JAMB UTME - Mathematics - 2003

Ajụjụ 1 Ripọtì

x varies directly as the product of u and v and inversely as their sum. If x = 3 when u = 3 and v = 1, what is the value of x if u = 3 and v = 3?

3

4

6

9

Akọwa Nkọwa

x \infty \frac{U V}{U + V} x = \frac{K U V}{U + V} K = \frac{x (U + V)}{U V} \frac{3 (3 + 1)}{3 \times 1} \frac{3 \times 4}{3} K = 4 x = \frac{4 U V}{U + V} x = \frac{4 \times 3 \times 3}{3 + 3} x = \frac{36}{6} x = 6

Gụọ akwụkwọ nkuzi na Variation (WAEC)

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Variation

Lee Azịza

Ajụjụ 2 Ripọtì

A cinema halll contains a certain number of people. If 25/2% are children, 95/2% are men and 84 are women, find the number of men in the hall.

63

84

113

133

Akọwa Nkọwa

Let's use algebra to solve this problem. Let the total number of people in the hall be x. Then we can write two equations based on the given information: - (25/2)% of x are children: (25/2)% * x = 0.25x - (95/2)% of x are men: (95/2)% * x = 0.95x We know that 84 people in the hall are women. So the total number of men and children in the hall is: x - 84 We can set up an equation using this expression and the two equations we found above: 0.25x + 0.95x = x - 84 Simplifying this equation, we get: 1.2x = x - 84 0.2x = -84 x = -420 This answer doesn't make sense because we can't have a negative number of people in the hall. We made an error somewhere. Let's check our work. We can see that our equation for the total number of people in the hall doesn't take into account the 84 women. We need to add 84 to our equation: x = 0.25x + 0.95x + 84 Simplifying this equation, we get: x = 1.2x + 84 0.2x = 84 x = 420 This is the correct total number of people in the hall. Now we can find the number of men: 0.95x = 0.95 * 420 = 399 So there are 399 men in the hall. Therefore, the answer is option (D) 133.

Lee Azịza

Ajụjụ 3 Ripọtì

Triangle OPQ is the solution of the inequalities

x - 1 < 0, y +

\geq

0, y - x

\geq

0

x + 1

\leq

0, y + x

\geq

0, y - x

\leq

0

x - 1

\geq

0, y - x

\geq

0, y + x

\geq

0

y + x

\geq

0, y - x

\leq

0

Linear Inequalities

Inequalities

Lee Azịza

Ajụjụ 4 Ripọtì

Evaluate $ \int_{3}^{2}(x^2-2x)\,dx $

4

2

4/3

1/3

Akọwa Nkọwa

To evaluate the integral ?23(x^2 - 2x)dx, we need to find the antiderivative of the integrand and then apply the definite integral between the limits of integration, 2 and 3. First, we find the antiderivative of x^2 - 2x: ∫(x^2 - 2x)dx = (1/3)x^3 - x^2 + C where C is the constant of integration. Next, we apply the limits of integration, 2 and 3: ∫2^3(x^2 - 2x)dx = [(1/3)3^3 - 3^2] - [(1/3)2^3 - 2^2] = (27/3 - 9) - (8/3 - 4) = 9 - (8/3) = (27/3) - (8/3) = 19/3 Therefore, the value of the integral ?23(x^2 - 2x)dx is 19/3. So the answer is 4/3.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Representation Of Data

Integration

Lee Azịza

Ajụjụ 5 Ripọtì

Determine the maximum value of y = 3x2 - x3

zero

2

4

6

Akọwa Nkọwa

To find the maximum value of y = 3x^2 - x^3, we can take the derivative of y with respect to x, set it equal to zero, and solve for x. First, we find the derivative of y: dy/dx = 6x - 3x^2 Setting this equal to zero gives: 6x - 3x^2 = 0 Factor out x: x(6 - 3x) = 0 Solving for x, we get two solutions: x = 0 or x = 2 To determine which value of x gives the maximum value of y, we can use the second derivative test. Taking the second derivative of y: d^2y/dx^2 = 6 - 6x Plugging in x = 0 and x = 2: d^2y/dx^2 |x=0 = 6 > 0 d^2y/dx^2 |x=2 = -6 < 0 Since the second derivative is positive at x = 0 and negative at x = 2, this means that x = 0 corresponds to a minimum value of y, while x = 2 corresponds to a maximum value of y. Therefore, the maximum value of y = 3x^2 - x^3 occurs when x = 2, and is equal to: y = 3(2)^2 - (2)^3 = 4. So the answer is 4.

