JAMB UTME - Mathematics - 2009

Ajụjụ 1 Ripọtì

A binary operation $plus$ defined on the set of integers is such that m $plus$ n = n + mn for all integers m and n. Find the inverse of -5 under this operation, if the identity element is 0.

\frac{- 5}{4}

\frac{- 5}{6}

o

5

Binary Operations

Lee Azịza

Ajụjụ 2 Ripọtì

In how many ways can a delegation of 3 be chosen from 5 men and 3 women. If at least 1 man and 1 woman must be included?

15

28

30

45

Probability

Permutation And Combination

Lee Azịza

Ajụjụ 3 Ripọtì

Evaluate $\frac{81.81+99.44}{20.09+36.16}$ correct to 3 significant figures.

6.24

3.22

2.78

2.13

Akọwa Nkọwa

To evaluate the expression: $$\frac{81.81 + 99.44}{20.09 + 36.16}$$ We simply substitute the numbers in the formula and calculate: $$\frac{81.81 + 99.44}{20.09 + 36.16} = \frac{181.25}{56.25} \approx \boxed{3.23}$$ Therefore, the expression evaluates to approximately 3.23.

Lee Azịza

Ajụjụ 4 Ripọtì

The distance travelled by a particle from a fixed point is given as s = (t3 - t2 - t + 5)cm. Find the minimum distance that the particle can cover from the fixed point.

2.3 cm

4.0 cm

5.2 cm

6.0 cm

Akọwa Nkọwa

The distance travelled by the particle is given by the function s = t^3 - t^2 - t + 5. To find the minimum distance that the particle can cover from the fixed point, we need to find the minimum value of this function. Taking the derivative of s with respect to t, we get s' = 3t^2 - 2t - 1. Setting s' equal to zero and solving for t, we get t = (2 ± √10)/3. To determine whether this critical point is a minimum or maximum, we need to check the second derivative of s. Taking the derivative of s' with respect to t, we get s'' = 6t - 2. When t = (2 + √10)/3, we have s'' > 0, which means that this critical point corresponds to a minimum. Therefore, the minimum distance that the particle can cover from the fixed point is achieved when t = (2 + √10)/3. Substituting this value of t into the function s, we get s_min = s((2 + √10)/3) = (16 - 8√10 + 27√10 - 25)/27 = (-9 + 19√10)/27 ≈ 0.463 cm. Therefore, the correct option is not listed as it is approximately 0.463 cm.

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 5 Ripọtì

Substract 164189 from 186309

11219

21129

21139

22119

Akọwa Nkọwa

To subtract 164189 from 186309, we need to perform column-wise subtraction starting from the rightmost column. In the ones column, we have 9 - 9 which gives us 0. We borrow 1 from the tens column and write it above the tens column. In the tens column, we have 0 - 8 (borrowed from the hundreds column) which gives us 2. In the hundreds column, we have 1 - 1 which gives us 0. In the thousands column, we have 6 - 4 which gives us 2. In the ten thousands column, we have 8 - 1 which gives us 7. So, the result of subtracting 164189 from 186309 is 22120. Therefore, the correct option is 22119.

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Number Bases

Lee Azịza

Ajụjụ 6 Ripọtì

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

15o

30o

45o

60o

Akọwa Nkọwa

x

\leq

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

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

= 30o

Gụọ akwụkwọ nkuzi na Coordinate Geometry (JAMB)

Coordinate Geometry

Lee Azịza

Ajụjụ 7 Ripọtì

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}

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Lee Azịza

Ajụjụ 8 Ripọtì

A cliff on the bank of a river is 300 meter high. if the angle of depression of a point on the opposite side of the river is $60^\circ$, find the width of the river.

100m

75√3 m

100√3m

200√3m

Akọwa Nkọwa

We can solve this problem using trigonometry. Let's draw a diagram of the situation:

    A
   /|
  / |
 /  | h = 300 m
/   |
-----------
  x   B

Where point A is the top of the cliff, point B is the unknown point on the opposite side of the river, and x is the width of the river.

We know that the angle of depression from A to B is 60 degrees. This means that the angle of elevation from B to A is also 60 degrees.

Using trigonometry, we can set up the following equation:

tan(60) = h / x

where h is the height of the cliff and x is the width of the river. We can solve for x:

x = h / tan(60)
x = 300 / √3
x = 100√3 meters

Therefore, the width of the river is 100√3 meters. Answer is correct.

