JAMB UTME - Mathematics - 2001

Swali 1 Ripoti

⊗	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

Tazama Jibu

Swali 2 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Probability (WAEC)

Probability

Tazama Jibu

Swali 3 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Positive And Negative Integers, Rational Numbers (WAEC)

Soma maelezo ya somo kwenye Modular Arithmetic (WAEC)

Soma maelezo ya somo kwenye Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Modular Arithmetic

Surds (radicals)

Tazama Jibu

Swali 4 Ripoti

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)

Maelezo ya Majibu

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)

Soma maelezo ya somo kwenye Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Tazama Jibu

Swali 5 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Circles (WAEC)

Circles

Tazama Jibu

Swali 6 Ripoti

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

4

3

2

1

Maelezo ya Majibu

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.

Tazama Jibu

Swali 7 Ripoti

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

6

5

√6

√5

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Measures Of Dispersion (JAMB)

Measures Of Dispersion

Tazama Jibu

Swali 8 Ripoti

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

Tazama Jibu

Swali 9 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Integration (JAMB)

Soma maelezo ya somo kwenye Linear Inequalities (WAEC)

Integration

Linear Inequalities

Tazama Jibu

Swali 10 Ripoti

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

Maelezo ya Majibu

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.

Tazama Jibu

Swali 11 Ripoti

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

6

2

3

8

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Triangles And Polygons (WAEC)

Triangles And Polygons

Tazama Jibu

Swali 12 Ripoti

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

0.039

0.358

0.033

0.308

Maelezo ya Majibu

(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

Soma maelezo ya somo kwenye Fractions, Decimals And Approximations (WAEC)

Soma maelezo ya somo kwenye Percentages (WAEC)

Fractions, Decimals And Approximations

Percentages

Tazama Jibu

Swali 13 Ripoti

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)

Maelezo ya Majibu

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)

Soma maelezo ya somo kwenye Algebraic Expressions (WAEC)

Algebraic Expressions

Tazama Jibu

Swali 14 Ripoti

Find the value of $ \theta $ in the diagram

60

^{\circ}

100

^{\circ}

120

^{\circ}

30

^{\circ}

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Circles (WAEC)

Circles

Tazama Jibu

Swali 15 Ripoti

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

Tazama Jibu

Swali 16 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Positive And Negative Integers, Rational Numbers (WAEC)

Soma maelezo ya somo kwenye Sets (WAEC)

Positive And Negative Integers, Rational Numbers

Sets

Tazama Jibu

Swali 17 Ripoti

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

Maelezo ya Majibu

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

Tazama Jibu

Swali 18 Ripoti

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

4a

1/8a

8a

1/4a

Number Bases

Tazama Jibu

Swali 19 Ripoti

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

490

495

496

498

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Permutation And Combination (JAMB)

Permutation And Combination

Tazama Jibu

Swali 20 Ripoti

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}

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Quadratic Equations (WAEC)

Quadratic Equations

Tazama Jibu

Swali 21 Ripoti

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

Maelezo ya Majibu

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)

Soma maelezo ya somo kwenye Areas (WAEC)

Areas

Tazama Jibu

Swali 22 Ripoti

The mean score is

7.0

8.7

9.5

11.0

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Loci (WAEC)

Soma maelezo ya somo kwenye Measures Of Dispersion (JAMB)

Soma maelezo ya somo kwenye Statistics (WAEC)

Loci

Measures Of Dispersion

Statistics

Tazama Jibu

Swali 23 Ripoti

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

Maelezo ya Majibu

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.

Tazama Jibu

Swali 24 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Representation Of Data (JAMB)

Soma maelezo ya somo kwenye Integration (JAMB)

Representation Of Data

Integration

Tazama Jibu

Swali 25 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Number Bases (WAEC)

Number Bases

Tazama Jibu

Swali 26 Ripoti

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

30

33

35

15

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Tazama Jibu

Swali 27 Ripoti

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

Maelezo ya Majibu

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

Tazama Jibu

Swali 28 Ripoti

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

Maelezo ya Majibu

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."