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 6 Ripọtì

The length a person can jump is inversely proportional to his weight. If a 20 kg person can jump 1.5 m, find the constant of proportionality

60

30

20

15

Akọwa Nkọwa

Inverse proportionality means that as one quantity increases, the other decreases in a proportional manner. In this case, the length a person can jump is inversely proportional to his weight. Let's assume that the constant of proportionality is represented by 'k'. We can use the following formula to solve for 'k': Jump length ∝ 1/Weight Jump length = k/Weight We know that a 20 kg person can jump 1.5 m. So we can substitute these values into the formula and solve for 'k': 1.5 = k/20 k = 30 Therefore, the constant of proportionality is 30. This means that if we multiply the weight of a person by 30, we will get the length of the jump they can make. For example, a person who weighs 60 kg will be able to jump half the distance of a person who weighs 30 kg.

Gụọ akwụkwọ nkuzi na Logical Reasoning (WAEC)

Gụọ akwụkwọ nkuzi na Ratio, Proportions And Rates (WAEC)

Logical Reasoning

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 7 Ripọtì

Factorize completely 4abx - 2axy -12b2x + 6bxy

2x(a - 3b)(2b - y)

2x(3b - a)(2b - y)

2x(a - 3b)(y - 2b)

2x(2b - a)(3b - y)

Akọwa Nkọwa

To factorize the given expression completely, we need to look for the common factors among the terms. First, we can take out 2 as a common factor from all the terms. So the expression becomes: 2(2abx - axy - 6b^2x + 3bxy) Next, we can take out ax as a common factor from the first two terms and 3bx as a common factor from the last two terms. So we get: 2ax(2b - y) - 3bx(2b - y) Now, we can see that (2b - y) is a common factor in both terms, so we can take it out. This gives us: (2b - y)(2ax - 3bx) So, the expression is completely factorized as (2b - y)(2ax - 3bx). Thus, the correct option is: 2x(a - 3b)(2b - y)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 8 Ripọtì

Simplify 1 - (7/3 x 5/4) + 2/5

-16/15

-79/60

-37/15

-151/60

Akọwa Nkọwa

To simplify the expression 1 - (7/3 x 5/4) + 2/5, we first need to evaluate the product of the fractions inside the parentheses: 7/3 x 5/4 = (7 x 5)/(3 x 4) = 35/12 Substituting this value back into the original expression, we get: 1 - 35/12 + 2/5 To combine the fractions, we need to find a common denominator. The least common multiple of 12 and 5 is 60, so we can convert the fractions as follows: 1 = 60/60 35/12 = (35 x 5)/(12 x 5) = 175/60 2/5 = 24/60 Substituting these values back into the expression, we get: 60/60 - 175/60 + 24/60 = -91/60 Therefore, the simplified form of the expression is -91/60. Therefore, the correct answer is: -79/60.

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

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Fractions, Decimals And Approximations

Number Bases

Lee Azịza

Ajụjụ 9 Ripọtì

PQRS is a cyclic quadrilateral. If ∠QPS = 75o, what is the size of ∠QRS?

180o

150o

105o

75o

Akọwa Nkọwa

Since PQRS is a cyclic quadrilateral, the opposite angles of the quadrilateral add up to 180 degrees. That is, ∠PQS + ∠QRS = 180° and ∠QPS + ∠RQS = 180° We are given that ∠QPS = 75°, so we can use the above equation to find ∠RQS. ∠QPS + ∠RQS = 180° 75° + ∠RQS = 180° ∠RQS = 180° - 75° = 105° Therefore, the size of ∠QRS is 105°. Option (C) "105" is the correct answer.

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 10 Ripọtì

A woman buys 270 oranges for ₦1800.00 and seels at 5 for ₦40.00. What is her profit?

