JAMB UTME - Mathematics - 1993

Question 1 Report

Integrate with respect to x

$\frac{x - x^{2}}{x^{4}}$ + k

\frac{4}{x^{4}} - \frac{3 + k}{x^{3}}

$\frac{1}{x} - \frac{1}{2 x^{2}}$ + k

$\frac{1}{3 x^{2}} - \frac{1}{2 x}$ + k

Answer Details

$\int \frac{1 - x}{x^{3}}$
= $\int_{x^{3}}^{1} - \int_{x^{3}}^{x}$
= x-3 dx - x-2dx
= $\frac{1}{2 x^{2}} + \frac{1}{x}$

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Integration

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Question 2 Report

A man's initial salary is ₦540.00 a month and increases after each period of six months by ₦36.oo a month. Find his salary in the eighth month of the third year

₦828.00

₦756.00

₦720.00

₦684.00

Answer Details

Initial salary = ₦540
increment = ₦36 (every 6 months)
Period of increment = 2 yrs and 6 months
amount(increment) = ₦36 x 5 = ₦180
The man's new salary = ₦540 = ₦180
= ₦720.00

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Sequence And Series

Ratio, Proportions And Rates

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Question 3 Report

A binary operation $ \ast $ is defined on a set of real numbers by $x \ast y = x^y$ for all real values of $x$ and $y$. If $x \ast 2 = x$. Find the possible values of $x$

0, 1

1, 2

2, 2

0, 2

Answer Details

x

*

y = xy
x

*

2 = x2
x

*

2 = x
∴ x2 - x = 0
x(x - 1) = 0
x = 0 or 1

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Binary Operations

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Question 4 Report

find the radius of a sphere whose surface area is 154cm² $ \left(\pi=\frac{22}{7}\right) $

7.00cm

3.50cm

3.00cm

1.75cm

Answer Details

Surface area = 154cm2 (area of sphere)
4

π

r2 = 154
r

\sqrt{\frac{154}{4 π}}

= 3.50cm

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Question 5 Report

The three sides of an isosceles triangle are length of lengths (x + 3), (2x + 3), (2x - 3) respectively. Calculate x.

5

1

3

6

Answer Details

2x + 3

\neq

2x - 3 for any value of x
∴ for the

△

to be isosceles, either
2x - 3 = x + 3 or 2x + 3 = x + 3
solve the two equations we arrive at
x = 6 or x = 0
When x = 6, the sides are 9, 15, 9
When x = 0, the sides are 3, 4, -3 since lengths of a

△

can never be negative then the value of x = 6

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Algebraic Expressions

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Question 6 Report

The shaded portion in the Venn diagram is

x

\cap

z

x

^{\circ}

\cap

y

\cap

z

x

\cap

y

^{\circ}

\cap

z

x

\cap

y

\cap

z

^{\circ}

Answer Details

The shaded part exists on x

\cap

z but not in y

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Logical Reasoning

Integration

Matrices And Determinants

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Question 7 Report

Find the area of the sector of a circle with radius 3m, if the angle of the sector is 60o

4.0m2

1m2

4.7m2

5.0m2

Answer Details

To find the area of a sector, we use the formula: Area of sector = (θ/360) x πr² Where θ is the angle of the sector in degrees, r is the radius of the circle, and π (pi) is a mathematical constant approximately equal to 3.14. In this problem, the radius is given as 3m and the angle of the sector is 60°. Substituting these values into the formula, we get: Area of sector = (60/360) x π(3)² = (1/6) x π(9) = (1/6) x 28.27 = 4.71m² (rounded to two decimal places) Therefore, the area of the sector of the circle with radius 3m and angle 60° is approximately 4.71m². The answer is option C.

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Circles

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Question 8 Report

Quantities in the proportions 1, 4, 6, 7 are to be represented in a pie chart. Calculate the angle of the sector with proportion 7

20o

80o

120o

140o

Answer Details

To calculate the angle of the sector with proportion 7, we first need to find the total of all the quantities in the proportion. Total quantity = 1 + 4 + 6 + 7 = 18 Next, we need to find the proportion of quantity 7 in relation to the total quantity. Proportion of quantity 7 = 7/18 To find the angle of the sector that represents this proportion, we need to use the formula: Angle of sector = Proportion of quantity * 360 degrees Angle of sector = (7/18) * 360 degrees Angle of sector = 140 degrees Therefore, the answer is option D, which is 140 degrees.

