JAMB UTME - Mathematics - 2014

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}

Bayanin Amsa

Q = (P

\cup

Q) - P
{6,7}

Karanta bayanin darasi akan Sets (JAMB)

Sets

Duba Amsa

Tambaya 2 Rahoto

Solve for x in 8x-2 = 2/25

4

6

8

10

Bayanin Amsa

8x-2 = 2/25
x-2 = 2/25 x 1/8
x-2 = 2/200
x-2 = 1/100
1/x2 = 1/100
x2 = 100
x = 10

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 3 Rahoto

Find the mid point of S(-5, 4) and T(-3, -2)

-4, 2

4, -2

-4, 1

4, -1

Bayanin Amsa

To find the midpoint of the line segment between two points, we need to average the x-coordinates and the y-coordinates of the two points separately. So, to find the midpoint of S(-5, 4) and T(-3, -2), we take the average of their x-coordinates and the average of their y-coordinates: Midpoint x-coordinate = (S x-coordinate + T x-coordinate) / 2 = (-5 + (-3)) / 2 = -4 Midpoint y-coordinate = (S y-coordinate + T y-coordinate) / 2 = (4 + (-2)) / 2 = 1 Therefore, the midpoint of S(-5, 4) and T(-3, -2) is (-4, 1).

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (JAMB)

Coordinate Geometry

Duba Amsa

Tambaya 4 Rahoto

Express the product of 0.00043 and 2000 in standard form.

8.6 x 10-3

8.3 x 10-2

8.6 x 10-1

8.6 x 10

Bayanin Amsa

0.00043 x 2000
= 43 x 10-5 x 2 x 103
= 43 x 2 x 10-5+3
= 86 x 10-2
= 8.6 x 101 x 10-2
= 8.6 x 10-1

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 5 Rahoto

A binary operation * is defined by x * y = xy. If x * 2 = 12 - x, find the possible values of x

3,4

3,-4

-3,4

-3,-4

Bayanin Amsa

x * y = xy
x * 2 = 12 - x
Thus by comparison,
x = x, y = 2
But x * y = x * 2
xy = 12 - x
x2 = 12 - x
x2 + x - 12 = 0
x2 + 4x - 3x - 12 = 0
x(x + 4) - 3(x + 4) = 0
(x - 3)(x + 4) = 0
x - 3 = 0 or x + 4 = 0
So x = 3 or x = -4

Karanta bayanin darasi akan Binary Operations (JAMB)

Binary Operations

Duba Amsa

Tambaya 6 Rahoto

What is the solution of x-5/x+3<-1?

-3 < x < 1

x < -3 or x > 1

-3 < x < 5

x < -3 or x > 5

Bayanin Amsa

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Linear Inequalities

Duba Amsa

Tambaya 7 Rahoto

If cos(x + 40)o = 0.0872, what is the value of x?

85o

75o

65o

45o

Bayanin Amsa

We are given that cos(x + 40) = 0.0872. To find the value of x, we need to use the inverse cosine function, also known as arccosine or cos^-1. Taking the inverse cosine of both sides, we get: arccos(cos(x + 40)) = arccos(0.0872) The inverse cosine and cosine functions are inverses of each other, so they "cancel out" on the left-hand side, leaving us with: x + 40 = arccos(0.0872) Using a calculator or a table of trigonometric values, we can find that arccos(0.0872) is approximately 84.74 degrees. Subtracting 40 from both sides, we get: x = 84.74 - 40 x = 44.74 So the value of x is approximately 44.74 degrees. None of the given options is an exact match, but the closest one is 45 degrees.

