JAMB UTME - Mathematics - 2000

Tambaya 1 Rahoto

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

Bayanin Amsa

Karanta bayanin darasi akan Polynomials (JAMB)

Polynomials

Duba Amsa

Tambaya 2 Rahoto

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}

Bayanin Amsa

\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}

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Duba Amsa

Tambaya 3 Rahoto

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

x√2

2x2

Bayanin Amsa

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

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Measures Of Dispersion

Duba Amsa

Tambaya 4 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Representation Of Data

Duba Amsa

Tambaya 5 Rahoto

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

Bayanin Amsa

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 6 Rahoto

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°

Bayanin Amsa

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)

Karanta bayanin darasi akan Loci (WAEC)

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Loci

Euclidean Geometry

Duba Amsa

Tambaya 7 Rahoto

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?

Bayanin Amsa

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Permutation And Combination

Duba Amsa

Tambaya 8 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 9 Rahoto

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}

Bayanin Amsa

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}.

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

Sets

Duba Amsa

Tambaya 10 Rahoto

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

-π

π/2

-π/2

Bayanin Amsa

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)

Karanta bayanin darasi akan Differentiation (JAMB)

Differentiation

Duba Amsa

Tambaya 11 Rahoto

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

2n+1

2n-1

1/4

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 12 Rahoto

Evaluate 5-3log52 x 22log23

11/8

2/5

1/8

Bayanin Amsa

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

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 13 Rahoto

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

Bayanin Amsa

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.

Duba Amsa

Tambaya 14 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Functions And Relations (WAEC)

Functions And Relations

Duba Amsa

Tambaya 15 Rahoto

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

π/2

-π/2

-π

Bayanin Amsa

\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

Karanta bayanin darasi akan Integration (JAMB)

Integration

Duba Amsa

Tambaya 16 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Sets

Quadratic Equations

Duba Amsa

Tambaya 17 Rahoto

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

Bayanin Amsa

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".

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Circles

Euclidean Geometry

Duba Amsa

Tambaya 18 Rahoto

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)

p/(p+1)

Bayanin Amsa

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)

Karanta bayanin darasi akan Matrices And Determinants (JAMB)

Karanta bayanin darasi akan Binary Operations (JAMB)

Matrices And Determinants

Binary Operations

Duba Amsa

Tambaya 19 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 20 Rahoto

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

Bayanin Amsa

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).

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 21 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 22 Rahoto

In how many ways can the word MATHEMATICS be arranged?

11!/(9! 2!)

11!/(9! 2! 2!)

11!/(2! 2! 2!)

11!/(2! 2!)

Bayanin Amsa

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

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Permutation And Combination

Duba Amsa

Tambaya 23 Rahoto

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

0.056

0.055

0.054

0.54

Bayanin Amsa

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 24 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Binary Operations (JAMB)

Binary Operations

Duba Amsa

Tambaya 25 Rahoto

Solve the inequality 2 - x > x2.

x < -2 or x > 1

x > 2 or x< -1

-1 < x < 2

-2 < x < 1

Bayanin Amsa

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

Karanta bayanin darasi akan Statistics (WAEC)

Statistics

Duba Amsa

Tambaya 26 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Differentiation (JAMB)

Algebraic Expressions

Differentiation

Duba Amsa

Tambaya 27 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Karanta bayanin darasi akan Statistics (WAEC)

Coordinate Geometry

Statistics

Duba Amsa

Tambaya 28 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 29 Rahoto

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

Bayanin Amsa

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

Duba Amsa

Tambaya 30 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Polynomials (JAMB)

Polynomials

Duba Amsa

Tambaya 31 Rahoto

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°

Bayanin Amsa

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.

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Probability

Linear Inequalities

Duba Amsa

Tambaya 32 Rahoto

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

Bayanin Amsa

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.

Duba Amsa

Tambaya 33 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Differentiation (JAMB)

Differentiation

Duba Amsa

Tambaya 34 Rahoto

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?

Bayanin Amsa

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

Karanta bayanin darasi akan Sets (JAMB)

Sets

Duba Amsa

Tambaya 35 Rahoto

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

zero

Bayanin Amsa

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.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Number Bases (WAEC)

Representation Of Data

Number Bases

Duba Amsa

Tambaya 36 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Areas (WAEC)

Volumes

Areas

Duba Amsa

Tambaya 37 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Triangles And Polygons

Duba Amsa

Tambaya 38 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 39 Rahoto

If 31410 - 2567 = 340x, find x.

Bayanin Amsa

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.

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 40 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Sets (WAEC)

Karanta bayanin darasi akan Measures Of Location (JAMB)

Karanta bayanin darasi akan Statistics (WAEC)

Representation Of Data

Sets

Measures Of Location

Statistics

Duba Amsa

Tambaya 41 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Trigonometry (JAMB)

Karanta bayanin darasi akan Differentiation (JAMB)

Karanta bayanin darasi akan Simple Operations On Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Trigonometry

Differentiation

Simple Operations On Algebraic Expressions

Triangles And Polygons

Duba Amsa

Tambaya 42 Rahoto

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

Bayanin Amsa

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.

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Circles

Coordinate Geometry

Duba Amsa

Tambaya 43 Rahoto

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

Bayanin Amsa

A =

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

ydx =

\int_{0}^{4}

x2dx = [

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

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

Duba Amsa

Tambaya 44 Rahoto

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

Bayanin Amsa

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Representation Of Data

Algebraic Expressions

Duba Amsa

Tambaya 45 Rahoto

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

2/3

zero

-2/3

-1

Bayanin Amsa

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.

Karanta bayanin darasi akan Fractions, Decimals And Approximations (WAEC)

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Number Bases (WAEC)

Fractions, Decimals And Approximations

Representation Of Data

Number Bases

Duba Amsa

Tambaya 46 Rahoto

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

160o

140o

120o