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Positive And Negative Integers, Rational Numbers

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Question 9 Report

If 2log3 y + log3 x2 = 4, then y is

4 - log3

\frac{4}{l o g_{3} x}

\frac{4}{x}

\pm

\frac{9}{x}

Answer Details

The given equation is 2log3 y + log3 x2 = 4. We can simplify this equation using the laws of logarithms. First, we can use the power rule of logarithms to rewrite x2 as (x)2. Then, we can use the product rule of logarithms to combine the two logarithms on the left-hand side of the equation: 2log3 y + log3 (x)2 = log3 y2 + log3 (x)2 Now, we can use the sum rule of logarithms to combine the two logarithms on the right-hand side of the equation: log3 y2(x)2 = log3 (x2y2) We can now rewrite the original equation as: log3 (x2y2) = 4 Using the definition of logarithms, we know that log3 (x2y2) = 4 is equivalent to 3^4 = x2y2. Therefore, we have: x2y2 = 81 Taking the square root of both sides, we get: xy = ±9 Since y is a positive real number, we can discard the negative solution. Therefore, we have: xy = 9 Finally, we can solve for y: y = 9/x So, the answer is "±9/x".

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Sets

Logarithms

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Question 10 Report

Evaluate $ \frac{3524}{0.05} $ correct to 3 significant figures

705

70,000

70, 480

70, 500

Answer Details

To evaluate the given expression to 3 significant figures, we need to round the result to the nearest third digit after the decimal point. To do this, we can perform the division and then round the result. The expression is: 3524 / 0.05 = 70,480 To round this to 3 significant figures, we need to consider the digits up to the third digit after the decimal point, which is 4. Since 4 is less than 5, we round down the previous digit, which is 8. Therefore, the final answer is: 70,500 So, is the correct answer.

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Fractions, Decimals And Approximations

Number Bases

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Question 11 Report

Solve for y in the equation 101 x 5(2x - 2) x 4(y - 1) = 1

\frac{3}{4}

\frac{5}{4}

\frac{2}{3}

5

Answer Details

10y x 5(2y - 2) x 4(y - 1) = 1
but 10y - (5 x 2)y = 5y x 2y
= (Law of indices)
5y x 2y x 5(2y - 2) x 4(y - 1) = 1
but 4(y - 1) = 22(y - 1)
= 2y - 2 (Law of indices)
5y x 5(2y -2) x 2(- 2) = 1
5(3y -2) x 2y x 2(2y -2) = 1
= 5(3y -2) x 2(3y -2) = 1
But ao = 1
10(3y -2) = 10o
3y - 2 = 0
? y =

\frac{2}{3}

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Question 12 Report

Solve the following equation (3x - 2)(5x - 4) = (3x - 2)2

\frac{- 2}{3}

, 1

1

\frac{2}{3}

, 1

\frac{2}{3}

,

\frac{4}{5}

Answer Details

(3x - 2)(5x - 4) = (3x - 2)2 = 5x2 - 22x + 6
= 9x2 = 12x + 4
6x2 - 10x + 4 = 0
6x2 - 6x - 4x + 4 = 0
6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0
x = 1 or

\frac{2}{3}

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Algebraic Expressions

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Question 13 Report

Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR

11o

73o

107o

138o

149o

Answer Details

QPS - QRK = 31o
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o

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Inequalities

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Question 14 Report

Which of the following is a factor of 15 + 7x - 2x2

x + 3

x - 3

x - 5

x + 5

Answer Details

Factorize 15 + 7x - 2x2
(5 - x)(3 + 2x); suppose 15 + 7x = 2x2 = 0
∴ (5 - x)(3 + 2x) = 0
x = 5 or x = -

\frac{3}{2}

Since 5 is a root, then (x - 5) is a factor

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Algebraic Expressions

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Question 15 Report

The bar chart shows the distribution of marks in a class test. How many students took the test?

