JAMB UTME - Mathematics - 2000

A function f(x) passes through the origin and its first derivative is 3x + 2. What is f(x)?

y = (3x2/)2 + 2x

y = (3x2)/2 + x

y = 3x2 + (x/2)

y = 3x2 +2x

Akọwa Nkọwa

Hints:
1. Integrate the given first derivative of f(x) at the boundaries, (0,0)
Then solve accordingly to get f(x) = y = (3x2/)2 + 2x

Gụọ akwụkwọ nkuzi na Functions And Relations (WAEC)

Functions And Relations

Lee Azịza

Ajụjụ 2 Ripọtì

if P and Q are fixed points and X is a point which moves so that XP = XQ, the locus of X is

A straight line

a circle

the bisector of angle PXQ

the perpendicular bisector of PQ

Akọwa Nkọwa

Given two fixed points P and Q, the locus of points X which are equidistant from P and Q is the perpendicular bisector of PQ. This is because any point on the perpendicular bisector of PQ will be equidistant from both P and Q by definition. Therefore, if X is a point which moves so that XP = XQ, it must lie on the perpendicular bisector of PQ. Therefore, the correct answer is "the perpendicular bisector of PQ".

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Circles

Euclidean Geometry

Lee Azịza

Ajụjụ 3 Ripọtì

If α and β are the roots of the equation 3x2 + 5x - 2 = 0, find the value of 1/α + 1/β

-5/3

-2/3

1/2

5/2

Akọwa Nkọwa

We know that if α and β are the roots of the quadratic equation ax^2 + bx + c = 0, then α + β = -b/a and α × β = c/a. In this equation, 3x^2 + 5x - 2 = 0, we have a = 3, b = 5, and c = -2. So, α + β = -b/a = -5/3 And, α × β = c/a = -2/3 We need to find the value of 1/α + 1/β = (α + β)/(α × β) = (-5/3) / (-2/3) = 5/2. Therefore, the value of 1/α + 1/β is 5/2.

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 4 Ripọtì

In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon.

8

6

4

3

Akọwa Nkọwa

2x + x = 180°, => 3x = 180°, and thus x = 60°
Each exterior angle = 60° but size of ext. angle = 360°/n
Therefore 60° = 360°/n
n = 360°/60° = 6 sides

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Triangles And Polygons

Lee Azịza

Ajụjụ 5 Ripọtì

3y = 4x - 1 and Ky = x + 3 are equations of two straight lines. If the two lines are perpendicular to each other, find K.

-4/3

-3/4

3/4

4/3

Akọwa Nkọwa

Grad of 3y = 4x - 1
y = 4x/3 - 1/3
Grad = 4/3
Grad of Ky = x + 3
y = x/k + 3/4
Grad = 1/k
Since two lines are perpendicular,
1/k = -3/4
-3k = 4
k = -4/3

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Algebraic Expressions

Differentiation

Lee Azịza

Ajụjụ 6 Ripọtì

A predator moves in a circle of radius $ \sqrt{2} $ centre $(0,0)$, while a prey moves along the line $y = x$. If $0 \le x \le 2$, at which point(s) will they meet?

(1,1) only

(1,1) and (1,2)

(0,0) and (1,1)

(√2,√2) only

Akọwa Nkọwa

The predator moves in a circular path with a radius of √2 centered at the origin (0,0), while the prey moves along the line y = x. The prey's movement is restricted between x = 0 and x = 2. At some point, the predator will catch up to the prey, and they will meet. The predator will catch the prey when the distance between them is less than or equal to √2. We can find the equation of the circle centered at (0,0) with a radius of √2: x^2 + y^2 = (√2)^2 x^2 + y^2 = 2 We can substitute y = x into the equation to get: x^2 + x^2 = 2 2x^2 = 2 x^2 = 1 x = ±1 Therefore, the possible points of intersection are (1,1) and (-1,-1), but we know that the prey can only be at x values between 0 and 2, so the only possible point of intersection is (1,1). Therefore, the answer is (1,1) only.

