JAMB UTME - Mathematics - 2002

The range of the data: k+2, k-3, k+4, k-2, k, k-5, k+3, k-1, and k+6 is

10

11

6

8

Akọwa Nkọwa

To find the range of a data set, we need to subtract the smallest value from the largest value. In this case, the smallest value is k-5 and the largest value is k+6. Therefore, the range is (k+6) - (k-5) = k+6 - k+5 = 11. So the answer is 11.

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Sets

Lee Azịza

Ajụjụ 2 Ripọtì

A trader bought goats for N4000 each. He sold them for N180,000 at a loss of 25%. How many goats did he buy?

60

50

45

36

Akọwa Nkọwa

Let's start by using a bit of math to understand the problem. We know that the trader bought each goat for N4000, and sold them for N180,000. We also know that he sold them at a loss of 25%. If the trader sold the goats at a loss of 25%, it means that he sold them for only 75% of their original value. We can calculate the original value of the goats by dividing the selling price by 75% or multiplying it by 4/3. So, the original value of the goats = 180,000 * (4/3) = 240,000. Since the trader bought each goat for N4000, we can find out how many goats he bought by dividing the total value of the goats (N240,000) by the price he paid for each goat (N4000). Number of goats = 240,000 / 4000 = 60. Therefore, the trader bought 60 goats. To summarize, the trader bought 60 goats and sold them for N180,000 at a loss of 25%. We found the answer by calculating the original value of the goats and dividing it by the price the trader paid for each goat.

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

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

Fractions, Decimals And Approximations

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 3 Ripọtì

Using the graph find the values of $p$ and $q$ if $px + qy \ge 4$

p = 2, q = -1

p = -1, q = 2

p = 2, q = 2

p = 1, q = 2

Akọwa Nkọwa

m =

\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 - 0}{0 - (4)} = \frac{2}{4} = \frac{1}{2}

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} \geq m$

$\frac{y - 0}{x + 4} \geq \frac{1}{2}$

2y $\geq$ x + 4, -x + 2y $\geq$ 4 = px + qy $\geq$ 4

p = -1, q = 2

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

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

Solution Of Linear Equations

Algebraic Expressions

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 4 Ripọtì

Simplify (√0.7 + √70)2

84.7

70.7

217.7

168.7

Akọwa Nkọwa

To simplify (√0.7 + √70)², we can use the formula (a + b)² = a² + 2ab + b², where a = √0.7 and b = √70. So, we have: (√0.7 + √70)² = (√0.7)² + 2(√0.7)(√70) + (√70)² Simplifying, we get: 0.7 + 2√(0.7 × 70) + 70 = 0.7 + 2√49 + 70 = 0.7 + 2(7) + 70 = 84.7 Therefore, the answer is 84.7.

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 5 Ripọtì

Make $r$ subject of the formula given that $\frac{x}{r+a}=\frac{a}{r}$

\frac{a^{2}}{(x - a)}

\frac{a^{2}}{(x + a)}

\frac{a}{x - a}

\frac{a}{x + a}

Introductory Calculus

Change Of Subject Of A Formula/Relation

Lee Azịza

Ajụjụ 6 Ripọtì

How many three-digit numbers can be formed from 32564 without repeating any of the digits?

120

10

20

60

Permutation And Combination

Lee Azịza

Ajụjụ 7 Ripọtì

A circle with radius 5cm has its radius increasing at the rate of 0.2m/s. What will be the corresponding increase in the area?

2π

5π

π

4π

Circles

Lee Azịza

Ajụjụ 8 Ripọtì

In a school, 220 students offer Biology or Mathematics or both, 125 offer Biology and 110 Mathematics. How many offer Biology but not Mathematics?

