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JAMB UTME - Mathematics - 1988

Pergunta 1 Relatório

Simplify \( \frac{x+2}{x+1} - \frac{x-2}{x+2} \)

3

x + 1

3x + 1

(x + 1) (x + 2)

5x + 1

(x + 1) (x + 2)

2x²+ 5x + 2

(x + 1) (x + 2)

Detalhes da Resposta

$\frac{x + 2}{x + 1}$ - $\frac{x - 2}{x + 2}$ = $\frac{(x + 2) (x + 2) - (x - 2) - (x - 2) (x + 1)}{(x + 1) (x + 2)}$
= $\frac{(x^{2} + 4 x + 4) - (x^{2} - x - 2)}{(x + 1) (x + 2)}$ = $\frac{x^{2} + 4 x + 4 - x^{2} + x + 2}{(x + 1) (x + 2)}$
= $\frac{5 x + 6}{(x + 1) (x + 2)}$

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Pergunta 2 Relatório

Two sisters, Taiwo and Keyinde, own a store. The ratio of Taiwo's share to Kehinde's is 11:9. Later, Keyinde sells \( \frac{2}{3} \) of her share to Taiwo for ₦720.00. Find the value of the store.

₦1,080.00

₦2,400.00

₦3,000.00

₦3,600.00

Detalhes da Resposta

Let value of store = X
Ratio of Taiwo's share to kehine's is 11:9 Keyinde sells $\frac{2}{3}$ of her share to Taiwo for ₦720
$\frac{2}{3}$ of 9 = 6
∴ Sum of the ratio = 11 + 9 = 20
$\frac{6}{20}$ of x = ₦720
$\frac{6 x}{20}$ = 720
∴ x = $\frac{720 \times 20}{6}$
x = ₦24,000

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Pergunta 4 Relatório

The solution of the quadratic equation px2 + qx + b = 0 is

\sqrt{\frac{- b \pm b^{2} - 4 a c}{2 a}}

\frac{- b \pm \sqrt{p^{2} - 4 p b}}{2 a}

\frac{- q \pm \sqrt{q^{2} - 4 b p}}{2 p}

\frac{- q \pm \sqrt{p^{2} - 4 b p}}{2 p}

Detalhes da Resposta

px2 + qx + b = 0
Using almighty formula
$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .........(i)
Where a = p, b = q and c = b
substitute for this value in equation (i)
= $\frac{- q \pm \sqrt{q^{2} - 4 b p}}{2 p}$

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Pergunta 5 Relatório

If \( \cos^2 \theta + \frac{1}{8} = \sin^2 \theta \), find \( \tan \theta \)

3

\frac{3 \sqrt{7}}{7}

3

\sqrt{7}

\sqrt{7}

Detalhes da Resposta

cos2 $θ$ + $\frac{1}{8}$ = sin2 $θ$ ..........(i)
from trigometric ratios for an acute angle, where cos $θ$ + sin2 $θ$ = 1 - cos $θ$ ........(ii)
Substitute for equation (i) in (i) = cos2 $θ$ + $\frac{1}{8}$ = 1 - cos2 $θ$
= cos2 $θ$ + cos2 $θ$ = 1 - $\frac{1}{8}$
2 cos2 $θ$ = $\frac{7}{8}$
cos2 $θ$ = $\frac{7}{2 \times 3}$
$\frac{7}{16}$ = cos $θ$
$\sqrt{\frac{7}{16}}$ = $\frac{\sqrt{7}}{4}$
but cos $θ$ = $\frac{adj}{hyp}$
opp2 = hyp2 - adj2
opp2 = 42 ( $\sqrt{7}$ )2
= 16 - 7
opp = $\sqrt{9}$ = 3
than $θ$ = $\frac{opp}{hyp}$
= $\frac{3}{\sqrt{7}}$
$\frac{3}{\sqrt{7}}$ x $\frac{7}{\sqrt{7}}$ = $\frac{3 \sqrt{7}}{7}$

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Trigonometry

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Pergunta 6 Relatório

Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is

(

\frac{- 1}{8}, \frac{19}{24}

)

8,

\frac{24}{19}

-8,

\frac{24}{19}

(

\frac{19}{24}, \frac{- 1}{8}

)