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Introductory Calculus

Trigonometry

Duba Amsa

Tambaya 8 Rahoto

If log7.5 = 0.8751, evaluate 2 log75 + log750

6.6252

6.6253

66.252

66.253

Bayanin Amsa

If log 7.5 = 0.8751
Then 2log75 + log750
= 2(1.8751) + 2.8751
= 3.7502 + 2.8751
= 6.6253

Karanta bayanin darasi akan Indices, Logarithms And Surds (JAMB)

Karanta bayanin darasi akan Logarithms (WAEC)

Indices, Logarithms And Surds

Logarithms

Duba Amsa

Tambaya 9 Rahoto

If ∣∣∣−x12−14∣∣∣=−12, find x

-6

-2

3

6

Bayanin Amsa

Explanation:

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

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

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 10 Rahoto

ind the value of k if y - 1 is a factor of y3 + 4y2 + ky - 6

-6

-4

O

1

Bayanin Amsa

if y - 1 is a factor of y3 + 4y2 + ky - 6, then
f(1) = (1)3 + 4(1)2 + k(1) - 6 = 0 (factor theorem)
1 + 4 + k - 6 = 0
5 - 6 + k = 0
-1 + k = 0
k = 1

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 11 Rahoto

If y = cos 3x, find $ \frac{\delta y}{\delta x} $

\frac{1}{3} \sin 3 x

- \frac{1}{3} \sin 3 x

3 sin 3x

-3 sin 3x

Bayanin Amsa

To find the derivative of y = cos 3x, we need to use the chain rule of differentiation. The chain rule states that if y = f(g(x)), then the derivative of y with respect to x is given by the product of the derivative of f with respect to g multiplied by the derivative of g with respect to x. In other words, δy/δx = δf/δg * δg/δx. Using the chain rule, we have: δy/δx = δ(cos 3x)/δ(3x) * δ(3x)/δx The derivative of cos 3x with respect to 3x can be found using the chain rule again: δ(cos 3x)/δ(3x) = -sin(3x) The derivative of 3x with respect to x is simply 3. Substituting these values in the original equation, we get: δy/δx = -sin(3x) * 3 Simplifying, we have: δy/δx = -3 sin(3x) Therefore, the correct option is -3 sin 3x. In summary, the derivative of y = cos 3x is -3 sin 3x, which is obtained using the chain rule of differentiation.

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Duba Amsa

Tambaya 12 Rahoto

Find the minimum value of y = x2 - 2x - 3

4

1

-1

-4

Bayanin Amsa

To find the minimum value of the function y = x^2 - 2x - 3, we can start by completing the square. First, let's add and subtract the value (-2/2)^2 = 1 to the expression inside the parentheses: y = x^2 - 2x + 1 - 1 - 3 Next, we can group the first three terms and write them as a perfect square: y = (x - 1)^2 - 4 Now we can see that the minimum value of the function occurs when (x - 1)^2 is zero, which happens when x = 1. Therefore, the minimum value of the function is y = -4, which occurs when x = 1. So the answer is -4, and we can explain it by completing the square to find the vertex of the parabolic function. The vertex of the parabola y = x^2 - 2x - 3 is (1, -4), and the minimum value of the function occurs at this point.

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 13 Rahoto

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

Duba Amsa

Tambaya 14 Rahoto

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

√2

\sqrt{3}

√6

√10

Bayanin Amsa

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.

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Measures Of Dispersion

Duba Amsa

Tambaya 15 Rahoto

How many sides has a regular polygon whose interior angle is 135o

12

10

9

8

Bayanin Amsa

The formula to find the interior angle of a regular polygon is: Interior angle = (n - 2) × 180° / n Where "n" is the number of sides of the polygon. We are given that the interior angle of the regular polygon is 135°, so we can substitute this value into the formula and solve for "n": 135 = (n - 2) × 180° / n Multiplying both sides by "n": 135n = (n - 2) × 180° Distributing on the right-hand side: 135n = 180n - 360° Subtracting 135n from both sides: 0 = 45n - 360° Adding 360° to both sides: 360° = 45n Dividing both sides by 45: 8 = n So the regular polygon has 8 sides. Looking at the given answer options, we see that the answer is (D) 8.

Duba Amsa

Tambaya 16 Rahoto

Values	0	1	2	3	4
Frequency	1	2	2	1	9

Find the mode of the distribution above

1

2

3

4

Bayanin Amsa

To find the mode of the distribution, we look for the value that appears most frequently in the dataset. From the given frequency table, we can see that the value "4" appears 9 times, which is more than any other value. Therefore, the mode of this distribution is "4". In other words, the mode is the value that occurs most frequently in the data set. It is a measure of central tendency that can be useful in describing a dataset.