95

80

125

110

120

Sets

Surds (radicals)

Lee Azịza

Ajụjụ 9 Ripọtì

The inverse function f(x) = 3x + 4 is

(x-4)/3

(x-5)/5

(x+3)/4

(x+4)/3

Logarithms

Lee Azịza

Ajụjụ 10 Ripọtì

The triangle PQR above is

an obtuse-angled triangle

a scalene triangle

an isosceles triangle

an equilateral triangle

Akọwa Nkọwa

∠PQR = 180 - 128 (∠s on a straight line)
= 52

^{\circ}

∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52

^{\circ}

∴ΔPQR is an isosceles triangle

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Lee Azịza

Ajụjụ 11 Ripọtì

Find the value of a if the line 2y - ax + 4 = 0 is perpendicular to the line y + (x/4) - 7 = 0

-4

4

8

-8

Algebraic Expressions

Application Of Differentiation

Lee Azịza

Ajụjụ 12 Ripọtì

The slope of the tangent to the curve $y = 2x^2 - 2x + 5$ at the point $(1, 6)$ is

4

1

6

5

3

Introductory Calculus

Trigonometry

Lee Azịza

Ajụjụ 13 Ripọtì

Simplify 52.4 - 5.7 - 3.45 - 1.75

41.4

41.5

42.1

42.2

Akọwa Nkọwa

To simplify the given expression, we can simply subtract the given values as follows: 52.4 - 5.7 - 3.45 - 1.75 = 41.5 Therefore, the answer is 41.5.

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

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 14 Ripọtì

The ninary operation * is defined on the set of integers p and q by p*q = pq + p + q. Find 2 * (3 * 4).

59

19

67

38

Binary Operations

Lee Azịza

Ajụjụ 15 Ripọtì

The time taken to do a piece of work is inversely proportional to the number of men employed. if it takes 45 men to do a piece of work in 5 days, how long will it take 25 men?

15 days

12 days

5 days

9 days

Akọwa Nkọwa

The time taken to do a piece of work is inversely proportional to the number of men employed, which means that the more men there are, the less time it will take to complete the work. In this case, we are told that 45 men can do the work in 5 days. If we let x be the number of days it takes for 25 men to do the work, then we can set up a proportion: 45 men x 5 days = 25 men x (number of days) Solving for x, we get: x = (45 x 5) / 25 = 9 days Therefore, it would take 25 men 9 days to do the same amount of work that 45 men can do in 5 days. The answer is 9 days.

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Variation

Lee Azịza

Ajụjụ 16 Ripọtì

If the 9th term of an A.P is five times the 5th term, find the relationship between a and d.

2a + 2 = 0

3a + 5d = 0

a + 3d = 0

a + 2d = 0

Akọwa Nkọwa

Let the fifth term of the AP be 'a + 4d'. Then, the ninth term will be 'a + 8d'. The problem states that the ninth term is five times the fifth term, which can be represented mathematically as: a + 8d = 5(a + 4d) Simplifying the equation gives: a + 8d = 5a + 20d Subtracting 'a' and 20d from both sides gives: -12d = -4a Dividing both sides by -4 gives: 3d = a So the relationship between 'a' and 'd' is a = 3d. Thus, the answer is a + 3d = 0.

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Solution Of Linear Equations

Sequence And Series

Lee Azịza

Ajụjụ 17 Ripọtì

Find the coordinates of the mid-point of x and y intercepts of the line 2y = 4x - 8

(2, 0)

(1, -2)

(-1, -2)

(1, 2)

Akọwa Nkọwa

To find the x-intercept of the line 2y = 4x - 8, we set y = 0 and solve for x. So, we have: 2(0) = 4x - 8 4x = 8 x = 2 This means the x-intercept is (2, 0). To find the y-intercept of the line, we set x = 0 and solve for y. So, we have: 2y = 4(0) - 8 2y = -8 y = -4 This means the y-intercept is (0, -4). The midpoint of two points (x1, y1) and (x2, y2) is: ((x1 + x2)/2, (y1 + y2)/2) Using this formula, we can find the midpoint of the x and y intercepts: ((2 + 0)/2, (0 + (-4))/2) = (1, -2) Therefore, the coordinates of the midpoint of the x and y intercepts of the line 2y = 4x - 8 is (1, -2). Answer option number 2.

