JAMB UTME - Mathematics - 1986

Pergunta 1 Relatório

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

Detalhes da Resposta

In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)

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Pergunta 2 Relatório

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

131

125

123

120

Detalhes da Resposta

Arrange in ascending order
89, 100, 108, 119, |120, 130|, 131, 131, 141, 161
Median =

\frac{120 + 130}{2}

= 125

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Pergunta 3 Relatório

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

Detalhes da Resposta

16x - 5 x 4x + 4 = 0
(4x)2 - 5(4x) + 4 = 0
let 4x = y
y2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
y = 4 or 1
4x = 4
x = 1
4x = 1
i.e. 4x = 4o, x = 0
∴ x = 1 or 0

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Pergunta 4 Relatório

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

4

5

4

2

5

6

7

Detalhes da Resposta

\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}

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Pergunta 5 Relatório

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

Detalhes da Resposta

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

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Pergunta 6 Relatório

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

Detalhes da Resposta

\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

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Pergunta 7 Relatório

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

Detalhes da Resposta

\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}

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Pergunta 8 Relatório

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}

Detalhes da Resposta

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}

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Pergunta 9 Relatório

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

Detalhes da Resposta

Total surface area(s) = 2(4 x 3) + 2(4 x 4)
= 2(12) + 2(16)
= 24 + 32
= 56cm2
1m2 costs ₦2.00
∴ 56m∴ will cost 56 x ₦2.00
= ₦112.00

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Pergunta 10 Relatório

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

Detalhes da Resposta

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

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Pergunta 11 Relatório

At what points does the straight line y = 2x + 1 intersect the curve y = 2x2 + 5x - 1?

(-2, -3) and(

\frac{1}{2}

, 0)

(1, 0), (1, 3)

(4, 0) and (0,1)

(2, 0) and (0,1)

Graphs Of Linear And Quadratic Functions

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Pergunta 12 Relatório

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

(a + 1)(a + 5)

(a - 5)(7a - 1)

(a + 5)(7a + 1)

a(7a + 1)

Detalhes da Resposta

(4a + 3)2 - (3a - 2)2 = a2 - b2
= (a + b)(a - b)
= [(4a + 3) + (3a - 2)][(4a + 3) + (3a - 2)]
= [(4a + 3 + 3a - 2)][(4a + 3 - 3a + 2)]
= (7a + 1)(a + 5)
∴ (a + 5)(7a + 1)

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Pergunta 13 Relatório

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

Detalhes da Resposta

\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

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Pergunta 14 Relatório

Factorize x2 + 2a + ax + 2x

(x + 2a)(x + 1)

(x + 2a)(x - 1)

(x2 - 1)(x - a)

(x + 2)(x + a)

Algebraic Expressions

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Pergunta 15 Relatório

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

(313)3

(1020)3

(1222)3

(1002) 3

Number Bases

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Pergunta 16 Relatório

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

2

11

5

11

3

11

4

11

Detalhes da Resposta

From 40 to 50 = 11 & number are prime i.e. 41, 43, 47
prob. of selecting a prime No. is

\frac{3}{11}

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Pergunta 17 Relatório

In the diagram, PQ and RS are chords of a circle centre O which meet at T outside the circle. If TP = 24cm. TQ = 8cm and TS = 12cm, find TR.

16cm

14cm

12cm

8cm

Detalhes da Resposta

PT x QT = TR x TS

24 x 8 = TR x 12

TR = $\frac{24 \times 8}{12}$

= = 16cm

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Pergunta 18 Relatório

PQ and PR are tangents from P to a circle centre O as shown in the figure. If QRP = $34^\circ$, find the angle marked x

34

^{\circ}

56

^{\circ}

68

^{\circ}

112

^{\circ}

Detalhes da Resposta

From the circle centre 0, if PQ & PR are tangents from P and QRP = 34

^{\circ}

Then the angle marked x i.e. QOP

34 $^{\circ}$ x 2 = 68 $^{\circ}$

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Pergunta 19 Relatório

Simplify $ \frac{1}{x-2} + \frac{1}{x+2} + \frac{2x}{x^2-4} $

2x

(x−2)(x+2)(x²−4)

2x

x²−4

x

x²−4

4x

x²−4

Detalhes da Resposta

\frac{1}{x - 2}

+

\frac{1}{x + 2}

+

\frac{2 x}{x^{2} - 4}

=

\frac{(x + 2) + (x - 2) + 2 x}{(x + 2) (x - 2)}

=

\frac{4 x}{x^{2} - 4}

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Pergunta 20 Relatório

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

Detalhes da Resposta

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

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Pergunta 21 Relatório

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}

Detalhes da Resposta

Prob. of purple shirt is

\frac{7}{18}

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Pergunta 22 Relatório

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)

Detalhes da Resposta

8a + 125ax3 = 23a + 53ax3
= a(23 + 53x3)
∴a[23 + (5x)3]
a3 + b3 = (a + b)(a2 - ab + b2)
∴ a(23 + (5x)3)
= a(2 + 5x)(4 - 10x + 25x2)

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Pergunta 23 Relatório

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

3

39

29

36

14

5

7

5

Detalhes da Resposta

cos z =

\frac{y^{2} + x^{2} - z^{2}}{2 y x}

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

= $\frac{29}{36}$

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Pergunta 24 Relatório

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}

%

Detalhes da Resposta

.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}

%

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Pergunta 25 Relatório

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}}

Detalhes da Resposta

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}}

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Pergunta 26 Relatório

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

Detalhes da Resposta

QY = 452 + 242 = 2025 + 576
= 2601
QY =

\sqrt{2601}

= 51

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Pergunta 27 Relatório

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

2

3

5

7

Detalhes da Resposta

Let the smallest number be x and the perfect square be y 252x = y.
By trial and error method, 252 x 9 = 1764
Check if y = 1764
y2 = 42
x = 7