Karanta bayanin darasi akan Variation (JAMB)

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Variation

Probability

Linear Inequalities

Modular Arithmetic

Sets

Duba Amsa

Tambaya 17 Rahoto

The pie chart above shows the monthly distribution of a man's salary on food items. If he spent ₦8,000 on rice, how much did he spent on yam?

₦42,000

₦18,000

₦16,000

₦12,000

Bayanin Amsa

The man's salary was divided into four food items: Rice, Yam, Beans, and Others. The chart shows that rice takes up 20% of his salary, and yam takes up 40% of his salary. Since rice takes up 20% of his salary, and he spent ₦8,000 on it, we can calculate the total salary of the man by dividing his spend on rice by 20%. ₦8,000 / 20% = ₦40,000 Since the total salary is ₦40,000 and yam takes up 40% of the salary, we can calculate how much he spent on yam by multiplying the total salary by 40%. ₦40,000 * 40% = ₦16,000 Therefore, the man spent ₦16,000 on yam.

Duba Amsa

Tambaya 18 Rahoto

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

Bayanin Amsa

\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

Karanta bayanin darasi akan Inequalities (JAMB)

Inequalities

Duba Amsa

Tambaya 19 Rahoto

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

\sqrt{2}

√5

3

5

Bayanin Amsa

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}

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Karanta bayanin darasi akan Sequence And Series (WAEC)

Surds (radicals)

Sequence And Series

Duba Amsa

Tambaya 20 Rahoto

y varies directly as w2. When y = 8, w = 2. Find y when w = 3

18

12

9

6

Bayanin Amsa

The given statement "y varies directly as w^2" can be written as an equation: y = k w^2 where k is a constant of proportionality. We are also given that when y = 8, w = 2. We can use this information to solve for k: 8 = k (2^2) 8 = 4k k = 2 Now that we know the value of k, we can use the equation to find y when w = 3: y = 2 (3^2) y = 18 Therefore, when w = 3, y is equal to 18. In other words, the problem is asking us to find the value of y when the value of w is changed from 2 to 3, given that y varies directly with w^2. We can use the equation y = k w^2 and the given information to solve for the constant of proportionality k. Once we have found k, we can use the equation to find y when w = 3.

Karanta bayanin darasi akan Variation (WAEC)

Variation

Duba Amsa

Tambaya 21 Rahoto

Find the value of

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

12

10

-1

-2

Bayanin Amsa

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

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 22 Rahoto

A woman bought a grinder for ₦60,000. She sold it at a loss of 15%. How much did she sell it?

₦53,000

₦52,000

₦51,000

₦50,000

Bayanin Amsa

The woman bought the grinder for ₦60,000 and sold it at a loss of 15%. This means that she sold it for 100% - 15% = 85% of its original price. To find out how much she sold it for, we can calculate 85% of ₦60,000: 85% of ₦60,000 = 0.85 x ₦60,000 = ₦51,000 Therefore, the woman sold the grinder for ₦51,000. The answer is option C.

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Financial Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Financial Arithmetic

Duba Amsa

Tambaya 23 Rahoto

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}

Bayanin Amsa

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 $^{\circ}$

Karanta bayanin darasi akan Angles And Intercepts On Parallel Lines (WAEC)

Angles And Intercepts On Parallel Lines

Duba Amsa

Tambaya 24 Rahoto

Find the median of 5,9,1,10,3,8,9,2,4,5,5,5,7,3 and 6

6

5

4

3

Bayanin Amsa

To find the median of a set of numbers, we need to arrange the numbers in order from smallest to largest and then find the middle number. Arranging the given numbers in order from smallest to largest, we get: 1, 2, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 9, 9, 10 There are 15 numbers in this set, so the median will be the average of the two middle numbers: the 7th and 8th numbers. The 7th number is 5, and the 8th number is also 5, so the median of this set of numbers is: (median) = (5 + 5) / 2 = 10 / 2 = 5 So the median of the given set of numbers is 5. Looking at the given answer options, we see that the answer is (B) 5.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Probability (WAEC)

