JAMB UTME - Mathematics - 1985

Question 1 Report

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

\frac{49}{24}

\frac{49}{12}

\frac{49}{13}

\frac{49}{3}

\frac{49}{36}

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}

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Quadratic Equations

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Question 2 Report

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

N180.00

n165.00

N150.00

N250.00

N200.00

Answer Details

S.I =

\frac{P T R}{100}

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

330 - P =

\frac{P T R}{100}

, A = P + I

i.e. A = P +

\frac{P T R}{100}

330 = P +

\frac{P (4) (8)}{100}

33000 = 32P + 100p

132P = 33000

P = N250.00

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Positive And Negative Integers, Rational Numbers

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Question 3 Report

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

[

\frac{7}{3}

+

\frac{5}{a}

(v - 1)] + b

[

\frac{7 b}{3}

+

\frac{5}{a}

(v - 1)]

\frac{7}{3}

[b +

\frac{5}{a}

(v - 1)]

\frac{7}{3}

[b +

\frac{5 v - 1}{a}

]

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

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Question 4 Report

In the figure, the area of the shaded segment is

3

π

9

\frac{\sqrt{3}}{4}

3

π - 3 \frac{\sqrt{3}}{4}

\frac{(\sqrt{3 - π)}}{4}

π + \frac{9 \sqrt{3}}{4}

Answer Details

Area of sector =

\frac{120}{360} \times π \times (3)^{2} = 3 π

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 $π - \frac{9 \sqrt{3}}{4}$

= 3 $π - 3 \frac{\sqrt{3}}{4}$

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Volumes

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Question 5 Report

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

101

11

110

111

10

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

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Number Bases

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Question 6 Report

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

₦45.00

₦48.00

₦52.00

₦60.00

₦53.00

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

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Fractions, Decimals And Approximations

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Question 7 Report

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

\frac{4}{\sqrt{3}}

\sqrt{\frac{3}{2}}

\sqrt{\frac{76}{2}}

4

\sqrt{\frac{76}{2}}

Answer Details

cos

\frac{90 - θ}{s i n^{2} θ}

=

\frac{\tan θ}{s i n^{2} θ}

Sin2

θ

=

\frac{1}{4}

\frac{c o s (90 - θ)}{s i n^{2} θ}

=

\frac{1}{\sqrt{3}}

=

\frac{4}{\sqrt{3}}

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Question 8 Report

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

(6, -8): (-2, 8)

(3, -4): (-1, 4)

(8, -4): (-1, 4)

(-8, 6): (8, -2)

(-4, 3): (4, -1)

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)

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Solution Of Linear Equations

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Question 9 Report

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

1

(

\sqrt{2} + 4 \sqrt{2}

)

(6

\sqrt{2}

8

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

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Question 10 Report

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

(21, 0)

(0, 12)

(1, 2)

(3, 4)

(4, 5)

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

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Quadratic Equations

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Question 11 Report

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

\sqrt{2}

\frac{1}{2}

22

1

2 - 1

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}

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Statistics

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Question 12 Report

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?

\frac{2}{3}

\frac{2}{15}

\frac{1}{2}

\frac{1}{3}

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}

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Probability

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Question 13 Report

Find the missing value in the table below

x	-4	-3	-2	-1	0	1	2	3
y=4?3x?x^3	80		18	8	4	0	-10	-32

-32

-14

40

22

37

Answer Details

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

= 4 + 9 + 27

= 13 + 27

= 40

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Question 14 Report

If f(x - 2) = 4x2 + x + 7, find f(1)

12

27

7

46

17

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

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Question 15 Report

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

190 sq.cm

20 sq.cm

210 sq.cm

160sq.cm

320sq.cm

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

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Question 16 Report

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?

15k

20k

50k

40k

45k

Answer Details

C = a + k

\frac{1}{N}

= c
=

\frac{a N + 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

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Question 17 Report

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

(

\frac{5}{2}

, 1)

(5, -4)

(2, 1)

(1,

\frac{- 5}{2}

)

(

\frac{- 5}{2}

, 1)

Answer Details

\frac{2}{2 r - 1}

-

\frac{5}{3}

=

\frac{1}{r + 2}

\frac{2}{2 r - 1}

-

\frac{1}{r + 2}

=

\frac{5}{3}

\frac{2 r + 4 - 2 r + 1}{2 r - 1 (r + 2)}

=

\frac{5}{3}

\frac{5}{(2 r + 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}

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Solution Of Linear Equations

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Question 18 Report

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

3

2

3

2

3

18

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 $\frac{3 p - q}{4 q + r}$
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 $\frac{15}{21}$ = $\frac{14}{21}$
= $\frac{2}{3}$

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Question 19 Report

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 ($x^\circ$, $y^\circ$)

28

^{\circ}

, 36

^{\circ}

36

^{\circ}

, 28

^{\circ}

43

^{\circ}

, 61

^{\circ}

61

^{\circ}

, 43

^{\circ}

36

^{\circ}

, 43

^{\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 $^{\circ}$ )

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Question 20 Report

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?