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Coordinate Geometry

Statistics

Lee Azịza

Ajụjụ 7 Ripọtì

Evaluate (1/2 - 1/4 - 1/8 - 1/16 + ...) - 1

2/3

zero

-2/3

-1

Akọwa Nkọwa

The given expression is a geometric series with the first term as 1/2 and the common ratio as -1/2. Using the formula for the sum of an infinite geometric series with |r| < 1, we can find that the sum of the series is: S = a/(1-r) = (1/2)/(1-(-1/2)) = (1/2)/(3/2) = 1/3 Therefore, the expression (1/2 - 1/4 - 1/8 - 1/16 + ...) - 1 simplifies to: 1/3 - 1 = -2/3 So, the answer is -2/3.

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

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Fractions, Decimals And Approximations

Representation Of Data

Number Bases

Lee Azịza

Ajụjụ 8 Ripọtì

Find the value of $ \displaystyle \int_{0}^{\pi}\frac{\cos^{2}\theta-1}{\sin^{2}\theta}\,d\theta $

π

π/2

-π/2

-π

Akọwa Nkọwa

\int_{0}^{π} \frac{c o s^{2} θ - 1}{s i n^{2} θ} d θ = \int_{0}^{π} \frac{- s i n^{2} θ}{s i n^{2} θ} = \int_{0}^{π} d θ = - π

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Integration

Lee Azịza

Ajụjụ 9 Ripọtì

P is a point on one side of the straight line UV and P moves in the same direction as UV. If the straight line ST is on the locus of P and angle VUS = 60°, find angle UST.

310°

130°

80°

50°

Akọwa Nkọwa

Make a good sketch of the question, and follow this steps.
Since ST is parallel to UV => angle USR = 50°(alternate to VUS.
Angle UST = 180° - 50° = 130° (angle on a straight line)

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Loci

Euclidean Geometry

Lee Azịza

Ajụjụ 10 Ripọtì

If the volume of a hemisphere is increasing at a steady rate of 18π m/s, at what rate is its radius changing when its is 6m?

2.30m/s

2.00 m/s

0.25 m/s

0.20 m/s

Akọwa Nkọwa

We can solve this problem using the formula for the volume of a hemisphere, which is given by V = (2/3)πr^3. To find the rate of change of the radius, we need to differentiate this equation with respect to time t. Thus, we have: dV/dt = (2/3)π(3r^2)(dr/dt) where dV/dt is the rate of change of volume (18π m/s), and r = 6 m is the radius of the hemisphere. Solving for dr/dt, we get: dr/dt = (3dV/dt)/(4πr^2) = (3(18π))/(4π(6)^2) = 0.25 m/s Therefore, the rate of change of the radius when the volume of the hemisphere is increasing at a steady rate of 18π m/s and its radius is 6 m is 0.25 m/s. Thus, the correct option is: 0.25 m/s.

Lee Azịza

Ajụjụ 11 Ripọtì

Find the inverse of p under the binary operation * defined by p*q = p + q - pq, where p and q are real numbers and zero is the identity

p

p -1

p/(p-1)

p/(p+1)

Akọwa Nkọwa

If P-1 is the inverse of P and O is the identity, Then P-1 * P = P * P-1 = 0
i.e. P-1 + P - P-1.P = 0
P-1 - P-1.P = -P
P-1(1 - P) = -P
P-1 = -P/(P-1)
= P/(P-1)

Gụọ akwụkwọ nkuzi na Matrices And Determinants (JAMB)

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Matrices And Determinants

Binary Operations

Lee Azịza

Ajụjụ 12 Ripọtì

The variance of x, 2x, 3x, 4x and 5x is

x√2

2x2

x2

3x

Akọwa Nkọwa

Hint: prepare a three-columned table, one for (x), another for the (deviation), and the last for the (squared-deviation).
Sum of (x) = 15x, algebraic sum of (deviation) = zero (0)
Sum of (squared-deviation) = 10x2
Variance = ∑(squared-deviation)/n = 10x2/5 = 2x2

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Measures Of Dispersion

Lee Azịza

Ajụjụ 13 Ripọtì

A frustrum of pyramid with square base has its upper and lower sections as squares of sizes 2m and 5m respectively and the distance between them 6m. Find the height of the pyramid from which the frustrum was obtained.

