JAMB UTME - Mathematics - 1984

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

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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?

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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

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

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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

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

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

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

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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$△$

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

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

= $\frac{r^2}{4}(2\pi -3\sqrt{3})$

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?

Answer Details

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

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

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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 $\frac{3}{7}$
= $\frac{1}{7}$

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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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?

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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

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

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

Using $△$XYZ in the figure, find XYZ.

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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'

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

Factorize $6x^2-14x-12$

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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

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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

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

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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

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

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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
= $\frac{6}{36}$
= $\frac{1}{6}$

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

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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

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

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by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY2 = 62 + 32

XY2 = 36 + 9 = 45

XY = $\sqrt{45}=3\sqrt{3}$

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

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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

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

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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

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

Answer Details

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

Answer Details

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 = $\frac{120}{2}={60}^o$

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?

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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

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

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

Answer Details

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

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

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

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

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

Answer Details

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

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.

Answer Details

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

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

Answer Details

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

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

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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

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

Answer Details

< 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)

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

Answer Details

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

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

Answer Details

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

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

Answer Details

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 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

Answer Details

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

What is the volume of this regular three dimensional figure?

Answer Details

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?

Answer Details

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 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

Answer Details

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 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?

Answer Details

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

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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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.

Answer Details

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

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

Answer Details

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)

Answer Details

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

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

Answer Details

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