JAMB UTME - Mathematics - 1984

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.

Maelezo ya Majibu

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

What is the volume of this regular three dimensional figure?

Maelezo ya Majibu

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

A = $\frac{1}{2}$ x 4 x 3

= 6cm2

V = 6 x 8

= 48cm2

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?

Maelezo ya Majibu

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

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

Maelezo ya Majibu

< 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)

Factorize $6x^2-14x-12$

Maelezo ya Majibu

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.

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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$△$

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

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

= $\frac{r^2}{4}(2\pi -3\sqrt{3})$

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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)

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

Maelezo ya Majibu

< 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

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?

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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)?

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

Maelezo ya Majibu

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

Using $△$XYZ in the figure, find XYZ.

Maelezo ya Majibu

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

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'

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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 = $\frac{120}{2}={60}^o$

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

Maelezo ya Majibu

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

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?

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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?

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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

PS = $\frac{12\times 8}{6}$

= 16cm

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

Maelezo ya Majibu

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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?

Maelezo ya Majibu

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

Maelezo ya Majibu

by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY2 = 62 + 32

XY2 = 36 + 9 = 45

XY = $\sqrt{45}=3\sqrt{3}$

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

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

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?

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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

Maelezo ya Majibu

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

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.

Maelezo ya Majibu

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

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

Maelezo ya Majibu

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