8.0 m

8.4 m

9.0 m

10.0 m

Polynomials

Lee Azịza

Ajụjụ 14 Ripọtì

X and Y are two events. The probability of X or Y is 0.7 and that of X is 0.4. If X and Y are independent, find the probability of Y.

0.30

0.50

0.57

1.80

Akọwa Nkọwa

P (X or Y) = P(X) + P(Y), when they are independent as given.
0.7 = 0.4 + P(Y)
P(Y) = 0.7 - 0.4 = 0.30

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 15 Ripọtì

In the diagram, POS and ROT are straight lines, OPQR is a parallelogram. |OS| = |OT| and ∠OST = 50o. Calculate ∠OPQ.

160o

140o

120o

100o

Lee Azịza

Ajụjụ 16 Ripọtì

Evaluate (2.813 x 10-3 x 1.063) (5.637 x 10-2)

0.056

0.055

0.054

0.54

Number Bases

Lee Azịza

Ajụjụ 17 Ripọtì

A man wishes to keep his money in a savings deposit at 25% compound interest so that after three years he can buy a car for ₦150,000. How much does he need to deposit?

₦112,000.50

₦96,000.00

₦85,714.28

₦76,800.00

Akọwa Nkọwa

Amount A = P(1+r)n;
A = ₦150,000, r = 25%, n = 3.
150,000 = P(1+0.25)3 = P(1.25)3
P = 150,000/1.253 =₦76,800.00

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Sets

Quadratic Equations

Lee Azịza

Ajụjụ 18 Ripọtì

In a youth club with 94 members, 60 like modern music, and 50 like traditional music. The number of members who like both traditional and modern music is three times those who do not like any type of music. How many members like only one type of music?

8

24

62

86

Akọwa Nkọwa

Use a venn diagram:
60-3x+3x+50-3x = 94-x.
110-3x+x = 94
-2x = 94-110
=>-2x = -16, this x = -8.
Members that like only one game:
= 60 - 3x + 50 - 3x
= 60 - 3x8 + 50 - 3x8
= 60 - 24 + 50 - 24
= 36 + 26 = 62

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Lee Azịza

Ajụjụ 19 Ripọtì

If P3446 - 23P26 = 2PP26, find the value of the digit P.

2

3

4

5

Akọwa Nkọwa

Covert everything to base 10 and collect like terms, such that:
210P - 42P = 434 + 406
168P = 840
P = 840/168 = 5

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 20 Ripọtì

In how many ways can the word MATHEMATICS be arranged?

11!/(9! 2!)

11!/(9! 2! 2!)

11!/(2! 2! 2!)

11!/(2! 2!)

Akọwa Nkọwa

Number of letter in MATHEMATICS = 11
Number of letter M = 2
Number of letter E = 2
Number of letter A = 2
Arrangement = 11!/(2! 2! 2!) ways

Gụọ akwụkwọ nkuzi na Permutation And Combination (JAMB)

Permutation And Combination

Lee Azịza

Ajụjụ 21 Ripọtì

The expression ax2 + bx + c equals 5 at x = 1. If its derivative is 2x + 1, what are the values of a, b, c respectively?

1, 3, 1

1, 2, 1

2, 1, 1

1, 1, 3

Akọwa Nkọwa

At x = 1, substituting x = 1 in the equation: ax2 + bx + c = 5;
f(1) => a + b + c = 5 .....(1)
Taking the first derivative of f(x) in the original equation gives dy/dx = 2ax + b = 2x + 1 (given)....(2)
From (2),=> b = 1, and 2ax = 2x, => a = 1.
substituting into (1) 1 + 1 + c = 5, => c = 5 - 2 = 3
Thus a = 1, b = 1 and c = 3

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 22 Ripọtì

A bowl is designed by revolving completely the area enclosed by y = x2 - 1, y = 3 and x ≥ 0 around the axis. What is the volume of this bowl?

