JAMB UTME - Mathematics - 2001

Tambaya 1 Rahoto

Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. Find the square of the mode.

49

121

25

64

Bayanin Amsa

Mode = score with highest frequency = 11.
Square of 11 = 121

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Integration

Linear Inequalities

Duba Amsa

Tambaya 2 Rahoto

If the gradient of the curve y = 2kx2 + x + 1 at x = 1 is 9, find k.

4

3

2

1

Bayanin Amsa

The given curve is y = 2kx² + x + 1, and we are asked to find the value of k when the gradient at x = 1 is 9. The gradient of a curve at a particular point is the slope of the tangent to the curve at that point. To find the gradient of this curve at x = 1, we need to differentiate it with respect to x: dy/dx = 4kx + 1 Now we can substitute x = 1 and set the result equal to 9, which gives: 4k(1) + 1 = 9 Simplifying this equation, we get: 4k = 8 Dividing both sides by 4, we obtain: k = 2 Therefore, the value of k that satisfies the given conditions is 2.

Duba Amsa

Tambaya 3 Rahoto

Factorize 4x2 - 9y2 + 20x + 25

(2x -3y + 5)(2x - 3y - 5)

(2x - 3y)(2x + 3y)

(2x - 3y +5)(2x + 3y + 5)

(2x + 5)(2x - 9y +5)

Bayanin Amsa

Given: 4x2 - 9y2 + 20x + 25
Collect like terms: 4x2 + 20x + 25 - 9y2
(2x + 5)(2x + 5) - 9y2
(2x + 5)2 - (3y)2
(2x - 3y +5)(2x + 3y + 5)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 4 Rahoto

Evaluate 21.05347 - 1.6324 x 0.43 to 3 decimal places

20.980

20.351

20.981

20.352

Bayanin Amsa

Hint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first.

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Fractions, Decimals And Approximations

Algebraic Expressions

Duba Amsa

Tambaya 5 Rahoto

⊗	k	l	m
k	l	m	k
l	m	k	l
m	k	l	m

The identity element with respect to the multiplication shown in the table above is

o

m

l

k

Binary Operations

Duba Amsa

Tambaya 6 Rahoto

Divide: ax3x - 26x2x + 156ax - 216 by a2x - 24ax + 108

ax - 2

ax + 2

ax - 8

ax - 6

Bayanin Amsa

To divide the given expression, we can use the long division method.

                  ax^2 - 26x + 156a - 216b
     ___________________________________
ax | ax^3 + 3x^2 - 26x^2 + 156ax - 216bx^2 - 24ax + 108
      -ax^3
      -----
           3x^2 - 216bx^2
           3x^2 - 3ax^2
           --------------
                     -216bx^2 + 3ax^2
                     -216bx^2 + 36ax^2
                     ------------------
                                    -33ax^2 + 156a - 216bx^2 - 24ax + 108
                                    -33ax^2 + 33bx^2
                                    ------------------
                                                156a - 33ax^2 - 24ax + 108
                                                156a - 33ax^2 - 156ax + 33ax
                                                -------------------------
                                                             -24ax + 108
                                                             -24ax + 24bx
                                                             -------------
                                                                       108 - 24bx
                                                                       108 - 2ax
                                                                       --------
                                                                                2ax - 24bx + 108

Therefore, the quotient is ax - 2 with a remainder of 2ax - 24bx + 108.

Hence, the answer is ax - 2.

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 7 Rahoto

If y = x sinx, find dy/dx when x = ?/2.

-?/2

-1

1

?/2

Bayanin Amsa

To find dy/dx, we need to take the derivative of y with respect to x. Using the product rule of differentiation, we have: dy/dx = (sinx) + (x)(cosx) So, when x = ?/2, we can substitute this value into the expression to get: dy/dx = (sin(?/2)) + (?/2)(cos(?/2)) We know that sin(?/2) = 1 and cos(?/2) = 0, so we can simplify the expression: dy/dx = (1) + (?/2)(0) dy/dx = 1 Therefore, the answer is 1.