Detalhes da Resposta

3x - 5y = 3, 2y - 6x = -5
-5y + 3x = 3........{i} x 2
2y - 6x = -5.........{ii} x 5
Substituting for x in equation (i)
-5y + 3( $\frac{19}{24}$ ) = 3
-5y + 3 x $\frac{19}{24}$ = 3
-5y = $\frac{3 - 19}{8}$
-5 = $\frac{24 - 19}{8}$
= $\frac{5}{8}$
y = $\frac{5}{8 \times 5}$
y = $\frac{- 1}{8}$
(x, y) = ( $\frac{19}{24}, \frac{- 1}{8}$ )

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Pergunta 8 Relatório

In the figure, a solid consists of a hemisphere surmounted by a right circular cone, with radius 3.0cm and height 6.0cm. Find the volume of the solid

36

π

cm 3

54

π

cm 2

18

π

cm 2

108

π

cm 2

Detalhes da Resposta

The volume of the solid = vol. of cone + vol. of hemisphere

volume of cone = $\frac{1}{2} π^{2} h$

= $\frac{1 π}{3} \times (3)^{2} x 6 = 18 π c m^{2}$

vol. of hemisphere = $\frac{4 π r^{3}}{6} = \frac{2 π r^{3}}{3}$

= $\frac{2 π}{3} \times (3)^{3} = 18 π c m^{3}$

vol. of solid = 18 $π$ + 18 $π$

= 36 $π$ cm3

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Pergunta 9 Relatório

A basket contain green, black and blue balls in the ratio 5 : 2 : 1. If there are 10 blue balls. Find the corresponding new ratio when 10 green and 10 black balls are removed from the basket

1 : 1 : 1

4 : 2 : 1

5 : 1 : 1

4 : 1 : 1

Detalhes da Resposta

Let x represent total number of balls in the basket.
If there are 10 blue balls, $\frac{1}{8}$ of x = 10
x = 10 x 8 = 80 balls
Green balls will be $\frac{5}{8}$ x 80 = 50 and black balls = $\frac{2}{8}$ x 80 = 20
Ratio = Green : black : blue
50 : 20 : 10
-10 : 10 : -
------------------
New Ratio 40 : 10 : 10
4 : 1 : 1

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Pergunta 11 Relatório

If x and y represent the mean and the median respectively of the following set of numbers 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, find the \( \frac{x}{y} \) correct to one decimal place

18

15

1.0

5

Detalhes da Resposta

Mean $\bar{x}$ = $\frac{156}{10}$ = 15.6
Median = $\bar{y}$ = $\frac{15 + 16}{2}$
$\frac{31}{2}$ = 15.5
$\frac{x}{y}$ = $\frac{15.6}{15.5}$ = 1.0065
1.0(1 d.p)

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Pergunta 13 Relatório

A tax player is allowed \( \frac{1}{8} \)th of his income tax-free, and pays 20% on the remainder. If he pays ₦490.00 tax, what is his income?

₦560.00

₦2,450.00

₦2,800.00

₦3,920.00

Detalhes da Resposta

He pays tax on 1 - $\frac{7}{8}$ = $\frac{1}{7}$ th of his income
20% is 490, 100% is $\frac{1000}{20}$ x 490, ₦2,450.00
= $\frac{7}{8}$ of his income = ₦2,450.00

$\frac{1}{\frac{7}{8}}$ x 2450
= $\frac{8 \times 2450}{7}$
= $\frac{19600}{7}$
= ₦2800.00

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Pergunta 16 Relatório

If \( \cos \theta = \frac{x}{y} \), find \( \cosec \theta \)

\frac{1}{y}

(x2 + y2)

\frac{x}{y}

\frac{1}{y}

\sqrt{x^{2} + y^{2}}

\frac{x - y}{y}

Detalhes da Resposta

Cos $θ$ = $\frac{x}{y}$ = $\frac{adj}{opp}$
(hyp2) = opp2 + adj2
(hyp2) = x2 + y2
hyp = $\sqrt{x^{2} + y^{2}}$
Cosec $θ$ = hyp
= x2 + y2
= $\frac{1}{y}$ $\sqrt{x^{2} + y^{2}}$

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Pergunta 17 Relatório

Evaluate \( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \) = \( \frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}} \)

