JAMB UTME - Mathematics - 1984

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$\frac{2}{3}-\frac{1}{5}$ = $\frac{10-3}{15}$
= $\frac{7}{15}$
$\frac{1}{3}$ Of $\frac{2}{5}$ = $\frac{1}{3}$ x $\frac{2}{5}$
= $\frac{2}{5}$
($\frac{2}{3}-\frac{1}{5}$) - $\frac{1}{3}$ of $\frac{2}{5}$
= $\frac{7}{15}-\frac{2}{15}$ = $\frac{1}{3}$
3 - $\frac{1}{1\frac{1}{2}}$ = 3 - $\frac{2}{3}$
= $\frac{7}{3}$
$\frac{\frac{2}{3}-\frac{1}{5}\text{of}\frac{2}{15}}{3-\frac{1}{1\frac{1}{2}}}$
= $\frac{\frac{1}{3}}{\frac{7}{3}}$
= $\frac{1}{3}$ x $\frac{3}{7}$
= $\frac{1}{7}$

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(y - 1)2 = 4y - 7
y2 - 2xy + 1 = 4y - 7
y2 - 6y + 8 = 0
(y - 4)(y - 2)
y = 4 or 2 = 4

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Pq + 1 = q2......(i)
t = $\frac{1}{p}$ - $\frac{1}{pq}$.........(ii)
p = $\frac{q^2-1}{q}$
Sub for p in equation (ii)
t = $\frac{1}{q^2-\frac{1}{q}}$ - $\frac{1}{\frac{q^2-1}{q}\times q}$
t = $\frac{q}{q^2-1}$ - $\frac{1}{q^2-1}$
t = $\frac{q-1}{q^2-1}$
= $\frac{q-1}{(q+1)(q-1)}$
= $\frac{1}{q+1}$

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$\frac{1}{tan\theta }$ = $\frac{cos\theta }{sin\theta }$
sin$\theta$ = $\frac{x}{y}$
cos$\theta$ = $\frac{\sqrt{y^2-x^2}}{y}$

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$\begin{array}{ccc}\text{x(mid point)}& f& fx\\ 3.4& 3& 10.2\\ 3.5& 6& 21.0\\ 3.6& 7& 25.2\\ 3.7& 4& 14.8\end{array}$
$\sum f$ = 20
$\sum fx$ = 71.2
mean = $\frac{\sum fx}{\sum f}$
= $\frac{71.2}{20}$
= 3.56

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f(x0 = 2(x - 3)2 + 3(x - 3) + 4
= (2 + 3) + (x - 3) + 4
5(x - 3) + 4
5x - 15 + 4
= 5x - 11
f(3) = 5 x 3 - 11
= 4

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x3 + px2 + qx + 1 = (x - 1) Q(x) + R
x - 2 = 0, x = 2, R = 0,
4p + 2p = -9........(i)
x3 + px2 + qx + 1 = (x - 1)Q(x) + R
-1 + p - q + 1 = 0
p - q = 0.......(ii)
Solve the equation simultaneously
p = $\frac{-3}{2}$
q = $\frac{-3}{2}$
p + q = $\frac{3}{2}$ - $\frac{3}{2}$
= $\frac{-6}{2}$
= -3

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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

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$\frac{3^n-3^{n-1}}{3^3\times 3^n-27\times 3^{n-1}}$ = $\frac{3^n-3^{n-1}}{3^3(3^n-3^{n-1})}$
= $\frac{3^n-3^{n-1}}{27(3^n-3^{n-1})}$
= $\frac{1}{27}$

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0.00014321940000 $\frac{0.0001432}{1940000}$ = k x 10n

where 1 ≤ $\le$ k ≤ $\le$ 10 and n is a whole number. Using four figure tables, the eqn. gives 7.38 x 10-11

k = 7.381, n = -11

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by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY2 = 62 + 32

XY2 = 36 + 9 = 45

XY = $\sqrt{45}=3\sqrt{3}$

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Total cost of production = ₦120.00
Labour Cost = $\frac{70}{120}$ x $\frac{{360}^o}{1}$
= 120o

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$\frac{PQ}{6+RT}=\frac{6}{12}=\frac{6}{PS}$(Similar triangles)

$\frac{6}{12}=\frac{8}{PS}$

PS = $\frac{12\times 8}{6}$

= 16cm

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Total cost of 1035 oranges at ₦145 each
= 1035 x 145
= 20025
Total selling price at ₦245 each
= (103)5 x 245
= 30325
Hence his gain = 30325 - 20025
= 10305