Representation Of Data

Probability

Duba Amsa

Tambaya 25 Rahoto

Evaluate Log28 + Log216 - Log24

3

4

5

6

Bayanin Amsa

= 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

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 26 Rahoto

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

Bayanin Amsa

\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

Karanta bayanin darasi akan Integration (JAMB)

Integration

Duba Amsa

Tambaya 27 Rahoto

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

5

3

-3

-5

Bayanin Amsa

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

Duba Amsa

Tambaya 28 Rahoto

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})

Bayanin Amsa

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}]

Duba Amsa

Tambaya 29 Rahoto

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

Bayanin Amsa

= \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

Karanta bayanin darasi akan Functions And Relations (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Functions And Relations

Surds (radicals)

Duba Amsa

Tambaya 30 Rahoto

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}

Bayanin Amsa

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

Karanta bayanin darasi akan Areas (WAEC)

Karanta bayanin darasi akan Simple Operations On Algebraic Expressions (WAEC)

Areas

Simple Operations On Algebraic Expressions

Duba Amsa

Tambaya 31 Rahoto

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}

Bayanin Amsa

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}

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Probability (WAEC)

Representation Of Data

Probability

Duba Amsa

Tambaya 32 Rahoto

Find the equation of the straight line through (-2, 3) and perpendicular to 4x + 3y - 5 = 0

3x - 4y + 18 = 0

3x + 2y - 18 = 0

4x + 5y + 3 = 0

5x - 2y - 11 = 0

Bayanin Amsa

4x + 3y - 5 = 0 (given)
The equation of the line perpendicular to the given line takes the form 3x - 4y = k
Thus, substitution x = -2 and y = 3 in 3x - 4y = k gives;
3(-2) - 4(3) = k
-6 - 12 = k
k = -18
Hence the required equation is 3x - 4y = -18
3x - 4y + 18 = 0

Duba Amsa

Tambaya 33 Rahoto

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}

Bayanin Amsa

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}

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Probability

Surds (radicals)

Duba Amsa

Tambaya 34 Rahoto

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!}

Bayanin Amsa

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

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Permutation And Combination

Duba Amsa

Tambaya 35 Rahoto

Factorize 2y2 - 15xy + 18x2

(2y - 3x) (y + 6x)

(2y - 3x) (y - 6x)

(2y + 3x) (y - 6x)

(3y + 2x) (y - 6x)

Bayanin Amsa

The expression 2y^2 - 15xy + 18x^2 can be factored as (2y - 3x)(y - 6x). To factor this expression, we look for two binomials that multiply to the given expression and have a common factor. In this case, (2y - 3x) and (y - 6x) are two binomials that multiply to 2y^2 - 15xy + 18x^2 and have a common factor of y - 6x. So, the factorization of 2y^2 - 15xy + 18x^2 is (2y - 3x)(y - 6x).

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 36 Rahoto

Find the value of 1101112 + 101002

11010112

10010012

10010112

10011112

Number Bases

Duba Amsa

Tambaya 37 Rahoto

In the figure above, what is the equation of the line that passes the y-axis at (0,5) and passes the x-axis at (5,0)?

y = x + 5

y = -x + 5

y = x - 5

y = -x - 5

Bayanin Amsa

The equation of the line is given by y = x + 5. To understand why, let's consider the two points the line passes through: (0,5) and (5,0). The first point (0,5) means that when x = 0, y = 5. The second point (5,0) means that when y = 0, x = 5. Using these two points, we can write an equation for the line that passes through them. The slope of the line is the difference in y values divided by the difference in x values, or (5 - 0) / (0 - 5) = -1. So the equation of the line is y = -x + b, where b is the y-intercept, or the point where the line crosses the y-axis. To find b, we use the first point (0,5), plug in x = 0, and solve for b: 5 = -0 + b, so b = 5. Putting it all together, the equation of the line is y = -x + 5.

