JAMB UTME - Mathematics - 1987

Question 1 Rapport

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

Détails de la réponse

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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Question 2 Rapport

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

Détails de la réponse

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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Question 3 Rapport

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}}

Détails de la réponse

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}

Voir la réponse

Question 4 Rapport

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

8

3

0

-1

5

Quadratic Equations

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Question 5 Rapport

Convert 241 in base 5 to base 8,

718

1078

1768

2418

Détails de la réponse

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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Number Bases

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Question 6 Rapport

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

625

8

625

1

8

Détails de la réponse

(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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Fractions, Decimals And Approximations

Number Bases

Surds (radicals)

Graphs Of Linear And Quadratic Functions

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Question 7 Rapport

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

Détails de la réponse

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

Voir la réponse

Question 8 Rapport

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

Détails de la réponse

\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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Positive And Negative Integers, Rational Numbers

Surds (radicals)

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Question 9 Rapport

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

Détails de la réponse

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

Voir la réponse

Question 10 Rapport

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

Détails de la réponse

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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Question 11 Rapport

Find the values of y which satisfy the simultaneous equations x + y = 5, x2 - 2y2 = 1

-12, -2

-12, 12

12, -2

2, -2

Détails de la réponse

x + y = 5.......(i)
x2 - 2y2 = 1.......(ii)
x = 5 - y.........(iii)
Subst. for x in eqn.(ii) = (5 - y)2 - 2y2 = 1
25- 10y + y2 - 2y2 = 1
25 - 1 = y2 + 10y
y2 + 10y2 - 24 = 0
(y + 12)(y - 2) = 0
Then Either y + 12 = 0 or y - 2 = 0
= (12, -2)

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Solution Of Linear Equations

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Question 12 Rapport

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}

Détails de la réponse

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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Variation

Sequence And Series

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Question 13 Rapport

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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Question 14 Rapport

Find the least length of a rod which can be cut into exactly equal strips, each of either 40cm or 48cm in length

120cm

240cm

360cm

480cm

Détails de la réponse

The least length is

\frac{40}{48}

=

\frac{5}{8}

for the rod to be cut in exactly equal trips
Ratio

\frac{5}{6}

:

\frac{48}{40}

\frac{\frac{5}{6}}{\frac{40}{48}}

= 1

\frac{5}{6}

x

\frac{48}{40}

=

\frac{240}{240}

= 1
The least length = 240cm

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Number Bases

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Question 15 Rapport

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

Détails de la réponse

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 =

\frac{adj}{hyp}

=

\frac{15}{17}

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Circles

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Question 16 Rapport

In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40o. Calculate ∠QSR.

30o

40o

45o

50o

Circles

Voir la réponse

Question 17 Rapport

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

Détails de la réponse

\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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Question 18 Rapport

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

Détails de la réponse

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

π

R cos

θ

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Circles

Coordinate Geometry

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Question 19 Rapport

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

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

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

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

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

Détails de la réponse

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

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

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

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

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

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

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Solution Of Linear Equations

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Question 20 Rapport

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}}

Détails de la réponse

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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Question 21 Rapport

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

x > -3

x < -3

2 < x < 3

-3 < x < -2

Détails de la réponse

x - 1 > 4(x + 2) = x - 1 > 4x + 8
4x + 8 < x - 1 = 4x - x < -1 -8

= 3x < -9

∴ x < -3

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Question 22 Rapport

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

Détails de la réponse

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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Question 23 Rapport

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

104

103

102

101

10-1

Détails de la réponse

1/2log10P
log10P1/2 = 1
P1/2 = 10
p = 102

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Logarithms

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Question 24 Rapport

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

Détails de la réponse

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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Quadratic Equations

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Question 25 Rapport

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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Question 26 Rapport

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

9

15

27

33

Détails de la réponse

a = u2 - 3y2 = (1)2 - 3(-1) = -2
b = 2uu + v2 + v2
= 2(1)(-1) + (-1)2 = -1
∴ 2(2a - b)(a - b∴) = [2(-2) - 1](-2 - (-1)2)
= -3 - 3
= 9

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Question 27 Rapport

Factorize 62x + 1 + 7(6x) - 5

[3(6x) - 5][2(6x)m + 1]