17

20

25

30

Answer Details

Number of students that took the test =

\sum f

. Where

f

is the frequencies
= 2 + 5 + 0 + 3 + 4 + 0 + 0 + 0 + 1 + 0 + 2 = 17

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Integration

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Question 16 Report

In the figure, the line segment ST is tangent to two circles at S and T. O and Q are the centres of the circles with OS = 5cm. QT = 2cm and OR = 14cm. Find ST

7

\sqrt{3}

cm

12.9cm

\sqrt{87}

cm

7cm

Answer Details

SQ2 + OS2 = OQ2 + 52 = 142

SQ2 = 142 - 52

196 - 25 = 171

ST2 + TQ2 = SQ2

ST2 + 22 = 171

ST2 = 171 - 4

= 167

ST = $\sqrt{167}$

= 12.92 = 12.9cm

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Modular Arithmetic

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Question 17 Report

In the diagram, O is the centre of the circle and POQ a diameter. If POR = $96^\circ$, find the value of ORQ.

84

^{\circ}

48

^{\circ}

45

^{\circ}

42

^{\circ}

Answer Details

OQ = OR = radii

< ROQ = 180 - 86 = 84 $^{\circ}$

$△$ OQR = Isosceles

R = Q

R + Q + 84 = 180(angle in a $△$ )

2R = 96 since R = Q

R = 48 $^{\circ}$

ORQ = 48 $^{\circ}$

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Circles

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Question 18 Report

Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ

35o

37o

49o

59o

Answer Details

In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)

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Circles

Triangles And Polygons

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Question 19 Report

The chances of three independent events X, Y, Z occurring are $ \frac{1}{2} $, $ \frac{2}{3} $, $ \frac{1}{4} $ respectively. What are the chances of Y and Z only occurring?

\frac{1}{8}

\frac{1}{24}

\frac{1}{12}

\frac{1}{4}

Answer Details

Chance of x =

\frac{1}{2}

Change of Y =

\frac{2}{3}

Chance of Z =

\frac{1}{4}

Chance of Y and Z only occurring
= Pr (Y ? Z ? Xc)
where Xc = 1 - Pr(X)
1 =

\frac{1}{2}

= 1

\frac{1}{2}

= Pr(Y) x Pr(Z) x Pr(Xc)
=

\frac{2}{3}

x

\frac{1}{4}

x

\frac{1}{2}

=

\frac{1}{12}

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Probability

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Question 20 Report

change 7110 to base 8

1078

1068

718

178

Answer Details

To convert the number 71 to base 8, we need to divide it repeatedly by 8 and note down the remainders until the quotient becomes 0. The remainders in reverse order will give us the equivalent number in base 8. Here's how we can do it: - 71 ÷ 8 = 8 remainder 7 - 8 ÷ 8 = 1 remainder 0 - 1 ÷ 8 = 0 remainder 1 So, the equivalent number in base 8 is 107. Therefore, the correct answer is option A: 1078.

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Number Bases

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Question 21 Report

simplify $ \frac{1}{\sqrt{3}-2} - \frac{1}{\sqrt{3}+2} $

3

\frac{2}{3}

7

-4

Answer Details

\frac{1}{\sqrt 3 - 2}

-

\frac{1}{\sqrt 3 + 2}

L.C.M = (3- 2) (3 + 2)
∴

\frac{1}{\sqrt{3 - 2}}

-

\frac{1}{\sqrt{3 - 2}}

=

\frac{\sqrt{3 + 2} - \sqrt{3 - 2}}{\sqrt{3 - 2} + \sqrt{3 - 2}}

$\frac{\sqrt 3 + 2 - \sqrt 3 + 2}{3 - 2 \sqrt 3 + 2 \sqrt 3 - 4}$ = $\frac{4}{3 - 2}$
= $\frac{4}{- 1}$
= -4

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Positive And Negative Integers, Rational Numbers

Surds (radicals)

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Question 22 Report

If $ \sin \theta = \cos \theta $, find $ \theta $ between 0^o and 360^o

45o, 225o

135o, 315o

45o, 315o

135o, 225o

Answer Details

sin

θ

= cos

θ

0

\leq

θ

\leq

360o
The acute angle where sin

θ

= cos

θ

= 45o
But at the fourth Quadrant Cos

θ

= +ve
at the 4th quadrant, value with respect to Q is
(360 - Q) where Q = acute angle
(360 - 45) = 315o
The two solution are 45o, 315o