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 9 Ripọtì

A man bought a second-hand photocopying machine for N34,000. He serviced is at a cost od N2,000 and then sold it i profit of 15%. What was the selling price?

N37,550

N40,400

N41,400

N42,400

Akọwa Nkọwa

The man bought a photocopying machine for N34,000 and then serviced it at a cost of N2,000. Therefore, his total cost becomes N34,000 + N2,000 = N36,000. He sold it at a profit of 15%. Profit is calculated as a percentage of the cost price. So, his profit is 15/100 x N36,000 = N5,400. To find the selling price, we add the profit to the cost price: Selling price = Cost price + Profit Selling price = N36,000 + N5,400 Selling price = N41,400 Therefore, the selling price is N41,400.

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Modular Arithmetic

Areas

Financial Arithmetic

Lee Azịza

Ajụjụ 10 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 11 Ripọtì

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

47

^{\circ}

52

^{\circ}

56

^{\circ}

99

^{\circ}

Akọwa Nkọwa

TXY = XTY = 47

^{\circ}

(Basa angle of an isosceles

△

)

< TYZ = 94 $^{\circ}$

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

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

= 52 $^{\circ}$

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Angles

Triangles And Polygons

Lee Azịza

Ajụjụ 12 Ripọtì

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}

Akọwa Nkọwa

p (pass at 5mark) =

\frac{12}{20}

=

\frac{3}{5}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Probability

Surds (radicals)

Lee Azịza

Ajụjụ 13 Ripọtì

Find the value of x in the diagram.

50o

60o

70o

80o

Akọwa Nkọwa

30 + x = 100

x = 100 - 30

= 70o

Lee Azịza

Ajụjụ 14 Ripọtì

If s = (2 + 3t)(5t - 4), find s/dt when t = 4/5 secs

0 units per sec

15 units per sec

22 unit per sec

26 units per sec

Akọwa Nkọwa

x = (2+3t)(5t-4)
Let u = 2+3t ∴du/dt = 3
and v = 5t-4 ∴dv/dt = 5
dx/dt = Vdu/dt + Udv/dt
= (5t-4)3 + (2+3t)5
= 15t - 12 + 10 + 15t
= 30t - 2
= 30x4/5 - 2
= 24 - 2
= 22

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 15 Ripọtì

A regular polygon has $150^\circ$ as the size of each interior angle. How many sides does it have?

12

10

9

8

Akọwa Nkọwa

A regular polygon has equal interior angles and equal sides. The sum of the interior angles of a polygon with n sides is (n-2) times 180 degrees. So, if a regular polygon has interior angles of 150 degrees, we can use this formula to find the number of sides: (n-2) x 180 = sum of interior angles of the polygon (n-2) x 180 = 150n Simplifying the equation: n - 2 = 150/180 * n n - 2 = 5/6 * n n = 12 Therefore, the regular polygon has 12 sides. Answer: 12.

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Polynomials

Angles

Triangles And Polygons

Lee Azịza

Ajụjụ 16 Ripọtì

In how many ways can 9 people be seated if 3 chairs are available

720

504

336

210

Akọwa Nkọwa

If there are 9 people and 3 chairs, the first person can choose any of the 9 chairs. After the first person has chosen their seat, there are only 8 chairs left for the second person to choose from. Once the second person has chosen their seat, there are only 7 chairs left for the third person to choose from. Therefore, the total number of ways in which 3 people can be seated in 9 chairs is: 9 x 8 x 7 = 504 Therefore, the answer is 504, which is option (B).

Gụọ akwụkwọ nkuzi na Permutation And Combination (JAMB)

Permutation And Combination

Lee Azịza

Ajụjụ 17 Ripọtì

Determine the value of x for which (x2 - 1) > 0

x < -1 or x > 1

-1 < x < 1

x > 0

x < -1

Akọwa Nkọwa

To find the value of x for which (x^2 - 1) > 0, we need to factorize the expression. (x^2 - 1) can be written as (x - 1)(x + 1). So now we have: (x - 1)(x + 1) > 0 For the product of two factors to be greater than zero, either both factors must be positive or both factors must be negative. If (x - 1) and (x + 1) are both positive, then x > 1. If (x - 1) and (x + 1) are both negative, then x < -1. Therefore, the solution is: x < -1 or x > 1 which means option (A) is the correct answer.

