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JAMB UTME - Mathematics - 1983

Question 1 Report

In a triangle PQT, QR = \( \sqrt{3}\text{ cm} \), PR = 3cm, PQ = \( 2\sqrt{3} \)cm and PQR = 30^o. Find angles P and R

P = 60o and R = 90o

P = 30o and R = 120o

P = 90o and R = 60o

P = 60o and R 60o

P = 45o and R = 105o

Answer Details

By using cosine formula, p2 = Q2 + R2 - 2QR cos p
Cos P = $\frac{Q^{2} + R^{2} - p^{2}}{2 Q R}$
= $\frac{(3)^{2} + 2 (\sqrt{3})^{2} - 3^{2}}{2 \sqrt{3}}$
= $\frac{3 + 12 - 9}{12}$
= $\frac{6}{12}$
= $\frac{1}{2}$
= 0.5
Cos P = 0.5
p = cos-1 0.5 = 60 $^{\circ}$
= < P = 60 $^{\circ}$
If < P = 60 $^{\circ}$ and < Q = 30

< R = 180 $^{\circ}$ - 90 $^{\circ}$
angle P = 60 $^{\circ}$ and angle R is 90 $^{\circ}$

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Question 3 Report

In the figure, find PRQ

66

\frac{1}{2}

^{\circ}

^{\circ}

125

^{\circ}

105

^{\circ}

65

^{\circ}

Answer Details

Angle subtended at any part of the circumference of the circle $\frac{125^{o}}{2}$ at centre = 360 $^{\circ}$ - 235 $^{\circ}$ = 125 $^{\circ}$

$\bar{P Q R}$ = $\frac{125}{2}$

= 62 $\frac{1}{2}$ $^{\circ}$

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Representation Of Data

Triangles And Polygons

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Question 4 Report

PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60^o. Find the perimeter of the figure PTRS. \( \pi = \frac{22}{7} \)

25cm

18cm

36cm

29cm

25.5cm

Answer Details

Circumference of PRS = $\frac{π}{2}$ = $\frac{22}{7}$ x $\frac{7}{1}$ x $\frac{1}{2}$
= 11cm
Side PT = 7cm, Side TR = 7cm
Perimeter(PTRS) = 11cm + 7cm + 7cm
= 25cm

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Question 5 Report

Simplify \( \frac{3}{2x-1} + \frac{2-x}{x-2} \)

2−x

(2x−1)(x−2)

5−x

(2x−1)(x−2)

4−2x

2x−1

6−3x

(2x−1)(x−2)

Answer Details

$\frac{3}{2 x - 1}$ + $\frac{2 - x}{x - 2}$
= $\frac{3}{2 x - 1}$ - $\frac{x - 2}{x - 2}$
= $\frac{3}{2 x - 1}$ - 1
= $\frac{3 - (2 x - 1)}{2 x - 1}$
= $\frac{3 - 2 x + 1}{2 x - 1}$
= $\frac{4 - 2 x}{2 x - 1}$

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Algebraic Expressions

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Question 10 Report

If \( \left(\frac{2}{3}\right)^m \left(\frac{3}{4}\right)^n = \frac{252}{729} \), find the values of m and n

m = 4, n = 2

m = -4, n = -2

m = -4, n =2

m = 4, n = -2

m = -2, n = 4

Answer Details

( $\frac{2}{3}$ )m ( $\frac{3}{4}$ )n = $\frac{252}{729}$
$\frac{2^{m}}{4^{n}}$ x $\frac{3^{n}}{3^{m}}$ = 283-6
2m - 2n x 3n - m = 28 x 3-6
m - 2n = 8........(i)
-m + n = -6........(ii)
Solving the equations simultaneously
m = 4, n = -2

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Number Bases

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Question 11 Report

y varies partly as the square of x and partly as the inverse of the square root of x. Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4.

y = $\frac{10 x^{2}}{31} + \frac{52}{31 \sqrt{x}}$

y = x2 + $\frac{1}{\sqrt{x}}$

y = x2 + $\frac{1}{x}$

y = $\frac{x^{2}}{31} + \frac{1}{31 \sqrt{x}}$

Answer Details

y = kx2 + $\frac{c}{\sqrt{x}}$
y = 2 when x = 1
2 = k + $\frac{c}{1}$
k + c = 2
y = 6 when x = 4
6 = 16k + $\frac{c}{2}$
12 = 32k + c
k + c = 2
32k + c = 12
= 31k + 10
k = $\frac{10}{31}$
c = 2 - $\frac{10}{31}$
= $\frac{62 - 10}{31}$
= $\frac{52}{31}$
y = $\frac{10 x^{2}}{31} + \frac{52}{31 \sqrt{x}}$