1M = 1

\frac{15}{57}

N

1M = 38

\frac{1}{4}

N

1M = 1

\frac{18}{57}

N

1M = 384

\frac{3}{4}

N

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}

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Question 21 Report

The factors of 9 - (x2 - 3x - 1)2 are

-(x - 4)(x + 1) (x - 1)(x - 2)

(x - 4)(x - 2) (x - 1)(x + 1)

-(x - 2)(x + 1) (x - 2) (x - 1)

(x - 2)(x + 2) (x - 1)(x + 1)

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)

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Question 22 Report

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

h = $\frac{a - b}{a d}$

h = $\frac{1 - b}{a d - b c}$

h = $\frac{(a - b)^{2}}{a d - b c}$

h =

\frac{a - b}{a d - b c}

h = $\frac{b - a}{a b - d c}$

Answer Details

a = $\frac{b (1 - c h)}{a - d h}$
a = $\frac{b - b c h}{1 - d h}$
= a - adh
= b - bch
a - b = bch + adn
a - b = adh
a - b = h(ad - bc)
h = $\frac{a - b}{a d - b c}$

View Answer

Question 23 Report

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

2

\frac{1}{3}

\frac{2}{15}

\frac{5}{3}

3

\frac{1}{2}

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}

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Question 24 Report

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

x = 2

\frac{1}{2}

, y = 3

\frac{1}{2}

x =

\frac{1}{2}

, y =

\frac{3}{2}

x = 2

\frac{1}{4}

, y = 3

\frac{1}{2}

x = 2

\frac{1}{3}

, y = 3

\frac{1}{2}

x = 2

\frac{1}{3}

, y = 2

\frac{1}{2}

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}

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Solution Of Linear Equations

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Question 25 Report

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

θ

^{\circ}

+

ϕ

^{\circ}

= 902

ϕ

^{\circ}

= 902 - 2

θ

^{\circ}

θ

^{\circ}

=

ϕ

^{\circ}

ϕ

^{\circ}

= 2

θ

^{\circ}

θ

^{\circ}

+ 2

ϕ

^{\circ}

Answer Details

180 -

ϕ

^{\circ}

=

θ

^{\circ}

+

ϕ

^{\circ}

(Sum of opposite interior angle equal to its exterior angle)

180 = 2 $ϕ$ + $θ$ $^{\circ}$

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Question 26 Report

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

1 -

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{3}}{2^{4} .3}

+ .........

1 -

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{4}}{2.4.3}

+ ..

1 +

\frac{x}{2}

+

\frac{x^{2}}{1.2}

+

\frac{x^{3}}{1.2.3}

+

\frac{x^{4}}{1.23.4}

+ .........

1 - x +

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{3}}{2^{4} .3}

+ .........

1 +

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{4}}{1.2.6}

+ ........

Answer Details

ex = 1 +

\frac{x}{2}

+

\frac{x^{2}}{1.2}

+

\frac{x^{3}}{1.2.3}

+

\frac{x^{4}}{1.23.4}

View Answer

Question 27 Report

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

$\frac{6}{7}$ < 0.865 < $\frac{13}{15}$

$\frac{13}{15}$ < $\frac{6}{7}$ < 0.865

$\frac{6}{7}$ < $\frac{13}{15}$ < 0.865

0.865 < $\frac{6}{7}$ < $\frac{13}{15}$

Answer Details

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

In ascending order, we have 0.8571, 0.8650, 0.8666

i.e. $\frac{6}{7}$ < 0.8650 < $\frac{13}{15}$

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Question 28 Report

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

No. of goals	0	1	2	3	4	5
No. of matches	3	5	7	3	1	0

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

(1.75, 5)

(1.75, 2)

(1.75, 1)

(65, 74)

Answer Details

\begin{array}{cc} x & f & f x \\ 0 & 3 & 0 \\ 1 & 5 & 5 \\ 2 & 7 & 14 \\ 3 & 4 & 12 \\ 4 & 1 & 4 \\ 5 & 0 & 0 \end{array}

