JAMB UTME - Mathematics - 1986

Ajụjụ 1 Ripọtì

Udoh deposited ₦150.00 in the bank. At the end of 5 years the simple interest on the principal was ₦55.00. At what rate per annum was the interest paid?

11%

7

\frac{1}{3}

%

5

3

\frac{1}{2}

%

Akọwa Nkọwa

.I = $\frac{P T R}{100}$
R = $\frac{100}{P T}$
$\frac{100 \times 50}{150 \times 6}$ = $\frac{22}{3}$
= 7 $\frac{1}{3}$ %

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

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

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

Sine, Cosine And Tangent Of An Angle

Introductory Calculus

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 2 Ripọtì

Find n if log24 + log27 - log2n = 1

10

14

27

28

Akọwa Nkọwa

log24 + log27 - log2n = 1
= log2(4 x 7) - log2n = 1
= log228 - log2n
= log $\frac{28}{2 n}$
$\frac{28}{2 n}$ = 21
= 2
2n = 28
∴ n = 14

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 3 Ripọtì

Factorize x2 + 2a + ax + 2x

(x + 2a)(x + 1)

(x + 2a)(x - 1)

(x2 - 1)(x - a)

(x + 2)(x + a)

Algebraic Expressions

Lee Azịza

Ajụjụ 4 Ripọtì

Find the smallest number by which 252 can be multiplied to obtain a perfect square

2

3

5

7

Introductory Calculus

Number Bases

Surds (radicals)

Lee Azịza

Ajụjụ 5 Ripọtì

In 1984, Ike was 24 yrs old and his father was 45 yrs old. In what year was Ike exactly half his father's age?

1981

1979

1982

1978

Akọwa Nkọwa

Let the no. of years be y
24 - y = $\frac{1}{2}$ (45 - y)
45 - y = 2(24 - y)
45 - y = 48 - 2y
2y - y = 48 - 45
∴ y = 3
The exact year = 1984
1984 - 3 = 1981

Lee Azịza

Ajụjụ 6 Ripọtì

In $ \triangle $XYZ, determine the cosine of angle Z.

3

39

29

36

14

5

7

5

Akọwa Nkọwa

cos z = $\frac{y^{2} + x^{2} - z^{2}}{2 y x}$

= $\frac{9 + 36 - 16}{2 (3) (6)}$

= $\frac{29}{36}$

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 7 Ripọtì

make U the subject of the formula $S = \sqrt{\frac{6}{u} - \frac{w}{2}}$

u =

\frac{12}{2 s^{2}}

u =

\frac{12}{2 s + w}

u =	12
	2s²−w

u =

\frac{12}{2 s^{2}}

+ w

Akọwa Nkọwa

S = $\sqrt{\frac{6}{u} - \frac{w}{2}}$
S = $\frac{12 - u w}{2 u}$
2us2 = 12 - uw
u(2s2 + w) = 12
u = $\frac{12}{2 s^{2} + w}$

Gụọ akwụkwọ nkuzi na Change Of Subject Of A Formula/Relation (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Change Of Subject Of A Formula/Relation

Quadratic Equations

Lee Azịza

Ajụjụ 8 Ripọtì

Evaluate (212)3 - (121)3 + (222)3

(313)3

(1020)3

(1222)3

Number Bases

Lee Azịza

Ajụjụ 9 Ripọtì

Of the nine hundred students admitted in a university in 1979, the following was the distribution by state: Anambrs 185, Imo 135, Kaduna 90, Kwara 110, Ondo 155, Oyo 225. In a pie chart drawn to represent this distribution, The angle subtended at the centre by anambra is