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 18 Ripọtì

If m * n = n - (m+2) for any real number m and n find he value of 3*(-5)

-6

-8

-10

-12

Akọwa Nkọwa

m * n = n - (m+2)
= -5 - (3+2)
= -5-5
= -10

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Binary Operations

Inequalities

Lee Azịza

Ajụjụ 19 Ripọtì

Find the locus of a particle which moves in the first quadrant so that it is equidistant from the lines x = 0 and y = 0 (where k is a constant)

x + y = 0

x - y = 0

x + y + k = 0

x - y - k = 0

Akọwa Nkọwa

The locus of a particle refers to the path that the particle follows in the coordinate plane. In this case, we want to find the path that a particle follows if it always remains at the same distance from the lines x = 0 and y = 0. Since the particle is equidistant from the x-axis and y-axis, it must lie on the perpendicular bisector of the line segment joining the origin to any point on the x-axis or y-axis. Let's consider the point (a,0) on the x-axis. The distance between this point and the particle is given by the distance formula: √[ (a-x)^2 + (0-y)^2 ] Similarly, for the point (0,b) on the y-axis, the distance between this point and the particle is given by: √[ (0-x)^2 + (b-y)^2 ] Since the particle is equidistant from both lines, we can set these two distances equal to each other: √[ (a-x)^2 + (0-y)^2 ] = √[ (0-x)^2 + (b-y)^2 ] Squaring both sides of the equation, we get: (a-x)^2 + y^2 = x^2 + (b-y)^2 Simplifying the equation yields: 2xy - 2bx - 2ay + a^2 + b^2 = 0 This is the equation of a hyperbola with the x-axis and y-axis as its asymptotes. However, we need to take into account that the particle is in the first quadrant, so we only consider the part of the hyperbola that lies in the first quadrant. Thus, the correct answer is x - y = 0.

Gụọ akwụkwọ nkuzi na Loci (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Loci

Statistics

Lee Azịza

Ajụjụ 20 Ripọtì

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

Akọwa Nkọwa

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.

Lee Azịza

Ajụjụ 21 Ripọtì

Find to infinity, the sum of the sequence $1, \frac{9}{10}, \left(\frac{9}{10}\right)^2, \left(\frac{9}{10}\right)^3, \ldots$

10

9

10/9

9/10

Akọwa Nkọwa

a = 1, r = \frac{9}{10} S_{n} = \frac{a}{1 - r} S_{n} = \frac{1}{1 - \frac{9}{10}} = 1 \div \frac{1}{10} = 1 \times \frac{10}{1} 10

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 22 Ripọtì

A binary operation ⊗ defined on the set of integers is such that m⊗n = m + n + mn for all integers m and n. Find the inverse of -5 under this operation, if the identity element is 0

-5/4

-5/6

zero

5

Akọwa Nkọwa

To find the inverse of -5 under this operation, we need to find an integer x such that -5⊗x = 0, where 0 is the identity element. Using the given binary operation, we know that m⊗n = m + n + mn. Substituting -5 for m and x for n, we get: -5⊗x = -5 + x - 5x = -4x - 5 We need to solve -4x - 5 = 0 for x. Adding 5 to both sides gives: -4x = 5 Dividing both sides by -4 gives: x = -5/4 Therefore, the inverse of -5 under this operation is -5/4. So, the correct option is: -5/4

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Binary Operations

Lee Azịza

Ajụjụ 23 Ripọtì

Simplify $ \frac{5+\sqrt{7}}{3+\sqrt{7}} $

17-√7

4-√7

15+√7

7-√7

Akọwa Nkọwa

To simplify the expression $\frac{5+\sqrt{7}}{3+\sqrt{7}}$, we need to rationalize the denominator. To do this, we can multiply the numerator and denominator by the conjugate of the denominator, which is $3-\sqrt{7}$. This gives us: $$\frac{5+\sqrt{7}}{3+\sqrt{7}}\cdot\frac{3-\sqrt{7}}{3-\sqrt{7}}=\frac{(5+\sqrt{7})(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}=\frac{15-5\sqrt{7}+3\sqrt{7}-7}{9-7}$$ Simplifying the numerator and denominator gives: $$\frac{8-2\sqrt{7}}{2}=\boxed{4-\sqrt{7}}$$ Therefore, $\frac{5+\sqrt{7}}{3+\sqrt{7}}$ simplifies to $4-\sqrt{7}$.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 24 Ripọtì

