JAMB UTME - Mathematics - 2016

Question 1 Report

Evaluate $ \frac{0.8 \times 0.43 \times 0.031}{0.05 \times 0.72 \times 0.021} $ correct to four significant figures.

14.09

14.1

14.12

14.11

Answer Details

\frac{0.8 \times 0.43 \times 0.031}{0.05 \times 0.72 \times 0.021}

x

\frac{10}{10}

= 14.11

View Answer

Question 2 Report

From the diagram above. Find the fraction of the shaded position?

\frac{1}{3}

\frac{1}{5}

\frac{1}{4}

\frac{1}{6}

Answer Details

θ

= 180o -(90 + 60)

θ

= 180o - 150o = 30o
Fraction of shaded position =

\frac{30}{360}

+

\frac{30}{360}

=

\frac{1}{12}

+

\frac{1}{12}

=

\frac{1}{6}

Read lesson note on Transformation In The Cartesian Plane (WAEC)

Read lesson note on Modular Arithmetic (WAEC)

Read lesson note on Indices, Logarithms And Surds (JAMB)

Transformation In The Cartesian Plane

Modular Arithmetic

Indices, Logarithms And Surds

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Question 3 Report

An arc of the length $16\pi$ cm subtends an angle of $80^o$ at the centre of the circle. Find the radius of the circle.

24cm

28cm

36cm

32cm

Answer Details

L =

\frac{θ}{360}

x 2

π

r
16

π

=

\frac{80}{360}

x 2

π

r

\frac{16 π \times 360}{80 \times 2 π}

= r
r = 36cm

Read lesson note on Circles (WAEC)

Circles

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Question 4 Report

Solve for x and y respectively
3x - 5y = 9
6x - 4y = 12

\frac{3}{4}

, 1

\frac{4}{3}

, 1

\frac{3}{4}

, -1

\frac{4}{3}

, -1

Answer Details

3x - 5y = 9 ------x2
6x - 4y = 12 -----x1
6x - 10y = 18
-6x - 4y = 12
____________

\frac{6 y}{- 6}

=

\frac{6}{6}

y = -1
in eq (1) 3x - 5y = 9
3x - 5(-1) = 9
3x + 5 = 9
3x = 4
x =

\frac{4}{3}

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

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Question 5 Report

Find $ \frac{dy}{dx} $, if $ y = \frac{2}{3}X^3 - \frac{4}{x} $

3x2 -

\frac{4}{x}

2x2 +

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

3x2 +

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

2x2 -

\frac{4}{x}

Answer Details

y =

\frac{2}{3}

x3 -

\frac{4}{x}

\frac{d y}{d x}

= 2x2 - Lx-2 = 2x2 +

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

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Read lesson note on Differentiation (JAMB)

Introductory Calculus

Differentiation

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Question 6 Report

If the mean of 4, y, 8 and 10 is 7. Find y?

6

10

7

9

Answer Details

To find the value of y, we need to use the formula for calculating the mean (also known as the average). The formula for calculating the mean of a set of numbers is: mean = (sum of numbers) / (number of numbers) In this case, we know that the mean of the numbers 4, y, 8, and 10 is 7. So we can plug in these values into the formula and solve for y: 7 = (4 + y + 8 + 10) / 4 Multiplying both sides by 4 gives us: 28 = 4 + y + 8 + 10 Simplifying the right side of the equation gives us: 28 = y + 22 Subtracting 22 from both sides gives us: 6 = y Therefore, the value of y that makes the mean of 4, y, 8, and 10 equal to 7 is 6. So the answer is 6.

Read lesson note on Linear Inequalities (WAEC)

Read lesson note on Progression (JAMB)

Linear Inequalities

Progression

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Question 7 Report

The bar chart above shows the marks obtained by students in a mathematics test.

How many students in all took the test?

40

30

10

20

Answer Details

3 + 2 + 5 + 5 + + 5 + 10 + 4 + 3 + 1 + 2 = 40

View Answer

Question 8 Report

If Q is a factor of 18 and T is prime numbers between 2 and 18. What is $Q \cap T$?