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Trigonometry

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Question 23 Report

If two angles of a triangle are 30o each and the longest side is 10cm. Calculate the length of each of the other sides

5cm

4cm

3√3 cm

10

\frac{3}{3}

cm

Answer Details

102 = 2x2 - 2x2 cos 120o (Cosine rule)
100 = 242 - 2x2 - 2x2 x -

\frac{1}{2}

100 = 3x2 + x2
= 3x2
x =

\sqrt{\frac{100}{3}}

=

\frac{10}{\sqrt{3}}

x

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

x = 10

\frac{3}{3}

cm

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Question 24 Report

If $9\left(x - \frac{1}{2}\right)^3x^2$

\frac{1}{2}

1

2

3

Answer Details

9(x -

\frac{1}{2}

) 3x2 = 32(x -

\frac{1}{2}

) = 3x2
∴ 2(x -

\frac{1}{2}

) = x2
2x - 1 = x2
hence x2 - 2x + 1 = 0
(x - 1)(x - 1) = 0
x = 1

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Number Bases

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Question 25 Report

Solve the following simultaneous equation for x. x2 + y - 5 = 0, y - 7x + 3 = 0

-2, 4

2, 4

-1,8

1, -8

Answer Details

x2 + y - 5 = 0.....(i)
y - 7x + 3 = 0.........(ii)
y = 7x - 3, substituting the value of y in equation (i)
x2 + (7x - 3) - 5 = 0
x2 + 7x + 3 = 0
(x + 8)(x - 1) = 0
x = -8 or 1

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Solution Of Linear Equations

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Question 26 Report

In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.

15o

25o

40o

50o

55o

Answer Details

Join PR
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o

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Algebraic Expressions

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Question 27 Report

Evaluate $ \left(x + \frac{1}{x} + 1\right)^2 - \left(x + \frac{1}{x} + 1\right)^2 $

4x2

(

\frac{2}{x}

+ 2)2

4

4(1 + x)

Answer Details

(x +

\frac{1}{x}

+ 1)2 - (x +

\frac{1}{x}

+ 1)2
= (x +

\frac{1}{x}

+ 1 + x +

\frac{- 1}{x}

- 1)(x -

\frac{1}{x}

+ 1 - x +

\frac{1}{x}

+ 1)
= (2x) (2 +

\frac{2}{x}

) = 2x x 2(1 +

\frac{1}{x}

)
4x (1 +

\frac{1}{x}

) = 4x + 4
= 4(1 + x)

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Quadratic Equations

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Question 28 Report

Make x the subject of the relation $ \frac{1+ax}{1-ax}=\frac{p}{q} $

\frac{p + q}{a (p - q)}

\frac{p - q}{a (p + q)}

\frac{p - q}{a p q}

\frac{p q}{a (p - q)}

Answer Details

To make x the subject of the given relation, we need to isolate x on one side of the equation. First, we can simplify the expression by finding the common denominator of the two fractions in the numerator: (1 + ax)/(1 - ax) = (1 - ax + ax + a^2x)/(1 - ax) = (1 + a^2x)/(1 - ax) Now we can cross-multiply to get rid of the denominator: (1 + a^2x) = pq - axpq Next, we can move the term with x to the left-hand side of the equation: a^2x + axpq = pq - 1 Finally, we can factor out x from the two terms on the left-hand side: x(a^2 + apq) = pq - 1 Therefore, we can isolate x by dividing both sides by (a^2 + apq): x = (pq - 1)/(a^2 + apq) So the correct option is: p-q / a(p+q) which is not among the given options.

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Solution Of Linear Equations

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Question 29 Report

The angle between latitudes 30oS and 13oN is

17o

33o

43o

53o

Answer Details

The angle between 2 latitudes one in northern hemisphere and the other in southern hemisphere and the other in southern hemisphere is the sum of their latitudes.
∴ Total angle difference = (30 + 13) = 43o

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Angles

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Question 30 Report

If x is negative, what is the range of values of x within which $ \frac{x+1}{3} > \frac{1}{X+3} $