50o

65o

74o

88o

Akọwa Nkọwa

Anambra = $\frac{185}{900}$ x $\frac{360}{1}$
= 74o

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Variation

Euclidean Geometry

Sets

Lee Azịza

Ajụjụ 10 Ripọtì

Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ

35o

37o

49o

59o

Polynomials

Euclidean Geometry

Lee Azịza

Ajụjụ 11 Ripọtì

An arc of a circle of radius 6cm is 8cm long. Find the area of the sector

5

\frac{1}{3}

cm2

24cm2

36cm2

84cm2

Akọwa Nkọwa

Radius of the circle r = 6cm, Length of the arc = 8cm
Area of sector = $\frac{θ}{360}$ x 2 $π$ r2........(i)
Length of arc = $\frac{θ}{360}$ x 2 $π$ r........(ii)
from eqn. (ii) $θ$ = $\frac{240}{π}$ , subt. for $θ$ in eqn (i)
Area x $\frac{240}{1}$ x $\frac{1}{360}$ x $\frac{π 6}{1}$
= 24cm2

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Coordinate Geometry

Lee Azịza

Ajụjụ 12 Ripọtì

An open rectangular box externally measures 4m x 3m x 4m. Find the total cost of painting the box externally if it costs ₦2.00 to paint one square meter

₦96.00

₦112.00

₦136.00

₦160.00

Lee Azịza

Ajụjụ 13 Ripọtì

The table below gives the scores of a group of students in a Mathematical test.

Scores	1	2	3	4	5	6	7	8
Frequency	2	4	7	14	12	6	4	1

If the mode in m and the number of students who scored 4 or less is s. What is (s, m)?

(27, 4)

(14,4)

(13,4)

(4,4)

Representation Of Data

Number Bases

Lee Azịza

Ajụjụ 14 Ripọtì

If 5(x + 2y) = 5 and 4(x + 3y) = 16, find 3(x + y)

7

1

3

27

Algebraic Expressions

Lee Azịza

Ajụjụ 15 Ripọtì

Simplify $\left(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\right)x\frac{1}{\sqrt{3}}$

√3

√5

2√3

7

-2

-1

Akọwa Nkọwa

( $\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}$ x $\frac{1}{\sqrt{3}}$ )
$\frac{1}{\sqrt{5} + \sqrt{3}} - \frac{1}{\sqrt{5} - \sqrt{3}}$ = $\frac{\sqrt{5} - \sqrt{3} - (\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3}) (5 \sqrt{3})}$
= $\frac{- 2 \sqrt{3}}{5 - 3}$
= $\frac{- 2 \sqrt{3}}{7}$

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Number Bases

Surds (radicals)

Lee Azịza

Ajụjụ 16 Ripọtì

Solve the equation 3x2 + 6x - 2 = 0

x = -1

\pm

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

x = -1

\pm

\frac{\sqrt{15}}{3}

x = -2

\pm

2

x = 3

\pm

\frac{\sqrt{3}}{15}

Akọwa Nkọwa

3x2 + 6x - 2 = 0
Using almighty formula i.e. x = $\frac{b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
a = 3, b = 6, c = -2
x = $\frac{- 6 \pm \sqrt{6^{2} - 4 (3) (- 2)}}{2 (3)}$
x = $\frac{6 \pm \sqrt{36 - 24}}{6}$
x = $\frac{6 \pm \sqrt{60}}{6}$
x = $\frac{- 6 \pm \sqrt{4 \times 15}}{6}$
x = $\frac{- 1 \pm \sqrt{15}}{3}$

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 17 Ripọtì

If $ y = \frac{x}{x - 3} + \frac{x}{x + 4} $ find y when $ x = -2 $

-

\frac{3}{5}

\frac{3}{5}

-

\frac{7}{5}

\frac{3}{5}

Akọwa Nkọwa

y = $\frac{x}{x - 3}$ + $\frac{x}{x + 4}$ when x = -2
y = $\frac{- 2}{- 5}$ + $\frac{(- 2)}{- 2 + 4}$
= $\frac{- 2}{5}$ + $\frac{- 2}{2}$
= $\frac{4 + - 10}{10}$
= $\frac{- 14}{10}$
= - $\frac{7}{5}$

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Algebraic Expressions

Measures Of Location

Quadratic Equations

Lee Azịza

Ajụjụ 18 Ripọtì

A man kept 6 black, 5 brown and 7 purple shirts in a drawer. What is the probability of his picking a purple shirt with his eyes closed?

