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JAMB UTME - Mathematics - 2014

Question 1 Rapport

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

Détails de la réponse

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.

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Surds (radicals)

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Question 3 Rapport

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)

Détails de la réponse

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.

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Probability

Linear Inequalities

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Question 4 Rapport

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

Détails de la réponse

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

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Voir la réponse

Question 7 Rapport

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}

Détails de la réponse

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}

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Probability

Surds (radicals)

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Question 9 Rapport

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

Détails de la réponse

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

\frac{δ y}{δ x}

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

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Algebraic Expressions

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Question 12 Rapport

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}

Détails de la réponse

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

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Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

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Question 13 Rapport

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

Détails de la réponse

= \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

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Functions And Relations

Surds (radicals)

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Question 15 Rapport

Evaluate Log28 + Log216 - Log24

3

4

5

6

Détails de la réponse

= 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

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Question 18 Rapport

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}

Détails de la réponse

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}

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Representation Of Data

Probability

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Question 19 Rapport

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

Détails de la réponse

\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

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Integration

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Question 20 Rapport

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!}

Détails de la réponse

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

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Permutation And Combination

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Question 21 Rapport

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

\sqrt{2}

√5

3

5

Détails de la réponse

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}

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Surds (radicals)

Sequence And Series

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Question 22 Rapport

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

cos 2x + k

\frac{1}{2}

cos 2x + k

- \frac{1}{2}

cos 2x + k

-cos 2x + k

Détails de la réponse

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.

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Integration

Trigonometry

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Question 23 Rapport

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})

Détails de la réponse

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}]

Voir la réponse

Question 26 Rapport

Find the value of

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

12

10

-1

-2

Détails de la réponse

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

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Number Bases

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Question 28 Rapport

If ∣∣∣−x12−14∣∣∣=−12, find x

-6

-2

3

6

Détails de la réponse

Explanation:

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

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

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Question 29 Rapport

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

t

-t

2

-2

Détails de la réponse

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

Voir la réponse

Question 33 Rapport

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

Détails de la réponse

(\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

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Solution Of Linear Equations

Modular Arithmetic

Application Of Differentiation

Quadratic Equations

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Question 34 Rapport

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}

Détails de la réponse

Q = (P

\cup

Q) - P
{6,7}

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Voir la réponse

Question 41 Rapport

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

Détails de la réponse

\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

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Inequalities

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Question 42 Rapport

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}

Détails de la réponse

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 $^{\circ}$

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Angles And Intercepts On Parallel Lines

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Question 43 Rapport

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}

Détails de la réponse

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
$P = \frac{12 Q}{R}$

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Simple Operations On Algebraic Expressions

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Question 45 Rapport

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

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Question 47 Rapport

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

5

3

-3

-5

Détails de la réponse

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

Voir la réponse

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