JAMB UTME - Mathematics - 1987

Vraag 1 Verslag

If the length of a square is increased by 20% while while its width is decreased by 20% to form a rectangle, what is the ratio of the area of the rectangle to the area of the square?

6 : 5

25 : 24

5 : 6

24 : 25

Antwoorddetails

Length and width of a square is 100%
Length increased by 20% and
Width decreased by 20% to form a rectangle
Length of rectangle = 120% to form a rectangle
Length of rectangle = 120% and
Width of rectangle = 80%
Area of rectangle = L x W
Area of square = W
Ratio of the area of the rectangle to the area of the square
A =

\frac{Area of rectangle}{Area of square}

\frac{120 \times 30}{100 \times 100}

=

\frac{96}{100}

= 24 : 25

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Vraag 2 Verslag

The sine, cosine and tangent of 210o are respectively

\frac{1}{2}

,

\frac{\sqrt{3}}{2}

,

\frac{\sqrt{3}}{2}

\frac{1}{2}

,

\frac{\sqrt{3}}{2}

,

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

\frac{1}{2}

,

\frac{\sqrt{3}}{2}

,

\frac{\sqrt{3}}{2}

$\frac{- 1}{2}$ , $\frac{\sqrt{- 3}}{2}$ , $\frac{\sqrt{3}}{3}$

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

Antwoorddetails

210o = 180o - 210o = 30o
From ratio of sides, sin -30o = -

\frac{1}{2}

Cos 210o = 180o - 210o = -30o
= cos -30o =

\frac{- 3}{2}

But tan 30o =

\frac{1}{3}

, rationalizing this
=

\frac{1}{3}

x

\frac{3}{3}

=

\frac{3}{3}

∴ =

\frac{- 1}{2}

,

\frac{\sqrt{- 3}}{2}

,

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

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Vraag 3 Verslag

A man invests a sum of money at 4% per annum simple interest. After 3 years, the principal amounts to ₦7,000.00. Find the sum invested.

₦7,840.00

₦6,250.00

₦6,160.00

₦5,833.33

Modular Arithmetic

Areas

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Vraag 4 Verslag

Four boys and ten girls can cut a field in 5 hours if the boys work at $ \frac{5}{4} $ the rate at which the girls work. How many boys will be needed to cut the field in 3 hours?

180

60

25

20

Antwoorddetails

Let x represents number of boys that can work at

\frac{5}{4}

the rate at which the 10 girls work
For 1hr. x boys will work for

\frac{\frac{1}{5}}{4}

x 10
x =

\frac{4}{5}

x 10 = 8 boys
8 boys will do the work of ten girls at the same rate 4 + 8 = 12 boys cut the field in 5 hrs
For 3 hrs,

\frac{12 \times 5}{3}

boys will be needed = 20 boys

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Vraag 5 Verslag

By selling 20 oranges for ₦1.35 a trader makes a profit of 8%. What is his percentage gain or loss if he sells the same 20 oranges for ₦1.10?

8%

10%

12%

15%

Antwoorddetails

profit 8% of ₦1.35 =

\frac{8}{100}

x ₦1.35 = ₦0.08
Cost price = ₦1.35 - ₦0.10 = ₦1.25
If he sells the 20 oranges for ₦1.10 now
%loss =

\frac{actual loss}{Cost price}

x 100

\frac{125 - 1.10}{1.25}

x 100
=

\frac{0.15 \times 100}{1.25}

=

\frac{15}{1.25}

= 12%

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Modular Arithmetic

Ratio, Proportions And Rates

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Vraag 6 Verslag

Simplify without using tables $ \frac{\log_2 6}{\log_2 8} - \frac{\log_2 3}{\log_2 \frac{1}{2}} $

