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JAMB UTME - Mathematics - 2014

Ibeere 1 Ìròyìn

The locus of a dog tethered to a pole with a rope of 4m is a

circle with diameter 4m

circle with radius 4m

semi-circle with diameter 4m

semi-circle with radius 4m

Awọn alaye Idahun

The locus of a dog tethered to a pole with a rope of 4m is a circle with radius 4m. When a dog is tethered to a pole with a rope, it can move around the pole within the radius of the rope. Therefore, the dog's possible positions form a circle centered at the pole, with the radius equal to the length of the rope, which in this case is 4 meters. Since the circle has a fixed radius of 4m, it is not a semi-circle, but a full circle. Therefore, the correct answer is "circle with radius 4m."

Ka ìwé ẹ̀kọ́ nípa Loci (JAMB)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Triangles And Polygons

Ibeere 3 Ìròyìn

Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis.

(\begin{matrix} 3 & - 3 \\ 8 & 2 \end{matrix})

(\begin{matrix} 3 & - 3 \\ 8 & 2 \end{matrix})

(\begin{matrix} 3 & 3 \\ 8 & 2 \end{matrix})

(\begin{matrix} - 2 & 2 \\ 2 & 2 \end{matrix})

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

Awọn alaye Idahun

y - 4x + 3 = 0
When y = 0, 0 - 4x + 3 = 0
Then -4x = -3
x = 3/4
So the line cuts the x-axis at point (3/4, 0).
When x = 0, y - 4(0) + 3 = 0
Then y + 3 = 0
y = -3
So the line cuts the y-axis at the point (0, 3)
Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;

[\frac{1}{2} (x_{1} + x_{2}), \frac{1}{2} (y_{1} + y_{2})]

[\frac{1}{2} (\frac{3}{4} + 0), \frac{1}{2} (0 + - 3)]

[\frac{1}{2} (\frac{3}{4}), \frac{1}{2} (- 3)]

[\frac{3}{8}, \frac{- 3}{2}]

Ibeere 4 Ìròyìn

In the figure above, KL//NM, LN bisects < KNM. If angles KLN is \(54^\circ\) and angle MKN is \(35^\circ\), calculate the size of angle KMN.

91

^{\circ}

89

^{\circ}

37

^{\circ}

19

^{\circ}

Awọn alaye Idahun

In the diagram above,

α

= 54

^{\circ}

(alternate angles; KL||MN) < KNM = 2

α

(LN is bisector of < KNM) = 108

^{\circ}

35 $^{\circ}$ + < KMN + 108 $^{\circ}$ = 180 $^{\circ}$ (sum of angles of $△$ )

< KMN + 143 $^{\circ}$ = 180 $^{\circ}$

< KMN = 180 $^{\circ}$ - 143 $^{\circ}$

= 37 $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Angles And Intercepts On Parallel Lines (WAEC)

Angles And Intercepts On Parallel Lines

Ibeere 7 Ìròyìn

Find the value of x in the figure above.

20

\sqrt{3}

cm

10

\sqrt{3}

cm

5

\sqrt{3}

cm

4

\sqrt{3}

cm

Awọn alaye Idahun

The figure above is a right triangle with sides x, 3 cm, and x√3 cm. Since it is a right triangle, we can use the Pythagorean theorem to find the value of x. The theorem states that the sum of the squares of the two shorter sides of a right triangle is equal to the square of the longest side, which is the hypotenuse. So, we have: x^2 + (3)^2 = (x√3)^2 Expanding the square on the right side: x^2 + 9 = x^2 * 3 Simplifying the equation: x^2 + 9 = 3x^2 Solving for x: 2x^2 = 9 x^2 = 4.5 Taking the square root of both sides: x = ±√(4.5) Since x has to be positive, we choose the positive square root: x = √(4.5) = 2√3 Finally, multiplying by 10: x = 10√3 So, the value of x in the figure is 10√3 cm.