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

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

Coordinate Geometry

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 18 Ripọtì

Simplify $ \frac{3}{5} \div \left(\frac{2}{7} \times \frac{4}{3} \div \frac{4}{9}\right) $

\frac{4}{5}

\frac{7}{10}

\frac{6}{6}

\frac{21}{6}

Akọwa Nkọwa

35÷(27×43÷49)=35÷(27×43×94)=35÷67=35×76=710

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 19 Ripọtì

(0.21 x 0.072 x 0.00054) (0.006 x 1.68 x 0.063), correct to 4.s.f

0.01286

0.01285

0.1286

0.1285

Akọwa Nkọwa

To find the product of the two numbers, we can simply multiply them together: (0.21 x 0.072 x 0.00054) x (0.006 x 1.68 x 0.063) = 0.000007728 x 0.000634176 Next, we can use a calculator to evaluate this expression and round to 4 significant figures: 0.000007728 x 0.000634176 = 0.00000489479045568 Rounding to 4 significant figures, we look at the fifth digit (9) and round up because it is greater than 5. Therefore, the final answer is: 0.01286 So the answer is 0.01286, correct to 4 significant figures.

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

Gụọ akwụkwọ nkuzi na Percentages (WAEC)

Fractions, Decimals And Approximations

Percentages

Lee Azịza

Ajụjụ 20 Ripọtì

Evaluate ∫sin 3x dx

(2/3) cos 3x + c

(1/3) cos 3x + c

(-1/3) cos 3x + c

(-2/3) cos 3x + c

Akọwa Nkọwa

To integrate the given function ∫sin 3x dx, we use the trigonometric identity: sin (2a) = 2sin a cos a Using this identity, we can write: sin 3x = sin (2x + x) = sin 2x cos x + cos 2x sin x Therefore, ∫sin 3x dx = ∫(sin 2x cos x + cos 2x sin x) dx Integrating each term separately, we get: = (-cos 2x / 2) cos x + (sin 2x / 2) sin x + c where c is the constant of integration. Simplifying this expression, we get: = -(1/2) cos 2x cos x + (1/2) sin 2x sin x + c = -(1/2) (cos 2x cos x - sin 2x sin x) + c = -(1/2) cos (2x - x) + c = -(1/2) cos x + c Therefore, the answer is: ∫sin 3x dx = -(1/2) cos x + c Hence, the option is: (-1/3) cos 3x + c.

Lee Azịza

Ajụjụ 21 Ripọtì

Find the mean of the data: 7, -3, 4, -2, 5, -9, 4, 8, -6, 12

3

4

2

1

Akọwa Nkọwa

To find the mean (average) of a set of data, you sum up all the numbers in the set and then divide by the total count of numbers. Let's calculate the mean for the given data set:

\[ \text{Data Set: } 7, -3, 4, -2, 5, -9, 4, 8, -6, 12 \]

1. Sum of all numbers:

\[ 7 + (-3) + 4 + (-2) + 5 + (-9) + 4 + 8 + (-6) + 12 = 20 \]

2. Total count of numbers: There are 10 numbers in the data set.

3. Mean (Average):

\[ \text{Mean} = \frac{20}{10} = 2 \]

Therefore, the mean of the given data set is 2

Gụọ akwụkwọ nkuzi na Probability (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Probability

Statistics

Lee Azịza

Ajụjụ 22 Ripọtì

The probability of a student passing any examination is 2/3. If the students takes three examination, what is the probability that he will not pass any of them

2/3

4/9

8/27

1/27

Akọwa Nkọwa

The probability of passing any one of the three examinations is 2/3. Therefore, the probability of not passing any one of the exams is 1 - 2/3 = 1/3. Since the student takes three examinations, the probability of not passing any of them is the probability of not passing the first exam multiplied by the probability of not passing the second exam multiplied by the probability of not passing the third exam. So, the probability of not passing any of the three exams is (1/3) x (1/3) x (1/3) = 1/27. Therefore, the answer is option (D), which is 1/27.