1

5

1

2

-1

2

log²3

log²7

Antwoorddetails

log26 - log23 = log2 (

\frac{6}{3}

) = log22
log28 - 2

\frac{1}{2}

log28 - log2

\frac{1}{2}

log28 - log2

\frac{1}{4}

\frac{l o g_{2} 2}{l o g_{2} 32} = \frac{l o g_{2} 2}{5 l o g_{2} 2}

=

\frac{1}{5}

N.B. log22 = 1

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Logarithms

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Vraag 7 Verslag

Solve the inequality x - 1 > 4(x + 2)

x > -3

x < -3

2 < x < 3

-3 < x < -2

Linear Inequalities

Areas

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Vraag 8 Verslag

The numbers 3, 2, 8, 5, 7, 12, 9 and 14 are the marks scored by a group of students in a class test. If P is scored by an group of students in a class test. If P is the mean and Q the median, the P + Q is

18

17

\frac{1}{2}

16

15

Antwoorddetails

Re-arranging them 2, 3, 5, 7, 8, 9, 12, 14
Mean = P and median = Q
P =

\frac{2 + 2 + 5 + 7 + 8 + 9 + 12 + 14}{8}

\frac{60}{8}

= 7.5
median (Q) =

\frac{7 + 8}{2}

\frac{15}{2}

= 7.5
P + Q = 7.5 + 7.5 = 15

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Number Bases

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Vraag 9 Verslag

Simplify $ \frac{x^2-y^2}{2x^2+xy-y^2} $

$\frac{x + y}{2 x + y}$

$\frac{x + y}{2 x - y}$

$\frac{x - y}{2 x - y}$

$\frac{x - y}{2 x + y}$

Antwoorddetails

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

$2 x^{2} + x y - y^{2} = 2 x^{2} - x y + 2 x y - y^{2}$

= $x (2 x - y) + y (2 x - y)$

= $(x + y) (2 x - y)$

$\frac{x^{2} - y^{2}}{2 x^{2} + x y - y^{2}} = \frac{(x + y) (x - y)}{(x + y) (2 x - y)}$

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

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Solution Of Linear Equations

Algebraic Expressions

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Vraag 10 Verslag

If a u2 - 3v2 and b = 2uv + v2 evaluate (2a - b)(a - b2), when u = 1 and v = -1

9

15

27

33

Algebraic Expressions

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Vraag 11 Verslag

From two points x and y, 8m apart, and in line with a pole, the angle of elevation of the top of the pole are 30o and 60o respectively. Fins the height of the pole assuming that x, y and the foot of the pole are the same horizontal plane and x and y are on the same side of the pole.

\sqrt{6}

4

\sqrt{3}

√3

\frac{12}{\sqrt{3}}

Antwoorddetails

From the diagram, WYZ = 60o, XYW = 180o - 60o
= 120o
LX = 30o
XWY = 180o - 120o + 30o
XWY = 30o
WXY = XYW = 30o
Side XY = YW
YW = 8m, sin 60o =

\frac{3}{2}

? sin 60o =

\frac{h}{Y W}

, sin 60o =

\frac{h}{8}

h = 8 x

\frac{3}{2}

= 4

\sqrt{3}

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Vraag 12 Verslag

Reduce each number to two significant figures and then evaluate $ \frac{0.021741 \times 1.2047}{0.023789} $

0.8

0.9

1.1

1.2

Antwoorddetails

\frac{0.021741 \times 1.2047}{0.023789}

=

\frac{0.022 \times 1.2}{0.024}

(to 216)
=

\frac{0.0264}{0.024}

= 1.1

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Fractions, Decimals And Approximations

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Vraag 13 Verslag

Find the values of m which make the following quadratic function a perfect square. x2 + 2(m + 1)x + m + 3

-1, 1

-1, 2

1, -2

2, -2

Antwoorddetails

x2 + 2(m + 1)x + m + 3 from (A + B)2 = a2 + 2AB + b2
2B =

\frac{2 (m + 1)}{2}

= m + 1 .......(i)
Also B2 = m + 3..........(ii)
Subst. for B in equation ..........(iii)
= (m + 1)2 = m + 3 = m2 + 2m + 1 = m + 3
= m2 + 2m - m + 1 - 3 = 0
m2 + m - 2 = 0
(m + 2)(m - 1) = 0 or m - 1 = 0
m = -2 or 1

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Vraag 14 Verslag

Find the eleventh term of the progression 4, 8, 16.....