₦2,160.00

₦1, 620.00

₦630.00

₦360.00

Akọwa Nkọwa

The cost of one orange can be found by dividing the total cost by the number of oranges: Cost of one orange = 1800/270 = ₦6.67 (rounded to two decimal places) The selling price of 5 oranges is ₦40.00. So the selling price of one orange is: Selling price of one orange = 40/5 = ₦8.00 Profit per orange = Selling price - Cost price = ₦8.00 - ₦6.67 = ₦1.33 Total profit = Profit per orange x number of oranges sold = ₦1.33 x 270 = ₦359.10 Therefore, the woman's profit is ₦360.00 (rounded up to the nearest naira). So the answer is: ₦360.00

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Financial Arithmetic

Lee Azịza

Ajụjụ 11 Ripọtì

Find the number of committees of three that can be formed consisting of two men and one woman from four men and three women

3

6

18

24

Akọwa Nkọwa

To find the number of committees of three that can be formed consisting of two men and one woman from four men and three women, we can use the combination formula. The number of ways to choose 2 men from 4 men is given by the combination formula as 4C2 = (4!)/(2!2!) = 6. Similarly, the number of ways to choose 1 woman from 3 women is 3C1 = 3. Therefore, the total number of committees of three that can be formed consisting of two men and one woman is the product of these two combinations: 6 x 3 = 18 So, there are 18 committees of three that can be formed consisting of two men and one woman from four men and three women. Therefore, the answer is 18.

Gụọ akwụkwọ nkuzi na Permutation And Combination (JAMB)

Permutation And Combination

Lee Azịza

Ajụjụ 12 Ripọtì

A triangle has vertices P(-1,6), Q(-3, -4) and R(1, -4).Find the midpoint of PQ and QR respectively.

(0,-2) and (-1, -4)

(-1,0) and (-1,-1)

(-2,1) and (0,1)

(-2,1) and (-1,-4)

Akọwa Nkọwa

To find the midpoint of a line segment, we can use the midpoint formula, which states that the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is ((x1 + x2)/2, (y1 + y2)/2). Using this formula, we can find the midpoint of PQ by plugging in the coordinates of P and Q: Midpoint of PQ = ((-1 - 3)/2, (6 - 4)/2) = (-2, 1) Similarly, we can find the midpoint of QR by plugging in the coordinates of Q and R: Midpoint of QR = ((-3 + 1)/2, (-4 - 4)/2) = (-1, -4) Therefore, the correct answer is: (-2,1) and (-1,-4)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (JAMB)

Coordinate Geometry

Lee Azịza

Ajụjụ 13 Ripọtì

Find the value of x and y respectively if 3x - 5y + 5 = 0 and 4x - 7y + 8 = 0

-5, -4

-4,. -5

4, 5

5, 4

Akọwa Nkọwa

3x - 5y + 5 = 0 → eqn1
4x - 7y + 8 = 0 → eqn2
eqn1 * 4; 12x - 20y + 20 = 0 → eqn3
eqn2 * 3; 12x - 21y + 24 = 0 → eqn4
eqn3 - eqn4 = y - 4 = 0
∴ y = 4
From eqn1,
3x - 5y + 5 = 0
3x - 5(4) + 5 = 0
3x - 20 + 5 = 0
3x - 15 = 0
3x = 15
x = 5
x and y = 5, 4 respectively

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 14 Ripọtì

Find the mean of the distribution above

1

2

3

4

Akọwa Nkọwa

M e a n = \frac{(0 \times 1) + (1 \times 2) + (2 \times 2) + (3 \times 1) + (4 \times 9)}{1 + 2 + 2 + 1 + 9} = \frac{(0 + 2 + 4 + 3 + 36)}{15} = \frac{45}{15} = 3

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Linear Inequalities

Measures Of Location

Statistics

Lee Azịza

Ajụjụ 15 Ripọtì

The graph of the function y = x2 + 4 and a straight line PQ are drawn to solve the equation x2 - 3x + 2 = 0. What is the equation of PQ?

y = 3x - 2

y = 3x + 2

y = 3x - 4

y = 3x + 4

Akọwa Nkọwa

To solve the equation x^2 - 3x + 2 = 0, we can factor it into (x - 1)(x - 2) = 0. This means that x = 1 or x = 2. Now, we can plug these values of x into the equation y = x^2 + 4 to get the corresponding y values. When x = 1, y = 5 and when x = 2, y = 8. We also know that the line PQ intersects the graph of y = x^2 + 4 at two points, since the equation x^2 - 3x + 2 = 0 has two solutions. Since the line PQ is a straight line, we can find its equation by using the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. To find the slope, we can use the two points of intersection between PQ and the parabola, which are (1, 5) and (2, 8). The slope is the change in y over the change in x, which is (8 - 5)/(2 - 1) = 3. To find the y-intercept, we can plug in the coordinates of one of the points into the slope-intercept form and solve for b. Using (1, 5), we get: 5 = 3(1) + b b = 2 So the equation of PQ is y = 3x + 2, which is.