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 23 Ripọtì

In the diagram, a cylinder is surmounted by a hemispherical bowl. Calculate the volume of the solid

198

π c m^{3}

180

π c m^{3}

162

π c m^{3}

216

π c m^{3}

Akọwa Nkọwa

V =

π r^{2} h + \frac{2}{3} π r^{3}

N.B h = r

V = $π x (3)^{2}$ x 20 = $\frac{2}{3} \times π \times (3)^{3}$

180 $π + 18 π = 198 c m^{3}$

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Volumes

Euclidean Geometry

Lee Azịza

Ajụjụ 24 Ripọtì

A solid hemisphere has a radius of 7 cm. Find the total surface area.

400 cm2

462 cm2

66 cm2

308 cm2

Lee Azịza

Ajụjụ 25 Ripọtì

Find the maximum value of y in the equation y = 1 - 2x - 3x2

5/4

5/3

3/4

4/3

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Quadratic Equations

Lee Azịza

Ajụjụ 26 Ripọtì

The sum of the interior angles of a polygon is 20 right angles. How many sides does the polygon have?

12

20

40

10

Akọwa Nkọwa

The sum of the interior angles of a polygon is given by the formula (n-2) x 180 degrees, where n is the number of sides of the polygon. In this problem, the sum of the interior angles is given as 20 right angles. One right angle is equal to 90 degrees, so 20 right angles would be 20 x 90 = 1800 degrees. Therefore, we can set up the equation (n-2) x 180 = 1800 and solve for n: (n-2) x 180 = 1800 n-2 = 10 n = 12 So the polygon has 12 sides. To summarize, the sum of the interior angles of a polygon can be calculated using (n-2) x 180 degrees, where n is the number of sides. Using this formula, we can solve for n when the sum of the interior angles is given. In this case, the polygon has 12 sides.

Gụọ akwụkwọ nkuzi na Mensuration (JAMB)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Mensuration

Triangles And Polygons

Lee Azịza

Ajụjụ 27 Ripọtì

In the diagram are two concentric circles of radii r and R respectively with centre O. If $r = \frac{2}{5}R$, express the area of the shaded portion in terms of $\pi$ and R

\frac{5}{9} π R^{2}

\frac{21}{25} π R^{2}

\frac{21}{23} π R^{2}

\frac{9}{25} π R^{2}

Akọwa Nkọwa

A

shaded portion

=

π R^{2} - π r^{2}

r =

\frac{2}{5}

R

A $shaded portion$ = $π R^{2} - π (\frac{2}{3} R)^{2} = π R^{2} - \frac{4 π R^{2}}{25}$

= $π R^{2} (1 - \frac{2}{3}) = \frac{21}{25} π R^{2}$

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 28 Ripọtì

The locus of a point P which is equidistant from two given points S and T is

the perpendicular bisector of ST

the angle bisector of PS and ST

a perpendicular to ST

a line parallel to ST

Akọwa Nkọwa

The correct answer is "the perpendicular bisector of ST". The locus of a point that is equidistant from two points S and T is the set of all points that are at the same distance from S and T. This set of points is a line that is equidistant from S and T, and this line is the perpendicular bisector of the line segment ST. Therefore, the answer is "the perpendicular bisector of ST".

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Loci

Euclidean Geometry

Lee Azịza

Ajụjụ 29 Ripọtì

The venn diagram above shows the numbers of students offering music and history in a class of 80 students. If a student is picked at random from the class, what is the probability that he offers Music only?

