JAMB UTME - Mathematics - 1985

If two fair coins are tossed, what is the probability of getting at least one head?

Answer Details

Prob. of getting at least one head
Prob. of getting one head + prob. of getting 2 heads
= $\frac{1}{4}$ + $\frac{2}{4}$
= $\frac{3}{4}$

In a class of 120 students, 18 of them scored an A grade in mathematics. If the section representing the A grade students on a pie chart has angle Zo at the centre of the circle, what is Z?

Answer Details

Total students = 120
grade = 18
Zo = $\frac{18}{120}$ x $\frac{360}{1}$
= 54o

In the figure, PQ is the tangent from P to the circle QRS with SR as its diameter. If QRS = $\theta$${}^{\circ }$and RQP = $\varphi$${}^{\circ }$, which of the following relationships between $\theta$${}^{\circ }$ and $\varphi$${}^{\circ }$ is correct

Answer Details

180 - $\varphi$${}^{\circ }$ = $\theta$${}^{\circ }$ + $\varphi$${}^{\circ }$ (Sum of opposite interior angle equal to its exterior angle)

180 = 2$\varphi$ + $\theta$${}^{\circ }$

Two points X and Y both on latitude 60oS have longitude 147oE and 153oW respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude (Take 2$\pi$R = 4 x 104km, where R is the radius of the earth)

Answer Details

Length of an area = $\frac{\theta }{360}$ × 2$\pi$r
Longitude difference = 147 + 153 = 300oN
distance between xy = $\frac{\theta }{360}$ × 2$\pi$R cos60o
= $\frac{300}{360}$ × 4 × 104 × $\frac{1}{2}$
= 1.667 × 104km (1667 km)

The factors of 9 - (x2 - 3x - 1)2 are

Answer Details

9 - (x2 - 3x - 1)2 = [3 - (x2 - 3x - 1)] [3 + (x2 - 3x - 1)]
= (3 - x2 + 3x + 1)(3 + x2 - 3x - 1)
= (4 + 3x - x2)(x2 - 3x + 2)
= (4 - x)(1 + x)(x - 1)(x - 2)
= -(x - 4)(x + 1) (x - 1)(x - 2)

In $△$ XYZ, XKZ = 90O, XK = 15cm, XY = 35cm and YK = 8cm. Find the area of $△$XYZ

Answer Details

By Pythagoras, KZ2 = 252 - 52

KZ2 = (25 + 15)(25 - 15) = 400

KZ = $\sqrt{400}$ = 20

area of XYZ = $\frac{1}{2}\times 28\times 15$

= 210 sq. cm

In the equation below, Solve for x if all the numbers are in base 2: $\frac{11}{x}$ = $\frac{1000}{x+101}$

Answer Details

$\frac{11}{x}$ = $\frac{1000}{x+101}$ = 11(x + 101)
1000x = 11x + 1111
1000x - 11x = 1111
101x = 1111
x = $\frac{1111}{101}$
x = 11

The number of goals scored by a football team in 20 matches is shown below:

$\begin{array}{ccccccc}\text{No. of goals}& 0& 1& 2& 3& 4& 5\\ \text{No. of matches}& 3& 5& 7& 3& 1& 0\end{array}$

What are the values of the mean and the mode respectively?

Answer Details

$\begin{array}{ccc}x& f& fx\\ 0& 3& 0\\ 1& 5& 5\\ 2& 7& 14\\ 3& 4& 12\\ 4& 1& 4\\ 5& 0& 0\end{array}$
$\sum f$ = 20
$\sum fx$ = 35
Mean = $\frac{\sum fx}{\sum f}$
= $\frac{35}{20}$
= $\frac{7}{4}$
= 1.75
Mode = 2
= 1.75, 2

Find the area of a regular hexagon inscribed in a circle of radius 8cm

Answer Details

Area of a regular hexagon = 8 x 8 x sin 60o
= 32 x $\frac{\sqrt{3}}{2}$ =
Area = 16$\sqrt{3}$ x 6 = 96 $\sqrt{3}$cm2