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A dice has 6 faces, 2 dice has 6 x 6 = 36 combined face
Prob. of both showing the same number of points
= $\frac{6}{36}$
= $\frac{1}{6}$

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P(x, y), P(0, 3) If x increases by 1 unit and y by 4 units, then ratio of x : y = 1 : 4
$\frac{x}{1}$ = $\frac{y}{4}$
y = 4x
Hence the sign of the graph is y + 3 = 4x

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log327 = 3log33
=3 log3$\frac{1}{81}$
= -4 log33 = -4
let log$\frac{1}{4}$64 = ($\frac{1}{4}$)x
= 64
4-x = 43
$\frac{lo{g}_{3}27-lo{g}_{\frac{1}{4}}64}{lo{g}_3\frac{1}{81}}$ = $\frac{3-(-3)}{-4}$
= $\frac{-6}{4}$
= $\frac{-3}{2}$

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$\frac{r}{h}$ = tan yo
h = $\frac{r}{tan{y}^o}$
= r cot yo

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Let x represent the price of an orange and
y represent the number of oranges that can be bought
xy = 240k, y = $\frac{240}{x}$.....(i)
If the price of an oranges is raised by $\frac{1}{2}$k per orange, number that can be bought for ?240 is reduced by 16
Hence, y - 16 = $\frac{240}{x+\frac{1}{2}}$
= $\frac{480}{2x+1}$
= $\frac{480}{2x+1}$.....(ii)
subt. for y in eqn (ii) $\frac{240}{x}$ - 16
= $\frac{480}{2x+1}$
= $\frac{240?16x}{x}$
= $\frac{480}{2x+1}$
= (240 - 16x)(2x + 1)
= 480x
= 480x + 240 - 32x2 - 16
480x = 224 - 32x2
x2 = 7
x = $\sqrt{7}$
= 2.5
= 2$\frac{1}{2}$k

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Asector = $\frac{60}{360}\times \pi r^2$

= $\frac{1}{6}\pi r^2$

A$△$ = $\frac{1}{2}{r}^2\sin {60}^o$

$\frac{1}{2}{r}^2\times \frac{\sqrt{3}}{2}=\frac{r^2\sqrt{3}}{4}$

A$\text{shaded portion}$ = Asector -
A$△$

= ($\frac{1}{6}\pi r^2-\frac{r^2\sqrt{3}}{4}{)}^3$

= $\frac{\pi r^2}{2}-\frac{3r^2\sqrt{3}}{4}$

= $\frac{r^2}{4}(2\pi -3\sqrt{3})$

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If the graph of y = x3 - 3x + 1 is plotted,the graph crosses the x-axis in the ranges -2 < x < -1 and 0 < x < 1

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When y = 5, y = x2 - 12x + 40, becomes
x2 - 12x + 40 = 5
x2 - 12x + 40 - 5 = 0
x2 + 12x + 35 = 0
x2 - 7x - 5x + 35 = 0
x(x - 7) - 5(x - 7) = 0
= (x - 5)(x - 7)
x = 5 or 7

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$\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}$ = $\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}$ x $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}$
= $\frac{(5\times 7)+(5\sqrt{7}\times 5)-(7\times \sqrt{5}\times 7)(-7\times 5)}{(\sqrt{7}{)}^2}$
= $\frac{5\sqrt{35}-7\sqrt{35}}{2}$
= $\frac{-2\sqrt{35}}{2}$
= - $\sqrt{35}$

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$P\propto \frac{Q^2}{R}$
$\therefore$ $P=\frac{K{Q}^2}{R}$
$36=\frac{K{Q}^2}{4}$
$K=\frac{36\times 4}{9}$
$K=16$
$P=\frac{16\times 5^2}{2}=200$

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2 x s = 280o(Angle at centre = 2 x < at circum)

S = $\frac{{280}^o}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

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From the diagram, PQRSTW is a regular hexagon.

Hexagon is a six sided polygon.

