\frac{3}{1}$
= $\frac{49}{12}$

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

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Fractions, Decimals, Approximations And Percentage

Quadratic Equations

Ajụjụ 25 Ripọtì

If a{ $\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}$ } = 6x. Find a in its simplest form

x2 - 1

x2 + 1

x2 + 4

x2 - 4

1

Akọwa Nkọwa

a{ $\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}$ } = a{ $\frac{(x + 1) (x + 2) - (x - 1) (x - 2)}{(x - 2) (x + 2)}$ }
= 6
$\frac{6 x}{x^{2} - 4}$ = 6x
a = x2 - 4

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Ajụjụ 26 Ripọtì

In the figure, the area of the shaded segment is

3

π

9

\frac{\sqrt{3}}{4}

3

π - 3 \frac{\sqrt{3}}{4}

\frac{(\sqrt{3 - π)}}{4}

π + \frac{9 \sqrt{3}}{4}

Akọwa Nkọwa

Area of sector = $\frac{120}{360} \times π \times (3)^{2} = 3 π$

Area of triangle = $\frac{1}{2} \times 3 \times 3 \times \sin 120^{o}$

= $\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9 \sqrt{3}}{4}$

Area of shaded portion = 3 $π - \frac{9 \sqrt{3}}{4}$

= 3 $π - 3 \frac{\sqrt{3}}{4}$

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Ajụjụ 27 Ripọtì

If the hypotenuse of right angled isosceles triangle is 2, what is the length of each of the other sides?

\sqrt{2}

\frac{1}{2}

22

1

2 - 1

Akọwa Nkọwa

45o = $\frac{x}{2}$ , Since 45o = $\frac{1}{\sqrt{2}}$
x = 2 x $\frac{1}{\sqrt{2}}$
= $\frac{2 \sqrt{2}}{2}$
= $\sqrt{2}$

Gụọ akwụkwọ nkuzi na Angles And Intercepts On Parallel Lines (WAEC)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Angles And Intercepts On Parallel Lines

Ajụjụ 30 Ripọtì

Without using table, calculate the value of 1 + sec2 30o

2

\frac{1}{3}

\frac{2}{15}

\frac{5}{3}

3

\frac{1}{2}

Akọwa Nkọwa

1 + sec2 30o = sec 30o
= $\frac{2}{\sqrt{3}}$
$\frac{(2)^{2}}{3}$
= $\frac{4}{3}$
1 + sec2 30o = sec 30o
= 1 + $\frac{4}{\sqrt{3}}$
= 2 $\frac{1}{3}$

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

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Loci (JAMB)

Gụọ akwụkwọ nkuzi na Matrices And Determinants (JAMB)

Sine, Cosine And Tangent Of An Angle

Introductory Calculus

Matrices And Determinants

Ajụjụ 31 Ripọtì

In $△$ XYZ, XY = 13cm, YZ = 9cm, XZ = 11cm and XYZ = $θ$ . Find cos $θ$ o

\frac{4}{39}

\frac{43}{39}

\frac{209}{3}

\frac{43}{78}

Akọwa Nkọwa

cos $θ$ = $\frac{13^{2} + 9^{2} - 11^{2}}{2 (13) (9)}$
= $\frac{169 + 81 - 21}{26 \times 9}$
cos $θ$ = $\frac{129}{26 \times 9}$
= $\frac{43}{78}$

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Introductory Calculus

Trigonometry

Ajụjụ 32 Ripọtì

In a class of 120 students, 18 of them scored an A grade in mathematics. If the section representing the A grade students on a pie chart has angle Zo at the centre of the circle, what is Z?

15

27

50

52

54

Akọwa Nkọwa

Total students = 120
grade = 18
Zo = $\frac{18}{120}$ x $\frac{360}{1}$
= 54o

Ajụjụ 33 Ripọtì

Solve the following equation $\frac{2}{2 r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$

(

\frac{5}{2}

, 1)

(5, -4)

(2, 1)

(1,

\frac{- 5}{2}

)

(

\frac{- 5}{2}

, 1)

Akọwa Nkọwa

$\frac{2}{2 r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$
$\frac{2}{2 r - 1}$ - $\frac{1}{r + 2}$ = $\frac{5}{3}$
$\frac{2 r + 4 - 2 r + 1}{2 r - 1 (r + 2)}$ = $\frac{5}{3}$
$\frac{5}{(2 r + 1) (r + 2)}$ = $\frac{5}{3}$
5(2r - 1)(r + 2) = 15
(10r - 5)(r + 2) = 15
10r2 + 20r - 5r - 10 = 15
10r2 + 15r = 25
10r2 + 15r - 25 = 0
2r2 + 3r - 5 = 0
(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)
(r - 1)(2r + 5) = 0
r = 1 or $\frac{- 5}{2}$

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Ajụjụ 36 Ripọtì

If two fair coins are tossed, what is the probability of getting at least one head?