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Algebraic Expressions

Quadratic Equations

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Question 12 Report

If \(0.0000152 \times 0.042 = A \times 10^8\), where \(1 \le A < 10\), find A and B

A = 9, B = 6.38

A = 6.38, B = -9

A = -6.38, B = -9

A = -9, B = -6.38

Answer Details

0.0000152 x 0.042 = A x 108

1 $\leq$ A < 10, it means values of A includes 1 - 9

0.0000152 = 1.52 x 10-5

0.00042 = 4.2 x 10-4

1.52 x 4.2 = 6.384

10-5 x 10-4

= 10-5-4

= 10-9

= 6.38 x 10-9

A = 6.38, B = -9

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Question 13 Report

A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours?

120km/hr

60km/hr

670km/hr

40km/hr

Answer Details

Speed = $\frac{d i s t a n c e}{t i m e}$
let x represent the speed, d represent distance
x = $\frac{d}{4}$
d = 4x
2x = $\frac{600 - d}{3}$
6x = 600 - d
6x = 600 - 4x
10x = 600
x = $\frac{600}{10}$
= 60km/hr

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Question 17 Report

Simplify \( \log_{10} a^{\frac{1}{3}} + \frac{1}{4}\log_{10} a - \frac{1}{12}\log_{10} a^{7} \)

1

\frac{7}{6}

log10 a

zero

10

Answer Details

log10 a $\frac{1}{3}$ + $\frac{1}{4}$ log10 a - $\frac{1}{12}$ log10a7 = log10 a $\frac{1}{3}$ + log10 $\frac{1}{4}$ - log10 a $\frac{7}{12}$
= log10 a $\frac{7}{12}$ - log10 a $\frac{7}{12}$
= log10 1 = 0

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Question 18 Report

Which of the following equations represents the graph above?

y = 2 + 7x + 4x2

y = 2 - 7x + 4x2

y = 2 + 7x - 4x2

y = 2 - 7x - 4x2

Answer Details

x = -2 or x = $\frac{1}{4}$

(x + 2)(4x - 1) = 0

y = 2 - 7x - 4x2

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Algebraic Expressions

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Question 19 Report

Find x if (xbase4)2 = 100100base2

6

12

100

210

110

Answer Details

(x2)4 to base10 gives (1 x 25) + (1 x 25) + (1 x 2)
32 + 4 = 36
x2 = 36, x = 6
610 to base 4 = $\begin{array}{cc} 4 & 6 \\ 1 r 2 \end{array}$
= 124

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Number Bases

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Question 20 Report

Make T the subject of the equation \( \frac{av}{1-v} = \sqrt{\frac{2v+T}{a+2T}} \)

T = $\frac{3 a v}{1 - v}$

T = $\frac{1 + v}{2 a^{2} v^{3}}$

T = $\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} + (1 - v)^{2}}$

\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} - (1 - v)^{3}}

Answer Details

$\frac{a v}{1 - v}$ = $\sqrt{\frac{2 v + T}{a + 2 T}}$
$\frac{(a v)^{3}}{(1 - v)^{3}}$ = $\frac{2 v + T}{a + 2 T}$
$\frac{a^{3} v^{3}}{(1^{3} - v)^{3}}$ = $\frac{2 v + T}{a + 2 T}$
= $\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} - (1 - v)^{3}}$

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Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Quadratic Equations

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Question 23 Report

In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per household:

No. of members per household	1	2	3	4	5	6	7	8	Total
No. of households	3	12	15	28	21	10	7	4	100

What is the median?

4

3

5

4.5

none

Answer Details

$\begin{array}{cc} x & f \\ 1 & 3 \\ 2 & 13 \\ 3 & 15 \\ 4 & 28 \\ 5 & 21 \\ 6 & 10 \\ 7 & 7 \\ 8 & 4 \end{array}$
Median is half of total frequency = 50th term
4 falls in the range = 4

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Question 26 Report

The figure FGHK is a rhombus. What is the value of angle X?