0.38

0.13

0.50

0.25

Akọwa Nkọwa

30 - x + x + 40 - x = 80 - 20
70 - x = 60
- x = 60 - 70
- x = - 10
∴ x = 10
Music only = 30 - x
= 30 - 10
20
P(music only) = 20/80
= 1/4
= 0.25

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Sets

Lee Azịza

Ajụjụ 30 Ripọtì

A bucket is 12 cm in diameter at the top, 8 cm in diameterat the bottom and 4 cm deep. Calculate its volume.

304π/3 cm3

144π cm3

128π cm3

72π cm3

Akọwa Nkọwa

To calculate the volume of the bucket, we need to use the formula for the volume of a frustum of a cone which is given by: V = (1/3) × π × h × (r₁² + r₂² + r₁ × r₂) where h is the height of the frustum, r₁ and r₂ are the radii of the top and bottom faces, respectively. In this case, the height of the frustum is 4 cm, the top radius is 6 cm (half of the diameter), and the bottom radius is 4 cm. Substituting these values into the formula, we get: V = (1/3) × π × 4 × (6² + 4² + 6 × 4) V = (1/3) × π × 4 × (36 + 16 + 24) V = (1/3) × π × 4 × 76 V = 304π/3 cubic centimeters Therefore, the volume of the bucket is 304π/3 cubic centimeters. Answer (1) is correct.

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Volumes

Areas

Angles

Lee Azịza

Ajụjụ 31 Ripọtì

Calculate the mean deviation of the set of numbers 7, 3, 14, 9, 7, and 8.

13/6

5/2

7/6

7/3

Akọwa Nkọwa

To find the mean deviation of a set of numbers, we first need to find the mean (average) of the set. Mean = (7+3+14+9+7+8) / 6 = 48/6 = 8 Next, we need to find the deviation of each number from the mean. To do this, we subtract the mean from each number: (7-8), (3-8), (14-8), (9-8), (7-8), (8-8) -1, -5, 6, 1, -1, 0 Notice that some of the deviations are negative and some are positive. To avoid canceling out the deviations that are opposite in sign, we take the absolute value of each deviation: 1, 5, 6, 1, 1, 0 Finally, we find the mean of these absolute deviations: Mean Deviation = (1+5+6+1+1+0) / 6 = 14/6 = 7/3 Therefore, the mean deviation of the given set of numbers is 7/3.

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Inequalities

Statistics

Lee Azịza

Ajụjụ 32 Ripọtì

If y = x2 - (1/x). find dy/dx

2x - (1/x2)

2x + x2

2x - x2

2x + (1/x2)

Akọwa Nkọwa

To find dy/dx, we need to differentiate y with respect to x using the rules of differentiation. Let's start by expressing y in terms of x: y = x^2 - (1/x) To differentiate this expression, we need to use the power rule and the quotient rule. The power rule states that if y = x^n, then dy/dx = nx^(n-1). The quotient rule states that if y = u/v, then dy/dx = (v du/dx - u dv/dx) / v^2. Using the power rule, we can find the derivative of x^2: d/dx (x^2) = 2x Using the quotient rule, we can find the derivative of (1/x): d/dx (1/x) = -1/x^2 Now we can use these results to find the derivative of y: dy/dx = d/dx (x^2) - d/dx (1/x) = 2x - (-1/x^2) = 2x + 1/x^2 Therefore, the answer is option D: 2x + (1/x^2).

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 33 Ripọtì

A chord of a circle subtends an angle of 12o degrees at the centre of a circle of diameter 4√3 cm. Calculate the area of the major sector.