Karanta bayanin darasi akan Differentiation (JAMB)

Differentiation

Duba Amsa

Tambaya 8 Rahoto

Find the locus of a point which moves such that its distance from the line y = 4 is a constant, k.

y = k + or - 4

y = 4 + or - k

y = 4 + k

y = k - 4

Bayanin Amsa

The locus of a point which moves such that its distance from the line y = 4 is a constant, k, is the set of all points which are equidistant from the line y = 4. This set of points is a pair of parallel lines, k units above and below the line y = 4. Therefore, the correct answer is: y = k + or - 4.

Karanta bayanin darasi akan Loci (JAMB)

Karanta bayanin darasi akan Loci (WAEC)

Loci

Duba Amsa

Tambaya 9 Rahoto

The graph shows the cumulative frequency of the distribution of masses of fertilizer for 48 workers in one institution. Which of the following gives the inter-quartile range?

\frac{1}{2} (Q_{3} - Q_{1})

Q3 - Q2

Q3 - Q3

Q3 - Q1

Probability

Linear Inequalities

Duba Amsa

Tambaya 10 Rahoto

A car dealer bought a second-hand car for ₦250,000 and spent ₦70,000 refurbishing it. He then sold the car for ₦400,000. What is the percentage gain?

60%

32%

25%

20%

Bayanin Amsa

The dealer bought the car for ₦250,000 and spent ₦70,000 refurbishing it, for a total cost of: Total cost = ₦250,000 + ₦70,000 = ₦320,000 The dealer then sold the car for ₦400,000. The profit made by the dealer is the difference between the selling price and the cost price: Profit = Selling price - Cost price Profit = ₦400,000 - ₦320,000 = ₦80,000 The percentage gain is the profit expressed as a percentage of the cost price: Percentage gain = (Profit / Cost price) x 100% Substituting the values we get: Percentage gain = (₦80,000 / ₦320,000) x 100% Percentage gain = 0.25 x 100% Percentage gain = 25% Therefore, the percentage gain made by the dealer is 25%, and the answer is.

Karanta bayanin darasi akan Percentages (WAEC)

Karanta bayanin darasi akan Fractions, Decimals, Approximations And Percentage (JAMB)

Percentages

Fractions, Decimals, Approximations And Percentage

Duba Amsa

Tambaya 11 Rahoto

If two graphs y = px2 + q and y = 2x2 -1 intersect at x = 2, find the value of p in terms q.

\frac{q - 8}{7}

\frac{7 - q}{4}

\frac{8 - q}{2}

\frac{7 + q}{8}

Bayanin Amsa

To find the value of p in terms of q, we can use the given information that the two graphs intersect at x = 2. This means that at x = 2, the values of y for both graphs are the same. So, we can substitute x = 2 into both equations and set them equal to each other: px^2 + q = 2x^2 - 1 Substituting x = 2: p(2)^2 + q = 2(2)^2 - 1 4p + q = 7 Solving for p in terms of q, we can isolate p: 4p = 7 - q p = (7 - q) / 4 Therefore, the value of p in terms of q is (7 - q) / 4.

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 12 Rahoto

Find the value of $ \theta $ in the diagram

60

^{\circ}

100

^{\circ}

120

^{\circ}

30

^{\circ}

Bayanin Amsa

Using cosine formula (t

\sqrt{3}

)2 = t2 + t2 - 2t2 cos

θ

3t2 = 2t2 - 2t2 cos $θ$ = 2t2(1 - cos $θ$ )

1 - cos $θ$ = $\frac{3 t^{2}}{2 t^{2}}$ = $\frac{3}{2}$

cos = 1 - $\frac{3}{2} = - \frac{1}{2}$

$θ$ = cos-1(- $\frac{1}{2}$ ) = 120 $^{\circ}$ and 240 $^{\circ}$

N.B 0 $\geq$ $θ$ 360

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 13 Rahoto

A sector of a circle of radius 7.2cm which subtends an angle of $300^\circ$ at the centre is used to form a cone. What is the radius of the base of the cone?

6cm

7cm

8cm

9cm

5cm

Loci

Duba Amsa

Tambaya 14 Rahoto

Simplify $ \left(\sqrt[3]{64a^3}\right)^{-1} $

4a

1/8a

8a

1/4a

Number Bases

Duba Amsa

Tambaya 15 Rahoto

Find the number of ways of selecting 8 subjects from 12 subjects for an examination

490

495

496

498

Bayanin Amsa

This question is asking us to find the number of ways to select 8 subjects out of 12 subjects. This is a combination problem since the order in which the subjects are selected does not matter. The formula for combination is nCr = n!/r!(n-r)!, where n is the total number of items, r is the number of items to be selected, and ! represents the factorial function. Applying this formula to the given problem, we have: 12C8 = 12!/8!(12-8)! = (12x11x10x9)/(4x3x2x1) = 495 Therefore, there are 495 ways of selecting 8 subjects out of 12 subjects. Hence, the correct option is 495.