(2,3)

(2,3,18)

(2,3,9)

(2,3,6)

Answer Details

Given that Q is a factor of 18, the possible values of Q are 1, 2, 3, 6, 9, and 18. We also know that T is a prime number between 2 and 18. The prime numbers between 2 and 18 are 2, 3, 5, 7, 11, 13, and 17. To find the intersection of Q and T, we look for the values that are common to both sets. The only values of Q that are also in the set of prime numbers between 2 and 18 are 2 and 3. Therefore, the intersection of Q and T is (2, 3). So the answer is: (2, 3).

Read lesson note on Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

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Question 9 Report

The curved surface area of a cylinder 5cm high is $110\text{cm}^2$. Find the radius of its base

$\pi = \frac{22}{7}$

2.6cm

3.5cm

3.6cm

7.0cm

Answer Details

The problem gives us the height and curved surface area of a cylinder and asks us to find the radius of its base. The formula for the curved surface area of a cylinder is 2πrh, where r is the radius and h is the height. We are given h = 5cm and curved surface area = 110cm². Substituting these values into the formula, we get: 2πr(5) = 110 Simplifying, we get: πr = 11 Dividing both sides by π, we get: r = 11/π The value of π is given as 22/7, so substituting this in the equation, we get: r = 11/(22/7) r = 3.5cm Therefore, the radius of the cylinder is 3.5cm. Option (B) is the correct answer.

Read lesson note on Areas (WAEC)

Areas

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Question 10 Report

Calculate the range of 20, -6, 25, 30, 21, 28, 32, 33, 34, 5, 3, 2, and 1.

32

36

33

40

Answer Details

To calculate the range of a set of numbers, you subtract the lowest number from the highest number. In this case, the lowest number is -6 and the highest number is 34. So the range is: 34 - (-6) = 40 Therefore, the correct answer is option D: 40.

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Read lesson note on Number Bases (WAEC)

Variation

Number Bases

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Question 11 Report

Evaluate $ \int (\sin x - 5x^2)\,dx $

-cosx - 10x + k

cosx -

\frac{5 x^{3}}{3}

+ k

-cosx -

\frac{5 x^{3}}{3}

+ k

cosx - 10x + k

Answer Details

\int

(sin x - 5x2) = -cosx -

\frac{5 x^{3}}{3}

+ k

Read lesson note on Representation Of Data (JAMB)

Read lesson note on Integration (JAMB)

Representation Of Data

Integration

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Question 12 Report

Calculate the perimeter of a sector of a circle of radius 9cm and angle 36o.

18cm

(18 +

\frac{9 π}{5}

)cm

(9 +

\frac{9 π}{5}

)cm

\frac{9 π}{5}

cm

Answer Details

The perimeter of a sector of a circle is the sum of the arc length and the two radii. To find the perimeter, we need to first find the arc length using the formula: arc length = (angle/360) x 2 x π x radius In this case, the angle is given as 36°, the radius is given as 9cm, and π is approximately equal to 3.14. Substituting these values into the formula, we get: arc length = (36/360) x 2 x 3.14 x 9 = 2.83 cm (rounded to two decimal places) Now, we can find the perimeter by adding the two radii to the arc length: perimeter = 2 x 9 + 2.83 = 20.83 cm Therefore, the answer is (B) (18 + 9π/5) cm, rounded to two decimal places.

Read lesson note on Circles (WAEC)

Circles

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Question 13 Report

OGIVE is constructed using

Third quartile range

Semi-quartile range

Cummulative frequency table

Inter-quartile table

Answer Details

An ogive is a graph used to show the cumulative frequency of a given dataset. It is constructed using a cumulative frequency table which contains the frequencies and the cumulative frequencies. The cumulative frequency is the total of the frequencies up to that point in the data set. Therefore, the correct answer is "Cumulative frequency table". The other options mentioned, third quartile range, semi-quartile range, and inter-quartile range, are statistical measures that can be used to describe the spread of a dataset, but they are not directly used in the construction of an ogive.