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Surds (radicals)

Ibeere 8 Ìròyìn

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

t

-t

2

-2

Awọn alaye Idahun

Mean x =

\frac{\sum x}{n}

= [(2 - t) + (4 + t) + (3 - 2t) + (2 + t) + (t - 1)

\div

] 5
= [11 - 1 + 3t - 3t]

\div

5
= 10

\div

5
= 2

Ibeere 10 Ìròyìn

If ∣∣∣−x12−14∣∣∣=−12, find x

-6

-2

3

6

Awọn alaye Idahun

Explanation:

| \begin{matrix} - x & 12 \\ - 1 & 4 \end{matrix} | = - 12

-4x - (-1)12 = -12
-4x + 12 = -12
-4x = -12 - 12
-4x = - 24
x = 6

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Ibeere 11 Ìròyìn

Evaluate \( \int (2x+3)^{\frac{1}{2}}\delta x \)

\frac{1}{12} (2 x + 3)^{6} + k

\frac{1}{3} (2 x + 3)^{\frac{1}{2}} + k

\frac{1}{3} (2 x + 3)^{\frac{3}{2}} + k

\frac{1}{12} (2 x + 3)^{\frac{3}{4}} + k

Awọn alaye Idahun

\int (2 x + 3)^{\frac{1}{2}} δ x

let u = 2x + 3,

\frac{δ y}{δ x} = 2

δ x = \frac{δ u}{2}

Now

\int (2 x + 3)^{\frac{1}{2}} δ x = \int u^{\frac{1}{2}} . \frac{δ x}{2}

= \frac{1}{2} \int u^{\frac{1}{2}} δ u

= \frac{1}{2} u^{\frac{3}{2}} \times \frac{2}{3} + k

= \frac{1}{3} u^{\frac{3}{2}} + k

= \frac{1}{3} (2 x + 3)^{\frac{3}{2}} + k

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Integration

Ibeere 12 Ìròyìn

If \(y = 4x^3 - 2x^2 + x\), find \(\frac{\delta y}{\delta x}\)

8x2 - 2x + 1

8x2 - 4x + 1

12x2 - 2x + 1

12x2 - 4x + 1

Awọn alaye Idahun

If y = 4x3 - 2x2 + x, then;

\frac{δ y}{δ x}

= 3(4x2) - 2(2x) + 1
= 12x2 - 4x + 1

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Ibeere 13 Ìròyìn

Evaluate Log28 + Log216 - Log24

3

4

5

6

Awọn alaye Idahun

= l o g_{2} \frac{8 \times 16}{4}

= l o g_{2} 32

= l o g_{2} 2^{5}

= 5 l o g_{2} 2

= 5 \times 1

= 5

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Ibeere 14 Ìròyìn

A number is chosen at random from 10 to 30 both inclusive. What is the probability that the number is divisible by 3?

\frac{2}{15}

\frac{1}{10}

\frac{1}{3}

\frac{2}{5}

Awọn alaye Idahun

Sample space S = {10, 11, 12, ... 30}
Let E denote the event of choosing a number divisible by 3
Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7
Prob (E) =

\frac{n (E)}{n (E)}

Prob (E) =

\frac{7}{21}

Prob (E) =

\frac{1}{3}

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Probability

Surds (radicals)

Ibeere 17 Ìròyìn

Simplify \( \frac{2\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} \)

3

\sqrt{6} - 7

3

\sqrt{6} - 7

3

\sqrt{6} - 1

3

\sqrt{6} + 1

Awọn alaye Idahun

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

= \frac{2 \sqrt{2} (\sqrt{2}) + (2 \sqrt{2}) (- \sqrt{3}) - \sqrt{3} (\sqrt{2}) - \sqrt{3} (- \sqrt{3})}{(\sqrt{2})^{2} - (\sqrt{3})^{2}}

= \frac{2 \times 2 - 2 \sqrt{6} - \sqrt{6} + 3}{2 - 3}

= \frac{4 - 3 \sqrt{6} + 3}{- 1}

= \frac{7 - 3 \sqrt{6}}{- 1}

= \frac{7}{- 1} - \frac{3 \sqrt{6}}{- 1}

= - 7 + 3 \sqrt{6}

= 3 \sqrt{6} - 7

Ka ìwé ẹ̀kọ́ nípa Functions And Relations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Functions And Relations

Surds (radicals)

Ibeere 18 Ìròyìn

From the Venn diagram above, the shaded parts represent

(P

\cap

Q)

\cup

(P

\cap

R)

(P

\cup

Q)

\cap

(P

\cap

R)

(P

\cup

Q)

\cup

(P

\cup

R)

(P

\cap

Q)

\cup

(P

\cup

R)

Awọn alaye Idahun

The Venn diagram above represents three sets: P, Q, and R. The shaded parts represent the elements that are in both P and Q (the intersection of P and Q), and the elements that are in both P and R (the intersection of P and R). Therefore, the shaded parts represent the set (P∩Q) and the set (P∩R). Option (A) (P∩Q)∪(P∩R) is the correct answer.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Probability