213

212

211

210

Antwoorddetails

a = 4, r =

\frac{4}{2}

\frac{8}{4}

= 2
n = 11
Tn = arn - 1
T11 = 4(2)11 - 1
4(2)10 = 212
since 4 = 22
= 212

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Sequence And Series

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Vraag 15 Verslag

The base of a pyramid is a square of side 8cm. If its vertex is directly above the centre, find the height, given that the edge is 4.3cm

6cm

5cm

4cm

3cm

Antwoorddetails

Base of pyramid of a square of side 8cm vertex directly above the centre edge =

4 \sqrt{3}

cm
From the diagram, the diagonal of one base is AC2 = 82 + 82
Ac2 = 64 + 64 = 128
AC =

8 \sqrt{2}

but OC =

\frac{1}{2}

AC = 8

\sqrt{\frac{2}{2}}

=

4 \sqrt{2}

cm
OE = h = height
h2 = (

4 \sqrt{3}

)2
16 x 2 - 16 x 2
48 - 32 = 16
h =

\sqrt{16}

= 4

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Vraag 16 Verslag

Find the probability of selecting a figure which is a parallelogram from a square, a rectangle, a rhombus, a kite and a trapezium

3

5

2

5

4

5

1

5

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Vraag 17 Verslag

A rectangular lawn has an area of 1815 square yards. if its length is 50 meters, find its width in meters given that 1 metre equals 1.1 yards

39.93

35.00

33.00

30.00

Antwoorddetails

1m = 1.1 yard, length(L)= 50m
= (50 x 1.1)yards
= 55 yards
Area(A) = length(L) y width (W)
1815 = 55y width (W)
width (w) =

\frac{1815}{55}

= 33 years

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Areas

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Vraag 18 Verslag

Below are the scores of group of students in a music test:

Scores	1	2	3	4	5	6	7	8	9
No of students	3	6	10	8	6	5	2	4	12

If CF(x) is the number of students with scores less than or equal to x, find CF(6)

40

38

33

5

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Vraag 19 Verslag

Taking the period of day light on a certain day to be from 5.30a.m to 7.00p.m. Calculate the angle of a pie chart designed to show the periods of the day light and of darkness on that day

187o30' 172o30'

135o225'

202o30' 157o30'

195o165'

Antwoorddetails

Period of day light is from 5.30 to 7.00p.m - 13hrs 30min. interval
24hrs make 1 day
∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins.
Angle of sector of daylight =

\frac{13 \times 60 + 30 \times 360}{(24 \times 60)}

=

\frac{810}{1440}

x 360o
=

\frac{291600}{1440}

= 202.5o
202.5o = 202.30o
darkness period = 360o - 202o 30'
= 157o30'
= 202o30' 157o 30'

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Vraag 20 Verslag

A train moves from P to Q at an average speed of 90km/h and immediately returns from Q to P through the same route at an average speed of 45km/h. Find the average speed for the entire journey

55.00km/h

60.00km/h

67.50km\h

75.00km\h

Antwoorddetails

Average speed from P to Q = 90km\h
Average speed from O to P = 45km/h
Average for the entire journey = 90 + 45

\frac{135}{2}

= 67.50 km/h

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Vraag 21 Verslag

In the diagram above, QR // TS, QR : TS = 2.3, Find the ratio of the area of triangle PQR to the area of the trapezium QRST.

4:9

4:5

1:3

2:3

Antwoorddetails

h1 =

\frac{h_{1} + h_{2}}{1.5}

=

\frac{h_{1}}{h_{2}} = 2

$\frac{A_{△ P Q R}}{A_{T P (Q R S T)}} = \frac{\frac{1}{2} \times 2 \times h_{1}}{\frac{1}{2} \times h_{2} (2 + 3)}$

$\frac{2}{5} \times \frac{h_{1}}{h_{2}} = \frac{2}{5} \times 2$

= $\frac{4}{5}$

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Vraag 22 Verslag

If P varies inversely as V and V varies directly as R2, find the relationship between P and R given that R = 7 when P = 2.