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Read lesson note on Integration (JAMB)

Representation Of Data

Integration

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Question 14 Report

Marks	2	3	4
Frequency	4	4	$y$

The table above shows the frequency distribution of marks obtained by a group of students. If the total mark is 48, find the value of y.

6

8

7

5

Answer Details

(2 x 4) + (3 x 4) + (4 x y) = 48
8 + 12 + 4y = 48
4y = 48 - 20

\frac{4 y}{4}

=

\frac{28}{4}

y = 7

View Answer

Question 15 Report

In the diagram above, l1 is parallel to l2, Find the value of < PMT

82o

36o

72o

118o

Answer Details

< MPT =

180^{\circ}

-

118^{\circ}

=

62^{\circ}

< PML =

62^{\circ}

(Alternating angles)
y + 2y +

10^{\circ}

+

62^{\circ}

=

180^{\circ}

(Angles on a straight line)
3y = 180 - 72

\frac{3 y}{3}

=

\frac{108}{3}

y =

36^{\circ}

< PMT = 2y + 10 = 2(36) + 10 =

82^{\circ}

Read lesson note on Triangles And Polygons (WAEC)

Triangles And Polygons

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Question 16 Report

The bar chart above shows the number of visitors received in a week. How many visitors were received on Friday, Tuesday and Sunday?

17

22

20

16

Answer Details

No of visitors = 10 + 5 + 1 = 16

Read lesson note on Representation Of Data (JAMB)

Read lesson note on Measures Of Location (JAMB)

Representation Of Data

Measures Of Location

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Question 17 Report

implify $1 - \left(\frac{1}{7} \times 3 \frac{1}{2}\right) \div \frac{3}{4}$

2

\frac{1}{3}

1

\frac{2}{3}

Answer Details

1 - (

\frac{1}{7}

x 3

\frac{1}{2}

)

\div

\frac{3}{4}

1 - (

\frac{1}{7}

x

\frac{7}{2}

) x

\frac{4}{3}

1 -

\frac{1}{2}

x

\frac{4}{3}

1 -

\frac{2}{3}

=

\frac{1}{3}

Read lesson note on Number Bases (WAEC)

Number Bases

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Question 18 Report

Integrate $ \frac{2x^3+2x}{x} $ with respect to x

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

- 2x + k

x3 + 2x + k

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

+ 2x + k

x3 - 2x + k

Answer Details

\int

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

=

\int

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

+

\frac{2 x}{x}

=

\int

(2x2 + 2) dx

\frac{2 x^{2} + 1}{2 + 1}

+ 2x + k
=

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

+ 2x + k

Read lesson note on Integration (JAMB)

Integration

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Question 19 Report

Find the derivative of $y = \left(\frac{1}{3}x + 6\right)$

2(

\frac{1}{3}

x + 6

\frac{2}{3}

(

\frac{1}{3}

x + 6)

\frac{2}{3}

(

\frac{1}{3}

x + 6)2

\frac{1}{3}

(

\frac{1}{3}

x + 6)2

Answer Details

Y = (

\frac{1}{3}

x + 6)2

\frac{d y}{d x}

= 2(

\frac{1}{3}

x + 6)

\frac{1}{3}

=

\frac{2}{3}

(

\frac{1}{3}

x + 6)

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Differentiation

Application Of Differentiation

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Question 20 Report

Make S the subject of the relation

$p = s + \frac{sm^2}{nr}$

s =

\frac{n r p}{n r + m^{2}}

s = nr +

\frac{m^{2}}{m r p}

s =

\frac{n r p}{m r}

+ m2

s =

\frac{n r p}{n r}

+ m2

Answer Details

p = s +

\frac{s m^{2}}{n r}

p = s + ( 1 +

\frac{m^{2}}{n r}

)
p = s (1 +

\frac{n r + m^{2}}{n r}

)
nr × p = s (nr + m2)
s =

\frac{n r p}{n r + m^{2}}

View Answer

Question 21 Report

The sum of the interior angles of a polygon is a given as 1080o. Find the number of the sides of the polygon.