Find the radius of a sphere whose surface area is 154 cm2

7.00 cm

3.50 cm

3.00 cm

1.75 cm

Akọwa Nkọwa

The surface area of a sphere is given by the formula: A = 4πr^2 where A is the surface area and r is the radius of the sphere. In this problem, we are given the surface area of the sphere, which is 154 cm^2. We can use this information to find the radius of the sphere. We start by rearranging the formula to solve for r: r = √(A/4π) Plugging in the value of A = 154 cm^2, we get: r = √(154/4π) ≈ 3.5 cm Therefore, the radius of the sphere is approximately 3.5 cm. So, the correct answer to this problem is option B: 3.50 cm.

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Volumes

Lee Azịza

Ajụjụ 25 Ripọtì

What is the locus or the mid-point of all the chords of length 6cm with circle of radius 5cm and with center 0?

A circle of radius 4 cm and the center 0

The perpendicular bisector of the chords

A straight line passing through 0

A circle of radius 6 cm and with center 0

Akọwa Nkọwa

x2 + 33 = 52
x2 + 9 = 25
x2 = 25 – 9
x2 = 16

x = √ 16
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Loci

Lee Azịza

Ajụjụ 26 Ripọtì

What is the mean deviation of x, 2x, x+1 and 3x. If their mean is 2?

0.5

1.0

1.5

2.0

Akọwa Nkọwa

The first step to finding the mean deviation is to calculate the mean of the given numbers. We are told that the mean of x, 2x, x+1, and 3x is 2. So, we can write an equation: (x + 2x + x+1 + 3x)/4 = 2 Simplifying this equation, we get: 7x/4 = 2 Multiplying both sides by 4/7, we get: x = 8/7 Now we can substitute this value of x back into the original numbers: x = 8/7 2x = 16/7 x+1 = 15/7 3x = 24/7 Next, we need to find the absolute deviation of each number from the mean: |x - 2| = |8/7 - 2| = 6/7 |2x - 2| = |16/7 - 2| = 2/7 |x+1 - 2| = |15/7 - 2| = 1/7 |3x - 2| = |24/7 - 2| = 10/7 The mean deviation is the average of these absolute deviations: (6/7 + 2/7 + 1/7 + 10/7)/4 = 19/28 Therefore, the mean deviation of the given numbers is 19/28. The answer closest to 19/28 among the options provided is 0.5. Therefore, the answer is (A) 0.5.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Representation Of Data

Measures Of Location

Financial Arithmetic

Lee Azịza

Ajụjụ 27 Ripọtì

If y = 3 cos 4x, dy/dx equals

6 sin 8x

-24 sin 4x

12 sin 4x

-12 sin 4x

Akọwa Nkọwa

Given, y = 3cos4x. To find dy/dx, we need to differentiate y with respect to x using the chain rule. dy/dx = -3 sin(4x) * d/dx(4x) = -3 sin(4x) * 4 = -12 sin(4x) Therefore, the correct option is "-12 sin 4x".

Lee Azịza

Ajụjụ 28 Ripọtì

The probability of a student passing any examination is $ \frac{2}{3} $. If the student takes three examinations, what is the probability that he will not pass any of them?

\frac{2}{3}

\frac{4}{9}

\frac{8}{27}

\frac{1}{27}

Akọwa Nkọwa

The probability of a student passing any examination is 2/3. Therefore, the probability of the student not passing any examination is 1 - 2/3 = 1/3. Since the student is taking three examinations, we can assume that each examination is independent of the others. Therefore, the probability of the student not passing any of the three examinations is (1/3) * (1/3) * (1/3) = 1/27. So, the answer is 1/27.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 29 Ripọtì

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}

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Linear Inequalities

Inequalities

Lee Azịza

Ajụjụ 30 Ripọtì

The histogram above represents the number of candidates that sat for Mathematics examination in a school. How many candidate scored more than 50 marks?

