JAMB UTME - Mathematics - 1984

Simplify $\frac{3^n-3^{n-1}}{3^3\times 3^n-27\times 3^{n-1}}$

Awọn alaye Idahun

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

The cost of production of an article is made up as follows: Labour ₦70, Power ₦15, Materials ₦30, Miscellaneous ₦5. Find the angle of the sector representing Labour in a pie chart

Awọn alaye Idahun

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

The quadratic equation whose roots are 1 - $\sqrt{13}$ and 1 + $\sqrt{13}$ is

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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

A straight lie y = mx meets the curve y = x2 - 12x + 40 in two distinct points. If one of them is (5, 5) find the other

Awọn alaye Idahun

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

The table below is drawn for a graph y = x3 - 3x + 1.

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

From x = -2 to x = 1, the graph crosses the x-axis in the range(s)

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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

0.00014321940000 $\frac{0.0001432}{1940000}$ = k x 10n where 1 ≤ $\le$ k < 10 and n is a whole number. The values K and n are

Awọn alaye Idahun

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

In the figure, PQRSTW is a regular hexagon. QS intersects RT at V. Calculate TVS.

Awọn alaye Idahun

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 = $\frac{120}{2}={60}^o$

Thirty boys and X girls sat for a test. The mean of the boys' scores and that of the girls were respectively 6 and 8. Find 8 if the total score was 468.

Awọn alaye Idahun

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

If sin $\theta$ = $\frac{x}{y}$ and 0o < 90o, then find $\frac{1}{tan\theta }$.

Awọn alaye Idahun

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

If pq + 1 = q2 and t = $\frac{1}{p}$ - $\frac{1}{pq}$ express t in terms of q

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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 = $\frac{q-1}{q^2-1}$
= $\frac{q-1}{(q+1)(q-1)}$
= $\frac{1}{q+1}$

What is the volume of this regular three dimensional figure?

Awọn alaye Idahun

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

A = $\frac{1}{2}$ x 4 x 3

= 6cm2

V = 6 x 8

= 48cm2

Find, without using logarithm tables, the value of $\frac{lo{g}_{3}27-lo{g}_{\frac{1}{4}}64}{lo{g}_3\frac{1}{81}}$

Awọn alaye Idahun

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

In a racing competition, Musa covered a distance 5x km in the first hour and (x + 10)km in the next hour. He was second to Nzozi who covered a total distance of 118km in the two hours. Which of the following inequalities is correct?

Awọn alaye Idahun

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

P varies directly as the square of Q and inversely as R. If P = 36, when Q = 3 and R = 4, find P when Q = 5 and R = 2.

Awọn alaye Idahun

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

A man invested a total of ₦50000 in two companies. If these companies pay dividends of 6% and 8% respectively, how much did he invest at 8% if the total yield is ₦3700?

Awọn alaye Idahun

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

A cone is formed by bending a sector of a circle Saving an angle or 210°. Find the radius of the base of the cone. If the diameter of the circle is 12cm.

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A trader in country where their currency 'MONI'(M) is in base five bought 1035 oranges at M145 each. If he sold the oranges at M245 each, what will be his gain?

Awọn alaye Idahun

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

If (x - 2) and (x + 1) are factors of the expression x3 + px2 + qx + 1, what is the sum of p and q

Awọn alaye Idahun

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

P sold his bicycle to Q at a profit of 10%. How much did the bicycle cost P?

Awọn alaye Idahun

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

The cumulative frequency function of the data below is given below by the equation y = cf(x). What is cf(5)?

$\begin{array}{cc}Score\left(n\right)& Frequency\left(f\right)\\ 3& 30\\ 4& 32\\ 5& 30\\ 6& 35\\ 8& 20\end{array}$

Awọn alaye Idahun

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

In the diagram, find the size of the angle marked ao

Awọn alaye Idahun

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

Using $△$XYZ in the figure, find XYZ.

Awọn alaye Idahun

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

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'

Tunde and Shola can do a piece of work in 18 days. Tunde can do it alone in x days, whilst Shola takes 15 days longer to do it alone. Which of the following equations is satisfied by x?

Awọn alaye Idahun

In the figure, 0 is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is

Awọn alaye Idahun

< 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)

Rationalize $\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}$

Awọn alaye Idahun

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

A variable point p(x, y) traces a graph in a two-dimensional plane. (0, 3) is one position of P. If x increases by 1 unit, y increases by 4 units. The equation of the graph is

Awọn alaye Idahun

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

If a = $\frac{2x}{1-x}$ and b = $\frac{1+x}{1-x}$, then a2 - b2 in the simplest form is

Awọn alaye Idahun

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

Find the area of the shaded portion of the semicircular figure.

Awọn alaye Idahun

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$△$

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

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

= $\frac{r^2}{4}(2\pi -3\sqrt{3})$

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

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In the figure, TSP = PRQ, QR = 8cm, PR = 6cm and ST = 12cm. Find the length SP.

Awọn alaye Idahun

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

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

PS = $\frac{12\times 8}{6}$

= 16cm

Find a factor which is common to all 3 binomial expressions $4a^2-9b^2,8a^3+27b^3,(4a+6b{)}^2$.

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The larger value of y for which (y - 1)2 = 4y - 7 is

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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

If 2x + 3y = 1 and x - y = 11, find (x + y).

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XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY.

Awọn alaye Idahun

by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY2 = 62 + 32

XY2 = 36 + 9 = 45

XY = $\sqrt{45}=3\sqrt{3}$

If f(x) = 2(x - 3)2 + 3(x - 3) + 4 and g(y) = 5 + y, find g [f(3)] and f[g(4)].

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

A right circular cone has a base radius r cm and a vertical angle 2yo. The height of the cone is

Awọn alaye Idahun

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

Find the x co-ordinates of the points of intersection of the two equations in the graph.

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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

Measurements of the diameters, in centimeters, in centimeters, of 20 copper spheres are distributed as shown below:

$\begin{array}{ccc}\text{Class boundary in cm}& \text{frequency}& \\ 3.35-3.45& 3& \\ 3.45-3.55& 6& \\ 3.55-3.65& 7& \\ 3.65-3.75& 4& \end{array}$

What is the mean diameter of the copper spheres?

Awọn alaye Idahun

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

Bola choose at random a number between 1 and 300. What is the probability that the number is divisible by 4?

Awọn alaye Idahun

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 = $\frac{72}{288}$
= $\frac{1}{4}$

If the price of oranges was raised by $\frac{1}{2}$k per orange. The number of oranges a customer can buy for ?2.40 will be less by 16. What is the present price of an orange?

Awọn alaye Idahun

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

Simplify $\frac{(\frac{2}{3}-\frac{1}{5})-\frac{1}{3}\text{of}\frac{2}{5}}{3-\frac{1}{1\frac{1}{2}}}$

Awọn alaye Idahun

$\frac{2}{3}-\frac{1}{5}$ = $\frac{10-3}{15}$
= $\frac{7}{15}$
$\frac{1}{3}$ 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 $\frac{3}{7}$
= $\frac{1}{7}$

Two fair dice are rolled. What is the probability that both show up the same number of points.

Awọn alaye Idahun

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

The sides of a triangle are(x + 4)cm, xcm and (x - 4)cm, respectively If the cosine of the largest angle is $\frac{1}{5}$, find the value of x

Awọn alaye Idahun

< 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

Factorize $6x^2-14x-12$

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