JAMB UTME - Mathematics - 1985

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

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

Detalhes da Resposta

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

The 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?

Detalhes da Resposta

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

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?

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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 $\frac{3}{1}$
= $\frac{49}{12}$

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

Detalhes da Resposta

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 missing value in the table below

Detalhes da Resposta

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

= 4 + 9 + 27

= 13 + 27

= 40

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

Detalhes da Resposta

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

Find x if log9x = 1.5

Detalhes da Resposta

If log9x = 1.5, 91.5 = x
∴ x = 27

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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$\frac{1}{2}$

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

Detalhes da Resposta

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

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?

Detalhes da Resposta

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

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

Detalhes da Resposta

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)

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

Detalhes da Resposta

-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]

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?

Detalhes da Resposta

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

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

Detalhes da Resposta

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 = $\frac{a?b}{ad?bc}$

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

Detalhes da Resposta

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 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${}^{\circ }$)

Detalhes da Resposta

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

Detalhes da Resposta

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

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?

Detalhes da Resposta

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 $\frac{18}{57}$

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?

Detalhes da Resposta

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

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)

Detalhes da Resposta

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 32y + 6(3y) = 27. Find y

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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)

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

Detalhes da Resposta

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

Detalhes da Resposta

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

Detalhes da Resposta

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

In the figure, the area of the shaded segment is

Detalhes da Resposta

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 $\pi -3\frac{\sqrt{3}}{4}$

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

Detalhes da Resposta

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

Detalhes da Resposta

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

Detalhes da Resposta

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 f(x - 2) = 4x2 + x + 7, find f(1)

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Detalhes da Resposta

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

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

Detalhes da Resposta

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

Factorize abx2 + 8y - 4bx - 2axy

Detalhes da Resposta

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

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

Detalhes da Resposta

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

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

Detalhes da Resposta