\frac{1}{4}

\frac{1}{2}

1

\frac{43}{78}

\frac{3}{4}

Akọwa Nkọwa

Prob. of getting at least one head
Prob. of getting one head + prob. of getting 2 heads
= $\frac{1}{4}$ + $\frac{2}{4}$
= $\frac{3}{4}$

Ajụjụ 38 Ripọtì

Write h in terms of a, b, c, d if a = $\frac{b (1 - c h)}{a - d h}$

h = $\frac{a - b}{a d}$

h = $\frac{1 - b}{a d - b c}$

h = $\frac{(a - b)^{2}}{a d - b c}$

h =

\frac{a - b}{a d - b c}

h = $\frac{b - a}{a b - d c}$

Akọwa Nkọwa

a = $\frac{b (1 - c h)}{a - d h}$
a = $\frac{b - b c h}{1 - d h}$
= a - adh
= b - bch
a - b = bch + adn
a - b = adh
a - b = h(ad - bc)
h = $\frac{a - b}{a d - b c}$

Ajụjụ 39 Ripọtì

Find the area of a regular hexagon inscribed in a circle of radius 8cm

16

\sqrt{3}

cm2

96

\sqrt{3}

cm2

192

\sqrt{3}

cm2

16

\sqrt{3}

cm2

33cm2

Akọwa Nkọwa

Area of a regular hexagon = 8 x 8 x sin 60o
= 32 x $\frac{\sqrt{3}}{2}$ =
Area = 16 $\sqrt{3}$ x 6 = 96 $\sqrt{3}$ cm2

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Ajụjụ 40 Ripọtì

In the equation below, Solve for x if all the numbers are in base 2: $\frac{11}{x}$ = $\frac{1000}{x + 101}$

101

11

110

111

10

Akọwa Nkọwa

$\frac{11}{x}$ = $\frac{1000}{x + 101}$ = 11(x + 101)
1000x = 11x + 1111
1000x - 11x = 1111
101x = 1111
x = $\frac{1111}{101}$
x = 11

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Ajụjụ 41 Ripọtì

In a restaurant, the cost of providing a particular type of food is partly constant and partially inversely proportional to the number of people. If cost per head for 100 people is 30k and the cost for 40 people is 60k, Find the cost for 50 people?

15k

20k

50k

40k

45k

Akọwa Nkọwa

C = a + k
$\frac{1}{N}$ = c
= $\frac{a N + k}{N}$
CN = aN + K
30(100) = a(100) + k
3000 = 100a + k.......(i)
60(40) = a(40) + k
2400 = 40a + k.......(ii)
eqn (i) - eqn (ii)
600 = 60a
a = 10
subt. for a in eqn (i) 3000 = 100(10) + K
3000 - 1000 = k
k = 2000
CN = 10N + 2000. when N = 50,
50C = 10(50) + 2000
50C = 500 + 2000
C = $\frac{2500}{50}$
= 50k

Gụọ akwụkwọ nkuzi na Variation (WAEC)

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Linear Inequalities

Ajụjụ 42 Ripọtì

22 $\frac{1}{2}$ % of the Nigerian Naira is equal to 17 $\frac{1}{10}$ % of a foreign currency M. What is the conversion rate of the M to the Naira?

1M = 1

\frac{15}{57}

N

1M = 38

\frac{1}{4}

N

1M = 1

\frac{18}{57}

N

1M = 384

\frac{3}{4}

N

Akọwa Nkọwa

N = 22 $\frac{1}{2}$ %, M = 17 $\frac{1}{10}$ %
M = $\frac{171}{10}$ %, N = $\frac{45}{2}$
$\frac{45}{2}$ x $\frac{10}{171}$
= $\frac{225}{171}$
= 1 $\frac{54}{171}$
= 1 $\frac{18}{57}$

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Percentages

Financial Arithmetic

Ajụjụ 43 Ripọtì

a sum of money was invested at 8% per annum simple interest. If after 4 years the money amounts to N330.00. Find the amount originally invested

N180.00

n165.00

N150.00

N250.00

N200.00

Akọwa Nkọwa

S.I = $\frac{P T R}{100}$

T = 4yrs, R = 8%, a = N330.00

330 - P = $\frac{P T R}{100}$ , A = P + I

i.e. A = P + $\frac{P T R}{100}$

330 = P + $\frac{P (4) (8)}{100}$

33000 = 32P + 100p

132P = 33000

P = N250.00

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

Positive And Negative Integers, Rational Numbers

Ajụjụ 44 Ripọtì

In the figure, PQ is the tangent from P to the circle QRS with SR as its diameter. If QRS = $θ$ $^{\circ}$ and RQP = $ϕ$ $^{\circ}$ , which of the following relationships between $θ$ $^{\circ}$ and $ϕ$ $^{\circ}$ is correct

θ

^{\circ}

+

ϕ

^{\circ}

= 902

ϕ

^{\circ}

= 902 - 2

θ

^{\circ}

θ

^{\circ}

=

ϕ

^{\circ}

ϕ

^{\circ}

= 2

θ

^{\circ}

θ

^{\circ}

+ 2

ϕ

^{\circ}

Akọwa Nkọwa

180 - $ϕ$ $^{\circ}$ = $θ$ $^{\circ}$ + $ϕ$ $^{\circ}$ (Sum of opposite interior angle equal to its exterior angle)

180 = 2 $ϕ$ + $θ$ $^{\circ}$

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Introductory Calculus

Trigonometry

Ajụjụ 46 Ripọtì

In the figure, MNQP is a cyclic quadrilateral. MN and Pq are produced to meet at X and NQ and MP are produced to meet at Y. If MNQ = 86 $^{\circ}$ and NQP = 122 $^{\circ}$ find (x $^{\circ}$ , y $^{\circ}$ )

28

^{\circ}

, 36

^{\circ}

36

^{\circ}

, 28

^{\circ}

43

^{\circ}

, 61

^{\circ}

61

^{\circ}

, 43

^{\circ}

36

^{\circ}

, 43

^{\circ}

Akọwa Nkọwa

y $^{\circ}$ = 180 $^{\circ}$ - (86 $^{\circ}$ + 58 $^{\circ}$ )

180 - 144 = 36 $^{\circ}$

x $^{\circ}$ = 180 - (94 + 58)

180 -152 = 28

(x $^{\circ}$ , y $^{\circ}$ ) = (28 $^{\circ}$ , 36 $^{\circ}$ )

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Polynomials

Differentiation

Inequalities

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