JAMB UTME - Mathematics - 1985

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

Antwoorddetails

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

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

Antwoorddetails

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

The factors of 9 - (x2 - 3x - 1)2 are

Antwoorddetails

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)

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

Antwoorddetails

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

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

Antwoorddetails

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

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

Antwoorddetails

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

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?

Antwoorddetails

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 $\frac{18}{57}$

If f(x - 2) = 4x2 + x + 7, find f(1)

Antwoorddetails

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

Antwoorddetails

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 = $\frac{6}{10}$
= $\frac{3}{5}$

Find x if log9x = 1.5

Antwoorddetails

If log9x = 1.5, 91.5 = x
∴ x = 27

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?

Antwoorddetails

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

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

Antwoorddetails

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

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

Antwoorddetails

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

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

Antwoorddetails

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

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

Antwoorddetails

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)

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

Antwoorddetails

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

Antwoorddetails

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

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?

Antwoorddetails

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

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

Antwoorddetails

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

Antwoorddetails

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

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?

Antwoorddetails

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

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

Antwoorddetails

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?

Antwoorddetails

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

Antwoorddetails

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

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

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

Antwoorddetails

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

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?

Antwoorddetails

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

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${}^{\circ }$)

Antwoorddetails

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${}^{\circ }$)

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

Antwoorddetails

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 = $\frac{a?b}{ad?bc}$

If 32y + 6(3y) = 27. Find y

Antwoorddetails

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

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

Antwoorddetails

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$\frac{1}{3}$

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

Antwoorddetails

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

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

Antwoorddetails

-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]

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

Antwoorddetails

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

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

Antwoorddetails

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

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)

Antwoorddetails

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)

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

Antwoorddetails

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

Find the missing value in the table below

Antwoorddetails

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

= 4 + 9 + 27

= 13 + 27

= 40

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?

Antwoorddetails

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

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

Antwoorddetails

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 $\frac{3}{1}$
= $\frac{49}{12}$

In the figure, the area of the shaded segment is

Antwoorddetails

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 $\pi -3\frac{\sqrt{3}}{4}$

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

Antwoorddetails

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

Antwoorddetails

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

Antwoorddetails

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

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

Antwoorddetails

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

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

Antwoorddetails

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

Factorize abx2 + 8y - 4bx - 2axy