5

7

6

8

Answer Details

The sum of the interior angles of a polygon is given by the formula (n-2)180, where n is the number of sides of the polygon. Setting this formula equal to 1080, we get (n-2)180 = 1080. Simplifying this equation, we get n-2 = 6, and therefore n = 8. Hence, the polygon has 8 sides. Therefore, the answer is 8.

Read lesson note on Triangles And Polygons (WAEC)

Triangles And Polygons

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Question 22 Report

If line p = 5x + 3 is parallel to line p = wx + 5. Find the value of w.

7

3

6

5

Answer Details

Two lines are parallel if they have the same slope. In this case, line p = 5x + 3 and line p = wx + 5 are parallel. Therefore, they have the same slope. The slope of p = 5x + 3 is 5. So, to find the value of w in p = wx + 5, we set the slope of this line equal to 5 and solve for w: w = 5 Therefore, the value of w is 5.

Read lesson note on Indices (WAEC)

Indices

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Question 23 Report

In a school of 150 students, 80 offer French while 60 offer Arabic and 20 offer neither. How many students offer both subjects?

45

10

35

30

Answer Details

80 - x + x + 60 - x + 20 = 150
160 - x = 150
x = 160 - 150 = 10

Read lesson note on Sets (JAMB)

Sets

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Question 24 Report

The venn diagram above shows a class of 40 students with the games they play. How many of the students play two games only?

19

16

15

4

Answer Details

To find the number of students who play two games only, we need to add up the number of students in the regions of the Venn diagram that represent playing two games only. Looking at the diagram, we can see that there are two regions that represent playing two games only: the intersection of the football and basketball circles, and the intersection of the basketball and volleyball circles. The number of students who play two games only is the sum of the number of students in these two regions. From the diagram, we can see that there are 7 students in the football and basketball region, and 8 students in the basketball and volleyball region. Therefore, the total number of students who play two games only is: 7 + 8 = 15 So, 15 of the 40 students play two games only.

Read lesson note on Positive And Negative Integers, Rational Numbers (WAEC)

Read lesson note on Sets (WAEC)

Positive And Negative Integers, Rational Numbers

Sets

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Question 25 Report

From the cyclic quadrilateral above, find < TVS

60o

80o

70o

50o

Answer Details

< TVS =

180^{\circ} - (80 + 20)

=

180 - 100 = 80^{\circ}

Read lesson note on Indices (WAEC)

Indices

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Question 26 Report

Scores	3	6	5	2
Frequency	2	3	4	6

From the table above, find the median.

3

5

4

6

Answer Details

Mean =

\frac{(2 \times 3) + (3 \times 6) + (4 \times 5) + (2 \times 6)}{2 + 3 + 4 + 6}

=

\frac{6 + 18 + 20 + 12}{15}

=

\frac{56}{15}

= 3.73
Median: Arrange numbers in ascending order from smallest to biggest to get: 2, 2, 2, 2, 2, 2, 3, 3, 5, 5, 5, 5, 6, 6, 6. The middle number is the 8th number which is 3.

Read lesson note on Linear Inequalities (WAEC)

Read lesson note on Measures Of Dispersion (JAMB)

Linear Inequalities

Measures Of Dispersion

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Question 27 Report

If $N = \frac{p}{2}\left(\frac{T_1 - T_2}{T_1}\right)$. Find P when $N = 12$, $T_1 = 27$ and $T_2 = 24$.

48

108

54

216

Answer Details

N =

\frac{p}{2}

(\frac{T_{1} - T_{2}}{T_{1}})

12 =

\frac{p}{2}

(\frac{27 - 24}{27})

24 = P

(\frac{3}{27})

P = 24 x 9 = 216

View Answer

Question 28 Report

If the 2nd term of a G.P is $ \frac{8}{9} $ and the 6th term is $ 4\frac{1}{2} $. Find the common ratio.