Linear Inequalities

Ibeere 21 Ìròyìn

If P = {1,2,3,4,5} and \( \cup \) Q = {1,2,3,4,5,6,7}, list the elements in Q

{6}

{7}

{6,7}

{5,6}

Awọn alaye Idahun

Q = (P

\cup

Q) - P
{6,7}

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Ibeere 22 Ìròyìn

Evaluate the inequality \( \frac{x}{2} + \frac{3}{4} \leq \frac{5x}{6} - \frac{7}{12} \)

x \geq 4

x≤3

x≥−3

x≤−4

Awọn alaye Idahun

\frac{x}{2} + \frac{3}{4} \leq \frac{5 x}{6} - \frac{7}{12}

12 \frac{x}{2} + 12 \frac{3}{4} \leq 12 \frac{5 x}{6} - 12 \frac{7}{12}

6x + 9

\leq

10x - 7
6x - 10x

\leq

- 7 - 9
-4x

\leq

-16
-4x/-4

\geq

-16/-4
x

\geq

4

Ka ìwé ẹ̀kọ́ nípa Inequalities (JAMB)

Inequalities

Ibeere 23 Ìròyìn

In how many ways can a team of 3 girls be selected from 7 girls?

\frac{7!}{3!}

\frac{7!}{4!}

\frac{7!}{3! 4!}

\frac{7!}{2! 5!}

Awọn alaye Idahun

A team of 2 girls can be selected from 7 girls in

^{7} C_{3}

= \frac{7!}{(7 - 3)! 3!}

= \frac{7!}{4! 3!} w a y s

Ka ìwé ẹ̀kọ́ nípa Permutation And Combination (JAMB)

Permutation And Combination

Ibeere 26 Ìròyìn

If \( \sin \theta = \frac{12}{13} \), find the value of \( 1 + \cos \theta \)

\frac{25}{13}

\frac{18}{13}

\frac{8}{13}

\frac{5}{13}

Sine, Cosine And Tangent Of An Angle

Number Bases

Ibeere 28 Ìròyìn

P varies directly as Q and inversely as R. When Q = 36 and R = 16, P = 27. Find the relation between P, Q and R.

P = \frac{Q}{12 R}

P = \frac{12 Q}{R}

P = 12 Q R

P = \frac{12}{Q R}

Awọn alaye Idahun

P \propto \frac{Q}{R}

P = K \frac{Q}{R}

When Q = 36, R = 16, P = 27
Then substitute into the equation

27 = K \frac{36}{16}

$K = \frac{27 \times 16}{36}$
$K = 12$
So the equation connecting P, Q and R is
$P = \frac{12 Q}{R}$

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Ka ìwé ẹ̀kọ́ nípa Simple Operations On Algebraic Expressions (WAEC)

Simple Operations On Algebraic Expressions

Ibeere 30 Ìròyìn

Numbers	1	2	3	4	5	6
Frequency	18	22	20	16	10	14

The table above represents the outcome of throwing a die 100 times. What is the probability of obtaining at least a 4?

\frac{1}{5}

\frac{1}{2}

\frac{2}{5}

\frac{3}{4}

Awọn alaye Idahun

Let E demote the event of obtaining at least a 4
Then n(E) = 16 + 10 + 14 = 40
Hence, prob (E) =

\frac{n (E)}{n (S)}

= \frac{40}{100}

= \frac{2}{5}

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Representation Of Data

Probability

Ibeere 32 Ìròyìn

Find the value of

\[ \begin{vmatrix} 0 & 3 & 2 \\ 1 & 7 & 8 \\ 0 & 5 & 4 \end{vmatrix} \]

12

10

-1

-2

Awọn alaye Idahun

0 | \begin{matrix} 7 & 8 \\ 5 & 4 \end{matrix} | - 3 | \begin{matrix} 1 & 8 \\ 0 & 4 \end{matrix} | + 2 | \begin{matrix} 1 & 7 \\ 0 & 5 \end{matrix} |

= 0(28 - 40) - 3(4 - 0) + 2(5 - 0)
= 0(-12) - 3(4) + 2(5)
= 0 - 12 + 10
= -2

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Ibeere 36 Ìròyìn

A cylindrical tank has a capacity of 6160m3. What is the depth of the tank if the radius of its base is 28cm?

