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JAMB UTME - Mathematics - 1988

Question 1 Report

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

Answer Details

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

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Number Bases

Modular Arithmetic

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Question 2 Report

Find correct to one decimal place, 0.24633 $\div$ 0.0306

0.8

1.8

8.0

8.1

Answer Details

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

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Fractions, Decimals And Approximations

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Question 3 Report

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

Answer Details

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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Indices, Logarithms And Surds

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Question 4 Report

If cos $θ$ = $\frac{x}{y}$ , find cosec $θ$

\frac{1}{y}

(x2 + y2)

\frac{x}{y}

\frac{1}{y}

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

\frac{x - y}{y}

Answer Details

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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Sine, Cosine And Tangent Of An Angle

Angles Of Elevation And Depression

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Question 5 Report

Solve the following equation equation for x2 + $\frac{2 x}{r^{2}}$ + $\frac{1}{r^{4}}$ = 0

r2

1

r⁴

-

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

1 - r

Answer Details

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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Number Bases

Quadratic Equations

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Question 6 Report

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

Answer Details

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

View Answer

Question 9 Report

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

Answer Details

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

View Answer

Question 10 Report

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

Answer Details

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

View Answer

Question 11 Report

Simplify $\frac{x - 7}{x^{2} - 9}$ x $\frac{x^{2} - 3 x}{x^{2} - 49}$

x

(x - 3) (x - 7)

x

(x + 3) (x - 7)

x

(x + 3) (x + 7)

x

(x - 3) (x + 7)

Answer Details

$\frac{x - 7}{x^{2} - 9}$ 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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Algebraic Expressions

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Question 12 Report

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)

Answer Details

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

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Algebraic Expressions

Quadratic Equations

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Question 13 Report

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

47

5

5

32

5

22

5

Answer Details

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

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

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

= $\frac{47}{5}$

View Answer

Question 14 Report

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}

Answer Details

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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Read lesson note on Angles (WAEC)

Angles And Intercepts On Parallel Lines

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Question 15 Report

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

Answer Details

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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Question 18 Report

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

Answer Details

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

View Answer

Question 19 Report

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}

Answer Details

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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Question 21 Report

Simplify $\frac{4 a^{2} - 49 b^{2}}{2 a^{2} - 5 a b - 7 b^{2}}$

a - b

2a + b

2a + 7b

a - b

2a - 7b

a + b

2a - 7b

a - b

Answer Details

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

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Solution Of Linear Equations

Algebraic Expressions

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Question 22 Report

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}

View Answer

Question 23 Report

Simplify $\frac{1 \frac{1}{2}}{2 + \frac{1}{4} of 32}$

3

256

3

32

6

85

Answer Details

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

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Number Bases

Number Bases

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Question 24 Report

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

Answer Details

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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Polynomials

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Question 25 Report

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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Question 26 Report

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

1

2

3

4

Answer Details

(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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Surds (radicals)

Lengths And Perimeters

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Question 27 Report

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

0.3010

0.4771

0.6532

0.9542

Answer Details

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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Question 28 Report

If cos2 $θ$ + $\frac{1}{8}$ = sin2 $θ$ , find tan $θ$

3

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

3

\sqrt{7}

\sqrt{7}

Answer Details

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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Sine, Cosine And Tangent Of An Angle

Trigonometry

Modular Arithmetic

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Question 29 Report

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

Answer Details

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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Probability

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Question 30 Report

If $\frac{1}{p}$ = $\frac{a^{2} + 2 a b + b^{2}}{a - b}$ and $\frac{l}{q}$ = $\frac{a + b}{a^{2} - 2 a b + b^{2}}$ Find $\frac{p}{q}$

a + b

a - b

1

a² - b²

a - b

a + b

a2 - b2

Algebraic Expressions

Algebraic Fractions

Simple Operations On Algebraic Expressions

Quadratic Equations

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Question 31 Report

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

Answer Details

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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Fractions, Decimals And Approximations

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Question 33 Report

$\begin{array}{cc} S c o r e s (x) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ F r e q u e n c y (f) & 7 & 11 & 6 & 7 & 7 & 5 & 3 \end{array}$

In the distribution above, the mode and median respectively are

1,3

1, 2

3, 3

0, 2

Answer Details

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

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Representation Of Data

Linear Inequalities

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Question 34 Report

If (g(y)) = $\frac{y - 3}{11}$ + $\frac{11}{y^{2} - 9}$ . what is g(y + 3)?

\frac{y}{11} + \frac{11}{y (y + 6)}

\frac{y}{11} + \frac{11}{y (y + 6)}

\frac{y}{11} + \frac{11}{y (y + 6)}

\frac{y}{11} + \frac{11}{y (y + 6)}

Solution Of Linear Equations

Algebraic Expressions

Number Bases

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Question 35 Report

If (IPO3)4 = 11510 find P

0

1

2

3

Answer Details

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

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Number Bases

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Question 36 Report

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

20

2

1.1

0.2

Answer Details

% 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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Percentages

Fractions, Decimals, Approximations And Percentage

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Question 40 Report

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}

Answer Details

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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Quadratic Equations

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Question 41 Report

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

Answer Details

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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Surds (radicals)

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Question 43 Report

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}

)

Answer Details

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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Algebraic Expressions

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Question 45 Report

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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Question 46 Report

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

Answer Details

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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Positive And Negative Integers, Rational Numbers

Ratio, Proportions And Rates

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Question 47 Report

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

Answer Details

$△$ 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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Question 48 Report

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

25o

40o

50o

65o

Answer Details

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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Polynomials

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Question 49 Report

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

Answer Details

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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Euclidean Geometry

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