\frac{1}{7}

\frac{11}{18}

\frac{7}{18}

\frac{7}{11}

Probability

Lee Azịza

Ajụjụ 19 Ripọtì

The ages of Tosan and Isa differ by 6 and the product of their ages is 187. Write their ages in the form (x, y), where x > y.

(12, 6)

(23, 17)

(17, 11)

(18, 12)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 20 Ripọtì

Find the probability that a number selected at random from 40 to 50 is a prime

2

11

5

11

3

11

4

11

Lee Azịza

Ajụjụ 21 Ripọtì

If x varies directly as y3 and x = 2 when y = 1, find x when y = 5

2

10

125

250

Akọwa Nkọwa

x $α$ y3
x = ky3
k = $\frac{x}{y^{3}}$
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Variation

Lee Azịza

Ajụjụ 22 Ripọtì

If $ \cos \theta = \frac{a}{b} $, find $ 1 + \tan^2 \theta $

\frac{b^{2}}{a^{2}}

\frac{a^{2}}{b^{2}}

\frac{a^{2} + b^{2}}{b^{2} - a^{2}}

\frac{2 a^{2} + b^{2}}{a^{2} + b^{2}}

Akọwa Nkọwa

cos $θ$ = $\frac{a}{b}$ , Sin $θ$ = $\sqrt{\frac{b^{2} - a^{2}}{a}}$
Tan $θ$ = $\sqrt{\frac{b^{2} - a^{2}}{a^{2}}}$ , Tan 2 = $\sqrt{\frac{b^{2} - a^{2}}{a^{2}}}$
1 + tan2 $θ$ = 1 + $\frac{b^{2} - a^{2}}{a^{2}}$
= $\frac{a^{2} + b^{2} - a^{2}}{a^{2}}$
= $\frac{b^{2}}{a^{2}}$

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

Lee Azịza

Ajụjụ 23 Ripọtì

A number of pencils were shared out among Bisi, Sola and Tunde in the ratio of 2 : 3 : 5 respectively. If Bisi got 5, how many were share out?

15

25

30

50

Akọwa Nkọwa

Let x r3epresent total number of pencils shared

B : S : T = 2 + 3 + 5 = 10

2 : 3 : 5

= $\frac{2}{10}$ x y

= 5

2y =5

2y = 50

∴ y = $\frac{50}{2}$

= 25

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

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Financial Arithmetic

Lee Azịza

Ajụjụ 24 Ripọtì

Find the median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119

131

125

123

120

Akọwa Nkọwa

Arrange in ascending order
89, 100, 108, 119, |120, 130|, 131, 131, 141, 161
Median = $\frac{120 + 130}{2}$
= 125

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Measures Of Location

Lee Azịza

Ajụjụ 25 Ripọtì

Factorize (4a + 3)2 - (3a - 2)2

(a + 1)(a + 5)

(a - 5)(7a - 1)

(a + 5)(7a + 1)

a(7a + 1)

Algebraic Expressions

Lee Azịza

Ajụjụ 26 Ripọtì

Find the reciprocal of $ \frac{\frac{2}{3}}{\frac{1}{2}+\frac{1}{3}} $

4

5

4

2

5

6

7

Akọwa Nkọwa

$\frac{2}{3}$ = $\frac{2}{3}$
= $\frac{2}{3}$ x $\frac{6}{5}$
= $\frac{4}{5}$
reciprocal of $\frac{4}{5}$ = $\frac{1}{\frac{4}{5}}$
= $\frac{5}{4}$

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

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

Fractions, Decimals And Approximations

Positive And Negative Integers, Rational Numbers

Lee Azịza

Ajụjụ 27 Ripọtì

Factorize completely 8a + 125ax3

(2a + 5x2)(4 + 26ax)

a(2 + 5x)(4 - 10x + 25x2)