7π cubic units

15π/2 cubic units

8π cubic units

17π/2 cubic units

Akọwa Nkọwa

∫30 π(y+1) dy = π[y2 + y30
= π(9/2 + 3) = 15π/2

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Volumes

Areas

Lee Azịza

Ajụjụ 23 Ripọtì

An equilateral triangle of side √3cm is inscribed in a circle. Find the radius of the circle.

2/3 cm

2 cm

1 cm

3 cm

Akọwa Nkọwa

In an equilateral triangle, all the sides are of equal length and each angle is 60 degrees. Let's draw an equilateral triangle with side √3 cm inscribed in a circle. Since the triangle is equilateral, we know that the circumcenter (center of the circle that passes through all the vertices of the triangle) is also the centroid (point of intersection of the medians). The median of an equilateral triangle is the line segment from a vertex to the midpoint of the opposite side. Therefore, the circumcenter is also the midpoint of any side of the triangle. Let's choose one side of the triangle and label its midpoint as point M. We can draw a perpendicular line from M to the opposite vertex of the triangle, which will bisect the side and form a right triangle with one leg being half of the side of the triangle and the other leg being the radius of the circle. Using the Pythagorean theorem, we can solve for the radius: (radius)^2 = (√3/2)^2 + (1/2)^2 (radius)^2 = 3/4 + 1/4 (radius)^2 = 1 radius = 1 cm Therefore, the radius of the circle is 1 cm. Answer: 1 cm.

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Gụọ akwụkwọ nkuzi na Simple Operations On Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Trigonometry

Differentiation

Simple Operations On Algebraic Expressions

Triangles And Polygons

Lee Azịza

Ajụjụ 24 Ripọtì

A binary operation * is defined by a*b = ab. If a*2 = 2-a, find the possible values of a.

1, -1

1, 2

2, -2

1, -2

Akọwa Nkọwa

a * b = ab
a * 2 = a2 = 2 - a
a2 + 2a - a - 2 = 0
a(a+2) - 1(a+2) = 0
(a-1)(a+2) = 0
a = 1, -2

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Binary Operations

Lee Azịza

Ajụjụ 25 Ripọtì

Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR

11o

73o

107o

138o

149o

Akọwa Nkọwa

QPS - QRK = 31o
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Lee Azịza

Ajụjụ 26 Ripọtì

Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 10, 9, 7, 10, 6, 5

16

14

12

10

Akọwa Nkọwa

To find the range, we need to subtract the smallest number from the largest number in the set. The smallest number in the set is 4 and the largest number is 10. So, the range is 10 - 4 = 6. To find the mode, we need to determine the number that occurs the most frequently in the set. In this case, the number 10 occurs the most frequently, appearing 7 times. So, the mode is 10. To find the sum of the range and the mode, we simply add the two values together. Thus, the sum of the range and the mode is 6 + 10 = 16. Therefore, the answer is option (A) 16.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Representation Of Data

Sets

Measures Of Location

Statistics

Lee Azịza

Ajụjụ 27 Ripọtì

If the population of a town was 240,000 in January 1998 and it increased by 2% each year, what would be the population of the town in January, 2000?