Make c the subject of formula v = 1 - $\frac{a}{5}$(b + $\frac{3c}{7}$)

Answer Details

Arrange the following numbers in ascending order of magnitude 67 $\frac{6}{7}$, 1315 $\frac{13}{15}$, 0.8650

Answer Details

67 $\frac{6}{7}$, 1315 $\frac{13}{15}$, 0.8650

In ascending order, we have 0.8571, 0.8650, 0.8666

i.e. 67 $\frac{6}{7}$ < 0.8650 < 1315

In the figure, the area of the shaded segment is

Answer Details

Area of sector = $\frac{120}{360}\times \pi \times (3{)}^2=3\pi$

Area of triangle = $\frac{1}{2}\times 3\times 3\times \sin {120}^o$

= $\frac{9}{2}\times \frac{\sqrt{3}}{2}=\frac{9\sqrt{3}}{4}$

Area of shaded portion = 3$\pi -\frac{9\sqrt{3}}{4}$

= 3 $\pi -3\frac{\sqrt{3}}{4}$

If a{$\frac{x+1}{x-2}-\frac{x-1}{x+2}$} = 6x. Find a in its simplest form

Answer Details

a{$\frac{x+1}{x-2}-\frac{x-1}{x+2}$} = a{$\frac{(x+1)(x+2)-(x-1)(x-2)}{(x-2)(x+2)}$}
= 6
$\frac{6x}{x^2-4}$ = 6x
a = x2 - 4

A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively?

Answer Details

$\frac{1\sqrt{3}}{(\frac{1}{2}{)}^2}$
= $\frac{4}{\sqrt{3}}$
= $\frac{\sqrt{3}}{\sqrt{3}}$
= $\frac{4\sqrt{3}}{\sqrt{3}}$
m = 64kg, V = $\frac{4\pi r^3}{3}$
= $\frac{4\pi (4{)}^3}{3}$
= $\frac{256\pi }{3}$ x 10-6m3
density(P) = $\frac{\text{Mass}}{\text{Volume}}$
= $\frac{64}{\frac{256\pi }{3\times {10}^{-6}}}$
= $\frac{64\times 3\times {10}^{-6}}{256}$
= $\frac{3}{4\times {10}^{-6}}$
m = PV = $\frac{3}{4\pi \times {10}^{-6}}$ x $\frac{4}{3}$ $\pi$[32 - 22] x 10-6
$\frac{3}{4\times {10}^{-6}}$ x $\frac{4}{3}$ x 5 x 10-6
= 5kg

If the quadratic function 3x2 - 7x + R is a perfect square, find R

Answer Details

3x2 - 7x + R. Computing the square, we have
x2 - $\frac{7}{3}$ = -$\frac{R}{3}$
($\frac{x}{1}?\frac{7}{6}$)2 = -$\frac{R}{3}$ + $\frac{49}{36}$
$\frac{?R}{3}$ + $\frac{49}{36}$ = 0
R = $\frac{49}{36}$ x $\frac{3}{1}$
= $\frac{49}{12}$

Write h in terms of a, b, c, d if a = $\frac{b(1?ch)}{a?dh}$

Answer Details

a = $\frac{b(1?ch)}{a?dh}$
a = $\frac{b?bch}{1?dh}$
= a - adh
= b - bch
a - b = bch + adn
a - b = adh
a - b = h(ad - bc)
h = $\frac{a?b}{ad?bc}$

PRSQ is a trapezium of area 14cm2 in which PQ||RS. If PQ = 4cm and SR = 3cm, Find the area of SQR in cm2

Answer Details

Area of trapezium = 14cm2
Area of trapezium = $\frac{1}{2}$(a + b)h
14 = $\frac{1}{2}$(4 + 3)h
14 = $\frac{7}{2}$h
h = $\frac{14\times 2}{7}$
= 4
Area of SQR = $\frac{1}{2}$(3 x 4)
= $\frac{12}{2}$
= 6.0

A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?