\sum f

= 20

\sum f x

= 35
Mean =

\frac{\sum f x}{\sum f}

=

\frac{35}{20}

=

\frac{7}{4}

= 1.75
Mode = 2
= 1.75, 2

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Question 29 Report

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

234.00 cm3

526.50 cm3

166.00 cm3

687cm3

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

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Question 30 Report

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

\sqrt{3 a}

3a

3a2

a

\sqrt{2}

a

\sqrt{3}

Answer Details

HJ2 = a2 + a2 = 2a2

HJ = $\sqrt{2 a^{2}} = a \sqrt{2}$

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

HN = $\sqrt{3 a^{2}}$

= a $\sqrt{3}$ cm

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Question 31 Report

Two points X and Y both on latitude $60^\circ\text{S}$ have longitude $147^\circ\text{E}$ and $153^\circ\text{W}$ respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude (Take $2\pi R = 4 \times 10^4\text{ km}$, where R is the radius of the earth)

16667km

28850km

8333km

2233km

Answer Details

Length of an area =

\frac{θ}{360}

× 2

π

r
Longitude difference = 147 + 153 = 300oN
distance between xy =

\frac{θ}{360}

× 2

π

R cos60o
=

\frac{300}{360}

× 4 × 104 ×

\frac{1}{2}

= 1.667 × 104km (1667 km)

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Circles

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Question 32 Report

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

\frac{1}{4}

\frac{1}{2}

1

\frac{43}{78}

\frac{3}{4}

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}

View Answer

Question 33 Report

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?

15

27

50

52

54

Answer Details

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

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Question 34 Report

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

16

\sqrt{3}

cm2

96

\sqrt{3}

cm2

192

\sqrt{3}

cm2

16

\sqrt{3}

cm2

33cm2

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

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Areas

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Question 35 Report

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

2, 3, 4, 5

2, 3, 4

2, 5

3, 4, 5

4, 5

Answer Details

-2 $\leq$ 2x - 6 < 4 = 2x - 6 < 4

= 2x < 10

= x < 5

2x $\geq$ -2 + 6 $\geq$

= x $\geq$ 2

∴ 2 $\leq$ x < 5 [2, 3, 4]

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Inequalities

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Question 36 Report

If 32y + 6(3y) = 27. Find y

3

-1

2

-3

1

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

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Solution Of Linear Equations

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Question 37 Report

What is the probability that a number chosen at random from the intergers between 1 and 10 inclusive is either a prime or a multiple of 3?

\frac{7}{10}

\frac{3}{5}

\frac{4}{5}

\frac{3}{10}

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}

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Probability

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Question 38 Report

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

100.02

1000.02

100.22

100.01

100.51

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

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Read lesson note on Areas (WAEC)

Number Bases

Areas

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Question 39 Report

Find x if log9x = 1.5

72.0

27.0

36.0

3.5

24.5

Answer Details

If log9x = 1.5, 91.5 = x
∴ x = 27

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Logarithms

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Question 40 Report

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

7.0

6.0

5.2

5.0

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

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Question 41 Report

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

115o

125o

135o

145o

Volumes

Angles

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Question 42 Report

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

-2

2

-1

9

1

Answer Details

y = x3 + x and y = x2 + 1

\begin{array}{cc} 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

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Read lesson note on Quadratic Equations (WAEC)

Coordinate Geometry

Quadratic Equations

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Question 43 Report

Factorize abx2 + 8y - 4bx - 2axy

(ax - 4)(bx - 2y)

(ax + b)(x - 8y)

(ax - 2y)(bx - 4)

(bx - 4)(ax - 2y)

(abx - 4)(x - 2y)

Answer Details

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

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Algebraic Expressions

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Question 44 Report

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

x2 - 1

x2 + 1

x2 + 4

x2 - 4

1

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{6 x}{x^{2} - 4}

= 6x
a = x2 - 4

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Algebraic Expressions

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Question 45 Report

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

\frac{4}{39}

\frac{43}{39}

\frac{209}{3}

\frac{43}{78}

Answer Details

cos

θ

=

\frac{13^{2} + 9^{2} - 11^{2}}{2 (13) (9)}

=

\frac{169 + 81 - 21}{26 \times 9}

cos

θ

=

\frac{129}{26 \times 9}

=

\frac{43}{78}

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Read lesson note on Trigonometry (JAMB)

Introductory Calculus

Trigonometry

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Question 46 Report

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?

S 18o W

S 72o W

S 72o E

S 27o E

S 27o W

Answer Details

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

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Bearings

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Question 47 Report

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?

5kg

16kg

19kg

6kg

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

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