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Karanta bayanin darasi akan Graphs Of Linear And Quadratic Functions (WAEC)

Coordinate Geometry

Graphs Of Linear And Quadratic Functions

Duba Amsa

Tambaya 38 Rahoto

Evaluate $ \int \sin 2x \, dx $

cos 2x + k

\frac{1}{2}

cos 2x + k

- \frac{1}{2}

cos 2x + k

-cos 2x + k

Bayanin Amsa

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.

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Trigonometry (JAMB)

Integration

Trigonometry

Duba Amsa

Tambaya 39 Rahoto

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

t

-t

2

-2

Bayanin Amsa

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

Duba Amsa

Tambaya 40 Rahoto

A man donates 10% of his monthly net earnings to his church. If it amounts to ₦4,500, what is his net monthly income?

₦40,500

₦45,000

₦52,500

₦62,000

Bayanin Amsa

We can begin by setting up an equation to represent the given situation. Let x be the man's net monthly income. Then, we know that he donates 10% of his net monthly income to his church, which amounts to ₦4,500. Mathematically, we can express this as: 10% of x = ₦4,500 To solve for x, we need to isolate the variable on one side of the equation. We can do this by dividing both sides of the equation by 10%, which is equivalent to multiplying both sides by 10/100 or 0.1: 10% of x ÷ 10% = ₦4,500 ÷ 10% x = ₦4,500 ÷ 0.1 x = ₦45,000 Therefore, the man's net monthly income is ₦45,000. Answer: ₦45,000

Duba Amsa

Tambaya 41 Rahoto

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

Bayanin Amsa

(\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

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Application Of Differentiation (JAMB)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Solution Of Linear Equations

Modular Arithmetic

Application Of Differentiation

Quadratic Equations

Duba Amsa

Tambaya 42 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Volumes (WAEC)

Volumes

Duba Amsa

Tambaya 43 Rahoto

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

Bayanin Amsa

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

\frac{δ y}{δ x}

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 44 Rahoto

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)

Bayanin Amsa

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.

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Karanta bayanin darasi akan Sets (WAEC)

Probability

Linear Inequalities

Sets

Duba Amsa

Tambaya 45 Rahoto

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}

Bayanin Amsa

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

Karanta bayanin darasi akan Change Of Subject Of A Formula/Relation (WAEC)

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Duba Amsa

Tambaya 46 Rahoto

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

Bayanin Amsa

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."

Karanta bayanin darasi akan Loci (JAMB)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Loci

Triangles And Polygons

Duba Amsa

Tambaya 47 Rahoto

The 4th term of an A.P is 13 while the 10th term is 31. Find the 21st term

175

85

64

45

Bayanin Amsa

Let's begin by recalling the formula for the nth term of an arithmetic progression (A.P): a_n = a_1 + (n - 1)d where a_n is the nth term of the A.P, a_1 is the first term, n is the number of the term, and d is the common difference between consecutive terms. We are given that the 4th term of the A.P is 13, so we can substitute these values into the formula to get: a_4 = a_1 + (4 - 1)d = 13 Simplifying this equation, we get: a_1 + 3d = 13 ---(1) We are also given that the 10th term of the A.P is 31, so we can use the formula again to get: a_10 = a_1 + (10 - 1)d = 31 Simplifying this equation, we get: a_1 + 9d = 31 ---(2) Now we need to solve for a_1 and d. We can do this by subtracting equation (1) from equation (2) to eliminate a_1: 6d = 18 d = 3 Substituting this value of d into equation (1), we get: a_1 + 3(3) = 13 a_1 = 4 So, the first term of the A.P is 4 and the common difference is 3. Now we can use the formula again to find the 21st term of the A.P: a_21 = a_1 + (21 - 1)d Substituting the values we found earlier, we get: a_21 = 4 + (20)(3) = 64 Therefore, the 21st term of the A.P is 64, and the correct answer is option (C).

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 48 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Circles

Surds (radicals)

Duba Amsa






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