P = 98R2

PR2 = 98

P =

\frac{1}{98 R^{2}}

P =

\frac{P R^{2}}{98}

Antwoorddetails

P =

\frac{1}{v}

and vR2 = P =

\frac{k}{v}

......(i)
and v KR2 .......(ii)
(where k is constant)
Subst. for v in equation (i) = p =

\frac{1^{2}}{K R}

.....(ii)
when r = 7, p = 2
2 =

\frac{k}{7^{2}}

k = 2 x 49
= 98
Subt. foe k in ....(iii)
P =

\frac{98}{R^{2}}

PR2 = 98

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Vraag 23 Verslag

Simplify without using tables $ \frac{2\sqrt{14}\times3\sqrt{21}}{7\sqrt{24}\times2\sqrt{98}} $

3√14

4

3√2

4

3√14

28

3√2

28

82

Antwoorddetails

\frac{2 \sqrt{14} \times 3 \sqrt{21}}{7 \sqrt{24} \times 2 \sqrt{98}}

=

\frac{6 \sqrt{14} \times 3 \times \sqrt{7} \times \sqrt{3}}{7 \times 2 \sqrt{6} \times \sqrt{7} \times \sqrt{14}}

=

\frac{3 \sqrt{3}}{14 \sqrt{6}}

=

\frac{3 \sqrt{3}}{14 \sqrt{2} \times \sqrt{3}}

=

\frac{\sqrt{3}}{\sqrt{2}}

x

\frac{\sqrt{2}}{\sqrt{2}}

=

\frac{3 \sqrt{2}}{28}

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Vraag 24 Verslag

Given that 1/2 log10 P = 1, find the value Of P

104

103

102

101

10-1

Logarithms

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Vraag 25 Verslag

The figure is an example of the construction of a

perpendicular bisector of a given straight line QR

perpendicular from a given point to a given line QR

perpendicular to a line from a given point P on that line

given angle

Euclidean Geometry

Simple Operations On Algebraic Expressions

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Vraag 26 Verslag

The goals scored by 40 football teams from three league divisions are recorded below:

Number of goals	0	1	2	3	4	5	6
frequency	4	3	15	16	1	0	1

What is the total number of goals scored by all the terms?

21

40

91

96

Antwoorddetails

Let x rep. number of goals

\begin{array}{cc} " f & " & f r e q u e n c y \\ x & f & f x \\ 0 & 4 & 0 \\ 1 & 3 & 3 \\ 2 & 15 & 30 \\ 3 & 16 & 48 \\ 4 & 1 & 4 \\ 5 & 0 & 0 \\ 6 & 1 & 6 \end{array}

\sum f x

= 91
Total number of goals scored is

\sum f x

= 91

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Vraag 27 Verslag

In the figure, find the value of x

110o

100o

90o

80o

Angles

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Vraag 28 Verslag

Convert 241 in base 5 to base 8,

718

1078

1768

2418

Antwoorddetails

2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o
50 + 20 + 1 = 7110
Convert 7110 to base 8
$\begin{array}{cc} 8 & 71 \\ 8 & 8 R 7 \\ 8 & 1 R 0 \\ 8 & 0 R 1 \end{array}$
= 1078

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Vraag 29 Verslag

An $ (n-2)^2 $ sided figure has n diagonals, find the number n of diagonals for a 25 sided figure

8

7

6

9

10

Variation

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Vraag 30 Verslag

Three angles of a nonagon are equal and the sum of six other angles is 1110o. Calculate the size of one of the equal angles.

