JAMB UTME - Mathematics - 1988

Question 1 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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Variation

Positive And Negative Integers, Rational Numbers

Ratio, Proportions And Rates

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Question 2 Report

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

47

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 3 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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Logarithms

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Question 4 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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Circles

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Question 5 Report

Factorize completely (x2 + x)2 - (2x + 2)2

(x + 1)(x + 2)(x - 2)

(x + 1) 2(x + 2) (x - 2)

(x + 1)2 (x + 2)2

(x + 1)2 (x + 2)(x - 2)

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Question 6 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 7 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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Angles And Intercepts On Parallel Lines

Angles

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Question 8 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 9 Report

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

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 10 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 11 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

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 12 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 13 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 14 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 15 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 16 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 17 Report

if a = -3, b = 2, c = 4, evaluate $ \frac{a^3-b^3-c^{\frac{1}{2}}}{b-a-c} $

37

-37

5

37

5

-37

View Answer

Question 18 Report

Tope bought X oranges at N5.00 each and some mangoes at N4.00 each. if she bought twice as many mangoes as oranges and spent at least N65.00 and at most N130.00, find the range of values of X.

4≤X≤5

5≤X≤8

5≤X≤10

8≤X≤10

Number Bases

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Question 19 Report

Simplify $ \frac{1\frac{1}{2}}{2+\frac{1}{4}\text{ 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

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Question 20 Report

In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45o and YZX = 55o, Find ZYQ

135o

125o

100o

90o

Answer Details

< RXZ = < ZYX = 45O(Alternate segment)

< ZYQ = 90 + 45

= 135O

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Circles

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Question 21 Report

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

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 22 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 23 Report

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

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 24 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 25 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 26 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 27 Report

If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3

1

3

2

3

8

27

4

9

Variation

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Question 28 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

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Question 29 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 30 Report

Four interior angles of a pentagon are 90o - xo, 90o + xo, 110o - 2xo, 110o + 2xo. Find the fifth interior angle

110o

120o

130o

140o

Answer Details

Let the fifth interior angle be y: sum of interior angle of a pentagon
= (2 x 5 - 4) x 90o
= 6 x 90o
= 540o
(90 - x) + (90 + x) + (110 - 2x) + (110 + 2x) + y = 540o
400o + y = 540o
y = 540 - 400o
y = 140o

View Answer

Question 31 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 32 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

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Question 33 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 34 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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Volumes

Indices, Logarithms And Surds

Areas

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Question 35 Report

What is the solution of the equation x2 - x - 1 + 0?

x = 1.6 and x = -0.6

x = -1.6 and x = 0.6

x = 1.6 and x = 0.6

x = -1.6 and x = -0.6

Quadratic Equations

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Question 36 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 37 Report

If $ \frac{1}{p} = \frac{a^2+2ab+b^2}{a-b} $ and $ \frac{l}{q} = \frac{a+b}{a^2-2ab+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 38 Report

If 7 and 189 are the first and fourth terms of a geometric progression respectively find the sum for the first
three terms of the progression

182

180

91

63

28

Sequence And Series

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Question 39 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 40 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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Volumes

Euclidean Geometry

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Question 41 Report

The solutions of x2 - 2x - 1 = 0 are the points of intersection of two graphs. if one of the graphs is y = 2 + x - x2, find the second graph

y = 1 - x

y = 1 + x

y = x - 1

y = 3x + 3

Algebraic Expressions

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Question 42 Report

if x is the addition of the prime numbers between 1 and 6; and y the H.C.F. of 6, 9, 15. Find the product of x and y

27

30

33

90

Answer Details

Prime numbers between 1 and 6 are 2, 3 and 5
x = 2 + 3
= 5 = 10
H.C.F. of 6, 9, 15 = 3
∴ y = 3
X x y = 10 x 3
= 30

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Probability

Number Bases

Sets

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Question 43 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