JAMB UTME - Mathematics - 1985

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

Awọn alaye Idahun

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

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)

Awọn alaye Idahun

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)

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

Awọn alaye Idahun

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 $\frac{15}{21}$ = $\frac{14}{21}$
= $\frac{2}{3}$

If 32y + 6(3y) = 27. Find y

Awọn alaye Idahun

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

Awọn alaye Idahun

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

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?

Awọn alaye Idahun

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

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?

Awọn alaye Idahun

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

In the figure, the area of the shaded segment is

Awọn alaye Idahun

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 $\pi -3\frac{\sqrt{3}}{4}$

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

Awọn alaye Idahun

$\frac{2}{2r-1}$ - $\frac{5}{3}$ = $\frac{1}{r+2}$
$\frac{2}{2r-1}$ - $\frac{1}{r+2}$ = $\frac{5}{3}$
$\frac{2r+4-2r+1}{2r-1(r+2)}$ = $\frac{5}{3}$
$\frac{5}{(2r+1)(r+2)}$ = $\frac{5}{3}$
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 $\frac{-5}{2}$

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

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-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]

Find x if log9x = 1.5

Awọn alaye Idahun

If log9x = 1.5, 91.5 = x
∴ x = 27

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

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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

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

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

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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 = $\frac{6}{10}$
= $\frac{3}{5}$

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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${}^{\circ }$)

Awọn alaye Idahun

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${}^{\circ }$)

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

Awọn alaye Idahun

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

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

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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 $\frac{3}{1}$
= $\frac{49}{12}$

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?

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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 = $\frac{a?b}{ad?bc}$

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

Awọn alaye Idahun

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$\frac{1}{2}$

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

Awọn alaye Idahun

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

Find the missing value in the table below

Awọn alaye Idahun

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

= 4 + 9 + 27

= 13 + 27

= 40

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

Awọn alaye Idahun

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

The factors of 9 - (x2 - 3x - 1)2 are

Awọn alaye Idahun

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)

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?

Awọn alaye Idahun

$\frac{1\sqrt{3}}{(\frac{1}{2}{)}^2}$
= $\frac{4}{\sqrt{3}}$
= $\frac{\sqrt{3}}{\sqrt{3}}$
= $\frac{4\sqrt{3}}{\sqrt{3}}$
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

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

Awọn alaye Idahun

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

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?

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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 a restaurant, the cost of providing a particular type of food is partly constant and partially inversely proportional to the number of people. If cost per head for 100 people is 30k and the cost for 40 people is 60k, Find the cost for 50 people?

Awọn alaye Idahun

C = a + k
$\frac{1}{N}$ = c
= $\frac{aN+k}{N}$
CN = aN + K
30(100) = a(100) + k
3000 = 100a + k.......(i)
60(40) = a(40) + k
2400 = 40a + k.......(ii)
eqn (i) - eqn (ii)
600 = 60a
a = 10
subt. for a in eqn (i) 3000 = 100(10) + K
3000 - 1000 = k
k = 2000
CN = 10N + 2000. when N = 50,
50C = 10(50) + 2000
50C = 500 + 2000
C = $\frac{2500}{50}$
= 50k

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

Factorize abx2 + 8y - 4bx - 2axy

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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$\frac{1}{3}$

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

Awọn alaye Idahun

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

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

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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

Awọn alaye Idahun

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

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?

Awọn alaye Idahun

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

Awọn alaye Idahun

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

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

Awọn alaye Idahun

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

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