80

90

100

115

Akọwa Nkọwa

MarkNo.ofCandidate6040702580159010100595

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Euclidean Geometry

Lee Azịza

Ajụjụ 31 Ripọtì

A polynomial in x whose roots are 4/3 and -3/5 is

15x2 - 11x – 12

15x2 + 11x – 12

12x2 - x – 12

12x2 + 11x – 15

Akọwa Nkọwa

If 4/3 and -3/5 are roots of a polynomial
Imply x = 4/3 and - 3/5
3x = 4 and 5x = -3
∴3x-4 = 0 and 5x+3 = 0 are factors
(3x-4)(5x+3) = 0 product of the factors
15x2 + 9x – 20x – 12 = 0 By expansion
15x2 - 11x – 12 = 0

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Lee Azịza

Ajụjụ 32 Ripọtì

What value of s will make the function x(4-x) a maximum

4

3

2

1

Akọwa Nkọwa

To find the value of s that will make the function x(4-x) a maximum, we need to take the derivative of the function with respect to x and set it equal to zero. The derivative of x(4-x) with respect to x is: d/dx (x(4-x)) = 4 - 2x Setting this derivative equal to zero and solving for x gives: 4 - 2x = 0 2x = 4 x = 2 Therefore, the maximum value of the function x(4-x) occurs when x = 2. To find the value of s, we substitute x = 2 into the equation s = 4 - x: s = 4 - 2 = 2 So the answer is s = 2.

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Integration

Quadratic Equations

Lee Azịza

Ajụjụ 33 Ripọtì

5, 8, 6 and k occur with frequency 3, 2, 4, and 1 respectively an d have a mean of 5.7. Find the value of k

4

3

2

1

Akọwa Nkọwa

The mean of the given numbers is 5.7. Mean = (sum of all numbers) / (total number of numbers) Therefore, (5*3 + 8*2 + 6*4 + k*1) / (3+2+4+1) = 5.7 Simplifying the above equation, we get: (15 + 16 + 24 + k) / 10 = 5.7 55 + k = 57 k = 2 Therefore, the value of k is 2.

Lee Azịza

Ajụjụ 34 Ripọtì

Find the infinity, the sum of the sequence $1, \frac{9}{10}, \left(\frac{9}{10}\right)^2, \left(\frac{9}{10}\right)^3, ...$

10

9

\frac{10}{9}

\frac{9}{10}

Akọwa Nkọwa

To find the sum of the infinite series, we can use the formula for the sum of an infinite geometric series, which is: S = a/(1-r) where a is the first term of the series and r is the common ratio between consecutive terms. In this series, the first term is 1 and the common ratio is (910/9)/(910/10) = 10/9. Therefore, the sum of the series is: S = 1/(1 - 10/9) = 1/(1/9) = 9 Therefore, the correct option is (B) 9.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 35 Ripọtì

Evaluate $ \int \sec^2 \theta \, d\theta $

sec

θ

tan

θ

+ k

tan

θ

+ k

2 sec

θ

+ k

sec

θ

+ k

Akọwa Nkọwa

\int

sec2

θ

d

θ

= tan

θ

+ k

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Integration

Lee Azịza

Ajụjụ 36 Ripọtì

The cumulative frequency curve above shoes the distribution of the scores of 50 students in an examination. Find the 36th percentile scores

18%

28%

36%

50%

Akọwa Nkọwa

To find the 36th percentile score, we need to locate the corresponding point on the cumulative frequency curve. The 36th percentile score is the score below which 36% of the data falls. First, we need to find the total frequency, which is 50 in this case. Then we calculate 36% of the total frequency: 36/100 x 50 = 18 This means that we are looking for the score below which 18 students have scored. We can locate this point on the cumulative frequency curve by drawing a horizontal line from 18 on the y-axis to where it intersects the curve. Then we draw a vertical line down to the x-axis to read off the corresponding score. From the curve, we can see that the 36th percentile score lies between 30 and 40. To estimate the exact score, we can use linear interpolation. The distance between 30 and 40 is 10, and the distance between the cumulative frequency at 30 (which is 16) and the cumulative frequency at 40 (which is 22) is 6. The distance from 30 to the 36th percentile score is: 10 x (18 - 16)/6 = 1.67 So the 36th percentile score is approximately: 30 + 1.67 = 31.67 Therefore, the answer is 28%.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Probability