2

5

5

3

3√5

3√2

Detalhes da Resposta

$\frac{8^{\frac{1}{3}} \times 5^{\frac{3}{2}}}{\frac{2}{10^{3}}}$ = $\frac{(2^{3})^{\frac{1}{3}} \times 5^{\frac{3}{2}}}{(2 \times 5)^{\frac{2}{3}}}$
= $\frac{2 \times 5}{2^{\frac{2}{3}} \times 5^{\frac{3}{2}}}$
= 21 - $\frac{2}{3}$
= 2 $\frac{1}{3}$
= 3 $\sqrt{2}$

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Pergunta 19 Relatório

In the figure, PS = 7cm and RY = 9cm. IF the area of parallelogram PQRS is 56cm2. Find the area of trapezium PQTS

56cm2

112cm2

120cm2

176cm2

Detalhes da Resposta

From the figure, PS = QR = YT = 7cm

Area of parallelogram PQRS = 56cm

56 = base x height, where base = 7

7 x h = 56cm,

h = $\frac{56}{7}$

= 8cm

Area of trapezium $\frac{1}{2}$ (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm

Area of trapezium PQTS = $\frac{1}{2}$ (23 + 7) x 8

$\frac{1}{2}$ x 30 x 8 = 120cmsq

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Pergunta 21 Relatório

The thickness of an 800 pages of book is 18mm. Calculate the thickness of one leaf of the book giving your answer in meters and in standard form

2.25 x 10-4m

50 x 10-4m

2.25 x 10-5m

4.50 x 10-5m

Detalhes da Resposta

Thickness of an 800 pages book = 18mm to meter
18 x 103m = 1.8 x 10-2m
One leaf = $\frac{1.8 \times 10^{- 2}}{800}$
= $\frac{1.8 \times 10^{- 2}}{8 \times 10^{2}}$
= $\frac{- 1.8}{8}$ x 10-4
= 0.225 x 10-4
= 2.25 x 10-5m

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Pergunta 22 Relatório

Solve the following equation equation for \(x\)

\[ x^2 + \frac{2x}{r^2} + \frac{1}{r^4} = 0 \]

r2

1

r⁴

-

\frac{1}{r^{2}}

1 - r

Detalhes da Resposta

x2 + $\frac{2 x}{r^{2}}$ + $\frac{1}{r^{4}}$ = 0
(x + $\frac{1}{r^{2}}$ ) = 0
x + $\frac{1}{r^{2}}$ = 0
x = $\frac{- 1}{r^{2}}$

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Pergunta 24 Relatório

If a metal pipe 10cm long has an external diameter of 12cm and a thickness of 1cm find the volume of the metal used in making the pipe

120

π

cm3

110

π

cm3

60

π

cm3

50

π

cm3

Detalhes da Resposta

The volume of the pipe is equal to the area of the cross section and length.
let outer and inner radii be R and r respectively.
Area of the cross section = (R2 - r2)
where R = 6 and r = 6 - 1
= 5cm
Area of the cross section = (62 - 52) $π$ = (36 - 25) $π$ cm sq
vol. of the pipe = $π$ (R2 - r2)L where length (L) = 10
volume = 11 $π$ x 10
= 110 $π$ cm3

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Pergunta 25 Relatório

Simplify \( \frac{x-7}{x^2-9} \, x \, \frac{x^2-3x}{x^2-49} \)

x

(x - 3) (x - 7)

x

(x + 3) (x - 7)

x

(x + 3) (x + 7)

x

(x - 3) (x + 7)

Detalhes da Resposta

$\frac{x - 7}{x^{2} - 9}$ x $\frac{x^{2} - 3 x}{x^{2} - 49}$
= $\frac{x - 7}{(x - 3) (x + 3)}$ x $\frac{x (x - 3)}{(x - 7) (x + 7)}$
= $\frac{x}{(x + 3) (x + 7)}$

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Pergunta 26 Relatório

Find \(m\) such that \((m + \sqrt{3})(1 - \sqrt{3})^2 = 6 - 2\sqrt{2}\)

1

2

3

4

Detalhes da Resposta

(m + $\sqrt{3}$ )(1 - $\sqrt{3}$ )2 = 6 - 2 $\sqrt{2}$
(m + $\sqrt{3}$ )(4 - 2 $\sqrt{3}$ ) = 6 - 2 $\sqrt{2}$
= 6 - 2 $\sqrt{3}$
4m - 6 + 4 - 2m $\sqrt{3}$ = 6 - 2 $\sqrt{3}$
comparing co-efficients,
4m - 6 = 6.......(i)
4 - 2m = -2.......(ii)
in both equations, m = 3