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Permutation And Combination

Duba Amsa

Tambaya 16 Rahoto

Find the range of 1/6, 1/3, 3/2, 2/3, 8/9 and 4/3

3/4

5/6

7/6

4/3

Bayanin Amsa

To find the range of a set of numbers, we need to find the difference between the largest and smallest numbers in the set. The given set of numbers are: 1/6, 1/3, 3/2, 2/3, 8/9, and 4/3. The smallest number in the set is 1/6 and the largest number in the set is 3/2. So, the range of the set is: 3/2 - 1/6 = 9/6 - 1/6 = 8/6 = 4/3 Therefore, the answer is 4/3.

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 17 Rahoto

An operation * is defined on the set of real numbers by a*b = a + b + 1. If the identity elements is -1, find the inverse of the element 2 under *.

4

zero

2

-2

Bayanin Amsa

First, let's find the inverse of the element 2 under the operation * on the set of real numbers. The inverse of an element "a" under the operation "*" is another element "b" such that a*b = b*a = identity element. The identity element under the operation "*" is given as -1, so we have: 2 * b = b * 2 = -1 Substituting the value of the identity element, we get: 2 * b = b * 2 + 1 = -1 Simplifying the equation, we get: 2b = -3 b = -3/2 Therefore, the inverse of the element 2 under the operation * is -3/2. To check if this is correct, we can substitute the values of 2 and -3/2 into the operation * and see if we get the identity element: 2 * (-3/2) = -3/2 * 2 = -1 Since we get the identity element, we can confirm that the inverse of the element 2 under the operation * is -3/2. So, the answer to the question is option (C) -2.

Duba Amsa

Tambaya 18 Rahoto

The sixth term of an A.P is half of its twelfth term. The first term of the A.P is equal to

zero

half of the common difference

double the common difference

the common difference

Bayanin Amsa

Let's assume that the first term of the A.P is 'a', and the common difference is 'd'. We know that for an A.P, the nth term is given by: a + (n-1)d According to the problem, the sixth term is equal to half of the twelfth term. So, we can write: a + 5d = 1/2 (a + 11d) Multiplying both sides by 2 to eliminate the fraction: 2a + 10d = a + 11d Simplifying, we get: a = d Therefore, the first term of the A.P is equal to the common difference. So, the correct answer is "the common difference."

Karanta bayanin darasi akan Sequence And Series (WAEC)

Sequence And Series

Duba Amsa

Tambaya 19 Rahoto

A point P moves such that it is equidistant from Points Q and R. Find QR when PR = 8cm and angle PRQ = 30°

4√3cm

8cm

8√3cm

4cm

Bayanin Amsa

Hint: Make a sketch of the moving points such that the hypotenuse is 8cm and the adjacent x cm.
Cos 30 = x/8
x = 8 cos 30 = 8√3cm

Karanta bayanin darasi akan Angles Of Elevation And Depression (WAEC)

Karanta bayanin darasi akan Sets (JAMB)

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

Angles Of Elevation And Depression

Sets

Simple Operations On Algebraic Expressions

Duba Amsa

Tambaya 20 Rahoto

If 6Pr = 6, find the value of 6Pr+1

30

33

35

15

Bayanin Amsa

The notation 6P_r means the number of permutations of r objects selected from a set of 6 distinct objects. Since 6P_r = 6! / (6 - r)!, we know that 6P_r = 6 when r = 1, since 6P₁ = 6! / (6 - 1)! = 6! / 5! = 6. To find the value of 6P_r+1, we can simply multiply 6P_r by (6 - r) since we are selecting one additional object. So, 6P_r+1 = 6P_r * (6 - r) = 6 * (6 - 1) = 30 when r = 1. Therefore, the answer is 30.