Answer Details

P(R1) = $\frac{6}{10}$
= $\frac{2}{3}$
P(R1 n R11) = P(both red)
$\frac{3}{5}$ x $\frac{5}{9}$
= $\frac{1}{3}$

Answer Details

Simple Space: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10)
Prime: (2, 3, 5, 7)
multiples of 3: (3, 6, 9)
Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6)
Probability = $\frac{6}{10}$
= $\frac{3}{5}$

At what real value of x do the curves whose equations are y = x3 + x and y = x2 + 1 intersect?

Answer Details

y = x3 + x and y = x2 + 1
$\begin{array}{cccccc}x& -2& -1& 0& 1& 2\\ Y=x^3+x& -10& -2& 0& 2& 10\\ y=x^2+1& 5& 2& 1& 2& 5\end{array}$
The curves intersect at x = 1

John gives one-third of his money to Janet who has ₦105.00. He then finds that his money is reduced to one-fourth of what Janet now has. Find how much money john has at first

Answer Details

Let x be John's money, Janet already had ₦105, $\frac{1}{3}$ of x was given to Janet
Janet now has $\frac{1}{3^2}$x + 105 = $\frac{x+315}{3}$
John's money left = $\frac{2}{3}$x
= $\frac{\frac{1}{4}(x+315)}{3}$
= $\frac{2}{3}$
24x = 3x + 945
∴ x = 45

If the hypotenuse of right angled isosceles triangle is 2, what is the length of each of the other sides?

Answer Details

45o = $\frac{x}{2}$, Since 45o = $\frac{1}{\sqrt{2}}$
x = 2 x $\frac{1}{\sqrt{2}}$
= $\frac{2\sqrt{2}}{2}$
= $\sqrt{2}$

Find the missing value in the table below

Answer Details

When x = -3, y = 4 - 3(3) - (-3)3

= 4 + 9 + 27

= 13 + 27

= 40

In the figure, MNQP is a cyclic quadrilateral. MN and Pq are produced to meet at X and NQ and MP are produced to meet at Y. If MNQ = 86${}^{\circ }$ and NQP = 122${}^{\circ }$ find (x${}^{\circ }$, y${}^{\circ }$)

Answer Details

y${}^{\circ }$ = 180${}^{\circ }$ - (86${}^{\circ }$ + 58${}^{\circ }$)

180 - 144 = 36${}^{\circ }$

x${}^{\circ }$ = 180 - (94 + 58)

180 -152 = 28

(x${}^{\circ }$, y${}^{\circ }$) = (28${}^{\circ }$, 36${}^{\circ }$)

List all integers satisfying the inequality -2 ≤ $\le$ 2 x -6 < 4

Answer Details

-2 ≤ $\le$ 2x - 6 < 4 = 2x - 6 < 4

= 2x < 10

= x < 5

2x ≥ $\ge$ -2 + 6 ≥ $\ge$

= x ≥ $\ge$ 2

∴ 2 ≤ $\le$ x < 5 [2, 3, 4]

In $△$ XYZ, XY = 13cm, YZ = 9cm, XZ = 11cm and XYZ = $\theta$. Find cos$\theta$o

Answer Details

cos$\theta$ = $\frac{{13}^2+9^2-{11}^2}{2\left(13\right)\left(9\right)}$
= $\frac{169+81-21}{26\times 9}$
cos$\theta$ = $\frac{129}{26\times 9}$
= $\frac{43}{78}$

Find correct to two decimals places 100 + $\frac{1}{100}$ + $\frac{3}{1000}$ + $\frac{27}{10000}$

Answer Details

100 + $\frac{1}{100}$ + $\frac{3}{1000}$ + $\frac{27}{10000}$
$\frac{1000,000+100+30+27}{10000}$ = $\frac{1,000.157}{10000}$
= 100.02

22$\frac{1}{2}$% of the Nigerian Naira is equal to 17$\frac{1}{10}$% of a foreign currency M. What is the conversion rate of the M to the Naira?