3 < x < 4

-4 < x < -3

-2 < x < -1

-3 < x < 0

Answer Details

\frac{x + 1}{3}

>

\frac{1}{X + 3}

=

\frac{x + 1}{3}

>

\frac{x + 3}{X + 3}

= (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3)
x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0
= x(x + 3)9x + 4) > 0
Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0
= x > -4 (solution only)
Case 2 (+, -, -) = x > 0, x + 4 < 0

= x > 0, x < -3, x < -4 = x < -3(solution only)

Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)

Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0
= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)

combining the solutions -4 < x < -3

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Fractions, Decimals And Approximations

Positive And Negative Integers, Rational Numbers

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Question 31 Report

Class	Frequency
1−5	2
6−10	4
11−15	5
16−20	2
21−25	3
26−30	2
31−35	1
36−40	1

Find the median of the observation in the table given.

11.5

12.5

14.0

14.5

Answer Details

Median = L1 + (

\frac{E f}{f m}

) - fo

\frac{\sum f}{2}

=

\frac{20}{2}

= 10, L1 = 10.5, fo = 6, fm = 5
Median = 10.5 +

\frac{(10 - 6)}{5}

5
= 10.5 + 4
= 14.5

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Representation Of Data

Measures Of Location

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Question 32 Report

If the function f is defined by f(x + 2) = 2x2 + 7x = 5, find f(-1)

-10

-8

4

10

Algebraic Expressions

Differentiation

Quadratic Equations

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Question 33 Report

Calculate the standard deviation of the following data: 7, 8, 9, 10, 11, 12, 13.

2

4

10

11

Answer Details

To calculate the standard deviation, we need to first find the mean (average) of the data. Mean = (7 + 8 + 9 + 10 + 11 + 12 + 13) / 7 = 10 Next, we need to find the difference between each data point and the mean, and square the differences. (7-10)^2 = 9 (8-10)^2 = 4 (9-10)^2 = 1 (10-10)^2 = 0 (11-10)^2 = 1 (12-10)^2 = 4 (13-10)^2 = 9 Then, we add up all of these squared differences and divide by the total number of data points. (9 + 4 + 1 + 0 + 1 + 4 + 9) / 7 = 28 / 7 = 4 Finally, we take the square root of this result to find the standard deviation. Standard deviation = √4 = 2 Therefore, the standard deviation of the given data is 2.

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Measures Of Dispersion

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Question 34 Report

Simplify $ \frac{1}{p} - \frac{1}{q} + \frac{p}{q} - \frac{q}{p} $

\frac{1}{p - q}

\frac{- 1}{p + q}

\frac{1}{p q}

\frac{1}{p q (p - q)}

Answer Details

\frac{1}{p}

-

\frac{1}{q}

+

\frac{p}{q}

-

\frac{q}{p}

=

\frac{q - p}{p q}

+

\frac{p^{2} - q^{2}}{p q}

\frac{q - p}{p q}

x

\frac{p q}{p^{2} q^{2}}

=

\frac{q - p}{p^{2} q^{2}}

\frac{- (p - q)}{(p + q) (p - q)}

=

\frac{- 1}{p + q}

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Algebraic Expressions

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Question 35 Report

A rectangular polygon has 150o as the size of each interior angle. How many sides has the polygon?

12

10

9

8

Answer Details

A rectangular polygon has each interior angle to be 150o
let the polygon has n-sides
therefore, Total interior angle 150 x n = 150n
hence 150n = (2n - 4)90
150n = 180n - 360
360 = (180 - 150)n
30n = 360
n = 12

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Polynomials

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Question 36 Report

PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP

18

^{\circ}

54

^{\circ}

90

^{\circ}

108

^{\circ}

Triangles And Polygons

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Question 37 Report

In the diagram, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.

6

\frac{3}{4}

cm

3

\frac{3}{8}

cm

4

\frac{3}{8}

cm

2

\frac{3}{8}

cm

Answer Details

Using angle bisector theorem: line PS bisects angle QPR

QS/QP = SR/PR

3/4 = SR/g

4SR = 27

SR = $\frac{27}{4}$

= 6 $\frac{3}{4}$ cm

View Answer

Question 38 Report

In the diagram, $QP // ST$: $PQR = 34^\circ$ $QRS = 73^\circ$ and $RS = RT$. Find $SRT$