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Pergunta 28 Relatório

For which of the following exterior angles is a regular polygon possible? i. 36o ii. 18o iii. 15o

i and ii

ii only

ii and iii

iii only

Detalhes da Resposta

for a regular polygon to be possible, it must have all sides angles equal. $\frac{360}{18}$ = 20 sides and $\frac{360}{15}$ = 24 sides
(ii) and (iii) are right

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Pergunta 30 Relatório

In the figure, PQRS is a circle. If chords QR and RS are equal, calculate the value of x.

80

^{\circ}

60

^{\circ}

45

^{\circ}

40

^{\circ}

Detalhes da Resposta

SRT is a straight line, where QRT = 120

SRQ = 180 $^{\circ}$ - 120 $^{\circ}$ = 60 $^{\circ}$ - (angle on a straight line)

also angle QRS = 180 $^{\circ}$ - 100 $^{\circ}$ (angle on a straight line) . In angles where QR = SR and angle SRQ = 60 $^{\circ}$

x = 100 - 60 = 40 $^{\circ}$

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Pergunta 32 Relatório

For what values of x is the curve \(y = \frac{x^2 + 3}{x + 4}\) decreasing?

-3 < x

\leq

0

-3

\geq

x < 0

0 < x < 3

0

\leq

x

\leq

3

Quadratic Equations

Graphs Of Linear And Quadratic Functions

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Pergunta 33 Relatório

Scores(\(x\))	0	1	2	3	4	5	6	7
Frequency(\(f\))	7	11	6	7	7	5	3

In the distribution above, the mode and median respectively are

1,3

1, 2

3, 3

0, 2

Detalhes da Resposta

From the distribution, Mode = 1 and
Median = $\frac{2 + 2}{2}$
= 2
= 1, 2

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Pergunta 34 Relatório

Simplify \( \frac{x-y}{x^{\frac{1}{3}}-x^{\frac{1}{3}}} \)

x2 + xy + y2

x

\frac{2}{3}

+ x

\frac{1}{3}

+ y

\frac{2}{3}

x

\frac{2}{3}

- x

\frac{1}{3}

y

\frac{2}{3}

y

\frac{2}{3}

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Pergunta 35 Relatório

If log102 = 0.3010 and log103 = 0.4771, evaluate; without using logarithm tables, log104.5

0.3010

0.4771

0.6532

0.9542

Detalhes da Resposta

If log102 = 0.3010 and log103 = 0.4771,
log104.5 = log10 $\frac{(3 \times 3)}{2}$
log103 + log103 - log102 = 0.4771 + 0.4771 - 0.0310
= 0.6532

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Pergunta 36 Relatório

In a class of 150 students, the sector in a pie chart representing the students offering Physics has angle 12o. How many students are offering Physics?

18

15

10

5

Detalhes da Resposta

No of students offering Physics are $\frac{12}{360}$ x 150
= 5

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Pergunta 37 Relatório

Simplify \( \frac{4a^2 - 49b^2}{2a^2 - 5ab - 7b^2} \)

a - b

2a + b

2a + 7b

a - b

2a - 7b

a + b

2a - 7b

a - b

Detalhes da Resposta

$\frac{4 a^{2} - 49 b^{2}}{2 a^{2} - 5 a b - 7 b^{2}}$ = $\frac{(2 a)^{2} - (7 b)^{2}}{(a - b) (2 a + 7 b)}$
= $\frac{(2 a + 7 b) (2 a - 7 b)}{(a - b) (2 a + 7 b)}$
= $\frac{2 a - 7 b}{a - b}$

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Pergunta 38 Relatório

In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle

\sqrt{3}

cm

\sqrt{7}

cm

3cm

7cm

Detalhes da Resposta

PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o
PR2 = 12 + 22 - 2(1)(2) x - cos 60o
= 5 - 2(1)(2) x - $\frac{1}{2}$
= 5 + 2 = 7
PR = $\sqrt{7}$ cm

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Pergunta 39 Relatório

In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR

47

5

5

32

5

22

5

Detalhes da Resposta

From similar triangle, $\frac{Q S}{Q P} = \frac{T Q}{Q R} = \frac{5}{12} = \frac{6}{Q R}$

QT = $\frac{6 \times 12}{5} = \frac{72}{5} = S R = Q R - Q S$

= $\frac{72}{5} - 5 = \frac{72 - 25}{5}$

= $\frac{47}{5}$

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Pergunta 40 Relatório

If two dice are thrown together, what is the probability of obtaining at least a score of 10?