4π cm2

32 π cm2

16 π cm2

8 π cm2

Circles

Lee Azịza

Ajụjụ 34 Ripọtì

In the diagram above, PST is a straight line, PQ = QS = RS. If ∠RST = $72^\circ$, find x

36

^{\circ}

18

^{\circ}

72

^{\circ}

24

^{\circ}

Akọwa Nkọwa

In Δ PQS, ∠PSQ = X(base ∠s of isoc Δ PQS)
In Δ QRS, ∠RQS = ∠PSQ + X(Extr ∠ = sum of two intr. opp ∠s)
∴ ∠RQS = X + X
= 2X
Also ∠QRS = 2X(base ∠s of isoc Δ QRS in Δ PRS,
72 = ∠RPS + ∠PRS(Extr, ∠ = sum of two intr. opp ∠s)
∴ 72 = x + 2x
72 = 3x
x = 72/3
x = 24

^{\circ}

Lee Azịza

Ajụjụ 35 Ripọtì

In the diagram, XZ is the diameter of the circle XYZW, with centre O and radius $ \frac{15}{2} $cm. If XY = 12cm, find the area of the triangle XYZ

45cm3

27cm3

75cm3

54cm3

Circles

Lee Azịza

Ajụjụ 36 Ripọtì

In the diagram above, a cylinder is summounted by a hemisphere bowl. Calculate the volume of the solid.

180 πcm3

162 πcm3

216 πcm3

198 πcm3

Akọwa Nkọwa

To find the volume of the solid, we need to find the volumes of the cylinder and the hemisphere bowl separately, and then add them together. The volume of the cylinder is given by the formula V_cylinder = πr^2h, where r is the radius of the base of the cylinder and h is the height of the cylinder. Looking at the diagram, we can see that the height of the cylinder is 12 cm and the radius of the base is also 12 cm (since the diameter of the cylinder is 24 cm). Therefore, the volume of the cylinder is: V_cylinder = π(12)^2(12) = 5,408π cm^3 The volume of the hemisphere bowl is given by the formula V_hemisphere = (2/3)πr^3, where r is the radius of the hemisphere. Again looking at the diagram, we can see that the radius of the hemisphere is also 12 cm (since it is the same as the radius of the base of the cylinder). Therefore, the volume of the hemisphere bowl is: V_hemisphere = (2/3)π(12)^3 = 9,216π/3 = 3,072π cm^3 To find the volume of the solid, we add the volume of the cylinder and the volume of the hemisphere bowl together: V_solid = V_cylinder + V_hemisphere = 5,408π + 3,072π = 8,480π cm^3 Therefore, the volume of the solid is 8,480π cm^3. : 198πcm^3 is not correct.

Gụọ akwụkwọ nkuzi na Volumes (WAEC)

Volumes

Lee Azịza

Ajụjụ 37 Ripọtì

In the diagram above are two concentric circles of radii r and R respectively with center O. If r = 2/3R, express the area of the shaded portion in terms of π and R

21/25πR2

9/25πR2

21/23πR2

5/9πR2

Akọwa Nkọwa

The area of the shaded region is equal to the difference between the areas of the two circles. The area of a circle is given by the formula A = πr^2, where r is the radius of the circle. Let's find the radius of the smaller circle: r = 2/3R Now we can express the area of the shaded region in terms of π and R: Area of shaded region = Area of larger circle - Area of smaller circle A = πR^2 - π(2/3R)^2 A = πR^2 - 4/9πR^2 A = (9/9πR^2 - 4/9πR^2) A = 5/9πR^2 Therefore, the answer is 5/9πR^2.

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 38 Ripọtì

If tan θ = 4/3, calculate sin2θ - cos2θ.

16/25

24/25

7/25

9/25

Trigonometry

Application Of Differentiation

Lee Azịza

Ajụjụ 39 Ripọtì

In the diagram PQ is parallel to RS. What is the value of $ \alpha + \beta + \gamma $

180

^{\circ}

90

^{\circ}

360

^{\circ}

200

^{\circ}

Angles And Intercepts On Parallel Lines

Lee Azịza

Ajụjụ 40 Ripọtì

The sum to infinity of the series: 1 + (1/3) + (1/9) + (1/27) + ... is

11/3

10/3

5/2

3/2

Akọwa Nkọwa

This is a geometric series with first term 1 and common ratio 1/3. The formula for the sum of an infinite geometric series is S = a/(1-r), where a is the first term and r is the common ratio. Plugging in the values, we get S = 1/(1 - 1/3) = 3/2. Therefore, the answer is 3/2.