30

^{\circ}

90

^{\circ}

150

^{\circ}

120

^{\circ}

60

^{\circ}

Answer Details

< HKF = 60 $^{\circ}$ , < KFG = 120 $^{\circ}$

< KFG = < KHG = x(opposite angles)

x = 120 $^{\circ}$

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Triangles And Polygons

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Question 27 Report

If w varies inversely as V and U varies directly as w3, Find the relationship between u and v given that u = 1, when v = 2

u =

\frac{8}{v^{3}}

v =

\frac{8}{u^{2} v^{3}}

u = 8v3

v = 8u2

Answer Details

W $α$ $\frac{1}{v}$ u $α$ w3
w = $\frac{k_{1}}{v}$
u = k2w3
u = k2( $\frac{k_{1}}{v}$ )3
= $\frac{k_{2} k_{1}^{2}}{v^{3}}$
k = k2k1k2
u = $\frac{k}{v^{3}}$
k = uv3
= (1)(2)3
= 8
u = $\frac{8}{v^{3}}$

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Change Of Subject Of A Formula/Relation

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Question 28 Report

PQRS is a desk of dimensions \(2\text{m} \times 0.8\) which is inclined at \(30^\circ\) to the horizontal. Find the inclination of the diagonal PR to the horizontal

30

^{\circ}

35'

30

^{\circ}

32'

15

^{\circ}

36'

10

10

^{\circ}

42'

Answer Details

tan $θ$ = $\frac{0.8}{2}$ = (0.4)

$θ$ = tan-1(0.4)

From the diagram, the inclination of the diagonal PR to the horizontal is 10 $^{\circ}$ 42'

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Measures Of Dispersion

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Question 29 Report

If O is the centre of the circle in the figure, find the value of x

50

260

100

65

130

Answer Details

From the diagram; The value of x = 360 $^{\circ}$ - 2(130 $^{\circ}$ )

= 360 - 260

= 100 $^{\circ}$

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Question 30 Report

The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart?

11

^{\circ}

19

\frac{1}{3}

^{\circ}

33

\frac{1}{3}

^{\circ}

91

\frac{1}{3}

^{\circ}

91

^{\circ}

Answer Details

Adding the values of all the items together, it gives 70

Okro sector = $\frac{145}{270}$ x $\frac{360^{o}}{1}$

= 19.33 $^{\circ}$

= 19 $\frac{1}{3}$ $^{\circ}$

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Question 31 Report

Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77

25.12

25.30

25.26

25.50

25.75

Answer Details

$\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}$
mean = $\frac{126.3}{5}$
= 25.26

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Integration

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Question 32 Report

One interior angle of a convex hexagon is 170o and each of the remaining interior angles is equal to xo. Find x

120o

110o

105o

102o

100o

Answer Details

An hexagon polygon is a six sided polygon
n = 6
sum of all interior angles of hexagon polygon will be (2n - 4) x 90o
= [2 x 6 - 4] x 90
= 720o
If one angle is 170o and each of the remaining five angles is 5xo
5xo + 170o = 720o
= 5x + 550
x = $\frac{550}{5}$
= 110o

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Polynomials

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Question 33 Report

The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.

47

50

48

\frac{1}{2}

48

49

Answer Details

By re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73
The median = $\frac{47 + 50}{2}$
$\frac{97}{2}$
= 48 $\frac{1}{2}$

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Question 34 Report

In the figure PT is a tangent to the circle with centre at O. If PQT = \(30^\circ\), find the value of PTO

30

^{\circ}

50

^{\circ}

24

^{\circ}

12

^{\circ}

60

^{\circ}

Answer Details

FROM the diagram, PQT = 50 $^{\circ}$

PTQ = 50 $^{\circ}$ (opposite angles are supplementary)

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Question 35 Report

Simplify \( \frac{3\sqrt{27a^{-9}}}{8} \)

9a²

2a³

3

2a³

2

3a²

3a²

2

Answer Details

$\frac{3 \sqrt{27 a^{- 9}}}{8}$ = $\frac{3 a^{- 3}}{2}$
= $\frac{3}{2}$ x $\frac{1}{a^{3}}$
= $\frac{3}{2 a^{3}}$

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Fractions, Decimals And Approximations

Surds (radicals)

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Question 37 Report

Find the angle of the sectors representing each item in pie chart of the following data 6, 10, 14, 16, 26

15o, 25o, 35o, 40o, 65o

60o, 100o, 140o, 160o, 260o

6o, 10o, 14o, 16o, 26o

30o, 50o, 70o, 80o, 130o

none of the above

Answer Details

6 + 10 + 14 + 16 + 26 = 72
$\frac{6}{72}$ x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively

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Polynomials

Measures Of Dispersion

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Question 38 Report

In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics?

15

10

18

12

20

Answer Details

2x - 24) $^{\circ}$ + (3x - 18) $^{\circ}$ + (2x + 12) $^{\circ}$ + (x + 12) $^{\circ}$ + x $^{\circ}$ = 360 $^{\circ}$

9x = 360 $^{\circ}$ + 18 $^{\circ}$

x = $\frac{378}{9}$

= 42 $^{\circ}$ , if x = 42 $^{\circ}$ , then add maths = $\frac{2 x - 42}{360}$ x 60

= $\frac{2 \times 42 - 24}{360}$ x 60

= $\frac{84 - 24}{6}$

= 10

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Question 39 Report

If a rod of length 250cm is measured as 255cm long in error, what is the percentage error in the measurement?

