Loading....






Press & Hold to Drag Around
Click Here to Close

JAMB UTME - Mathematics - 2009

Question 1 Report

i. \(S \cap T \cap W = S\)
ii. \(S \cup T \cup W = W\)
ii. \(T \cap W = S\)

If \(S \subset T \subset W\). Which of the above statements are true?

i and ii

i and iii

ii and iii

i, ii and iii

Answer Details

From the given statements, we know that: i. S∩T∩W = S: This means that the elements that are common to S, T, and W are only those in S. ii. S∪T∪W = W: This means that the union of S, T, and W includes all the elements in W. iii. T∩W = S: This means that the elements that are common to T and W are only those in S. Using these statements, we can conclude that: - Statement i is true because it tells us that the only elements that are common to S, T, and W are those in S. This means that S is a subset of both T and W, which is consistent with statement iii. However, statement ii doesn't provide any additional information about the relationship between S, T, and W. - Statement ii is false because it tells us that the union of S, T, and W includes all the elements in W. This means that W is a subset of S∪T, which is inconsistent with statement iii. We know that S is a subset of both T and W, but this doesn't necessarily mean that S is a subset of S∪T. - Statement iii is true because it tells us that the elements that are common to T and W are only those in S. This means that S is a subset of both T and W, which is consistent with statement i. Therefore, the correct answer is: i and iii are true.

View Answer

Question 2 Report

If \( p = \begin{pmatrix} x + 3 & x + 2 \\ x + 1 & x - 1 \end{pmatrix} \) evaluate x if \(|p| = -10\)

-5

-2

2

5

Answer Details

If p =

(\begin{matrix} x + 3 & x + 2 \\ x + 1 & x - 1 \end{matrix})

evaluate x if |p| = -10
|P| = (x + 3)(x - 1) - (x + 2)(x + 1)
x2 + 2x - 3 - (x2 + 3x + 2)
x2 + 2x - 3 - x2 - 3x - 2
-x - 5 = -10
-x = -10 + 5
x = 5

Read lesson note on Polynomials (JAMB)

Polynomials

View Answer

Question 8 Report

Solve 52(x - 1) x 5x + 1 = 0.04

\frac{1}{3}

\frac{1}{4}

-

\frac{1}{5}

-

\frac{1}{3}

Answer Details

52(x - 1) x 5x + 1 = 0.04
52x - 1 x 5x + 1 =

\frac{1}{25}

53x - 1 = 5-2
3x - 1= -2
3x = -1
therefore, x = -

\frac{1}{3}

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

View Answer

Question 14 Report

Find the range of values of \(x\) for which \(3x - 7 \le 0\) and \(x + 5 > 0\)

-5 < x <

\frac{7}{3}

-5

\leq

x

\leq

\frac{7}{3}

-5 < x

\leq

\frac{7}{3}

-5

\leq

x <

\frac{7}{3}

Answer Details

To solve this inequality, we need to work with each inequality separately and then find the values of x that satisfy both. Starting with the first inequality: 3x - 7 ≤ 0 Adding 7 to both sides: 3x ≤ 7 Dividing by 3: x ≤ 7/3 Now, let's work with the second inequality: x + 5 > 0 Subtracting 5 from both sides: x > -5 To satisfy both inequalities, x must be greater than -5 and less than or equal to 7/3. Therefore, the range of values for x is: -5 < x ≤ 7/3

Read lesson note on Linear Inequalities (WAEC)

Read lesson note on Inequalities (JAMB)

Linear Inequalities

Inequalities

View Answer

Question 16 Report

Simplify \(4\frac{3}{4} - 6\frac{1}{4}\)

-7

\frac{7}{8}

\frac{- 2}{7}

\frac{- 10}{21}

\frac{10}{21}

Answer Details

4

\frac{3}{4}

- 6

\frac{1}{4}

\frac{19}{4}

-

\frac{25}{4}

............(A)

\frac{21}{5}

-

\frac{5}{4}

.............(B)
Now work out the value of A and the value of B and then find the value

\frac{A}{B}

A =

\frac{19}{4}

-

\frac{25}{4}

=

\frac{- 6}{4}

B =

\frac{21}{5}

x

\frac{5}{20}

=

\frac{105}{20}

=

\frac{21}{4}

But then

\frac{A}{B}

=

\frac{- 6}{4}

\frac{21}{4}

=

\frac{- 6}{4}

\div

\frac{21}{4}

=

\frac{- 6}{4}

x

\frac{4}{21}

=

\frac{- 24}{84}

=

\frac{- 2}{7}

View Answer

Question 20 Report

If \(m * n = n-(m + 2)\) for any real numbers m and n, find the value of \(3 * (-5)\)

-6

-8

-10

-12

Answer Details

If m x n = n
= n - (m + 2)
3

*

(-5) = -5 - (3 + 2)
= -5 - 5
= -10

Read lesson note on Solution Of Linear Equations (WAEC)

Read lesson note on Differentiation (JAMB)

Read lesson note on Quadratic Equations (WAEC)

Solution Of Linear Equations

Differentiation

Quadratic Equations

View Answer

Question 21 Report

In the figure above, TS//XY and XY = TY,

47

^{\circ}

52

^{\circ}

56

^{\circ}

99

^{\circ}

Answer Details

TXY = XTY = 47

^{\circ}

(Basa angle of an isosceles

△

)

< TYZ = 94 $^{\circ}$

n $^{\circ}$ = 180 - (94 $^{\circ}$ + 34 $^{\circ}$ )

= 180 $^{\circ}$ - 128 $^{\circ}$

= 52 $^{\circ}$

Read lesson note on Angles (WAEC)

Read lesson note on Triangles And Polygons (WAEC)

Triangles And Polygons

View Answer

Question 22 Report

No. of days	1	2	3	4	5	6
No. of students	20	\(2x\)	60	40	\(x\)	50

The distribution above shows the number of days a group of 260 students were absent from school in a particular term. How many students were absent for at least four days in the term

180

120

110

40

Answer Details

3x + 170 = 260
3x = 260 - 170
3x = 90

\to

x = 30
therefore, 20 + 2x + 60 + 40
= 180

View Answer

Question 26 Report

If s = (2 + 3t)(5t - 4), find \( \frac{dy}{dx} \) when t = \( \frac{4}{5} \) sec

0 units per sec.