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 41 Ripọtì

Find the range of values of x for which $ \frac{(x+2)}{4}-\frac{2x-3}{3}<4 $

x > -6

x > -3

x < 8

x < 4

Lee Azịza

Ajụjụ 42 Ripọtì

In the diagram above, $XZ$ is the diameter of the circle $XZW$, with center $O$ and radius $15/2$ cm. If $XY = 12$ cm, find the area of the triangle $XYZ$

54 cm2

45 cm2

27 cm2

75 cm2

Circles

Lee Azịza

Ajụjụ 43 Ripọtì

A hunter 1.6 m tall, views a bird on top of a tree at an angle of 45 degrees. if the distance between the hunter and the tree is 10.4 m, find the height of the tree.

9.0 m

12. 0 m

8.8 m

10.4 m

Coordinate Geometry

Number Bases

Lee Azịza

Ajụjụ 44 Ripọtì

The mean of a set of six numbers is 60. If the mean of the first five is 50, find the sixth number in the set.

105

100

95

110

Akọwa Nkọwa

The mean of a set of six numbers is 60. Therefore, the sum of the six numbers is 6 x 60 = 360. The mean of the first five is 50. Therefore, the sum of the first five numbers is 5 x 50 = 250. To find the sixth number, we can subtract the sum of the first five numbers from the sum of all six numbers: 6th number = sum of all six numbers - sum of the first five numbers 6th number = 360 - 250 6th number = 110 Therefore, the sixth number in the set is 110.

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Integration

Euclidean Geometry

Lee Azịza

Ajụjụ 45 Ripọtì

Solve for x in the equation x3 - 5x2 - x + 5 = 0

1, - 1, or 5

1, 1, or -5

-1, 1, or -5

1, 1, or 5

Akọwa Nkọwa

To solve this equation, we need to use algebraic methods to isolate the variable x on one side of the equation. First, we can combine like terms on the left-hand side of the equation: x^3 - 5x^2 - x + 5 = 0 Next, we can try to factor the equation, looking for factors that will make the equation equal to zero. By trial and error, we can see that (x-1) is a factor of the equation: (x-1)(x^2 - 4x - 5) = 0 Now we can use the zero product property, which tells us that if the product of two factors is zero, then at least one of the factors must be zero. So we can set each factor equal to zero and solve for x: x-1 = 0 or x^2 - 4x - 5 = 0 Solving for x-1, we get x = 1. Solving for x in the quadratic equation, we can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a In this case, a = 1, b = -4, and c = -5. Plugging these values into the quadratic formula, we get: x = (-(-4) ± sqrt((-4)^2 - 4(1)(-5))) / 2(1) Simplifying, we get: x = (4 ± sqrt(36)) / 2 x = (4 ± 6) / 2 x = 5 or x = -1 Therefore, the solutions for x are 1, 5, and -1.

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 46 Ripọtì

Find the equation of the set of points which are equidistant from the parallel lines x = 1 and x = 7

y = 3

x = 3

x = 4

y = 4

Akọwa Nkọwa

To find the set of points which are equidistant from the parallel lines x=1 and x=7, we can begin by finding the midpoint of the segment connecting the two parallel lines. The midpoint of the segment joining the two parallel lines is ((1+7)/2, 0) = (4,0). Now, let (x,y) be any point that is equidistant from the two parallel lines. Then, the distance from (x,y) to the line x=1 is |x-1|, and the distance from (x,y) to the line x=7 is |x-7|. Since the point (x,y) is equidistant from the two lines, we have: |x-1| = |x-7| Solving for x, we get: x-1 = -(x-7) or x-1 = x-7 Solving each equation for x, we get: x = 4 or x = -6 Since the distance from (x,y) to x=1 is the same as the distance from (x,y) to x=7, it follows that the set of points that are equidistant from the two parallel lines is the vertical line passing through the midpoint of the segment joining the two parallel lines, namely, the line x=4. Therefore, the equation of the set of points which are equidistant from the parallel lines x=1 and x=7 is x = 4. Thus, the correct option is x = 4.