480,000

249,696

249,600

244,800

Akọwa Nkọwa

To solve this problem, we can use the formula for exponential growth, which is: A = P(1 + r)^t where A is the final amount, P is the initial amount, r is the annual growth rate as a decimal, and t is the time in years. In this problem, we are given P = 240,000, r = 0.02, and t = 2 (since we want to find the population after two years). We can plug these values into the formula and simplify: A = 240,000(1 + 0.02)^2 A = 240,000(1.02)^2 A = 240,000(1.0404) A = 249,696 Therefore, the population of the town in January 2000 would be 249,696. So the answer is 249,696.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Representation Of Data

Lee Azịza

Ajụjụ 28 Ripọtì

Evaluate 5-3log52 x 22log23

8

11/8

2/5

1/8

Akọwa Nkọwa

5-3log52 x 22log23i Let -3log52 = p => log52-3 = p
∴2-3 = 5p
∴5-3log52 = 5log52-3 = 5pii 22log23 = q => log532 = q
∴32 = 2q
∴222log23 = 2q
= 32 = 9
i x ii = 2-3 x 9 = 1/23 x 9
= 1/8 x 9
= 9/8 = 11/8

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 29 Ripọtì

The cumulative frequency curve represents the ages of ages of students in a school. What age group do 70% of the students belongs?

17.5 - 20.5

16.5 - 19.5

15.5 - 19.5

15.5 - 18.5

Representation Of Data

Algebraic Expressions

Lee Azịza

Ajụjụ 30 Ripọtì

If y = 2x - sin2x, find dy/dx when x = π/4

π

-π

π/2

-π/2

Akọwa Nkọwa

y = 2x cos2x - sin2x
dy/dx = 2 cos2x +(-2x sin2x) - 2 cos2x
= 2 cos2x - 2x sin2x - cos2x
= -2x sin2x
= -2 x (π/4) sin2 x (π/4)
= -(π/2) x 1 = -(π/2)

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 31 Ripọtì

A ship sails a distance of 50km in the direction S50°E and then sails a distance of 50km in the direction ₦40°E. FInd the bearing of the ship from the original position.

S90°E

₦40°E

S95°E

₦85°E

Algebraic Expressions

Lee Azịza

Ajụjụ 32 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

Permutation And Combination

Lee Azịza

Ajụjụ 33 Ripọtì

if (x - 1), (x + 1) and (x - 2) are factors of the polynomial ax3 + bx2 + cx - 1, find a, b, c in that order.

-1/2, 1., 1/2

1/2, 1, 1/2

1/2, 1, -1/2

1/2, -1, 1/2

Akọwa Nkọwa

This is a polynomial of the 3rd order, thus x should have three answers. Use the factors given to get values of x as 1, -1 and -2.
Form three equations, and carry out elimination and subsequent substitution to get a = -1/2, b = 1, and c = 1/2

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 34 Ripọtì

If the mean of the numbers 0, (x+2), (3x+6), and (4x+8) is 4, find their mean deviation.

zero

2

3

4

Akọwa Nkọwa

The mean deviation of a set of numbers is the average of the absolute deviations from their mean. To find the mean deviation of the given set of numbers, we need to first find their mean. Mean = (0 + (x+2) + (3x+6) + (4x+8))/4 Mean = (8x + 16)/4 Mean = 2x + 4 We are given that the mean is 4. Therefore, 2x + 4 = 4 2x = 0 x = 0 Substituting x = 0 in the set of numbers, we get: 0, 2, 6, 8 Mean = (0+2+6+8)/4 = 4 To find the mean deviation, we need to find the deviation of each number from the mean and then take the average of their absolute values. Deviations from the mean: 0-4 = -4 2-4 = -2 6-4 = 2 8-4 = 4 Mean deviation = (|-4| + |-2| + |2| + |4|)/4 Mean deviation = (12/4) Mean deviation = 3 Therefore, the answer is 3.

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Representation Of Data

Number Bases

Lee Azịza

Ajụjụ 35 Ripọtì

The 3rd term of an A.P is 4x - 2y and the 9th term is 10x - 8y. Find the common difference.

19x - 17y

8x - 4y

x - y

2x

Akọwa Nkọwa

In an Arithmetic Progression (A.P), the difference between any two adjacent terms is constant. Let d be the common difference of the A.P, then: - The 3rd term is the first term plus two common differences, i.e. a + 2d = 4x - 2y - The 9th term is the first term plus eight common differences, i.e. a + 8d = 10x - 8y We can now solve for d by subtracting the first equation from the second: a + 8d - (a + 2d) = 10x - 8y - (4x - 2y) 6d = 6x - 6y d = (x - y) Therefore, the common difference is x - y, which is option (C).