(2a + 5x)(4 - 10ax + 25x2)

a(2 + 5x)(4 + 10ax + 25x2)

Algebraic Expressions

Lee Azịza

Ajụjụ 28 Ripọtì

Divide the L.C.M of 48, 64, and 80 by their H.C.F

20

30

48

60

Akọwa Nkọwa

48 = 24 x 3, 64 = 26, 80 = 24 x 5
L.C.M = 26 x 3 x 5
H.C.F = 24
$\frac{2^{6} \times 3 \times 5}{2^{4}}$
= 22 x 3 x 5
= 4 x 3 x 5
= 12 x 5
= 60

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Number Bases

Sets

Lee Azịza

Ajụjụ 29 Ripọtì

The people in a city with a population of 0.9 million were grouped according to their ages. Use the diagram to determine the number of people in the 15 - 29 years group

29 x 104

26 x 104

16 x 104

13 x 104

Akọwa Nkọwa

15 - 29 years is represented by 104 $^{\circ}$

Number of people in the group is $\frac{104}{360}$ x 0.9m

= 260000 = 26 x 104

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 30 Ripọtì

A girl walks 45 meters in the direction 050o from a point Q to a point X. She then walks 24 meters in the direction 140o from X to a point Y. How far is she then from Q?

69m

57m

51m

21m

Akọwa Nkọwa

QY = 452 + 242 = 2025 + 576
= 2601
QY = $\sqrt{2601}$
= 51

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Euclidean Geometry

Lee Azịza

Ajụjụ 31 Ripọtì

Find the values of x which satisfy the equation 16x - 5 x 4x + 4 = 0

1 and 4

-2 and 2

0 and 1

-1 and 0

Solution Of Linear Equations

Lee Azịza

Ajụjụ 32 Ripọtì

If U and V are two distinct fixed points and W is a variable points such that UWV is a right angle, what is the locus of W?

The perpendicular bisector of UV

A circle with UV as radius

A line parallel to the line UV

A circle with the line UV as the diameter

Loci

Statistics

Lee Azịza

Ajụjụ 33 Ripọtì

Three boys shared some oranges. The first received 1/3 of the oranges and the second received 2/3 of the remaining. If the third boy received the remaining 12 oranges, how many oranges did they share

60

54

48

42

Number Bases

Modular Arithmetic

Lee Azịza

Ajụjụ 34 Ripọtì

Find the total surface area of solid cone of radius $2\sqrt{3}$ cm and slanting side $4\sqrt{3}$

8

\sqrt{3} π

cm2

24

π

cm2

15

\sqrt{3} π

cm2

36

π

cm2

Akọwa Nkọwa

Total surface area of a solid cone
r = 2 $\sqrt{3}$
= $π r^{2}$ + $π$ rH
H = 4 $\sqrt{3}$ , $π$ r(r + H)
∴ Area = $π$ 2 $\sqrt{3}$ [2 $\sqrt{3}$ + 4 $\sqrt{3}$ ]
= $π$ 2 $\sqrt{3}$ (6 $\sqrt{3}$ )
= 12 $π$ x 3
= 36 $π$ cm2

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 35 Ripọtì

If Musa scored 75 in biology instead of 57, his average mark in four subjects would have been 60. What was his total mark?

282

240

222

210

Akọwa Nkọwa

Let x represent Musa's total mark when he scores 57 in biology and Let Y represent Musa's total mark when he now scored 75 in biology, if he scored 75 in biology his new total mark will be $\frac{Y}{4}$ = 60, y = 4 x 60 = 240
To get his total mark when he scored 57, subtract 57 from 75 to give 18, then subtract this 18 from the new total mark(ie. 240)
= 240 - 18
= 222