8.0m

7.5m

5.0m

2.5m

Awọn alaye Idahun

Using

V = π r^{2} h

6160 = 22/7 x 28 x 28 x h

h = \frac{6160}{22 \times 4 \times 28}

h = 2.5 m

Ka ìwé ẹ̀kọ́ nípa Volumes (WAEC)

Ibeere 39 Ìròyìn

What is the common ratio of the G.P. \( (\sqrt{10}+\sqrt{5})+(\sqrt{10}+2\sqrt{5})+\ldots \)?

\sqrt{2}

√5

3

5

Awọn alaye Idahun

Common ratio r of the G.P is

r = \frac{T_{n} + 1}{T_{n}} = \frac{T_{2}}{T_{1}}

r = \frac{\sqrt{10} + 2 \sqrt{5}}{\sqrt{10} + \sqrt{5}}

r = \frac{\sqrt{10} + 2 \sqrt{5}}{\sqrt{10} + \sqrt{5}} \times \frac{\sqrt{10} - \sqrt{5}}{\sqrt{10} - \sqrt{5}}

= \frac{(\sqrt{10}) (\sqrt{10}) + (\sqrt{10}) (- \sqrt{5}) + (2 \sqrt{5}) (\sqrt{10}) + (2 \sqrt{5}) (- \sqrt{5})}{(\sqrt{10})^{2} - (\sqrt{5})^{2}}

\frac{10 - \sqrt{50} + 2 \sqrt{50} - 10}{10 - 5}

\frac{\sqrt{50}}{5}

\frac{\sqrt{25 \times 2}}{5}

\frac{5 \sqrt{2}}{5}

\sqrt{2}

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Surds (radicals)

Sequence And Series

Ibeere 43 Ìròyìn

Find y, if

\[ \begin{pmatrix} 5 & -6 \\ 2 & -7 \end{pmatrix} \begin{pmatrix} 5 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ -11 \end{pmatrix} \]

8

5

3

2

Awọn alaye Idahun

(\begin{matrix} 5 & - 6 \\ 2 & - 7 \end{matrix}) (\begin{matrix} 5 \\ 2 \end{matrix}) = (\begin{matrix} 7 \\ - 11 \end{matrix})

By matrices multiplication;
5x - 6y = 7 ........(1)
2x - 7y = -11 ......(2)
2 x (1): 10x - 12y = 14 .......(3)
5 x (2): 10x - 35y = -55 ......(4)
(3) - (4): 23y = 69
y = 69/23 = 3

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Ka ìwé ẹ̀kọ́ nípa Application Of Differentiation (JAMB)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Solution Of Linear Equations

Modular Arithmetic

Application Of Differentiation

Quadratic Equations

Ibeere 44 Ìròyìn

Evaluate \( \int \sin 2x \, dx \)

cos 2x + k

\frac{1}{2}

cos 2x + k

- \frac{1}{2}

cos 2x + k

-cos 2x + k

Awọn alaye Idahun

The value of the integral of sin(2x)dx is -(1/2)cos(2x) + k, where k is an arbitrary constant of integration. The integral of sin(2x) can be found using substitution or by recognizing that sin(2x) is the derivative of -(1/2)cos(2x). The constant of integration k is added to account for the fact that there are infinitely many functions that have the same derivative as sin(2x). The constant can take any value and is introduced to reflect the inherent uncertainty in finding an antiderivative.

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Integration

Trigonometry

Ibeere 45 Ìròyìn

If gt2 - k - w = 0, make g the subject of the formula

\frac{k + w}{t^{2}}

\frac{k - w}{t^{2}}

\frac{k + w}{t}

\frac{k - w}{t}

Awọn alaye Idahun

We can solve for g by manipulating the given equation, gt² - k - w = 0, to isolate g on one side of the equation. First, we can add k and w to both sides of the equation to obtain: gt² = k + w Next, we can divide both sides of the equation by t² to solve for g: g = (k + w)/t² Therefore, the solution is: g = (k + w)/t² Hence, the answer is: (k + w)/t²

Ka ìwé ẹ̀kọ́ nípa Change Of Subject Of A Formula/Relation (WAEC)

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Ibeere 46 Ìròyìn

The gradient of a line joining \((x,4)\) and \((1,2)\) is \(\frac{1}{2}\). Find the value of \(x\)

5

3

-3

-5

Awọn alaye Idahun

Gradient m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

\frac{1}{2} = \frac{2 - 4}{1 - x}

1 - x = 2(2 - 4)
1 - x = 4 - 8
1 - x = -4
-x = -4 - 1
x = 5

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