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 36 Ripọtì

If 31410 - 2567 = 340x, find x.

7

8

9

10

Akọwa Nkọwa

31410 - 2567 = 340x,
Convert 2567 and 340x to base 10, such that:
314 - 139 = 3x2 + 4x
=> 3x2 + 4x - 175 = 0 (quadratic)
Factorising, (x - 7) (3x + 25) = 0,
either x = 7 or x = -25/3 ( but x cannot be negative)
Therefore, x = 7.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 37 Ripọtì

In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40o. Calculate ∠QSR.

30o

40o

45o

50o

Circles

Lee Azịza

Ajụjụ 38 Ripọtì

Solve the inequality 2 - x > x2.

x < -2 or x > 1

x > 2 or x< -1

-1 < x < 2

-2 < x < 1

Akọwa Nkọwa

2 - x > x2
2 - x - x2 > 0
(x+2)(x-1) < 0

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Statistics

Lee Azịza

Ajụjụ 39 Ripọtì

Given that the various faces of a fair dice 1, 2, 3, 4, 5, 6 appeared 30, 43, 54, 40, 41, 32 times respectively in a single toss. Picture the figures as being represented in a simple table with number (X) against frequency (f).

If a pie chart is used to depict the data, the angle corresponding to 4 is?

10°

16°

40°

60°

Akọwa Nkọwa

To find the angle corresponding to 4 in the pie chart, we need to calculate the percentage of times 4 appeared in the toss. Total frequency = 30 + 43 + 54 + 40 + 41 + 32 = 240 Frequency of 4 = 40 Percentage of times 4 appeared = (Frequency of 4 / Total frequency) x 100% = (40/240) x 100% = 16.67% The pie chart represents a circle of 360°. Therefore, to find the angle corresponding to 4 in the pie chart, we need to calculate 16.67% of 360°. Angle corresponding to 4 = (16.67/100) x 360° = 60° Therefore, the angle corresponding to 4 in the pie chart is 60°. Option (d) is the correct answer.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Probability

Linear Inequalities

Lee Azịza

Ajụjụ 40 Ripọtì

In the diagram above, EFGH is a circle center O. Fh is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH.

16 cm

20 cm

26 cm

32 cm

Akọwa Nkọwa

To solve this problem, we need to use the property of a circle that states that a diameter is twice the radius of the circle. Since FH is a diameter, then FH = 2OH, where OH is the radius of the circle. Since GE is a chord that meets FH at right angle at the point N, then N is the midpoint of FH. Therefore, NH = NF = 8 cm. Let's use the Pythagorean theorem to find OH. Since GE is a chord of the circle, we can draw radii from the center O to the endpoints of the chord so that we have a right triangle. Let x be the length of OE. Then, HG = 2x, and OH = x + 12 (since FH = 2OH = 2(x + 12) = 2x + 24). Using the Pythagorean theorem, we have: x^2 + 12^2 = (2x)^2 x^2 + 144 = 4x^2 3x^2 = 144 x^2 = 48 x = sqrt(48) = 4sqrt(3) Therefore, OH = x + 12 = 4sqrt(3) + 12. Finally, we can calculate FH: FH = 2OH = 2(4sqrt(3) + 12) = 8sqrt(3) + 24 ≈ 32.39 cm. Therefore, the answer is 32 cm.