Lee Azịza

Ajụjụ 36 Ripọtì

A regular polygon of n sides has 160o as the size of each interior angle. Find n

18

14

16

12

Polynomials

Triangles And Polygons

Lee Azịza

Ajụjụ 37 Ripọtì

If $ \frac{a}{c} = \frac{c}{d} = k $, find the value of $ \frac{3a^2-ac+c^2}{3b^2-bd+d^2} $

3k2

2k - k2

\frac{3 b^{2} k^{2} - b k^{2} d + d k^{2}}{3 b^{2} - b d + d^{2}}

k2

Akọwa Nkọwa

$\frac{a}{c}$ = $\frac{c}{d}$ = k
∴ $\frac{a}{b}$ = bk
$\frac{c}{d}$ = k
∴ c = dk
= $\frac{3 a^{2} - a c + c^{2}}{3 b^{2} - b d + d^{2}}$
= $\frac{3 (b k)^{2} - (b k) (d k) + d k^{2}}{3 b^{2} - b d + a^{2}}$
= $\frac{3 b^{2} k^{2} - b k^{2} d + d k^{2}}{3 b^{2} - b d + d^{2}}$
k = $\frac{3 b^{2} k^{2} - b k^{2} d + d k^{2}}{3 b^{2} - b d + d^{2}}$

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Circles

Algebraic Expressions

Quadratic Equations

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 38 Ripọtì

Simplify $ \frac{0.0324 \times 0.00064}{0.48 \times 0.012} $

3.6 x 102

3.6 x 10-2

3.6 x 10-3

3.6 x 10-4

Akọwa Nkọwa

$\frac{0.0324 \times 0.00064}{0.48 \times 0.012}$ = $\frac{324 \times 10^{- 4} \times 10^{- 5}}{48 x 10^{- 2} x 12 x 10^{- 3}}$
$\frac{324 \times 64 \times 1 0^{- 9}}{48 \times 12 \times 10^{- 5}}$ = 36 x 10-4
= 3.6 x 10-3

Lee Azịza

Ajụjụ 39 Ripọtì

If P = 18, Q = 21, R = -6 and S = -4, Calculate $ \frac{(P-Q)^3+S^2}{R^3} + S^2 $

-11

216

11

216

-43

116

43

116

Akọwa Nkọwa

$\frac{(P - Q)^{3} + S^{2}}{R^{3}}$ = $\frac{(18 - 21)^{3} + (- 4)^{2}}{(- 6)^{3}}$
= $\frac{- 27 + 16}{R^{3}}$
= $\frac{- 11}{- 216}$
= $\frac{11}{216}$

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 40 Ripọtì

Simplify $ \frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}} $

1

3

1

3

9

Akọwa Nkọwa

\frac{9^{\frac{1}{3}} \times 27^{- \frac{1}{3}}}{3^{- \frac{1}{6}} \times 3^{\frac{2}{3}}}

=

\frac{(3^{2})^{\frac{1}{3}} \times (3^{3})^{- \frac{1}{3}}}{3^{- \frac{1}{6}} \times 3^{\frac{2}{3}}}

=

\frac{3^{\frac{2}{3}} \times 3^{\frac{2}{3}}}{3^{- \frac{1}{6}} \times 3^{\frac{2}{3}}}

=

\frac{3^{\frac{5}{6}}}{3^{- \frac{5}{6}}}

= 1

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Representation Of Data

Probability

Number Bases

Lee Azịza

Ajụjụ 41 Ripọtì

Find all real numbers $x$ which satisfy the inequality $\frac{1}{3}(x + 1) - 1 > \frac{1}{5}(x + 4)$

x < 11

x < -1

x > 6

x > 11

Akọwa Nkọwa

$\frac{1}{3}$ (x + 1) - 1 > $\frac{1}{5}$ (x + 4)
= $\frac{x + 1}{3}$ - 1 > $\frac{x + 4}{5}$
$\frac{x + 1}{3}$ - $\frac{x + 4}{5}$ - 1 > 0
= $\frac{5 x + 5 - 3 x - 12}{15}$
= 2x - 7 > 15
= 2x > 12
= x > 11