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 16 Ripọtì

BY how much is the mean of 30, 56, 31, 55, 43 and 44 less than the median?

0.75

0.50

0.33

0.17

Akọwa Nkọwa

M e a n = \frac{30 + 56 + 31 + 55 + 43 + 44}{6} = \frac{259}{6} = 43.167 M e d i a n = 30, 31, 43, 44, 55, 56 = \frac{43 + 44}{2} = \frac{87}{2} = 43.5 M e d i a n - m e a n = 43.3 - 43.17 = 0.33

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Measures Of Location

Statistics

Lee Azịza

Ajụjụ 17 Ripọtì

Find the slope of the curve y = 2x2n+ 5x - 3 at (1, 4).

4

6

7

9

Akọwa Nkọwa

To find the slope of the curve at a specific point, we need to find the first derivative of the curve and substitute the x-coordinate of the point of interest to obtain the slope. The given curve is y = 2x²n + 5x - 3, and its first derivative is obtained by differentiating each term with respect to x. dy/dx = 4xn + 5 Substituting x = 1 (as given) into the derivative, we get: dy/dx = 4(1)n + 5 = 4 + 5 = 9 Therefore, the slope of the curve at the point (1,4) is 9.

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Introductory Calculus

Integration

Lee Azịza

Ajụjụ 18 Ripọtì

A trapezium has two parallel sides of length 5cm and 9cm. If the area is 21cm2, find the distance between the parallel sides

3 cm

4 cm

6 cm

7 cm

Akọwa Nkọwa

The area of a trapezium is given by the formula A = (1/2)(a+b)h, where a and b are the lengths of the parallel sides, and h is the perpendicular distance between them. In this question, we are given that the parallel sides have lengths 5cm and 9cm, and the area is 21cm^2. We can plug these values into the formula and solve for h: 21 = (1/2)(5+9)h 21 = (1/2)(14)h 21 = 7h h = 3 Therefore, the distance between the parallel sides is 3cm. So the answer is option A: 3cm.

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 19 Ripọtì

An aeroplane flies due north from airport P to Q and then flies due east R. If Q is equidistant from P and R, find the bearing of P and R

90

^{\circ}

135

^{\circ}

225

^{\circ}

270

^{\circ}

Akọwa Nkọwa

Since PQ = QR, Δ PQR is a right angle isosceles triangle

∴ ∠PQR = 45

^{\circ}

=> The bearing of P from R = 180 + 45
= 225

^{\circ}

Gụọ akwụkwọ nkuzi na Bearings (WAEC)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Bearings

Circles

Lee Azịza

Ajụjụ 20 Ripọtì

Triangle OPQ above is the solution of the inequalities

x + 1 ≥ 0, y + x ≤ 0, y - x ≥ 0

y + x ≤ 0, y - x ≥ 0, x -1 ≥ 0

x - 1 ≤ 0, y - x ≥ o, y + x ≥ 0

x - 1 ≤ 0, y + x ≤ 0, y - x ≤ 0

Akọwa Nkọwa

Lines bounding Δ OPQ
OQ; y - x = 0
y - x ≥ 0
PQ; x + 1 = 0
x + 1 ≥ = 0
PO; y + x = 0
y + x ≤ 0
∴ x + 1 ≥ 0, y + x ≤ 0, y - x ≥ 0

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Inequalities

Lee Azịza

Ajụjụ 21 Ripọtì

Find the equation of the locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1).

4x + 2y = 5

4x - 2y = 5

2x + 2y = 5

2x + y = 5

Akọwa Nkọwa

Locus of a point P(x,y) which is equidistant from Q(0,0) and R(2,1) is the perpendicular bisector of the straight line joining Q and R
Mid point QR = (x2+x1)/2 . (y2+y1)/2
= 2+0/2 . 1+0/2

Gradient of Qr =	y2-y1
	x2-x1

=	1-0
	2-0

= 1/2

Gradient of PM(M)=	-1
	1/2

= -2
Equation of Pm = y - y1 = m(x-x1)
i.e y - 1/2 = -2(x-1)
2y - 1 = -4(x-1)
2y - 1 = -4x + 4
2y + 4x = 5

Gụọ akwụkwọ nkuzi na Loci (JAMB)

