JAMB UTME - Mathematics - 1983

If M represents the median and D the mode of the measurements 5, 9, 3, 5, 7, 5, 8, then (M, D) is

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If x + 2 and x - 1 are factors of the expression 1x3 + 2kx2 + 24, find the values of 1 and K

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f(x) = Lx3 + 2kx2 + 24
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
1 = -6

Which of the following equations represents the graph above?

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x = -2 or x = $\frac{1}{4}$

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

y = 2 - 7x - 4x2

Simplify $\frac{3}{2x-1}$ + $\frac{2-x}{x-2}$

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

The scores of 16 students in a mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65,
70. What is the sum of the median and modal scores?

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

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($\frac{2}{3}$)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

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

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

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

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

On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures.

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Area of the paper = area of square = L x B or S2
where s = S x k
Area of the paper = (2.524)2
area of the diagram = (0.524)2
area not covered = (2.524)2 - (0.524)2
= 6.370576 - 0.274576
= 6.096
= 6.10cm2 (2 s.f)

Make T the subject of the equation $\frac{av}{1?v}$ = $\sqrt{\frac{2v+T}{a+2T}}$

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

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

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< HKF = 60${}^{?}$, < KFG = 120${}^{?}$

< KFG = < KHG = x(opposite angles)

x = 120${}^{?}$

The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x

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(3x + 1)2 = (3x - 1)2 + 2 (Pythagoras's theorem)
9x2 + 6x + 1 = 9x - 6x2 + 1 + x2
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12

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.

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

In the figure PT is a tangent to the circle with centre at O. If PQT = 30${}^{?}$, find the value of PTO

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

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

A ship H leaves a port P and sails 30 km due south. Then it sails 50km due west. What is the bearing of H from P

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Simplify $\frac{3\sqrt{27a^{-9}}}{8}$

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

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

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

= 360 - 260

= 100${}^{\circ }$

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?

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Speed = $\frac{distance}{time}$
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

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

$\begin{array}{cccccccccc}\text{No. of members per household}& 1& 2& 3& 4& 5& 6& 7& 8& \text{Total}\\ \text{No. of households}& 3& 12& 15& 28& 21& 10& 7& 4& 100\end{array}$

What is the median?

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

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.

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$\frac{1}{2}$

A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received ₦5,000 more than X. What is the total profit of the company for the year?

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Total sharing ratio is 9
X has 4, Y has 4 + 1
If 1 is ₦5000
Total profit = 5000 x 9
= ₦45,000

The value of $(0.303{)}^3-(0.02{)}^3$ is

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The letters of the word MATRICULATION are cut and put into a box. One letter is drawn at random from the box. Find the probability of drawing a vowel

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Vowels of letters are 6 in numbers
prob. of vowel = $\frac{6}{13}$

Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0

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x2 + y - 8 = 0, y + 5x - 2 = 0
Rearranging, x2 + y = 8.....(i)
5x + y = 2.......(ii)
Subtract eqn(ii) from eqn(i)
x2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6, -1

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?

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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 }$

If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled?

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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

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W $\alpha$ $\frac{1}{v}$u $\alpha$ 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 = $\frac{8}{v^3}$

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

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tan$\theta$ = $\frac{0.8}{2}$ = (0.4)

$\theta$ = tan-1(0.4)

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

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

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

Find x if (xbase4)2 = 100100base2

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(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 = $\begin{array}{cc}4& 6\\ 1r2& \end{array}$
= 124

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

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$\frac{2}{3}$ + 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${}^{\circ }$

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

Answer Details

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

Without using tables find the numerical value of log749 + log7($\frac{1}{7}$)

Answer Details

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

Find the missing value in the following table:

$\begin{array}{ccccccc}x& -2& -1& 0& 1& 2& 3\\ y=x^3-x+3& & 3& 3& 3& 9& 27\end{array}$

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When x = -2, y = x3 - x + 3
= -23 - (-2) + 3
= -8 + 2 + 3
= -3

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

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

In the figure, find PRQ

Answer Details

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

$\overline{PQR}$ = $\frac{125}{2}$

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

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?

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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 = $\frac{2x-42}{360}$ x 60

= $\frac{2\times 42-24}{360}$ x 60

= $\frac{84-24}{6}$

= 10

PQRS is a cyclic quadrilateral. If ∠QPS = 75o, what is the size of ∠QRS?

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If 0.0000152 x 0.042 = A x 108, where 1 ≤ $\le$ A < 10, find A and B

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0.0000152 x 0.042 = A x 108

1 ≤ $\le$ 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

Simplify log10 a$\frac{1}{3}$ + $\frac{1}{4}$log10 a - $\frac{1}{12}$log10a7

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

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

Answer Details

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

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

Answer Details

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

Correct each of the numbers 59.81798 and 0.0746829 to three significant figures and multiply them, giving your answer to three significant figures

Answer Details

59.81798 = 59.8 (3 s.f)
0.0746829 = 0.0747
59.8 x 0.0747 = 4.46706
= 4.48 (3s.f)

Solve the following equations 4x - 3 = 3x + y = x - y = 3, 3x + y = 2y + 5x - 12

Answer Details

4x - 3 = 3x + y = x - y = 3.......(i)
3x + y = 2y + 5x - 12.........(ii)
eqn(ii) + eqn(i)
3x = 15
x = 5
substitute for x in equation (i)
5 - y = 3
y = 2

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

Answer Details

< SPQ = 80${}^{?}$

< SPQ + < SRQ = 180(Supplementary)

80 + < QRS = 180${}^{?}$ - 80${}^{?}$

= 100${}^{?}$

Factorize completely 81a4 - 16b4

Answer Details

81a4 - 16b4 = (9a2)2 - (4b2)2
= (9a2 + 4b2)(9a2 - 4b2)
N:B 9a2 + 4b2 = (3a - 2b)(3a - 2b)

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

Answer Details

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

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

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