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 21 Rahoto

The bar chart above shows different colours of passing a particular point of a certain street in two minutes. What fraction of the total number of cars

3/25

2/25

1/5

4/15

Bayanin Amsa

The total number of cars that passed the point on the street in two minutes is not given in the chart. Therefore, we cannot determine the fraction of the total number of cars from the information provided. The chart only shows the relative frequency or proportion of cars of different colors that passed the point during the two-minute observation period. For example, if the chart shows that 3 out of 25 cars were blue, then the proportion of blue cars passing the point during the two minutes was 3/25. To find the fraction of the total number of cars, we would need to know the total number of cars that passed the point during the two-minute observation period.

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

Karanta bayanin darasi akan Sets (WAEC)

Positive And Negative Integers, Rational Numbers

Sets

Duba Amsa

Tambaya 22 Rahoto

Evaluate (0.142 x 0.275) ÷ 7(0.02) to 3 decimal places.

0.039

0.358

0.033

0.308

Bayanin Amsa

(0.142 x 0.275) 7(0.02) = (0.14 x 0.14 x 0.275) ÷ (7 x 0.02) = (0.14 x 0.14 x 0.275)/(0.14)
=0.14 x 0.275 = 0.0385 Approx. 0.039

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

Karanta bayanin darasi akan Percentages (WAEC)

Fractions, Decimals And Approximations

Percentages

Duba Amsa

Tambaya 23 Rahoto

Find the principal which amounts to ₦5,500 at a simple interest in 5 years at 2% per annum.

₦4,900

₦5,000

₦4,700

₦4,800

Bayanin Amsa

Principal, P = Amount, A - Interest, I.
A = P + I
I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10
But A = P + I,
=> 5500 = P + (P/10)
=> 55000 = 10P + P
=> 55000 = 11P
Thus P = 55000/11 = ₦5,000

Karanta bayanin darasi akan Sine, Cosine And Tangent Of An Angle (WAEC)

Karanta bayanin darasi akan Areas (WAEC)

Sine, Cosine And Tangent Of An Angle

Areas

Duba Amsa

Tambaya 24 Rahoto

P(-6, 1) and Q(6, 6) are the two ends of the diameter of a given circle. Calculate the radius.

6.5 units

13.0 units

3.5 units

7.0 units

Bayanin Amsa

The diameter of a circle is the distance between any two points on the circle passing through the center of the circle. In this problem, we are given the two endpoints of the diameter of a circle: P(-6, 1) and Q(6, 6). We can find the distance between these two points using the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2) where d is the distance between the two points (P and Q in this case), and (x1, y1) and (x2, y2) are the coordinates of the two points. So, substituting the given values in the distance formula, we get: d = sqrt((6 - (-6))^2 + (6 - 1)^2) d = sqrt(12^2 + 5^2) d = sqrt(144 + 25) d = sqrt(169) d = 13 Therefore, the diameter of the circle is 13 units. The radius of the circle is half the diameter, so: radius = diameter / 2 = 13 / 2 = 6.5 Hence, the radius of the circle is 6.5 units. So, the answer is.

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 25 Rahoto

The chord ST of a circle is equal to the radius, r, of the circle. Find the length of arc ST.

πr/6

πr/2

πr/12

πr/3

Bayanin Amsa

In a circle, the length of an arc is proportional to the angle it subtends at the center of the circle. Since ST is a chord and its length is equal to the radius of the circle, then angle SOT, where O is the center of the circle, is 60 degrees (because it subtends an equilateral triangle). Therefore, the length of arc ST is 1/6 of the circumference of the circle (because 60 degrees is 1/6 of a full revolution, which is 360 degrees). The circumference of the circle is 2πr, so the length of arc ST is: Length of arc ST = 1/6 * 2πr = πr/3 Hence, the answer is option D: πr/3.

