JAMB UTME - Mathematics - 2000

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

2n+1

2n-1

4 Ìdáhùn tó tọ́

1/4

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 2 Ìròyìn

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

11o

73o

107o Ìdáhùn tó tọ́

138o

149o

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Polynomials

Wo Idahun

Ibeere 3 Ìròyìn

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

y = (3x2/)2 + 2x Ìdáhùn tó tọ́

y = (3x2)/2 + x

y = 3x2 + (x/2)

y = 3x2 +2x

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Functions And Relations (WAEC)

Functions And Relations

Wo Idahun

Ibeere 4 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Quadratic Equations

Wo Idahun

Ibeere 5 Ìròyìn

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

2/3

zero

-2/3 Ìdáhùn tó tọ́

-1

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals And Approximations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Fractions, Decimals And Approximations

Representation Of Data

Number Bases

Wo Idahun

Ibeere 6 Ìròyìn

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

zero

3 Ìdáhùn tó tọ́

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Representation Of Data

Number Bases

Wo Idahun

Ibeere 7 Ìròyìn

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?

62 Ìdáhùn tó tọ́

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Sets

Wo Idahun

Ibeere 8 Ìròyìn

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

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

Ìdáhùn tó tọ́

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

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

Awọn alaye Idahun

\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}

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Wo Idahun

Ibeere 9 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

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).

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Sequence And Series

Wo Idahun

Ibeere 10 Ìròyìn

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

6 Ìdáhùn tó tọ́

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Triangles And Polygons

Wo Idahun

Ibeere 11 Ìròyìn

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 Ìdáhùn tó tọ́

-3/4

3/4

4/3

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Algebraic Expressions

Differentiation

Wo Idahun

Ibeere 12 Ìròyìn

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 Ìdáhùn tó tọ́

1/2, 1, 1/2

1/2, 1, -1/2

1/2, -1, 1/2

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 13 Ìròyìn

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

-π

π/2

-π/2 Ìdáhùn tó tọ́

Awọn alaye Idahun

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)

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Differentiation

Wo Idahun

Ibeere 14 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

A =

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

ydx =

\int_{0}^{4}

x2dx = [

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

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

Wo Idahun

Ibeere 15 Ìròyìn

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 Ìdáhùn tó tọ́

0.50

0.57

1.80

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 16 Ìròyìn

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} Ìdáhùn tó tọ́

Awọn alaye Idahun

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}.

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Ka ìwé ẹ̀kọ́ nípa Sets (JAMB)

Sets

Wo Idahun

Ibeere 17 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (WAEC)

Circles

Coordinate Geometry

Wo Idahun

Ibeere 18 Ìròyìn

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

30o

40o

45o

50o Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 19 Ìròyìn

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 Ìdáhùn tó tọ́

16.5 - 19.5

15.5 - 19.5

15.5 - 18.5

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Representation Of Data

Algebraic Expressions

Wo Idahun

Ibeere 20 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Differentiation

Wo Idahun

Ibeere 21 Ìròyìn

If 31410 - 2567 = 340x, find x.

7 Ìdáhùn tó tọ́

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 22 Ìròyìn

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° Ìdáhùn tó tọ́

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Probability

Linear Inequalities

Wo Idahun

Ibeere 23 Ìròyìn

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 Ìdáhùn tó tọ́

0.20 m/s

Awọn alaye Idahun

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.

Wo Idahun

Ibeere 24 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Polynomials

Wo Idahun

Ibeere 25 Ìròyìn

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

0.056

0.055 Ìdáhùn tó tọ́

0.054

0.54

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 26 Ìròyìn

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

x√2

2x2 Ìdáhùn tó tọ́

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Measures Of Dispersion

Wo Idahun

Ibeere 27 Ìròyìn

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

5 Ìdáhùn tó tọ́

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Quadratic Equations

Wo Idahun

Ibeere 28 Ìròyìn

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 Ìdáhùn tó tọ́

8π cubic units

17π/2 cubic units

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Volumes (WAEC)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Volumes