2

\frac{3}{2}

\frac{2}{3}

3

Answer Details

T2 =

\frac{8}{9}

, T6 = 4

\frac{1}{2}

ar =

\frac{8}{9}

ar5 =

\frac{9}{2}

\frac{a r}{a r^{5}}

=

\frac{8}{9}

x

\frac{2}{9}

\frac{1}{r^{4}}

=

\frac{16}{81}

r =

\sqrt[4]{\frac{81}{16}}

=

\frac{3}{2}

Read lesson note on Sequence And Series (WAEC)

Sequence And Series

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Question 29 Report

Find $ \frac{dy}{dx} $. If $y = 3x^3 + 2x^2 + 3x + 1$

9x2 + 4x + 3

9x2 - 4x - 3

9x2 + 4x - 3

9x2 - 4x + 3

Answer Details

y = 3x3 + 2x2 + 3x + 1

\frac{d y}{d x}

= 9x2 + 4x + 3

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Read lesson note on Application Of Differentiation (JAMB)

Trigonometry

Differentiation

Application Of Differentiation

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Question 30 Report

Evaluate $ \frac{27^{\frac{1}{3}} - 8^{\frac{2}{3}}}{16^{\frac{2}{4}} \times 2} $

\frac{23}{5}

-

\frac{23}{5}

\frac{1}{8}

\frac{21}{7}

Answer Details

\frac{27^{\frac{1}{3}} - 8^{\frac{2}{3}}}{16^{\frac{2}{4}} \times 2}

\frac{27^{\frac{1}{3}} - 8^{\frac{2}{3}}}{16^{\frac{2}{4}} \times 2}

=

\frac{3 \sqrt{27} - 3 {\sqrt{8}}^{2}}{\sqrt{16} \times 2}

\frac{3 - 4}{4 \times 2}

=

\frac{- 1}{8}

Read lesson note on Surds (radicals) (WAEC)

Read lesson note on Simple Operations On Algebraic Expressions (WAEC)

Read lesson note on Quadratic Equations (WAEC)

Surds (radicals)

Simple Operations On Algebraic Expressions

Quadratic Equations

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Question 31 Report

In the cyclic quadrilateral above . Find < PRO

70o

20o

50o

30o

Answer Details

< PRO =

180^{\circ} - (100 + 50)

=

180 - 150 = 30^{\circ}

View Answer

Question 32 Report

If x10 = 235. Find x

15

12

14

13

Answer Details

x10 = 235
x10 = 2 x 51 + 3 x 50
= 10 + 3
= x10 = 13

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Read lesson note on Number Bases (WAEC)

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Solution Of Linear Equations

Number Bases

Quadratic Equations

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Question 33 Report

Evaluate $ \int_{0}^{\frac{\pi}{2}} \sin x\,dx $

2

1

-1

-2

Answer Details

\int_{0}^{\frac{π}{2}}

sin xdx = -cosx|

\frac{π}{2}

= -(cos

\frac{π}{2}

- cos0) = -(0-1) = 1

Read lesson note on Introductory Calculus (WAEC)

Read lesson note on Sequence And Series (WAEC)

Introductory Calculus

Sequence And Series

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Question 34 Report

Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45o

2

π

8 + 2

π

16 + 2

π

16 + 16

π

Answer Details

Perimeter = OP + OQ + PQ
= 8 + 8 + PQ
length PQ =

\frac{θ}{360 \times 2 π r}

=

\frac{45}{360}

x 2 x

π

x 8
= 2

π

Perimeter of sector 2r + L
Where l = length of arc and r = radius
∴ P = 2(8) + 2

π

= 16 + 2

π

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Coordinate Geometry

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Question 35 Report

If $y = 2x^3 + 6x^2 + 6x + 1$, Find $\frac{dy}{dx}$

6x2 + 12x + 1

6x2 + 6x + 1

6x2 + 6x + 6

6x2 + 12x + 6

Answer Details

The derivative of y with respect to x, denoted as dy/dx, is the rate at which y changes with respect to x. To find dy/dx, we need to apply the power rule and the sum rule for differentiation. The power rule states that d(x^n)/dx = nx^(n-1). The sum rule states that d(f(x) + g(x))/dx = df(x)/dx + dg(x)/dx. Given that y = 2x^3 + 6x^2 + 6x + 1, we can differentiate it term by term: d(2x^3)/dx = 6x^2 d(6x^2)/dx = 12x d(6x)/dx = 6 d(1)/dx = 0 Applying the sum rule, we have: dy/dx = 6x^2 + 12x + 6 So, the answer is 6x^2 + 12x + 6.