Karanta bayanin darasi akan Loci (WAEC)

Karanta bayanin darasi akan Circles (WAEC)

Loci

Circles

Duba Amsa

Tambaya 26 Rahoto

Evaluate ∫2(2x-3)2/3dx

3/5(2x-3)5/3 + k

6/5(2x-3)5/3 + k

2x-3+k

2(2x-3)+k

Bayanin Amsa

To evaluate the definite integral ∫2(2x-3)^2/3 dx, we can use the following formula: ∫uⁿ du = (uⁿ⁺¹)/(n+1) + C where C is the constant of integration. Using this formula, we can rewrite the integral as: ∫2(2x-3)^2/3 dx = 2 ∫(2x-3)^2/3 dx Let u = 2x - 3, then du/dx = 2 and dx = du/2. Substituting this into the integral, we get: 2 ∫(2x-3)^2/3 dx = 2 ∫u^2/3 (du/2) = ∫u^2/3 du Using the formula, we get: ∫u^2/3 du = (u^5/3)/(5/3) + C Substituting back for u, we get: ∫2(2x-3)^2/3 dx = (2(2x-3)^5/3)/(5/3) + C Simplifying, we get: ∫2(2x-3)^2/3 dx = (6(2x-3)^5/3)/5 + C Therefore, the correct answer is option A: 3/5(2x-3)^5/3 + k.

Duba Amsa

Tambaya 27 Rahoto

The distribution of colours of beads in a bowl is given above. What is the probability that a bead selected at random will be blue or white?

7/15

2/5

1/3

1/15

Bayanin Amsa

Total number of beads
= 1+2+4+5+3
=15
Number of blue beads = 1
P(Blue beads) = 1/15
Numbers of white beads = 4
P(white beads) = 4/15
∴P(Blue of white beads)
= P(Blue) + P(White)
= 1/15 + 4/15
= 5/15
= 1/3

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 28 Rahoto

The mean score is

7.0

8.7

9.5

11.0

Bayanin Amsa

Mean ((4*3) + (7*5) + (8*2) + (11*7) + (13*2) + (8*1))/20
(12 + 35 + 16 + 77 + 26 + 8)/20
174/20 = 8.7

Karanta bayanin darasi akan Loci (WAEC)

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Karanta bayanin darasi akan Statistics (WAEC)

Loci

Measures Of Dispersion

Statistics

Duba Amsa

Tambaya 29 Rahoto

Find the value of P if the line joining (P, 4) and (6, -2) is perpendicular to the line joining (2, P) and (-1, 3).

4

6

3

zero

Bayanin Amsa

Hint: If two lines a perpendicular, the gradient of one is equal to minus the reciprocal of the other (P, 4) and (6, -2)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

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

Coordinate Geometry

Simple Operations On Algebraic Expressions

Duba Amsa

Tambaya 30 Rahoto

If $x = \frac{y}{2}$, evaluate $\left(\frac{x^3}{y^3} + \frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^2}{y^2}\right)$

5/8

5/2

5/4

5/16

Algebraic Fractions

Quadratic Equations

Duba Amsa

Tambaya 31 Rahoto

Triangle SPT is the solution of the linear inequalities

2y - x - 2 ≤ 0, y + 2x + 2 ≥ 0, -2 ≤ x ≤ - 1

-2 ≤ x ≤ 2, y ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0

2y - x - 2 ≤ 0, y + 2x + 2 ≤ 0, y ≥ 0, x ≤ 0

2y - x - 2 ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0

Bayanin Amsa

Δ SPT is the solution of the inequalities
2y - x - 2 ≤ 0
2y ≤ 2 + x
y ≤ 2/2 + x/2
y ≤ 1 + 1/2x
y + 2x + 2 ≥ 0,
y ≥ -2 -2x
∴ the solution of the inequalities
2y - x - 2 ≤ 0,
y + 2x + 2 ≥ 0
= -2 ≤ x ≤ -1

Karanta bayanin darasi akan Linear Inequalities (WAEC)

Linear Inequalities

Duba Amsa

Tambaya 32 Rahoto

A straight line makes an angle of 30° with the positive x-axis and cuts the y-axis at y = 5. Find the equation of the straight line.

y = (x/10) + 5

y = x + 5

√3y = - x + 5√3

√3y = x + 5√3

Bayanin Amsa

Cos 30 = 5/x
x cos 30 = 5, => x = 5√3
Coordinates of P = -5, 3, 0
Coordinates of Q = 0, 5
Gradient of PQ = (y2 - y1) (X2 - X1) = (5 - 0)/(0 -5√3)
= 5/5√3 = 1/√3
Equation of PQ = y - y1 = m (x -x1)
y - 0 = 1/√3 (x -(-5√3))
Thus: √3y = x + 5√3

Duba Amsa

Tambaya 33 Rahoto

Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.