1

6

1

12

5

6

11

12

Detalhes da Resposta

The total sample space when two dice are thrown together is 6 x 6 = 36
$\begin{matrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1. & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 & 1.6 \\ 2 & 2.1 & 2.2 & 2.3 & 2.4 & 2.5 & 2.6 \\ 3 & 3.1 & 3.2 & 3.3 & 3.4 & 3.5 & 3.6 \\ 4 & 4.1 & 4.2 & 4.3 & 4.4 & 4.5 & 4.6 \\ 5 & 5.1 & 5.2 & 5.3 & 5.4 & 5.5 & 5.6 \\ 6 & 6.1 & 6.2 & 6.3 & 6.4 & 6.5 & 6.6 \end{matrix}$
At least 10 means 10 and above
P(at least 10) = $\frac{6}{36}$
= $\frac{1}{6}$

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Pergunta 41 Relatório

A 5.0g of salt was weighted by Tunde as 5.1g. What is the percentage error?

20

2

1.1

0.2

Detalhes da Resposta

% error = $\frac{actual error}{true value}$ x 100
Where actual error = 5.1 - 5.0 = 0.1
true value = 5.0g
% error = $\frac{0.1}{5.0}$ x 100
= $\frac{10}{5}$
= 2

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Pergunta 42 Relatório

If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference

4

8

6

\frac{2}{3}

9

\frac{1}{3}

Surds (radicals)

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Pergunta 43 Relatório

Simplify \( \frac{1\frac{1}{2}}{2+\frac{1}{4}\text{ of }32} \)

3

256

3

32

6

85

Detalhes da Resposta

$\frac{1 \frac{1}{2}}{2 + \frac{1}{4} of 32}$
= $\frac{\frac{3}{2}}{2 + \frac{1}{4} \times 32}$
$\frac{3}{2}$ x 4 = 6

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Pergunta 45 Relatório

In the figure, PQ is a parallel to ST and QRS = \(40^\circ\). Find the value of x.

55

^{\circ}

60

^{\circ}

65

^{\circ}

75

^{\circ}

Detalhes da Resposta

From the figure, 3x + x - 40 $^{\circ}$ = 180 $^{\circ}$

4x = 180 $^{\circ}$ + 40 $^{\circ}$

4x = 220 $^{\circ}$

x = $\frac{220}{4}$

= 55 $^{\circ}$

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Pergunta 46 Relatório

In the figure, PS = RS = QS and QRS = 50o. Find QPR

25o

40o

50o

65o

Detalhes da Resposta

In the figure PS = RS = QS, they will have equal base QR = RP

In angle SQR, angle S = 50O

In angle QRP, 65 + 65 = 130O

Since RQP = angle RPQ = $\frac{180 - 130}{2}$

= $\frac{50}{2} = 25^{o}$

QPR = 25O

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Pergunta 47 Relatório

PQR is a triangle in which PQ = 10cm and QPR = 60oS is a point equidistant from P and Q. Also S is a point equidistant from PQ and PR. If U is the foot of the perpendicular from S on PR, find the length SU in cm to one decimal place

2.7

4.33

3.1

3.3

Detalhes da Resposta

$△$ PUS is right angled
$\frac{U S}{5}$ = sin60o
US = 5 x $\frac{\sqrt{3}}{2}$
= 2.5 $\sqrt{3}$
= 4.33cm

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Pergunta 48 Relatório

Find correct to one decimal place, \(0.24633 \div 0.0306\)

0.8

1.8

8.0

8.1

Detalhes da Resposta

$\frac{0.24633}{0.03060}$ multiplying throughout by 100,000
= $\frac{24633}{3060}$
= 8.05
= 8.1

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Pergunta 49 Relatório

If (IPO3)4 = 11510 find P

0

1

2

3

Detalhes da Resposta

1 x 43 + P x 42 + 0 x 4 + 3 = 11510
16p + 67 = 115 p = $\frac{48}{16}$
= 3

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1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

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