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Pergunta 28 Relatório

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

18

14

16

12

Detalhes da Resposta

Interior + exterior = 360
160 + exterior = 360
Exterior = 360 - 160
Exterior = 20
n = 360/exterior
n = 360/20
n = 18

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Pergunta 29 Relatório

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)

Detalhes da Resposta

x - y = 6.......(i)
xy = 187.......(ii)
From equation (i), x(6 + y)
sub. for x in equation (ii) = y(6 + y)
= 187
y2 + 6y = 187
y2 + 6y - 187 = 0
(y + 17)(y - 11) = 0
y = -17 or y = 11
y cannot be negative, y = 11
Sub. for y in equation(i) = x - 11
= 16
x = 6 + 11
= 17
∴(x, y) = (17, 11)

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Pergunta 30 Relatório

Find n if log24 + log27 - log2n = 1

10

14

27

28

Detalhes da Resposta

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

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Pergunta 31 Relatório

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

Detalhes da Resposta

Anambra =

\frac{185}{900}

x

\frac{360}{1}

= 74o

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Pergunta 32 Relatório

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

Detalhes da Resposta

\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}}

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Pergunta 33 Relatório

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

Detalhes da Resposta

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

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Pergunta 34 Relatório

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

Detalhes da Resposta

(

\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}

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Pergunta 35 Relatório

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

20

30

48

60

Detalhes da Resposta

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

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Pergunta 36 Relatório

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

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Pergunta 37 Relatório

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

Detalhes da Resposta

Let x = the number of oranges
The 1st received 1/3 of x = 1/3x
∴Remainder = x - 1/3x = 2x/3
The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3
The 3rd received 12 oranges
∴1/3x + 4x/9 + 12 = x
(3x + 4x + 108)/9 = x
3x + 4x + 108 = 9x
7x + 108 = 9x
9x - 7x = 108
2x = 108
x = 54 oranges

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Pergunta 38 Relatório

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

Detalhes da Resposta

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}

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Pergunta 39 Relatório

In the figure, $ \triangle $PQT is isosceles. PQ = QT, SRQ = $35^\circ$, TPQ = $20^\circ$ and PQR is a straight line.Calculate TSR

20

^{\circ}

55

^{\circ}

75

^{\circ}

140

^{\circ}

Detalhes da Resposta

Given

△

isosceles PQ = QT, SRQ = 35

^{\circ}

TPQ = 20 $^{\circ}$

PQR = is a straight line

Since PQ = QT, angle P = angle T = 20 $^{\circ}$

Angle PQR = 180 $^{\circ}$ - (20 + 20) = 140 $^{\circ}$

TQR = 180 $^{\circ}$ - 140 $^{\circ}$ = 40 $^{\circ}$ < on a straight line

QSR = 180 $^{\circ}$ - (40 + 35) $^{\circ}$ = 105 $^{\circ}$

TSR = 180 $^{\circ}$ - 105 $^{\circ}$

= 75 $^{\circ}$

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Pergunta 40 Relatório

The figure is a solid with the trapezium PQRS as its uniform cross-section. Find its volume

120m2

576m3

816m3

1056m3

Detalhes da Resposta

Volume of solid = cross section x H

Since the cross section is a trapezium

= $\frac{1}{2} (6 + 11) \times 12 \times 8$

= 6 x 17 x 8 = 816m3

Ler a nota de aula em Volumes (WAEC)

Volumes

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Pergunta 41 Relatório

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

Detalhes da Resposta

\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

Ver Resposta

Pergunta 42 Relatório

Simplify $ \frac{1}{5x+5} + \frac{1}{7x+7} $

\frac{12}{35 x + 1}

\frac{1}{35 (x + 1)}

\frac{12 x}{35 (x + 7)}

\frac{12}{35 x + 35}

Detalhes da Resposta

\frac{1}{5 x + 5}

+

\frac{1}{7 x + 7}

=

\frac{1}{5 (x + 1)}

+

\frac{1}{7 (x + 1)}

=

\frac{7 + 5}{35 (x + 1)}

=

\frac{12}{35 (x + 1)}

Ler a nota de aula em Solution Of Linear Equations (WAEC)

Ler a nota de aula em Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

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Pergunta 43 Relatório

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

2

10

125

250

Detalhes da Resposta

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

Ler a nota de aula em Variation (JAMB)

Variation

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Pergunta 44 Relatório

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)

Detalhes da Resposta

M = mode = the number having the highest frequency = 4
S = Number of students with 4 or less marks
= 14 + 7 + 4 + 2
= 27
∴ (M,S) = (27, 4)

Ler a nota de aula em Representation Of Data (JAMB)

Ler a nota de aula em Number Bases (WAEC)

Representation Of Data

Number Bases

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Pergunta 45 Relatório

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

7

1

3

27

Detalhes da Resposta

5(x + 2y) = 5
∴ x + 2y = 1.....(i)
4(x + 3y) = 16 = 42
x + 3y = 2 .....(ii)
x + 2y = 1.....(i)
x + 3y = 2......(ii)
y = 1
Substitute y = 1 into equation (i) = x + 2y = 1
∴ x + 2(1) = 1
x + 2 = 1
∴ x = 1
∴ 3x + y = 3-1 + 1
= 3 = 1

Ler a nota de aula em Algebraic Expressions (WAEC)

Algebraic Expressions

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Pergunta 46 Relatório

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

Detalhes da Resposta

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

Ler a nota de aula em Sine, Cosine And Tangent Of An Angle (WAEC)

Ler a nota de aula em Financial Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Financial Arithmetic

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Pergunta 47 Relatório

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

Detalhes da Resposta

15 - 29 years is represented by 104