[3(6x) + 5][2(6x) - 1]

[(6x) - 5][3(6x) + 1]

[2(6x) + 5][3(6x) - 1]

Détails de la réponse

62x + 1 + 7(6x) - 5
Let 6x = y....(i)
6yx + 7y - 5 = 0
(3y + 5)(2y - 1) = 0.....(i)
Subt. for y from eqn. (i) in eqn. (ii)
= [3(6x) + 5][2(6x) - 1]

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Question 28 Rapport

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

Détails de la réponse

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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Solution Of Linear Equations

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Question 29 Rapport

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'

Détails de la réponse

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'

Voir la réponse

Question 30 Rapport

The formula Q = 1.5 + 0.5n gives the cost Q(in Naira)of feeding n people one additional person

350k

200k

150k

50k

Détails de la réponse

Q = 1.5 + 0.5n gives the cost 1(in Naira) of feeding n people for a week. Extra cost of feeding one additional person = n = 1
Subt. for n in the formula Q = 1.5 + 0.5(1)
= 15 + 0.5(2)
Q = ₦2
= 200k

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Question 31 Rapport

What is the locus or the mid-point of all the chords of length 6cm with circle of radius 5cm and with center 0?

A circle of radius 4 cm and the center 0

The perpendicular bisector of the chords

A straight line passing through 0

A circle of radius 6 cm and with center 0

Détails de la réponse

x2 + 33 = 52
x2 + 9 = 25
x2 = 25 – 9
x2 = 16

x = √ 16
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0

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Circles

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Question 32 Rapport

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

Détails de la réponse

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

Voir la réponse

Question 33 Rapport

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

Détails de la réponse

QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR

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Question 34 Rapport

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

Détails de la réponse

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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Lengths And Perimeters

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Question 35 Rapport

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

Détails de la réponse

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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Question 36 Rapport

If $ P = \frac{2}{3}\left(\frac{1-r^2}{n^2}\right) $, find n when r = $ \frac{1}{3} $ and p = 1

3

2

1

3

2

3

Détails de la réponse

If P =

\frac{2}{3}

(

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

), find n when r =

\frac{1}{3}

and p = 1
p =

\frac{2 (1 - r^{2})}{3 n^{2}}

when r =

\frac{1}{3}

and p = 1
1 =

\frac{2}{3}

\frac{(1 - (\frac{1}{3})^{2})}{n^{2}}

n2 =

\frac{2 (3 - 1)}{3 \times 3}

n2 =

\frac{2 \times 2}{3 \times 3}

=

\frac{4}{9}

n =

\frac{4}{9}

=

\frac{2}{3}

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Probability

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Question 37 Rapport

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}

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

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

Détails de la réponse

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}

Lisez la note de cours sur Sine, Cosine And Tangent Of An Angle (WAEC)

Lisez la note de cours sur Trigonometry (JAMB)

Lisez la note de cours sur Coordinate Geometry (WAEC)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Coordinate Geometry

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Question 38 Rapport

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

Détails de la réponse

A rectangle, a square and a rhombus are the only parallelogram
∴ Probability of a parallelogram =

\frac{3}{5}

Voir la réponse

Question 39 Rapport

In the figure, find the value of x

110o

100o

90o

80o

Détails de la réponse

Z = X = Y = 180o .........(i)

2Z + Y + Y = 190 = 2Z + 2Y = 180

Z + Y = 90o........(ii)

hence x + (z + y) = 180

x + 90o = 180o

x = 180o - 90o

= 90o

Lisez la note de cours sur Angles (WAEC)

Angles

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Question 40 Rapport

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

213

212

211

210

Détails de la réponse

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

Lisez la note de cours sur Sequence And Series (WAEC)

Sequence And Series

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Question 41 Rapport

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%

Détails de la réponse

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%

Lisez la note de cours sur Modular Arithmetic (WAEC)

Lisez la note de cours sur Ratio, Proportions And Rates (WAEC)

Modular Arithmetic

Ratio, Proportions And Rates

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Question 42 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Angles And Intercepts On Parallel Lines (WAEC)

Angles And Intercepts On Parallel Lines

Voir la réponse

Question 43 Rapport

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

Détails de la réponse

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