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Loci

Lee Azịza

Ajụjụ 22 Ripọtì

The range of 4, 3, 11, 9, 6, 15, 19, 23, 27, 24,21 and 16 is

16

21

23

24

Akọwa Nkọwa

In mathematics, the range of a set of numbers is the difference between the largest and smallest numbers in the set. To find the range of the given set of numbers, we need to first find the largest and smallest numbers in the set. The smallest number in the set is 3, and the largest number is 27. Therefore, the range of the set is: 27 - 3 = 24 Hence, the correct answer is: 24.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sets

Sequence And Series

Lee Azịza

Ajụjụ 23 Ripọtì

The result of tossing a fair die 120 times is summarized above. Find the value of x

19

20

21

22

Akọwa Nkọwa

12 + 20 + x + 21 + x-1 +28 = 120
2x + 80 = 120
2x = 120 - 80
2x = 40
x = 20

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Transformation In The Cartesian Plane (WAEC)

Probability

Transformation In The Cartesian Plane

Lee Azịza

Ajụjụ 24 Ripọtì

The sum of the first n terms of an arithmetic progresssion is 252. If the first term is -16 and the last term is 72, find the number of terms in the series

6

7

8

9

Akọwa Nkọwa

To find the number of terms in an arithmetic progression, we can use the formula: sum = (n/2)(first term + last term) where sum is the sum of the first n terms, n is the number of terms, and the first and last terms are given. Plugging in the given values, we get: 252 = (n/2)(-16 + 72) Simplifying the equation, we get: 252 = 28n Dividing both sides by 28, we get: n = 9 Therefore, the correct answer is: 9. The arithmetic progression has 9 terms.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 25 Ripọtì

If $ \frac{9^{2x-1}}{27^{x+1}} = 1 $ find the value of x.

8

5

3

2

Akọwa Nkọwa

Cross-multiplying, 92x-1 = 1 x 27x+1
32(2x-1) = 33(x+1)
Solving by equating index: 2(2x-1) = 3(x+1), => x = 5

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 26 Ripọtì

Given:

U = {Even numbers between 0 and 30}

P = {Multiples of 6 between 0 and 30}

Q = {Multiples of 4 between 0 and 30}

Find (P∪Q)c

{2, 10, 14, 22, 26}

{0, 10, 14, 22, 26}

{2,4, 14, 18, 26}

{0, 2, 6, 22, 26}

Akọwa Nkọwa

In set theory, the complement of a set A, denoted by A', is the set of all elements in the universal set that are not in A. In this question, we are given three sets: U, P, and Q. U is the set of even numbers between 0 and 30, P is the set of multiples of 6 between 0 and 30, and Q is the set of multiples of 4 between 0 and 30. To find (P∪Q)c, we first need to find P∪Q, which is the union of P and Q. The union of two sets is the set of all elements that are in either set. In this case, P∪Q is the set of all multiples of either 4 or 6 between 0 and 30. We can list out the elements of P∪Q: {0, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30} To find (P∪Q)c, we need to find the complement of P∪Q, which is the set of all even numbers between 0 and 30 that are not multiples of either 4 or 6. We can list out the even numbers between 0 and 30 that are not multiples of 4 or 6: {2, 10, 14, 22, 26} Therefore, the answer is option A: {2, 10, 14, 22, 26}.

Lee Azịza

Ajụjụ 27 Ripọtì

On the pie chart, there are four sectors of which three angles are $45^\circ$, $90^\circ$ and $135^\circ$. If the smallest sector represents N28.00, how much is the largest sector?

N96.00

N84.00

N48.00

N42.00

Akọwa Nkọwa

To solve this problem, we need to first find the total angle of the pie chart, which is always 360∘. Next, we need to determine the value of one degree of the pie chart. We can do this by dividing the value of the smallest sector (N28.00) by its corresponding angle (45∘): N28.00 ÷ 45∘ = N0.62 per degree Now that we know the value of one degree, we can use it to find the value of the largest sector, which has an angle of 135∘: N0.62 × 135∘ = N83.70 Therefore, the largest sector represents N83.70, which is closest to the option N84.00.