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Algebraic Expressions

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Question 36 Report

In the diagram MN is a chord of a circle KMN centre O and radius 10cm. If < MON = 140o, find, correct to the nearest cm, the length of the chord MN.

19cm

18cm

17cm

12cm

Answer Details

From the diagram

sin 70o = $\frac{x}{10}$

x = 10sin 70o

= 9.3969

Hence, length of chord MN = 2x

= 2 x 9.3969

= 18.7938

= 19cm (nearest cm)

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Circles

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Question 37 Report

Evaluate $ \frac{12.02 \times 20.06}{26.04 \times 60.06} $, correct to three significant figures.

0.157

0.154

0.155

0.158

Answer Details

Explanation:

\frac{12.02 \times 20.06}{26.04 \times 60.06}

=

\frac{241.1212}{15698.96}

= 0.154

Read lesson note on Positive And Negative Integers, Rational Numbers (WAEC)

Read lesson note on Modular Arithmetic (WAEC)

Positive And Negative Integers, Rational Numbers

Modular Arithmetic

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Question 38 Report

Given U = {x is a positive integer less than 15} and P = {x is even number from 1 to 14}. Find the compliment.

{1, 3, 5, 7, 9, 11, 13, 15}

{2, 3, 5, 7, 9, 11, 13}

{1, 3, 5, 7, 9, 11, 13}

{2, 3, 5, 7, 11, 15}

Answer Details

The complement of a set contains all the elements that are not in the set. In this problem, the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}, which is the set of positive integers less than 15. The set P = {2, 4, 6, 8, 10, 12, 14} is the set of even numbers from 1 to 14. To find the complement of P, we need to determine the set of all elements that are not in P. Since every positive integer less than 15 is either even or odd, we can determine the complement of P by finding the set of all odd numbers less than 15. This set is {1, 3, 5, 7, 9, 11, 13}. Therefore, the answer is {1, 3, 5, 7, 9, 11, 13}.

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Sets

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Question 39 Report

Rationalize $ \frac{\sqrt{6}-\sqrt{4}}{\sqrt{6}+\sqrt{4}} $

5 + 2

\sqrt{6}

5 - 4

\sqrt{6}

5 + 4

\sqrt{6}

5 - 2

\sqrt{6}

Answer Details

\frac{\sqrt{6} - \sqrt{4}}{\sqrt{6} + \sqrt{4}}

=

\frac{6 - 2 \sqrt{6} - 2 \sqrt{6} + 4}{6 - 4}

\frac{10 - 4 \sqrt{6}}{2}

= 5 - 2

\sqrt{6}

Read lesson note on Surds (radicals) (WAEC)

Surds (radicals)

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Question 40 Report

Simplify $ \frac{0.026 \times 0.36}{0.69} $. Leave your answer in standard form.

1.36 x 10-4

1.36 x 10-3

1.36 x 10-1

1.36 x 10-2

Answer Details

To simplify 0.026 × 0.360 / 0.69, we can follow the order of operations, which is to perform multiplication and division first, then addition and subtraction. Multiplying 0.026 and 0.360, we get: 0.00936 Dividing by 0.69, we get: 0.01356521739 We can express this answer in standard form by moving the decimal point to the right until we have a number between 1 and 10, then multiplying by an appropriate power of 10. In this case, we can move the decimal point three places to the right to get: 1.35652173913 × 10^-2 Rounding this to two decimal places, we get: 1.36 × 10^-2 Therefore, the answer is: 1.36 x 10^-2.

Read lesson note on Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

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Question 41 Report

If a car travels 120km on 45 litres of petrol, how much petrol is needed for a journey of 600km?

