JAMB UTME - Mathematics - 1984

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

Antwortdetails

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

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

Antwortdetails

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

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

Antwortdetails

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

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

Antwortdetails

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 = $\frac{120}{2}={60}^o$

Factorize $6x^2-14x-12$

Antwortdetails

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

Antwortdetails

by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY2 = 62 + 32

XY2 = 36 + 9 = 45

XY = $\sqrt{45}=3\sqrt{3}$

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?

Antwortdetails

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

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?

Antwortdetails

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

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

Antwortdetails

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

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

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

Antwortdetails

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

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.

Antwortdetails

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

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

Antwortdetails

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$△$

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

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

= $\frac{r^2}{4}(2\pi -3\sqrt{3})$

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.

Antwortdetails

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

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

Antwortdetails

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

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

Antwortdetails

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

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?

Antwortdetails

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

Antwortdetails

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

Antwortdetails

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

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

Antwortdetails

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

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

Antwortdetails

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

Antwortdetails

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

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

Antwortdetails

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

Antwortdetails

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

Antwortdetails

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

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?

Antwortdetails

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

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

Antwortdetails

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

What is the volume of this regular three dimensional figure?

Antwortdetails

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

A = $\frac{1}{2}$ x 4 x 3

= 6cm2

V = 6 x 8

= 48cm2

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

Antwortdetails

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 = $\frac{72}{288}$
= $\frac{1}{4}$

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?

Antwortdetails

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

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

Antwortdetails

< 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)

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

Antwortdetails

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)

Antwortdetails

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

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

Antwortdetails

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

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?

Antwortdetails

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

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.

Antwortdetails

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

Antwortdetails

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

Using $△$XYZ in the figure, find XYZ.

Antwortdetails

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

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'

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

Antwortdetails

< 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

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

Antwortdetails

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

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

Antwortdetails

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

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

PS = $\frac{12\times 8}{6}$

= 16cm

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

Antwortdetails

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

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

Antwortdetails

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

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

Antwortdetails

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

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

Antwortdetails

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

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

Antwortdetails

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