Lee Azịza

Ajụjụ 28 Ripọtì

Find the range of values of $x$ which satisfy the inequalities $4x - 7 \le 3x$ and $3x - 4 \le 4x$

-4

\leq

x

\leq

7

-7

\leq

x

\leq

4

x

\geq

-7

\leq

x

\leq

6

Akọwa Nkọwa

To solve the system of inequalities 4x - 7 ≤ y ≤ 3x and 3x - 4 ≤ y ≤ 4x, we can use the following steps: 1. For the first inequality, isolate y in the middle by adding 7 to both sides: 4x - 7 + 7 ≤ y + 7 ≤ 3x + 7, which simplifies to 4x ≤ y + 7 ≤ 3x + 7. 2. For the second inequality, isolate y in the middle by adding 4 to both sides: 3x - 4 + 4 ≤ y + 4 ≤ 4x + 4, which simplifies to 3x ≤ y + 4 ≤ 4x + 4. 3. We want to find the range of values of x that satisfy both inequalities. To do this, we need to find the intersection of the two intervals [4x, 3x + 7] and [3x, 4x + 4] for y. 4. We start by finding the maximum lower bound of the two intervals. The lower bound of [4x, 3x + 7] is 4x, and the lower bound of [3x, 4x + 4] is 4x. So the maximum lower bound is 4x. 5. We then find the minimum upper bound of the two intervals. The upper bound of [4x, 3x + 7] is 3x + 7, and the upper bound of [3x, 4x + 4] is 4x + 4. So the minimum upper bound is 3x + 7. 6. Therefore, the range of values of x that satisfy both inequalities is 4x ≤ y + 7 ≤ 3x + 7 and 3x - 4 ≤ y ≤ 4x, which simplifies to 4x ≤ y + 7 ≤ 3x + 7 and 3x - 4 ≤ y ≤ 4x. 7. To solve for x, we can use the second inequality: 3x - 4 ≤ y ≤ 4x. We can substitute the upper bound of y with 4x and the lower bound of y with 3x - 4 to get 3x - 4 ≤ 3x - 4 ≤ x ≤ 4x. Therefore, the range of values of x that satisfy both inequalities is -4 ≤ x ≤ 7. Therefore, the correct option is "-4 ≤ x ≤ 7".

Lee Azịza

Ajụjụ 29 Ripọtì

Simplify 2134 x 234

103114

103214

122314

132114

Number Bases

Lee Azịza

Ajụjụ 30 Ripọtì

A bag contains 5 blacks balls and 3 red balls. Two balls are picked at random without replacement. What is the probability that a black and red balls are picked?

15/28

13/28

5/14

3/14

Akọwa Nkọwa

There are 5 black balls and 3 red balls in the bag, so there are a total of 8 balls. When two balls are picked at random without replacement, there are (8 choose 2) = 28 ways of doing so. To pick a black ball and a red ball, we can pick one black ball from the 5 available, and one red ball from the 3 available. There are (5 choose 1) = 5 ways to pick one black ball, and (3 choose 1) = 3 ways to pick one red ball. By the multiplication principle, there are a total of 5 × 3 = 15 ways to pick a black ball and a red ball. Therefore, the probability of picking a black ball and a red ball is 15/28.

Lee Azịza

Ajụjụ 31 Ripọtì

The locus of a point P which moves on one side only of a straight line XY so that ∠XPY = $90^\circ$ is

a circle

a semicircle

an arc of a circle through X, Y

the perpendicular bisector of XY

Coordinate Geometry

Lee Azịza

Ajụjụ 32 Ripọtì

The histogram above shows the ages of the victims of a pollution. How many people were involved in the pollution?

15

18

20

21

Akọwa Nkọwa

age	People
10	3
20	4
30	5
40	6
50	2
	20

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Euclidean Geometry

Lee Azịza

Ajụjụ 33 Ripọtì

Find the range of the value of $x$ satisfying the inequalities $5 + x \le 8$ and $13 + x \ge 7$

-3 $\leq$ x $\leq$ 3

3 $\leq$ x $\leq$ 6

-6 $\leq$ x $\leq$ 3

-6 $\leq$ x $\leq$ -3

Akọwa Nkọwa

5 + x $\leq$ 8 and 13 + x $\geq$ 7
5 + x $\leq$ 8
X $\leq$ 8 – 5
X $\leq$ 3
And 13 + x $\geq$ 7
X $\geq$ 7 – 15
X $\geq$ -7
Combining the two together
-6 $\leq$ x $\leq$ 3

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Linear Inequalities

Inequalities

Lee Azịza

Ajụjụ 34 Ripọtì

In a class of 40 students, 32 offer mathematics, 24 offer Physics, and 4 offer neither Mathematics nor Physics. How many offer both Mat
ematics and Physics?