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Loci

Circles

Coordinate Geometry

Lee Azịza

Ajụjụ 47 Ripọtì

The acres for rice, pineapple, cassava, cocoa, and palm oil in a certain district are given respectively as 2, 5, 3, 11, and 9. What is the angle sector for cassava in a pie chart?

108°

180°

36°

60°

Akọwa Nkọwa

To find the angle sector for cassava in a pie chart, we need to calculate what percentage of the total area is occupied by cassava. First, we need to find the total area of the pie chart, which represents the total acres of all the crops in the district. The total acres can be found by adding up the acres for each crop: 2 + 5 + 3 + 11 + 9 = 30 So the total area of the pie chart is 30 acres. Next, we need to find what percentage of the total area is occupied by cassava. We can do this by dividing the acres of cassava by the total acres and then multiplying by 100: (3 / 30) x 100 = 10% Therefore, the angle sector for cassava in the pie chart is 10% of the total area of the pie chart. To find the angle in degrees, we need to multiply the percentage by 360 (the total number of degrees in a circle): 10% x 360 = 36° So the angle sector for cassava in the pie chart is 36°.

Lee Azịza

Ajụjụ 48 Ripọtì

Without using tables, evaluate (343)1/3 x (

0.14)-1 x (25)-1/2

10

12

8

7

Akọwa Nkọwa

We can simplify the expression using the laws of exponents and basic arithmetic operations. First, let's evaluate each exponent: - (343)^(1/3) = 7, since 7^3 = 343 - (0.14)^(-1) = 1/0.14 ≈ 7.14 - 25^(-1/2) = 1/√25 = 1/5 Next, let's substitute these values back into the original expression: (343)^(1/3) x (0.14)^(-1) x 25^(-1/2) = 7 x 7.14 x 1/5 Multiplying these three numbers gives us: 7 x 7.14 x 1/5 ≈ 10 Therefore, the answer is 10.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 49 Ripọtì

If dy/dx = 2x - 3 and y = 3 when x = 0, find y in terms of x.

2x2 - 3x

x2 - 3x

x2 - 3x - 3

x2 - 3x + 3

Akọwa Nkọwa

To solve this problem, we need to integrate both sides of the equation with respect to x. dy/dx = 2x - 3 Integrating both sides with respect to x gives: y = x^2 - 3x + c where c is the constant of integration. To find the value of c, we use the fact that y = 3 when x = 0: 3 = 0^2 - 3(0) + c c = 3 Therefore, the equation of y in terms of x is: y = x^2 - 3x + 3

Gụọ akwụkwọ nkuzi na Differentiation (JAMB)

Differentiation

Lee Azịza

Ajụjụ 50 Ripọtì

Find the derivative of y = sin2(5x) with respect to x.

10 sin 5x cos 5x

5 sin5x cos 5x

2 sin 5x cos 5x

15 sin 5x cos 5x

Akọwa Nkọwa

To find the derivative of y = sin^2(5x), we use the chain rule of differentiation, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function. In this case, the outer function is sin^2 and the inner function is 5x. Thus, y' = 2 sin(5x) cos(5x) * 5 Simplifying this expression, we get: y' = 10 sin(5x) cos(5x) Therefore, the answer is: 10 sin(5x) cos(5x).

Lee Azịza






Pịa ma Jide iji Dọkpụrụ Ya
Pịa Ebe a ka Imechi