Answer Details

N = 22$\frac{1}{2}$%, M = 17$\frac{1}{10}$%
M = $\frac{171}{10}$%, N = $\frac{45}{2}$
$\frac{45}{2}$ x $\frac{10}{171}$
= $\frac{225}{171}$
= 1 $\frac{54}{171}$
= 1 $\frac{18}{57}$

a sum of money was invested at 8% per annum simple interest. If after 4 years the money amounts to N330.00. Find the amount originally invested

Answer Details

S.I = PTR100 $\frac{PTR}{100}$

T = 4yrs, R = 8%, a = N330.00

330 - P = PTR100 $\frac{PTR}{100}$, A = P + I

i.e. A = P + PTR100 $\frac{PTR}{100}$

330 = P + P(4)(8)100 $\frac{P(4)(8)}{100}$

33000 = 32P + 100p

132P = 33000

P = N250.00

Solve the following equation $\frac{2}{2r-1}$ - $\frac{5}{3}$ = $\frac{1}{r+2}$

Answer Details

$\frac{2}{2r-1}$ - $\frac{5}{3}$ = $\frac{1}{r+2}$
$\frac{2}{2r-1}$ - $\frac{1}{r+2}$ = $\frac{5}{3}$
$\frac{2r+4-2r+1}{2r-1(r+2)}$ = $\frac{5}{3}$
$\frac{5}{(2r+1)(r+2)}$ = $\frac{5}{3}$
5(2r - 1)(r + 2) = 15
(10r - 5)(r + 2) = 15
10r2 + 20r - 5r - 10 = 15
10r2 + 15r = 25
10r2 + 15r - 25 = 0
2r2 + 3r - 5 = 0
(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)
(r - 1)(2r + 5) = 0
r = 1 or $\frac{-5}{2}$

The bearing of a bird on a tree from a hunter on the ground is ₦72oE. What is the bearing of the hunter from the birds?

Answer Details

Bearing of Hunter from Bird is
180 + 27 = 207o
Note that bearing is always taken from the north
207o = S 27o W

Solve the simultaneous equations 2x - 3y + 10 = 10x - 6y = 5

Answer Details

2x - 3y = -10; 10x - 6y = -5
2x - 3y = -10 x 2
10x - 6y = 5
4x - 6y = -20 .......(i)
10x - 6y = 5.......(ii)
eqn(ii) - eqn(1)
6x = 15
x = $\frac{15}{6}$
= $\frac{5}{2}$
x = 2$\frac{1}{2}$
Sub. for x in equ.(ii) 10($\frac{5}{2}$) - 6y = 5
y = 3$\frac{1}{2}$

If ex = 1 + $\frac{x}{2}$ + $\frac{x^2}{1.2}$ + $\frac{x^3}{\mathrm{1.2.3}}$ + .....Find $\frac{1}{e^x}$

Answer Details

ex = 1 + $\frac{x}{2}$ + $\frac{x^2}{1.2}$ + $\frac{x^3}{\mathrm{1.2.3}}$ + $\frac{x^4}{\mathrm{1.23.4}}$

In the figure, GHIJKLMN is a cube of side a. Find the length of HN.

Answer Details

HJ2 = a2 + a2 = 2a2

HJ = $\sqrt{2a^2}=a\sqrt{2}$

HN2 = a2 + (a$\sqrt{2}$)2 = a2 + 2a2 = 3a2

HN = $\sqrt{3a^2}$

= a$\sqrt{3}$cm

Without using table, calculate the value of 1 + sec2 30o

Answer Details

1 + sec2 30o = sec 30o
= $\frac{2}{\sqrt{3}}$
$\frac{(2{)}^2}{3}$
= $\frac{4}{3}$
1 + sec2 30o = sec 30o
= 1 + $\frac{4}{\sqrt{3}}$
= 2$\frac{1}{3}$

($\sqrt{2}+4\sqrt{2}$) ($\sqrt{3}+\sqrt{2}$) ($\sqrt{3}+\sqrt{2}$) is equal to

Answer Details

($\sqrt{2}+4\sqrt{2}$) ($\sqrt{3}+\sqrt{2}$) ($\sqrt{3}+\sqrt{2}$)
($\sqrt{3}+\sqrt{2}$ + ($\sqrt{3}+\sqrt{2}$) = 3 + ($\sqrt{6}-\sqrt{6}$) - 2
= 1