Areas

Wo Idahun

Ibeere 29 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Ka ìwé ẹ̀kọ́ nípa Measures Of Location (JAMB)

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Representation Of Data

Sets

Measures Of Location

Statistics

Wo Idahun

Ibeere 30 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Binary Operations (JAMB)

Binary Operations

Wo Idahun

Ibeere 31 Ìròyìn

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 Ìdáhùn tó tọ́

3/10

1/4

1/20

Awọn alaye Idahun

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

Wo Idahun

Ibeere 32 Ìròyìn

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 Ìdáhùn tó tọ́

(1,1) and (1,2)

(0,0) and (1,1)

(√2,√2) only

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (WAEC)

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Coordinate Geometry

Statistics

Wo Idahun

Ibeere 33 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

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".

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Euclidean Geometry (JAMB)

Circles

Euclidean Geometry

Wo Idahun

Ibeere 34 Ìròyìn

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

5 Ìdáhùn tó tọ́

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 35 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 36 Ìròyìn

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° Ìdáhùn tó tọ́

80°

50°

Awọn alaye Idahun

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)

Ka ìwé ẹ̀kọ́ nípa Loci (WAEC)

Ka ìwé ẹ̀kọ́ nípa Euclidean Geometry (JAMB)

Loci

Euclidean Geometry

Wo Idahun

Ibeere 37 Ìròyìn

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?

45 Ìdáhùn tó tọ́

Awọn alaye Idahun

Ka ìwé ẹ̀kọ́ nípa Permutation And Combination (JAMB)

Permutation And Combination

Wo Idahun

Ibeere 38 Ìròyìn

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 Ìdáhùn tó tọ́

\frac{- 2}{5}

, 1

2, 3/5

Awọn alaye Idahun

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.

Wo Idahun

Ibeere 39 Ìròyìn

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 -1

p/(p-1) Ìdáhùn tó tọ́

p/(p+1)

Awọn alaye Idahun

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)

Ka ìwé ẹ̀kọ́ nípa Matrices And Determinants (JAMB)

Ka ìwé ẹ̀kọ́ nípa Binary Operations (JAMB)

Matrices And Determinants

Binary Operations

Wo Idahun

Ibeere 40 Ìròyìn

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 Ìdáhùn tó tọ́

249,600

244,800

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Representation Of Data

Wo Idahun

Ibeere 41 Ìròyìn

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

160o

140o

120o

100o Ìdáhùn tó tọ́

Awọn alaye Idahun

Wo Idahun

Ibeere 42 Ìròyìn

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

π/2

-π/2

-π Ìdáhùn tó tọ́

Awọn alaye Idahun

\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

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Integration

Wo Idahun

Ibeere 43 Ìròyìn

In how many ways can the word MATHEMATICS be arranged?

11!/(9! 2!)

11!/(9! 2! 2!)

11!/(2! 2! 2!) Ìdáhùn tó tọ́

11!/(2! 2!)

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Permutation And Combination (JAMB)

Permutation And Combination

Wo Idahun

Ibeere 44 Ìròyìn

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

2/3 cm

2 cm

1 cm Ìdáhùn tó tọ́

3 cm

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Ka ìwé ẹ̀kọ́ nípa Simple Operations On Algebraic Expressions (WAEC)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Trigonometry

Differentiation

Simple Operations On Algebraic Expressions

Triangles And Polygons

Wo Idahun

Ibeere 45 Ìròyìn

Solve the inequality 2 - x > x2.

x < -2 or x > 1

x > 2 or x< -1

-1 < x < 2

-2 < x < 1 Ìdáhùn tó tọ́

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Statistics

Wo Idahun

Ibeere 46 Ìròyìn

Evaluate 5-3log52 x 22log23

11/8 Ìdáhùn tó tọ́

2/5

1/8

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Logarithms

Wo Idahun

Ibeere 47 Ìròyìn

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 Ìdáhùn tó tọ́

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Sets (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Sets

Quadratic Equations

Wo Idahun






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