720 litres

160 litres

225 litres

960 litres

Answer Details

To find out how much petrol is needed for a journey of 600 km, we need to use the given information about the car's fuel consumption. The car travels 120 km on 45 litres of petrol, so we can set up a proportion: 120 km / 45 litres = 600 km / x litres We can cross-multiply to solve for x: 120 km * x litres = 45 litres * 600 km x = (45 litres * 600 km) / 120 km x = 225 litres Therefore, the amount of petrol needed for a journey of 600 km is 225 litres. So the correct option is: 225 litres.

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Question 42 Report

From the diagram above, Find the value of < ROP

110o

70o

95o

85o

Answer Details

< ROP =

95^{\circ}

(Exterior angle of cyclic quadrilateral)

Read lesson note on Angles (WAEC)

Angles

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Question 43 Report

Factorize k2 - 2kp + p2.

(k +p)2

(k - p)2

k2 + p2

k2 - p2

Answer Details

k2 - 2kp + p2 = (k - p)2

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Polynomials

Algebraic Expressions

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Question 44 Report

Evaluate $ \int (\cos 4x + \sin 3x)\,dx $

sin4x - cos3x + k

sin4x + cos3x + k

\frac{1}{4}

sin4x -

\frac{1}{3}

cos3x + k

\frac{1}{4}

sin4x +

\frac{1}{3}

cos3x + k

Representation Of Data

Integration

Financial Arithmetic

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Question 45 Report

An arc subtends an angle of 30o at the centre of a circle radius 12cm. Calculate the length of the arc.

6

π

cm

2

π

cm

3

π

cm

9

π

cm

Answer Details

Length or arc =

\frac{θ}{360}

x 2

π

r
=

\frac{30}{360}

x 2 x

\frac{22}{7}

x 12
=

\frac{1}{12}

x 2 x

π

x 2 = 2

π

cm

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Circles

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Question 46 Report

Determine the mean score of the student that took the mathematics test?

4.5

4.3

4.2

4.6

Answer Details

(3 x 0) + (1 x 2) + (5 x 2) + (5 x 3) + (5 x 4) +
(10 x 5) + (4 x 6) + (3 x 7) + (1 x 8) + (2 x 9)

\div

40
=

\frac{168}{40}

= 4.2

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Integration

Linear Inequalities

Statistics

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Question 47 Report

From the diagram above, find the required roots/region of x.

x

\leq

-1, x

\geq

4

x

\leq

-1, x

\geq

4

-1

\leq

x

\leq

4

x

\leq

-1, o

\leq

x

\leq

4

Answer Details

The required roots/region of x is: - -1 ≤ x ≤ 4 The diagram shows a closed circle at -1 and an open circle at 4, indicating that -1 and 4 are included and excluded from the solution set, respectively. This means that any value of x that is greater than or equal to -1 and less than or equal to 4 is a solution to the inequality. Therefore, the required region of x is the interval between -1 and 4, including -1 and excluding 4.

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Linear Inequalities

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Question 48 Report

Find the mean of 10, 8, 5, 11, 12, 9, 6, 3, 15, and 23.

11.2

10

10.2

11.3

Answer Details

To find the mean of a set of numbers, you add up all the numbers and then divide by the total count of numbers. So, for the given set of numbers, we add them up: 10 + 8 + 5 + 11 + 12 + 9 + 6 + 3 + 15 + 23 = 102 There are 10 numbers in the set, so we divide the sum by 10: 102 ÷ 10 = 10.2 Therefore, the mean of the given set of numbers is 10.2. So the answer is 10.2.

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Integration

Statistics

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Question 49 Report

he mean of 2-t, 4+t, 3-2t, 2+t and t-1 is

2

t

-t

-2

Answer Details

Mean =

\frac{2 - t + 4 + t + 3 - 2 t + 2 + t + t - 1}{5}

=

\frac{10}{5}

= 2

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Probability

Euclidean Geometry

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Question 50 Report

The pie chart above shows the distribution of subjects offered by students in SSS III level. If 80 students enrolled in the class. What is the size of the angle of the sector in economics?

24o

39o

32o

36o

Answer Details

xo + 36o + 54o + 28o + 90o + 120o = 360o
x + 328o = 360o
x = 360o - 328o
x = 32o

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Representation Of Data

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