The ratio of the length of two similar rectangular blocks is 2 : 3. If the volume of the larger block is 351cm3, then the volume of the other block is

Answer Details

Let x represent total vol. 2 : 3 = 2 + 3 = 5
$\frac{3}{5}$x = 351
x = $\frac{351\times 5}{3}$
= 585
Volume of smaller block = $\frac{2}{3}$ x 585
= 234.00

Find the values of p for which the equation x2 - (p - 2)x + 2p + 1 = 0

Answer Details

Equal roots implies b2 - 4ac = 0
a = 1b = - (p - 2), c = 2p + 1
[-(p - 2)]2 - 4 x 1 x (2p + 1) = 0
p2 - 4p + 4 - 4(2p + 1) = 0
p2 - 4p = 4 - 8p - 4 = 0
p2 - 12p = 0
p(p - 12) = 0
p = 0 or 12

If f(x - 2) = 4x2 + x + 7, find f(1)

Answer Details

f(x - 2) = 4x2 + x + 7
x - 2 = 1, x = 3
f(x - 2) = f(1)
= 4(3)2 + 3 + 7
= 36 + 10
= 46

Solve for (x, y) in the equation 2x + y = 4: x^2 + xy = -12

Answer Details

2x + y = 4......(i)
x^2 + xy = -12........(ii)
from eqn (i), y = 4 - 2x
= x2 + x(4 - 2x)
= -12
x2 + 4x - 2x2 = -12
4x - x2 = -12
x2 - 4x - 12 = 0
(x - 6)(x + 2) = 0
sub. for x = 6, in eqn (i) y = -8, 8
=(6,-8); (-2, 8)

If cos $\theta$ = $\frac{\sqrt{3}}{2}$ and $\theta$ is less than 90o. Calculate cos$\frac{90-\theta }{si{n}^2\theta }$

Answer Details

cos$\frac{90-\theta }{si{n}^2\theta }$
= $\frac{\tan \theta }{si{n}^2\theta }$
Sin2$\theta$ =$\frac{1}{4}$
$\frac{cos(90-\theta )}{si{n}^2\theta }$
= $\frac{1}{\sqrt{3}}$
= $\frac{4}{\sqrt{3}}$

Which of these angles can be constructed using ruler and a pair of compasses only?

Answer Details

Factorize abx2 + 8y - 4bx - 2axy

Answer Details

abx2 + 8y - 4bx - 2axy = (abx2 - 4bx) + (8y - 2axy)
= bx(ax - 4) 2y(ax - 4) 2y(ax - 4)
= (bx - 2y)(ax - 4)

If 32y + 6(3y) = 27. Find y

Answer Details

32y + 6(3y) = 27
This can be rewritten as (3y)2 + 6(3y) = 27
Let 3y = x
x2 + 6x - 27 = 0
(x + 9)(x - 3) = 0
when x - 3 = 0, x = 3
sub. for x in 3y = x
3y = 3
log33 = y
y = 1

If three numbers P, Q, R are in ratio 6 : 4 : 5, find the value of $\frac{3p-q}{4q+r}$

Answer Details

P : Q : r = 6 : 4 : 5
5 = 6 + 4 + 5
= 15
P = $\frac{6}{15}$, q = $\frac{4}{15}$, r = $\frac{5}{15}$ = $\frac{1}{3}$
To find $\frac{3p-q}{4q+r}$
3p - q = 3 x $\frac{6}{15}$ - $\frac{4}{15}$
$\frac{18}{15}$ - $\frac{4}{15}$ = $\frac{14}{15}$
∴ 4q + r = 4 x $\frac{4}{15}$ + $\frac{5}{15}$
$\frac{16}{15}$ = $\frac{16}{15}$ + $\frac{5}{15}$
= $\frac{21}{15}$
$\frac{14}{15}$ x $\frac{15}{21}$ = $\frac{14}{21}$
= $\frac{2}{3}$