Statistics

Lee Azịza

Ajụjụ 37 Ripọtì

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

-5

-2

2

5

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Lee Azịza

Ajụjụ 38 Ripọtì

If $y = 3 \cos 4x$, $\frac{dy}{dx}$ equals

6 sin 8x

-24 sin 4x

12 sin 4x

-12 sin 4x

Akọwa Nkọwa

To differentiate y = 3 cos 4x, we will use the chain rule of differentiation. The chain rule states that if y = f(g(x)), then y' = f'(g(x))g'(x). Here, f(x) = 3 cos x, and g(x) = 4x. So, f'(x) = -3 sin x, and g'(x) = 4. Using the chain rule, we have: y' = f'(g(x))g'(x) = -3 sin (4x) * 4 = -12 sin (4x) Therefore, the derivative of y = 3 cos 4x is y' = -12 sin (4x).

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Modular Arithmetic

Lee Azịza

Ajụjụ 39 Ripọtì

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}

$\frac{2}{3}$

$\frac{4}{5}$

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Fractions, Decimals And Approximations

Number Bases

Surds (radicals)

Lee Azịza

Ajụjụ 40 Ripọtì

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

\frac{1}{3}

\frac{1}{4}

-

\frac{1}{5}

-

\frac{1}{3}

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 41 Ripọtì

The sum of the first n terms of the arithmetic progression 5, 11, 17, 23, 29, 35, ... is

n(3n - 0.5)

n(3n + 2)

n(3n + 2.5)

n(3n + 5)

Akọwa Nkọwa

To find the sum of the first n terms of an arithmetic progression, we use the formula: Sn = n/2[2a + (n-1)d] where Sn is the sum of the first n terms, a is the first term, and d is the common difference between the terms. In this case, the first term is 5 and the common difference is 6 (since each term is 6 more than the previous one). So, using the formula: Sn = n/2[2(5) + (n-1)(6)] Simplifying, we get: Sn = n/2[10 + 6n - 6] Sn = n/2[6n + 4] Sn = 3n^2 + 2n Therefore, the correct answer is n(3n + 2).

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 42 Ripọtì

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

-5

-2

2

5

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Angles Of Elevation And Depression (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Angles Of Elevation And Depression

Surds (radicals)

Lee Azịza

Ajụjụ 43 Ripọtì

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

Akọwa Nkọwa

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.

Lee Azịza

Ajụjụ 44 Ripọtì

A chord drawn 5 cm away from the center of a circle of radius 13 cm. Calculate the length of the chord

7cm

9cm

12cm

24cm

Akọwa Nkọwa

x2 + 52 = 132
x2 + 25 = 169
x2 = 144

x = √ 144
= 12
Length of the chord AB = 2x
2*12= 24cm

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Lee Azịza

Ajụjụ 45 Ripọtì

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

-6

-8

-10

-12

Akọwa Nkọwa

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

*

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

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Solution Of Linear Equations

Differentiation

Quadratic Equations

Lee Azịza

Ajụjụ 46 Ripọtì

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

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

210

40

120

160

Akọwa Nkọwa

20 + X + 50 + 40 + 2X + 60 = 260
3X + 170 = 260
3X = 260 - 170
3x = 90
x = 30
Absent for at least 4 days

4	5	6
40	2x	60

i.e 40 + 2x + 60 = 40 + (2 x 30) + 60
= 40 + 60 + 60
= 160

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Lee Azịza

Ajụjụ 47 Ripọtì

Simplify $4\frac{3}{4} - 6\frac{1}{4}$

-7

\frac{7}{8}

\frac{- 2}{7}

\frac{- 10}{21}

\frac{10}{21}

Akọwa Nkọwa

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}

Lee Azịza

Ajụjụ 48 Ripọtì

I.S∩T∩W=S II. S∪T∪W=S

III. T∩W=S

If S⊂T⊂W, which of the above statements are true?

I and II

I and III

II and III

I, II and III

Akọwa Nkọwa

S∩T∩W = S(This means that all the elements of s are in T and also in W.)
S∪T∪W = W (imply W is a universal set for S and T) T∩W = S imply all the elements of S are in T and W)
S ⊂ T ⊂ W
∴ I, II, and III

Lee Azịza

Ajụjụ 49 Ripọtì

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

Akọwa Nkọwa

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