68

^{\circ}

102

^{\circ}

107

^{\circ}

141

^{\circ}

Answer Details

Construction joins R to P such that SRP = straight line

R = 180 $^{\circ}$ - 107 $^{\circ}$

< p = 180 $^{\circ}$ - (107 $^{\circ}$ - 34 $^{\circ}$ )

108 - 141 $^{\circ}$ = 39 $^{\circ}$

Angle < S = 39 $^{\circ}$ (corr. Ang.) But in $△$ SRT

< S = < T = 39 $^{\circ}$

SRT = 180 - (39 $^{\circ}$ + 39 $^{\circ}$ )

= 180 $^{\circ}$ - 78 $^{\circ}$

= 102 $^{\circ}$

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Angles And Intercepts On Parallel Lines

Statistics

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Question 39 Report

In the diagram, PQRs is a circle with 0 as centre and PQ/RT. If RTS = $32^\circ$. Find PSQ

32

^{\circ}

45

^{\circ}

58

^{\circ}

90

^{\circ}

Answer Details

< PSO =

\frac{1}{2}

< SOQ =

\frac{1}{2}

(180) = 90

^{\circ}

< RTS = < PQS = 32 $^{\circ}$ (Alternative angle)

< PSQ = 90 - < PSQ = 90 $^{\circ}$ - 32 $^{\circ}$

= 58 $^{\circ}$

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Circles

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Question 40 Report

Solve without using tables $ \log_{5}(62.5) - \log_{5}\left(\frac{1}{2}\right) $

3

4

5

8

Answer Details

log5(62.5) - log5(

\frac{1}{2}

)
= log5

\frac{(62.5)}{\frac{1}{2}}

- log5(2 x 62.5)
= log5(125)
= log553 - 3log55
= 3

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Positive And Negative Integers, Rational Numbers

Logarithms

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Question 41 Report

Divide the expression x3 + 7x2 - x - 7 by -1 + x2

-x3 + 7x2 - x - 7

-x3 = 7x + 7

x - 7

x + 7

Algebraic Expressions

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Question 42 Report

Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of $ \frac{1}{2} $o at the centre of the circle

2

π

π

\frac{2}{3}

\frac{π}{2}

Answer Details

Diameter = 8cm
∴ Radius = 4cm
Length of arc =

\frac{θ}{360}

x 2

π

r but Q = 22

\frac{1}{2}

∴ Length

\frac{22 \frac{1}{2}}{360}

x 2 x

π

x 4
=

\frac{22 \frac{1}{2} \times 8 π}{360}

=

\frac{180}{360}

=

\frac{π}{2}

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Circles

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Question 43 Report

If $ \sqrt{x^2 + 9} = x + 1 $, solve for x

5

4

3

2

1

Answer Details

\sqrt{x^{2} + 9}

= x + 1
x2 + 9 = (x + 1)2 + 1
0 = x2 + 2x + 1 - x2 - 9
= 2x - 8 = 0
2(x - 4) = 0
x = 4

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Fractions, Decimals, Approximations And Percentage

Quadratic Equations

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Question 44 Report

From the figure, calculate TH in centimeters

\frac{5}{\sqrt{3} + 1}

\frac{5}{\sqrt{3} - 1}

\frac{5}{\sqrt{3}}

\frac{\sqrt{3}}{5}

Answer Details

TH = tan 45

^{\circ}

, TH = QH

$\frac{T H}{5 + Q H}$ = tan 30 $^{\circ}$

TH = (b + QH) tan 30 $^{\circ}$

QH = 56 (5 + QH) $\frac{1}{\sqrt{3}}$

QH(1 - $\frac{1}{\sqrt{3}}$ ) = $\frac{5}{\sqrt{3}}$

QH = $\frac{5 \sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}$

= $\frac{5}{\sqrt{3} - 1}$

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Surds (radicals)

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Question 45 Report

The mean of the ages of ten secondary school pupils is 16 but when the age of their teacher is added to it the men becomes 19. Find the age of the teacher

27

35

38

49

Answer Details

Let the age of the teacher be T. According to the problem, the mean age of the 10 pupils is 16. Therefore, the sum of their ages is 10 x 16 = 160. When the teacher's age is added to it, the sum becomes 160 + T. Also, it is given that the new mean age after adding the teacher's age is 19. So, we have: (160 + T) / 11 = 19 Multiplying both sides by 11, we get: 160 + T = 209 Subtracting 160 from both sides, we get: T = 49 Therefore, the age of the teacher is 49.