20

16

8

4

Akọwa Nkọwa

Using a venn diagram, let x = number who offer both Maths and Physics. So that (32-x) offer Maths and another (24-x) offer Physics.
(32-x) + (24-x) + (x) + (4 who offer neither) = 40
=> 60 - x = 40
=> -x = -20
Therefore x = 20.

Lee Azịza

Ajụjụ 35 Ripọtì

In the diagram, STUV is a straight line. < TSY = < UXY = 40o and < VUW = 110o. Calculate < TYW

150o

140o

130o

120o

Akọwa Nkọwa

< TUW = 110

^{\circ}

= 180

^{\circ}

(< s on a straight line)
< TUW = 180

^{\circ}

- 110

^{\circ}

= 70

^{\circ}

In

△

XTU, < XUT + < TXU = 180

^{\circ}

i.e. < YTS + 70

^{\circ}

= 180
< XTU = 180 - 110

^{\circ}

= 70

^{\circ}

Also < YTS + < XTU = 180 (< s on a straight line)
i.e. < YTS + < XTU - 180(< s on straight line)
i.e. < YTS + 70

^{\circ}

= 180
< YTS = 180 - 70 = 110

^{\circ}

in

△

SYT + < YST + < YTS = 180

^{\circ}

(Sum of interior < s)
SYT + 40 + 110 = 180
< SYT = 180 - 150 = 30
< SYT = < XYW (vertically opposite < s)
Also < SYX = < TYW (vertically opposite < s)
but < SYT + < XYW + < SYX + < TYW = 360
i.e. 30 + 30 + < SYX + TYW = 360
but < SYX = < TYW
60 + 2(< TYW) = 360
2(< TYW) = 360

^{\circ}

- 60
2(< TYW) = 300

^{\circ}

TYW =

\frac{300}{2}

= 150

^{\circ}

< SYT

Lee Azịza

Ajụjụ 36 Ripọtì

If y = 3 sin(-4x), dy/dx is

12x cos (4x)

-12x cos (-4x)

-12 cos (-4x)

12 sin (-4x)

Akọwa Nkọwa

To find the derivative of y with respect to x, we use the chain rule of differentiation, which states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). In this case, we have y = 3 sin(-4x), so f(g(x)) = 3 sin(g(x)) and g(x) = -4x. Therefore, f'(g(x)) = 3 cos(g(x)) and g'(x) = -4. Plugging these values into the chain rule, we get: dy/dx = f'(g(x)) * g'(x) = 3 cos(-4x) * (-4) = -12 cos(4x) Therefore, the correct answer is: -12x cos(-4x).

Lee Azịza

Ajụjụ 37 Ripọtì

Simplify (√98 -√50)/√32

3

1

1/2

1/4

Akọwa Nkọwa

To simplify the expression (√98 -√50)/√32, we can first simplify the square roots in the numerator and denominator. √98 can be written as √(49 × 2), which simplifies to 7√2. Similarly, √50 can be written as √(25 × 2), which simplifies to 5√2. Finally, √32 can be written as √(16 × 2), which simplifies to 4√2. Now, we can substitute these simplified values into the original expression to get: (7√2 - 5√2)/(4√2) Simplifying the numerator further by combining like terms, we get: 2√2/4√2 Which simplifies to: 1/2 Therefore, the answer is 1/2.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Number Bases

Modular Arithmetic

Surds (radicals)

Lee Azịza

Ajụjụ 38 Ripọtì

In the diagram above, PQ is parallel to RS. What is the value of ? + ? + ??

360

^{\circ}

200

^{\circ}

180

^{\circ}

90

^{\circ}

Akọwa Nkọwa

γ + S = 180(Allied angles)

S = 180 - γ
α + P = 180(Allied angles)
p = 180 - α
β = P + S
β = 180 - α + 180 - γ
β = 360 - α - γ
β + α + γ = 360

Gụọ akwụkwọ nkuzi na Angles And Intercepts On Parallel Lines (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Angles And Intercepts On Parallel Lines