6

2

3

8

Bayanin Amsa

Let the ext. angle = x
Thus int. angle = 2x
But sum of int + ext = 180 (angle of a straight line).
2x + x = 180
3x = 180
x = 180/3 = 60
Each ext angle = 360/n
=> 60 = 360/n
n = 360/60 = 6

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Triangles And Polygons

Duba Amsa

Tambaya 34 Rahoto

A cylindrical tank has a capacity of 3080 m3. What is the depth of the tank if the diameter of its base is 14 m?

(Take pi = 22/7)

23 m

25 m

20 m

22 m

Bayanin Amsa

The formula for the volume of a cylindrical tank is V = πr^2h, where V is the volume, r is the radius, and h is the height or depth of the tank. We are given that the capacity or volume of the tank is 3080 m^3, and we can find the radius by dividing the diameter by 2: d = 14 m r = d/2 = 7 m Substituting the given values in the formula, we get: 3080 = (22/7) x 7^2 x h Simplifying the expression, we get: 3080 = 22 x 7 x 7 x h/7 3080 = 22 x 7 x h 3080 = 154h h = 3080/154 h = 20 Therefore, the depth of the tank is 20 meters. Answer: 20 m

Duba Amsa

Tambaya 35 Rahoto

Given an isosceles triangle with length of 2 equal sides t units and opposite side √3t units with angle θ. Find the value of the angle θ opposite to the √3t units.

100°

120°

30°

60°

Bayanin Amsa

In an isosceles triangle, the two equal sides are opposite to the two equal angles. Therefore, the angle opposite to the side of length √3t would be equal to the other equal angle of the triangle. Let's call this angle x. Then, using the fact that the sum of the angles in a triangle is 180°, we can write: x + x + θ = 180 Simplifying this equation, we get: 2x + θ = 180 Now, we know that the two equal sides of the triangle have length t, so we can use the cosine formula to write: cos(x) = (t/2) / √(3t^2/4) Simplifying this expression, we get: cos(x) = 1/√3 Taking the inverse cosine of both sides, we get: x = 30° Substituting this value of x into the earlier equation, we get: 2(30°) + θ = 180° θ = 120° Therefore, the answer is option (B) 120°.

Duba Amsa

Tambaya 36 Rahoto

Triangle SPT is the solution of the linear inequalities

2y - x - 2

\leq

0, y + 2x + 2

\geq

0, x

\geq

0

2y - x - 2

\geq

0, y + 2x + 2

\leq

0, -2

\geq

x

\geq

-1

-2

\geq

x

\geq

2, y

\leq

0, y + 2x + 2

\geq

0, x

\geq

0

2y - x - 2

\geq

0, y + 2x + 2

\geq

0, y

\leq

0, x

\geq

0

Solution Of Linear Equations

Linear Inequalities

Duba Amsa

Tambaya 37 Rahoto

Differentiate (2x+5)2(x-4) with respect to x.

4(2x+5)(x-4)

4(2x+5)(4x-3)

(2x+5)(2x-13)

(2x+5)(6x-11)

Differentiation

Duba Amsa

Tambaya 38 Rahoto

In the figure above, PQR is a straight line segment, PQ = QT. Triangle PQT is an isosceles triangle, ?SQR is 75$^{\circ}$ and ?QPT is 25$^{\circ}$. Calculate the value of ?RST.

45

^{\circ}

55

^{\circ}

25

^{\circ}

50

^{\circ}

Bayanin Amsa

In ? PQT,
?PTQ = 25

^{\circ}

(base ?s of isosceles ?)
In ? QSR,
?RQS = ?QPT + ?QTP
(Extr = sum of interior opposite ?s)
?RQS = 25 + 25
= 50

^{\circ}

Also in ? QSR,
75 + ?RQS + ?QSR = 180

^{\circ}

(sum of ?s of ?)
?75 + 50 + ?QSR = 180
125 + ?QSR = 180
?QSR = 180 - 125
?QSR = 55

^{\circ}

But ?QSR and ?RST are the same
?RST = 55

^{\circ}

Duba Amsa

Tambaya 39 Rahoto

Solve the equations
m2 + n2 = 29
m + n = 7

(2, 3) and ( 3, 5)

(2, 5) and (5, 2)

(5, 2) and ( 5, 3)