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Circles

Coordinate Geometry

Lee Azịza

Ajụjụ 41 Ripọtì

Simplify $ \frac{1}{1+\sqrt{5}} - \frac{1}{1-\sqrt{5}} $

-

\frac{1}{2} \sqrt{5}

\frac{1}{2} \sqrt{5}

--

\frac{1}{4} \sqrt{5}

5

Akọwa Nkọwa

\frac{1}{1 + \sqrt{5}}

-

\frac{1}{1 - \sqrt{5}}

=

\frac{3 - \sqrt{5} - 3 - \sqrt{5}}{(3 + \sqrt{5}) (3 - \sqrt{5}}

=

\frac{- 2 \sqrt{5}}{9 - 5}

=

\frac{- 2 \sqrt{5}}{4}

= -

\frac{1}{2} \sqrt{5}

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 42 Ripọtì

If $U = \{x : x\text{ is an integer and }1 \le x \le 20\}$

$E_1 = \{x : x\text{ is a multiple of }3\}$

$E_2 = \{x : x\text{ is a multiple of }4\}$

and an integer is picked at random from $U$, find the probability that it is not in $E_2$

3/4

3/10

1/4

1/20

Akọwa Nkọwa

U = {1, 2, 3, 4, 5,..., 20}
E1 = {3, 6, 9, 12, 15, 18}
E2 = {4, 8, 12, 16, 20}
P(E1) = 5/20
P(not E1) = 1 - (5/20) = 15/20 = 3/4

Lee Azịza

Ajụjụ 43 Ripọtì

If the diagram is the graph of y = x2, the shaded area is

64 square units

\frac{126}{4}

square units

32 square units

\frac{64}{3}

square units

Akọwa Nkọwa

A =

\int_{x = 0}^{x = 4}

ydx =

\int_{0}^{4}

x2dx = [

\frac{x^{3}}{3} + c]_{0}^{4}

= $\frac{64}{3}$ square units

Lee Azịza

Ajụjụ 44 Ripọtì

Let P = {1, 2, u, v, w, x}; Q = {2, 3, u, v, w, 5, 6, y} and R = {2, 3, 4, v, x, y}.

Determine (P-Q) ? R

{1, x}

{x y}

{x}

ɸ

Akọwa Nkọwa

To determine (P-Q) ? R, we need to first find the set difference (P-Q), which is the set of elements that are in P but not in Q. So, we need to remove all the elements in Q from P. (P-Q) = {1, w, x} Then, we take the intersection of (P-Q) and R, which means we need to find all the elements that are common to both (P-Q) and R. (P-Q) ? R = {x} Therefore, the answer is {x}.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Lee Azịza

Ajụjụ 45 Ripọtì

Simplify 3(2n+1) - 4(2n-1) 2n+1 - 2n

2n+1

2n-1

4

1/4

Akọwa Nkọwa

Start by expanding 3(2n+1) - 4(2n-1) 2n+1 - 2n:
3 x 2n x 21 - 22 x 2n x 2-1 2n x 2 -2n
Solving the equation above gives;
2n x 21 x 2 2n = 2 x 2 = 4

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 46 Ripọtì

If $\frac{(2\sqrt{3}-\sqrt{2})}{(\sqrt{3}+2\sqrt{2})}=m+n\sqrt{6}$, find the values of m and n respectively.

1, -2

-2, 1

\frac{- 2}{5}

, 1

2, 3/5

Akọwa Nkọwa

Rationalize

\frac{(2 \sqrt{3} - \sqrt{2})}{(\sqrt{3} + 2 \sqrt{2})}

and equate to

m + n \sqrt{6}

. Such that m = -2, and n = 1.

Lee Azịza

Ajụjụ 47 Ripọtì

A trader realizes 10x - x2 naira profit from the sale of x bags on corn. How many bags will give him the desired profit?

4

5

6

7

Akọwa Nkọwa

The trader's profit function can be represented as P(x) = 10x - x^2, where x is the number of bags sold and P(x) is the profit obtained from selling x bags of corn. To find the number of bags that will give him the desired profit, we need to maximize the profit function P(x). We can do this by finding the derivative of P(x) with respect to x, and then setting it to zero to find the critical point(s) of the function. P'(x) = 10 - 2x Setting P'(x) to zero gives: 10 - 2x = 0 Solving for x gives: x = 5 Therefore, the trader will make the maximum profit when he sells 5 bags of corn. This means that the answer is, 5.

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza






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