15 units per seconds

22 units per seconds

26 units per seconds

Answer Details

s = (2 + 3t)(5t - 4), find

\frac{d y}{d x}

when t =

\frac{4}{5}

sec
= 10t - 8 + 15t2 - 12t

\frac{d s}{d t}

= 15t2 - 2t - 8
= 30t - 2 ; t =

\frac{4}{5}

= 30t - 2
= 30 x

\frac{4}{5}

- 2

\to

24 - 2
= 22 sec

Read lesson note on Differentiation (JAMB)

Differentiation

View Answer

Question 32 Report

Evaluate \( \int \sec^2 \theta \, d\theta \)

sec

θ

tan

θ

+ k

tan

θ

+ k

2 sec

θ

+ k

sec

θ

+ k

Answer Details

\int

sec2

θ

d

θ

= tan

θ

+ k

Read lesson note on Integration (JAMB)

Integration

View Answer

Question 34 Report

Find the acute angle between the straight lines \(y = x\) and \(y = \sqrt{3x}\)

15o

30o

45o

60o

Answer Details

x

\leq

90o (acute angle)
x2 = 3x
3x = 90o
x =

\frac{90^{o}}{3}

= 30o

Read lesson note on Coordinate Geometry (JAMB)

Coordinate Geometry

View Answer

Question 35 Report

Find the value of \( \sin 45^\circ - \cos 30^\circ \)

\frac{2 + \sqrt{3}}{4}

\frac{\sqrt{2} + \sqrt{3}}{4}

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

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

Answer Details

S i n 45 - c o s 30 \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} = \frac{\sqrt{2} - \sqrt{3}}{2}

Read lesson note on Sine, Cosine And Tangent Of An Angle (WAEC)

Read lesson note on Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

View Answer

Question 39 Report

A student spent \( \frac{1}{5} \) of his allowances on books, \( \frac{1}{3} \) of the remainder on food and kept the rest for contingences. What fraction was kept?

$\frac{7}{16}$

\frac{8}{15}

$\frac{2}{3}$

$\frac{4}{5}$

Answer Details

Spent

\frac{1}{5}

on books remaining

\frac{4}{5}

spent

\frac{1}{3}

of

\frac{4}{5}

on food

\frac{4}{5}

x

\frac{1}{3}

=

\frac{4}{15}

therefore,

\frac{4}{5}

-

\frac{4}{15}

=

\frac{12 - 4}{15}

=

\frac{8}{15}

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

Read lesson note on Number Bases (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Fractions, Decimals And Approximations

Number Bases

Surds (radicals)

View Answer

Question 43 Report

If the hypotenuse of a right-angled isosceles triangle is 2cm , what is the area of the triangle?

\frac{1}{\sqrt{2}}

cm2

1 cm2

\sqrt{2}

cm2

2

s q r t 2

cm2

Answer Details

An isosceles triangle has two equal sides and two equal angles opposite those sides. In a right-angled isosceles triangle, the two equal angles are 45 degrees each, since the sum of all angles in a triangle is 180 degrees and the right angle is 90 degrees. Let's use the Pythagorean theorem to find the length of the legs of the triangle. If the hypotenuse is 2cm, then each leg is x cm. Applying the Pythagorean theorem, we have: x^2 + x^2 = 2^2 2x^2 = 4 x^2 = 2 x = sqrt(2) Therefore, the length of each leg of the triangle is sqrt(2) cm. To find the area of the triangle, we can use the formula: Area = (base * height) / 2 Since the triangle is isosceles, the base and height are equal. Therefore, we can use either leg as the base and the other leg as the height. Area = (sqrt(2) * sqrt(2)) / 2 Area = 2 / 2 Area = 1 cm^2 So, the area of the right-angled isosceles triangle is 1 cm^2.

View Answer

Question 46 Report

If \(P = \begin{bmatrix} x+3 & x+2 \\ x+1 & x-1 \end{bmatrix}\) evaluate x if \(|P| = -10\)

-5

-2

2

5

Answer Details

P = [\begin{array}{cc} x + 3 & x + 2 \\ x + 1 & x - 1 \end{array}]

evaluate x if |P| = -10
(x+3)(x-1) - {(x+1)(x+2)} = -10
x2 - x + 3x - 3 - {x2 + 2x + x + 2} = -10
x2 + 2x - 3 - {x2 + 3x + 2} = -10
-x - 5 = -10
-5 + 10 = x
5 = x
∴x = 5

Read lesson note on Solution Of Linear Equations (WAEC)

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Angles Of Elevation And Depression (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Angles Of Elevation And Depression

Surds (radicals)

View Answer

Question 48 Report

Marks	2	3	4	5	6	7	8	9
No. of students	3	4	1	0	4	5	2	1

The table above shows the distribution of marks of students in a test. Find the probability of passing the test if the pass mark is 5.

\frac{3}{5}

\frac{2}{5}

\frac{7}{20}

\frac{1}{5}

Answer Details

p (pass at 5mark) =

\frac{12}{20}

=

\frac{3}{5}

Read lesson note on Probability (WAEC)

Read lesson note on Surds (radicals) (WAEC)

Probability

Surds (radicals)

View Answer

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50