(5, 3) and (3, 5)

Bayanin Amsa

m2 + n2 = 29 .......(1)
m + n = 7 ............(2)
From (2),
m = 7 - n
but m2 + n2 = 29, substituting;
(7-n)2 + n2 = 29
49 - 14n + n2 + n2 = 29
=> 2n2 -14n + 20 = 0
Thus n2 -7n + 10 = 0
Factorizing;
(n-5)(n-2) = 0
n - 5 = 0, => n = 5
n - 2 = 0, => n = 2.
When n = 5,
m + n = 7, => m = 2,
When n = 2,
m + n = 7, => m = 5.
Thus (m,n) = (5,2) and (2,5)

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Duba Amsa

Tambaya 40 Rahoto

The bearings of P and Q from a common point N are 020° and 300° respectively. If P and Q are also equidistant from N, find the bearing of P from Q.

040°

070°

280°

320°

Bayanin Amsa

Hint: Simply make a sketch of the question and find the bearing to be 070°

Karanta bayanin darasi akan Bearings (WAEC)

Bearings

Duba Amsa

Tambaya 41 Rahoto

Find the dimensions of a rectangle of greatest area which has a fixed perimeter p.

square of sides p

square of sides 2p

square of sides (p/2)

square of sides (p/4)

Bayanin Amsa

Let the rectangle be a square of sides p/4.
So that perimeter of square = 4p
4 x (p/4) = p.

Karanta bayanin darasi akan Integration (JAMB)

Karanta bayanin darasi akan Areas (WAEC)

Integration

Areas

Duba Amsa

Tambaya 42 Rahoto

Find the area bounded by the curves y = 4 - x2 and y = 2x + 1

20(1/3) sq. units

20(2/3) sq. units

10(2/3) sq. units

10(1/3) sq. units

Bayanin Amsa

Hint:
y = 4 - x2 and y = 2x + 1
=> 4 - x2 = 2x + 1
=> x2 + 2x - 3 = 0
(x+3)(x-1) = 0
thus x = -3 or x = 1.
Integrating x2 + 2x - 3 from (-3, to 1) w.r.t x will give 31/3 = 10(1/3)

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 43 Rahoto

Given distribution of color beads: blue, black, yellow, white and brown with frequencies 1, 2, 3, 4, and 5 respectively. Find the probability that a bead picked at random will be blue or white.

7/15

2/5

1/3

1/15

Bayanin Amsa

There are a total of 1+2+3+4+5 = 15 beads. The probability of picking a blue bead is 1/15, since there is only one blue bead out of the total of 15 beads. The probability of picking a white bead is 4/15, since there are 4 white beads out of the total of 15 beads. To find the probability of picking a blue or white bead, we add the probabilities of picking a blue bead and a white bead: P(blue or white) = P(blue) + P(white) P(blue or white) = 1/15 + 4/15 P(blue or white) = 5/15 P(blue or white) = 1/3 Therefore, the probability of picking a blue or white bead is 1/3.

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Probability

Surds (radicals)

Duba Amsa

Tambaya 44 Rahoto

The histogram above shows the distribution of passengers in taxis of a certain motor park. How many taxis have more than 4 passengers

16

17

14

15

Bayanin Amsa

Looking at the histogram, we can see that the height of the bars represents the number of taxis with a particular number of passengers. We can count the number of taxis with more than 4 passengers by adding up the heights of the bars for 5, 6, 7, and 8 passengers. Counting these bars, we can see that there are a total of 17 taxis with more than 4 passengers. Therefore, the answer is 17.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Integration (JAMB)

Representation Of Data

Integration

Duba Amsa

Tambaya 45 Rahoto

Given that $p = 1 + \sqrt{2}$ and $q = 1 - \sqrt{2}$, evaluate $\frac{(p^2 - q^2)}{2pq}$.

2(2+√2)

-2(2+√2)

2√2

-2√2

Bayanin Amsa

First, let's evaluate p^2 and q^2: p^2 = (1+√2)^2 = 1 + 2√2 + 2 = 3 + 2√2 q^2 = (1-√2)^2 = 1 - 2√2 + 2 = 3 - 2√2 Next, let's substitute the values of p^2 and q^2 into the expression (p^2 - q^2)/(2pq): (p^2 - q^2)/(2pq) = ((3 + 2√2) - (3 - 2√2))/(2(1+√2)(1-√2)) = (4√2)/(2(1-2)) = (4√2)/(-2) = -2√2 Therefore, the answer is -2√2.