210o

150o

105o

50o

Antwoorddetails

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Monagon = 9
Where 3 are equal and 6 other angles = 1110o
(2 x 9 - 4)90o = (18 - 4)90o
14 x 90o = 1260o
9 angles = 12600, 6 angles = 110o
Remaining 3 angles = 1260o - 1110o = 150o
Size of one of the 3 angles

\frac{150}{3}

= 50o

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Vraag 31 Verslag

If $ \tan \theta = \frac{m^2-n^2}{2mn} $ find $ \sec \theta $

m²+n²

(m²−n²)

m²+n²

2mn

mn

2(m²+n²⁾

m²n²

2m²−n²

Antwoorddetails

Tan

θ

=

\frac{m^{2} - n^{2}}{2 m n}

\frac{Opp}{Adj}

by pathagoras theorem
= Hyp2 = Opp2 + Adj2
Hyp2 = (m2 - n2) + (2mn)2
Hyp2 = m4 - 2m2n4 - 4m2 - n2
Hyp2 = m4 + 2m2 + n2n
Hyp2 = (m2 - n2)2
Hyp2 =

\frac{m^{2} + n^{2}}{2 m n}

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Vraag 32 Verslag

Make $y$ the subject of the formula $Z = x^2 + \frac{1}{y^3}$

y =

\frac{1}{(z - x^{2})^{3}}

y =

\frac{1}{(z + x^{2}) \frac{1}{3}}

y =

\frac{1}{(z x^{2}) \frac{1}{3}}

y =

3 \sqrt{z - x^{2}}

Antwoorddetails

Z = x2 +

\frac{1}{y^{3}}

Z - x2 =

\frac{1}{y^{3}}

y3 =

\frac{1}{Z x^{2}}

y =

3 \sqrt{\frac{1}{2} x^{2}}

∴ y =

3 \sqrt{Z - x^{2}}

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Vraag 33 Verslag

Two brothers invested a total of ₦5,000.00 on a farm project, the farm yield was sold for ₦15,000.00 at the end of the season. If the profit was shared in the ratio 2 : 3, what is the difference in the amount to profit received by the brothers?

₦2,000.00

₦4,000.00

₦6,000.00

₦10,000.00

Antwoorddetails

Total amount invested by A and B = ₦5,000
farm yield was sold for ₦15,000.00
profit = 15,000.00 - 5,000.00
= ₦10,000.00
Profit was shared in ratio 2 : 3
2 + 3 = 5
A received

\frac{2}{5}

of profit =

\frac{2}{5}

x 10,000 = ₦4,000.00
A receive

\frac{3}{5}

of profit =

\frac{3}{5}

x 10,000 = ₦6,000.00
Difference in profit received = ₦6,0000 - ₦4,000.00
= ₦2,000.00

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Vraag 34 Verslag

What is the circumference of latitude 0oS if R is the radius of the earth?

R cos

θ

2

π

R cos

θ

R sin

θ

2

π

R sin

θ

Antwoorddetails

Circumference of Latitude 0oS where R is the radius of the earth is = 2

π

R cos

θ

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Vraag 35 Verslag

Evaluate without using tables $ (0.008)^{\frac{1}{3}} \times (0.16)^{-\frac{3}{2}} $

625

8

625

1

8

Antwoorddetails

(0.008) -

\frac{1}{3}

x (0.16) -

\frac{3}{2}

= (8 x 10-3)10-3 x (16 x 10-2)-

\frac{3}{2}

=

\frac{(2^{3})}{10^{3}}

- 3 x

\frac{(2^{4})}{10}

-

\frac{- 3}{2}

=

\frac{625}{8}

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Vraag 36 Verslag

The minimum value of $y$ in the equation $y=x^2-6x+8$

8

3

0

-1

5

Quadratic Equations

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Vraag 37 Verslag

A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room

5

17

8

17

8

15

12

17

Antwoorddetails

ABCD is the floor, by pythagoras
AC2 = 144 + 81 =

\sqrt{225}

AC = 15cm
Height of room is 8m, diagonal of floor is 15m
∴ The cosine of the angle which a diagonal of the room makes with the floor is EC2 = 152
cosine =