55

10

5

4

2

Answer Details

% error = $\frac{Actual error}{real value}$ x 100
= $\frac{5}{250}$ x 100
= 2

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Percentages

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Question 40 Report

Given that cos z = L , whrere z is an acute angle, find an expression for \( \frac{\cot z - \cosec z}{\sec z + \tan z} \)

1-L

1+L

L²√3

1+L

1+L³

L²

L(L−1)

1−L+1√1−L²

Answer Details

Given Cos z = L, z is an acute angle
$\frac{cot z - cosec z}{sec z + tan z}$ = cos z
= $\frac{cos z}{sin z}$
cosec z = $\frac{1}{sin z}$
cot z - cosec z = $\frac{cos z}{sin z}$ - $\frac{1}{sin z}$
cot z - cosec z = $\frac{L - 1}{sin z}$
sec z = $\frac{1}{cos z}$
tan z = $\frac{sin z}{cos z}$
sec z = $\frac{1}{cos z}$ + $\frac{sin z}{cos z}$
= $\frac{1}{l}$ + $\frac{sin z}{L}$
the original eqn. becomes
$\frac{cot z - cosec z}{sec z + tan z}$ = $\frac{L - \frac{1}{sin z}}{1 + s i n \frac{z}{L}}$
= $\frac{L (L - 1)}{sin z (1 + sin z)}$
= $\frac{L (L - 1)}{sin z + 1 - c o s^{2} z}$
= sin z + 1
= 1 + $\sqrt{1 - L^{2}}$
= $\frac{L (L - 1)}{1 - L + 1 \sqrt{1 - L^{2}}}$

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Matrices And Determinants

Modular Arithmetic

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Question 41 Report

In the figure, PQR is a straight line. Find the values of x and y.

^{\circ}

x = 15

^{\circ}

, y = 52.5

^{\circ}

x = 22.5

^{\circ}

, y = 45.0

^{\circ}

x = 56.25

^{\circ}

, y = 11.5

^{\circ}

Answer Details

$\frac{2}{3}$ + 3y + 45 $^{\circ}$ = 180 $^{\circ}$

3x + 6y + 90 $^{\circ}$ = 360 $^{\circ}$

3x + 67 = 270.......(i) x 2, 5x + y + y = 180 $^{\circ}$

5x + 2y = 180 $^{\circ}$ .......(ii) x 6

6 x 12y = 540..........(iii)

30x + 12y = 1080.........(iv)

egn(iv) - eqn(iii)

24x = 540

x = 22.5 and y = 33.75 $^{\circ}$

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Question 42 Report

Without using tables find the numerical value of \( \log_{7} 49 + \log_{7}\left(\frac{1}{7}\right) \)

1

2

3

7

9

Answer Details

log749 + log7 $\frac{1}{7}$
= log77
= 1

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Positive And Negative Integers, Rational Numbers

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Question 44 Report

PQRS is a cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of \(50^\circ\) with the tangent as shown in the figure. What is the size of QRS?

50

^{\circ}

40

^{\circ}

110v

80

^{\circ}

^{\circ}

Answer Details

< SPQ = 80 $^{\circ}$

< SPQ + < SRQ = 180(Supplementary)

80 + < QRS = 180 $^{\circ}$ - 80 $^{\circ}$

= 100 $^{\circ}$

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Indices, Logarithms And Surds

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Question 45 Report

Given a regular hexagon, calculate each interior angle of the hexagon

60o

30o

120o

45o

135o

Answer Details

Sum of interior angles of polygon = (2n - 4)rt < s

sum of interior angles of an hexagon
(2 x 6 - 4) x 90o = (12 - 4) x 90o
= 8 x 90o
= 720o
each interior angle will have $\frac{720^{o}}{6}$
= 120o

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Question 48 Report

In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS

2

\sqrt{3}

cm

4

\sqrt{6}

cm

2

\sqrt{6}

cm

8

\sqrt{6}

cm

8cm

Answer Details

From the diagram, sin 60o = $\frac{P R}{8}$
PR = 8 sin 60 = $\frac{8 \sqrt{3}}{2}$
= 4 $\sqrt{3}$
Cos 45o = $\frac{P R}{P S}$ = $\frac{4 \sqrt{3}}{P S}$
PS Cos45o = 4 $\sqrt{3}$
PS = 4 $\sqrt{3}$ x 2
= 4 $\sqrt{6}$

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Angles Of Elevation And Depression

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