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

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Modular Arithmetic

Surds (radicals)

Duba Amsa

Tambaya 46 Rahoto

What is the rate of change of the volume v of a hemisphere with respect to its radius r when r = 2?

8π

16π

2π

4π

Bayanin Amsa

To solve this problem, we need to use the formula for the volume of a hemisphere, which is V = (2/3)πr^3. We want to find the rate of change of V with respect to r, which is given by dV/dr = 2πr^2. When r = 2, we can substitute this value into the formula to get dV/dr = 2π(2)^2 = 8π. Therefore, the rate of change of the volume of a hemisphere with respect to its radius when r = 2 is 8π.

Karanta bayanin darasi akan Introductory Calculus (WAEC)

Karanta bayanin darasi akan Integration (JAMB)

Introductory Calculus

Integration

Duba Amsa

Tambaya 47 Rahoto

The bar chart shows different colours of cars passing a particular point of a certain street in two minutes. What fraction of the cars is yellow

\frac{1}{5}

\frac{2}{25}

\frac{4}{15}

\frac{3}{25}

Bayanin Amsa

\begin{matrix} colour of cars & Number (frequency) \\ y e l l o w & 3 \\ w h i t e & 4 \\ r e d & 8 \\ g r e e n & 2 \\ b l u e & 6 \\ b l a c k & 2 \\ 25 \end{matrix}

Thus, the fraction of the total numbers that are yellow is $\frac{3}{25}$

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Number Bases (WAEC)

Representation Of Data

Number Bases

Duba Amsa

Tambaya 48 Rahoto

Team P and Q are involved in a game of football.What is the probability that the game ends in a draw?

2/3

1/2

1/3

1/4

Bayanin Amsa

P(Games ends in draw)
This implies that Team P and Q wins
∴ P(P wins) = 1/2
P(Q wins) = 1/2
∴ P(game ends in a draw) = 1/2*1/2 = 1/4

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 49 Rahoto

Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. The mean score is

7.0

8.7

9.5

11.0

Bayanin Amsa

To find the mean score, we need to calculate the sum of all the scores, and then divide it by the total number of scores. The scores are given as 4, 7, 8, 11, 13, 8 with corresponding frequencies of 3, 5, 2, 7, 2, 1 respectively. So, the sum of all the scores is: 4 × 3 + 7 × 5 + 8 × 2 + 11 × 7 + 13 × 2 + 8 × 1 = 12 + 35 + 16 + 77 + 26 + 8 = 174 The total number of scores is: 3 + 5 + 2 + 7 + 2 + 1 = 20 Therefore, the mean score is: mean = sum of scores / total number of scores = 174 / 20 = 8.7 So, the answer is.

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Loci (WAEC)

Karanta bayanin darasi akan Statistics (WAEC)

Representation Of Data

Loci

Statistics

Duba Amsa

Tambaya 50 Rahoto

Find the variance of 2, 6, 8, 6, 2 and 6

6

5

√6

√5

Bayanin Amsa

To find the variance of a set of numbers, you need to follow these steps: 1. Find the mean (average) of the set. 2. Subtract the mean from each number in the set. 3. Square each of the differences. 4. Find the mean (average) of the squared differences. Let's apply these steps to the set of numbers given: 2, 6, 8, 6, 2, and 6. 1. The mean of the set is (2+6+8+6+2+6)/6 = 5. 2. Subtracting the mean from each number gives us the following differences: -3, 1, 3, 1, -3, 1. 3. Squaring each difference gives us: 9, 1, 9, 1, 9, 1. 4. Finding the mean of these squared differences gives us (9+1+9+1+9+1)/6 = 5. Therefore, the variance of the set is 5. In summary, variance is a measure of how spread out a set of numbers is. A high variance means that the numbers in the set are far from the mean, while a low variance means that the numbers are closer to the mean.

Karanta bayanin darasi akan Measures Of Dispersion (JAMB)

Measures Of Dispersion

Duba Amsa






Latsa & Riƙe don Ja Shi Gabaɗaya
Danna nan don rufewa