Triangles And Polygons

Lee Azịza

Ajụjụ 39 Ripọtì

If nP3 - 6(nC4) = 0, find the value of n

5

6

7

8

Akọwa Nkọwa

The given expression is an equation that involves the binomial coefficients nP3 and nC4, where n is an integer. We are asked to find the value of n that satisfies the equation: nP3 - 6(nC4) = 0 Using the formula for the binomial coefficients, we can simplify the expression: nP3 = n(n-1)(n-2) nC4 = n(n-1)(n-2)(n-3)/4! Substituting these expressions into the given equation, we get: n(n-1)(n-2) - 6(n(n-1)(n-2)(n-3)/4!) = 0 Simplifying further: n(n-1)(n-2)(1 - 6(n-3)/4!) = 0 Since n is an integer, we must have either n = 0, n = 1, n = 2 or 1 - 6(n-3)/4! = 0. However, n cannot be 0, 1, or 2 because nP3 and nC4 are defined only for n >= 4. Therefore, we need to solve the equation: 1 - 6(n-3)/4! = 0 6(n-3)/4! = 1 n-3 = 4! / 6 n-3 = 4 n = 7 Hence, the value of n that satisfies the given equation is 7.

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 40 Ripọtì

The sum of four numbers is 12145. What is the average expressed in base five?

114

141

401

411

Number Bases

Lee Azịza

Ajụjụ 41 Ripọtì

Evaluate $ \log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 $

-5.5

-2.5

2.5

5.5

Akọwa Nkọwa

To solve this equation, we can use the logarithmic identity log(a) + log(b) = log(ab) and log(a) - log(b) = log(a/b). Using this identity, we can simplify the equation: log√24 + log1216 − log432 = log(24^(1/2)) + log(1216) - log(432) = log(4^(1/2) * 6^(1/2)) + log(1216) - log(432) = log(2*3) + log(1216) - log(432) = log(2*3*1216/432) = log(16) Therefore, the answer is 2.5. In summary, we use logarithmic identities to simplify the equation and find that the answer is 2.5.

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 42 Ripọtì

Find $\left(\frac{1}{0.06} \div \frac{1}{0.042}\right)^{-1}$

1.43

1.53

3.14

4.42

Modular Arithmetic

Lee Azịza

Ajụjụ 43 Ripọtì

Three consecutive terms of a geometric progression are given as n-2, n and n=3. Find the common ratio

1/4

1/2

2/3

3/2

Akọwa Nkọwa

In a geometric progression, each term is obtained by multiplying the previous term by a fixed constant called the common ratio (r). Therefore, if we divide any two consecutive terms in a geometric progression, we get the same value for the common ratio. In this problem, the three consecutive terms are given as n-2, n, and n+3. We can divide the second term (n) by the first term (n-2) to get: n / (n-2) We can also divide the third term (n+3) by the second term (n) to get: (n+3) / n Since these two values must be equal to the common ratio (r), we can set them equal to each other and solve for r: n / (n-2) = (n+3) / n n^2 = n(n-2) + (n+3)(n-2) n^2 = n^2 - 2n + n^2 - 5n + 6 0 = n^2 - 7n + 6 0 = (n-1)(n-6) This equation has two solutions, n=1 and n=6. However, n cannot be equal to 1 because that would make the first term (n-2) equal to -1, which is not allowed in a geometric progression. Therefore, n must be equal to 6. Plugging n=6 back into the equation for the common ratio, we get: r = (n+3) / n = 9 / 6 = 3/2 Therefore, the correct answer is: 3/2.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 44 Ripọtì

XYZ is a circle enter o and a radius 7cm. Find the area of the shaded region. θ = 45 x 2 = 90

84cm2

77cm2

38cm2

14cm2

Akọwa Nkọwa

θ = 45 x 2 (∠ at center twice ∠ at θ)
θ = 90
Area of sector = θ/360 x πr2
= 90/360 x 22/7 x 7 x 7
= 77/2cm2
Area of triangle = 1/2 ab sin θ
= 1/2 x 7 x 7 sin 90
= 1/2 x 7 x 7 x 1
= 49/2
Area of shaded region = 77/2 - 49/2

1	1
	2

= 28/2
= 14cm2

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 45 Ripọtì

An arc of a circle subtends an angle of $30^\circ$ on the circumference of a circle of radius 21cm. Find the length of the arc.

11cm

22cm

44cm

66cm

Akọwa Nkọwa

θ = 30 x 2(∠ at center twice ∠ at θ)
∴ θ = 60

^{\circ}

Length of arc =

\frac{θ}{360} \times 2 π r = \frac{60}{360} \times 2 \times \frac{22}{7} \times \frac{21}{1} = 22 c m

Lee Azịza






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