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Integration

Statistics

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Question 46 Report

If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

0, 8

-1	5
	3

2, 3

1, -1

Number Bases

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Question 47 Report

If ₦225.00 yields ₦27.00 in x years simple interest at the rate of 4% per annum, find x

3

4

12

17

Answer Details

To solve this problem, we use the formula for simple interest: I = P*r*t Where: I = Interest earned P = Principal amount r = Rate of interest per year (as a decimal) t = Time period in years In this case, we are given: P = ₦225.00 (the principal amount) I = ₦27.00 (the interest earned) r = 4% = 0.04 (the rate of interest per year) Substituting these values into the formula, we get: 27 = 225 * 0.04 * t Simplifying the equation, we have: 27 = 9t Dividing both sides by 9, we get: t = 3 Therefore, the answer is 3 years. Explanation: The interest earned on the principal amount of ₦225.00 is ₦27.00. This means that for every year, the interest earned is 27/225 = 0.12 or 12% of the principal amount. We know that the rate of interest per year is 4%, which is much lower than the interest earned per year of 12%. This means that it will take more than 1 year for the interest earned to reach the rate of interest. In fact, it takes 3 years for the interest earned to reach the rate of interest of 4%, which is why the answer is 3.

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Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

Financial Arithmetic

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Question 48 Report

A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime

\frac{2}{11}

\frac{5}{11}

\frac{6}{11}

\frac{8}{11}

Answer Details

To solve this problem, we need to first determine the set of possible numbers that could be selected at random between 20 and 30. Since the problem states that both 20 and 30 are inclusive, the set of possible numbers is: {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} To find the probability that the selected number is prime, we need to determine the number of prime numbers in the set of possible numbers, and then divide that by the total number of possible numbers. Prime numbers are numbers that are only divisible by 1 and themselves. The prime numbers in the set of possible numbers are: { 23, 29 } Therefore, the probability that the selected number is prime is 2/11 or approximately 0.18.

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Question 49 Report

$Weight(s)$	$0-10$	$10-20$	$20-30$	$40-50$
Number of coconuts	$10$	$27$	$19$	$6$	$2$

Estimate the mode of the frequency distribution above.

13.2g

15.0g

16.8g

17.5g

Answer Details

Mode = a +

\frac{(b - a) (F_{m} - F_{b})}{2 F_{m} - F_{a} - F_{b}}

=

L_{1} + \frac{Δ_{1} x^{c}}{Δ_{1} + Δ_{2}}

=

10 + \frac{(20 - 10) (27 - 10)}{2 (27) - 10 - 19}

= 10 +

\frac{170}{25}

= 10 + 6.8
= 16.8

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Representation Of Data

Probability

Linear Inequalities

Statistics

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Question 50 Report

If sin x = cos 50o, then x equals

40o

45o

50o

90o

130o

Answer Details

We know that sin x = cos 50°. To solve for x, we need to find the angle whose sine is equal to cos 50°. We can use the trigonometric identity that sin (90° - θ) = cos θ. So, sin x = cos 50° can be rewritten as sin x = sin (90° - 50°). Using the identity, we have: sin x = sin 40° Now, we need to find the angle x whose sine is equal to sin 40°. Since sine is a periodic function, there are multiple angles whose sine is equal to a given value. One such angle is x = 40°. However, sine is also negative in the third and fourth quadrants. In the third quadrant, x = 180° - 40° = 140° and in the fourth quadrant, x = 360° - 40° = 320° also satisfy the equation sin x = sin 40°. However, since x has to be between 0° and 360°, we can eliminate the30°. Therefore, the possible values of x are 40°, 140°, and 320°. However, since we know that sin x = cos 50° and cos is positive in the first quadrant, x cannot be in the third or fourth quadrants. Therefore, the only possible value of x is x = 40°. Hence, the answer is x = 40°.

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Introductory Calculus

Trigonometry

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\(Weight(s)\)	\(0-10\)	\(10-20\)	\(20-30\)	\(40-50\)
Number of coconuts	\(10\)	\(27\)	\(19\)	\(6\)	\(2\)






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