Soma maelezo ya somo kwenye Sequence And Series (WAEC)

Sequence And Series

Tazama Jibu

Swali 29 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Sine, Cosine And Tangent Of An Angle (WAEC)

Soma maelezo ya somo kwenye Areas (WAEC)

Sine, Cosine And Tangent Of An Angle

Areas

Tazama Jibu

Swali 30 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Angles Of Elevation And Depression (WAEC)

Soma maelezo ya somo kwenye Sets (JAMB)

Soma maelezo ya somo kwenye Simple Operations On Algebraic Expressions (WAEC)

Angles Of Elevation And Depression

Sets

Simple Operations On Algebraic Expressions

Tazama Jibu

Swali 31 Ripoti

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)

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Integration (JAMB)

Soma maelezo ya somo kwenye Areas (WAEC)

Integration

Areas

Tazama Jibu

Swali 32 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Probability (WAEC)

Soma maelezo ya somo kwenye Surds (radicals) (WAEC)

Probability

Surds (radicals)

Tazama Jibu

Swali 33 Ripoti

Evaluate 21.05347 - 1.6324 x 0.43 to 3 decimal places

20.980

20.351

20.981

20.352

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Fractions, Decimals And Approximations (WAEC)

Soma maelezo ya somo kwenye Algebraic Expressions (WAEC)

Fractions, Decimals And Approximations

Algebraic Expressions

Tazama Jibu

Swali 34 Ripoti

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

Tazama Jibu

Swali 35 Ripoti

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}

Maelezo ya Majibu

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}

Tazama Jibu

Swali 36 Ripoti

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

Tazama Jibu

Swali 37 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Representation Of Data (JAMB)

Soma maelezo ya somo kwenye Loci (WAEC)

Soma maelezo ya somo kwenye Statistics (WAEC)

Representation Of Data

Loci

Statistics

Tazama Jibu

Swali 38 Ripoti

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

-?/2

-1

1

?/2

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Differentiation (JAMB)

Differentiation

Tazama Jibu

Swali 39 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Loci (WAEC)

Soma maelezo ya somo kwenye Circles (WAEC)

Loci

Circles

Tazama Jibu

Swali 40 Ripoti

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

Tazama Jibu

Swali 41 Ripoti

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

ax - 2

ax + 2

ax - 8

ax - 6

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Algebraic Expressions (WAEC)

Algebraic Expressions

Tazama Jibu

Swali 42 Ripoti

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°

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Bearings (WAEC)

Bearings

Tazama Jibu

Swali 43 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Coordinate Geometry (WAEC)

Soma maelezo ya somo kwenye Simple Operations On Algebraic Expressions (WAEC)

Coordinate Geometry

Simple Operations On Algebraic Expressions

Tazama Jibu

Swali 44 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Introductory Calculus (WAEC)

Soma maelezo ya somo kwenye Integration (JAMB)

Introductory Calculus

Integration

Tazama Jibu

Swali 45 Ripoti

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

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Loci (JAMB)

Soma maelezo ya somo kwenye Loci (WAEC)

Loci

Tazama Jibu

Swali 46 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Linear Inequalities (WAEC)

Linear Inequalities

Tazama Jibu

Swali 47 Ripoti

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%

Maelezo ya Majibu

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.

Soma maelezo ya somo kwenye Percentages (WAEC)

Soma maelezo ya somo kwenye Fractions, Decimals, Approximations And Percentage (JAMB)

Percentages

Fractions, Decimals, Approximations And Percentage

Tazama Jibu

Swali 48 Ripoti

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

Maelezo ya Majibu

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

Soma maelezo ya somo kwenye Probability (WAEC)

Probability

Tazama Jibu

Swali 49 Ripoti

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°

Maelezo ya Majibu

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°.

Tazama Jibu

Swali 50 Ripoti

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}

Maelezo ya Majibu

\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 $\frac{3}{25}$

Soma maelezo ya somo kwenye Representation Of Data (JAMB)

Soma maelezo ya somo kwenye Number Bases (WAEC)

Representation Of Data

Number Bases

Tazama Jibu






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