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JAMB UTME - Mathematics - 2014

Vraag 1 Verslag

Find the standard deviation of 5, 4, 3, 2, 1

√2

\sqrt{3}

√6

√10

Antwoorddetails

To find the standard deviation of the numbers 5, 4, 3, 2, and 1, we need to follow these simple steps: Step 1: Calculate the mean (average) of the given numbers. - Add the numbers together: 5 + 4 + 3 + 2 + 1 = 15. - Divide the sum by the total number of values: 15 ÷ 5 = 3. Therefore, the mean of the numbers is 3. Step 2: Calculate the variance of the given numbers. - Subtract the mean from each number: 5 - 3 = 2, 4 - 3 = 1, 3 - 3 = 0, 2 - 3 = -1, 1 - 3 = -2. - Square each of the differences: 2^2 = 4, 1^2 = 1, 0^2 = 0, (-1)^2 = 1, (-2)^2 = 4. - Add up the squared differences: 4 + 1 + 0 + 1 + 4 = 10. - Divide the sum by the total number of values: 10 ÷ 5 = 2. Therefore, the variance of the numbers is 2. Step 3: Calculate the standard deviation of the given numbers. - Take the square root of the variance: √2 = 1.41421356. Therefore, the standard deviation of the numbers 5, 4, 3, 2, and 1 is approximately 1.41421356.

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Measures Of Dispersion

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Vraag 2 Verslag

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}

Antwoorddetails

Q = (P

\cup

Q) - P
{6,7}

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Vraag 3 Verslag

Evaluate Log28 + Log216 - Log24

3

4

5

6

Antwoorddetails

= 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

Lees lesnotitie op Logarithms (WAEC)

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Vraag 4 Verslag

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

cos 2x + k

\frac{1}{2}

cos 2x + k

- \frac{1}{2}

cos 2x + k

-cos 2x + k

Antwoorddetails

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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Vraag 9 Verslag

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

5

3

-3

-5

Antwoorddetails

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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Vraag 10 Verslag

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}

Antwoorddetails

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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Vraag 12 Verslag

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

Antwoorddetails

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

\frac{δ y}{δ x}

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

Lees lesnotitie op Algebraic Expressions (WAEC)

Algebraic Expressions

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Vraag 14 Verslag

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

Antwoorddetails

(\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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Lees lesnotitie op Quadratic Equations (WAEC)

Solution Of Linear Equations

Modular Arithmetic

Application Of Differentiation

Quadratic Equations

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Vraag 15 Verslag

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

Antwoorddetails

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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Vraag 17 Verslag

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)

Antwoorddetails

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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Lees lesnotitie op Linear Inequalities (WAEC)

Lees lesnotitie op Sets (WAEC)

Probability

Linear Inequalities

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Vraag 18 Verslag

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}

Antwoorddetails

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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Vraag 22 Verslag

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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Vraag 25 Verslag

Find the value of

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

12

10

-1

-2

Antwoorddetails

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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Vraag 29 Verslag

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})

Antwoorddetails

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}]

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Vraag 31 Verslag

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

Antwoorddetails

= \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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Vraag 32 Verslag

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

t

-t

2

-2

Antwoorddetails

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

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Vraag 34 Verslag

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!}

Antwoorddetails

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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Vraag 35 Verslag

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}

Antwoorddetails

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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Vraag 38 Verslag

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

\sqrt{2}

√5

3

5

Antwoorddetails

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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Vraag 39 Verslag

If ∣∣∣−x12−14∣∣∣=−12, find x

-6

-2

3

6

Antwoorddetails

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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Vraag 41 Verslag

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}

Antwoorddetails

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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Vraag 42 Verslag

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

Antwoorddetails

\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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Vraag 44 Verslag

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

Antwoorddetails

\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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Vraag 45 Verslag

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}

Antwoorddetails

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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Vraag 46 Verslag

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

Antwoorddetails

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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