Sum of interior angles of polygon = (2n - 4)90${}^{\circ }$ = [2 x 6 - 4] x 90 = 8 x 90 = 720${}^{\circ }$

each angle = $\frac{{720}^o}{6}={120}^o$ and TVS = $\frac{120}{2}={60}^o$

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Let the selling price(SP from P to Q be represented by x
i.e. SP = x
When SP = x at 10% profit
CP = $\frac{100}{100}$ + 10 of x = $\frac{100}{110}$ of x
when Q sells to R, SP = ₦209 at loss of 5%
Q's cost price = Q's selling price
CP = $\frac{100}{95}$ x 209
= 220.00
x = 220
= $\frac{2200}{11}$
= 200
= ₦200.00

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Numbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c. To get the number of figures divisible by 4, We solve by method of A.P
Let x represent numbers divisible by 4, nth term = a + (n - 1)d
a = 4, d = 4
Last term = 4 + (n - 1)4
288 = 4 + 4n - n
= $\frac{288}{4}$
= 72
rn(Note: 288 is the last Number divisible by 4 between 1 and 300)
Prob. of x = $\frac{72}{288}$
= $\frac{1}{4}$

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Volume of the three dimensional figures = v = A x h

A = $\frac{1}{2}$ x 4 x 3

= 6cm2

V = 6 x 8

= 48cm2

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< B is the largest since the side facing it is the largest, i.e. (x + 4)cm

Cosine B = $\frac{1}{5}$
= 0.2 given
b2 - a2 + c2 - 2a Cos B
Cos B = $\frac{a^2+c^2-b^2}{2ac}$
$\frac{1}{5}$ = $\frac{x^2+?(x-4{)}^2-(x+4{)}^2}{2x(x-4)}$
$\frac{1}{5}$= $\frac{x(x-16)}{2x(x-4)}$
$\frac{1}{5}$ = $\frac{x-16}{2x-8}$
= 5(x - 16)
= 2x - 8
3x = 72
x = $\frac{72}{3}$
= 24

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a2 - b2 = ($\frac{2x}{1-x}$)2 - ($\frac{1+x}{1-x}$)2
= ($\frac{2x}{1-x}+\frac{1+x}{1-x}$)($\frac{2x}{1-x}-\frac{1+x}{1-X}$)
= ($\frac{3x+1}{1-x}$)($\frac{x-1}{1-x}$)
= $\frac{3x+1}{x-1}$

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1 - $\sqrt{13}$ and 1 + $\sqrt{13}$
Product of roots = (1 - $\sqrt{13}$) (1 + $\sqrt{13}$) = -12
x2 - (sum of roots) x + (product of roots) = 0
x2 - 2x + 12 = 0

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Total yield = ₦3,700
Total amount invested = ₦50,000
Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest
then yield on x = $\frac{6}{100}$ x and yield on y = $\frac{8}{100}$y
Hence, $\frac{6}{100}$x + $\frac{8}{100}$y
= 3,700.........(i)
x + y = 50,000........(ii)
6x + 8y = 370,000 x 1
x + y = 50,000 x 6
6x + 8y = 370,000.........(iii)
6x + 6y = 3000,000........(iv)
Eqn (iii) - Eqn (ii)
2y = 70,000
y = $\frac{70,000}{2}$
= 35,000
Money invested at 8% is ₦35,000

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If y = 2x + 1 and y = x2 - 2x + 1

then x2 - 2x + 1 = 2x + 1

x2 - 4x = 0

x(x - 4) = 0

x = 0 or 4

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$\frac{\sin y}{3}=\frac{\sin {120}^o}{5}$

sin 120${}^{\circ }$ = sin 60${}^{\circ }$

5 sin y = 3 sin 60${}^{\circ }$

sin y = $\frac{3\sin {60}^o}{5}$

$\frac{3\times 0.866}{5}$

= $\frac{2.598}{5}$

y = sin-1 0.5196 = 30${}^{\circ }$ 18'

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Total distance covered by Musa in 2 hrs
= x + 10 + 5x
= 6x + 10
Ngozi = 118 km
If they are equal, 6x + 10 = 188
but 6x + 10 < 118

6x < 108

= x < 18

0 < x < 18 = 0 $\le$ x < 18

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If y = cf(x)
cf(5) = 30 + 32 + 30
= 92

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Le the number of girls = A; Total score of boys = 30 $\times$ 6 = 180
Total score of girls = A $\times$ 8 = 8A
$\therefore$ 180 + 8A = 468
8A = 288
A = 36

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< R = 180${}^{\circ }$ - 45${}^{\circ }$ (sum of angles on a straight line)

< R = < P = 45${}^{\circ }$ (corresponding angles)

< PSO = < P = 45${}^{\circ }$ ($△$PSO is a right angle)