WAEC SSCE - Further Mathematics - 2021

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To find the value of p, we need to use the concept of perpendicular vectors. Two vectors are perpendicular if and only if the dot product of the two vectors is equal to zero. The dot product of two vectors can be calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. If the angle between the two vectors is 90 degrees, the cosine of the angle is zero and the dot product is also zero. Therefore, to find the value of p, we need to calculate the dot product of the two vectors, 2i + pj and 4i - 2j, and set it equal to zero. The dot product of two vectors (a, b) and (c, d) is given by: (a, b) * (c, d) = ac + bd So, the dot product of 2i + pj and 4i - 2j is: (2i + pj) * (4i - 2j) = (2 * 4) + (p * -2) = 8 - 2p = 0 Therefore, 2p = 8 and p = 4. So, the value of p is 4.

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A prime number is a number greater than 1 that is only divisible by 1 and itself. The prime numbers on a die are 2, 3, and 5. We know that the die is fair, which means that each number has an equal probability of being rolled. There are a total of 60 rolls, so we can use the frequency table to count the number of times each prime number appears on the die. The frequency of the number 2 is 10, the frequency of the number 3 is 14, and the frequency of the number 5 is 8. Therefore, the total number of times a prime number was rolled is: 10 + 14 + 8 = 32 The probability of rolling a prime number on a fair die is the total number of times a prime number was rolled divided by the total number of rolls: 32/60 = 8/15 Therefore, the answer is option D, 8/15.

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To find the normal reaction to the plane, we can use the following approach: Step 1: Draw a diagram of the situation. In this case, we have a body of mass 15kg on a smooth plane inclined at 60° to the horizontal. The weight of the body acts vertically downwards and is equal to mg, where m is the mass of the body and g is the acceleration due to gravity. Step 2: Resolve the weight of the body into two components, one perpendicular to the plane and the other parallel to the plane. The component of the weight perpendicular to the plane is equal to mg cos 60°, which is equal to (15 kg) x (10 m/s\(^2\)) x cos 60° = 75 N. Step 3: The normal reaction to the plane is equal in magnitude but opposite in direction to the component of the weight perpendicular to the plane. Therefore, the normal reaction to the plane is 75 N in the upward direction. Step 4: Check the options given in the question and select the one that matches the value obtained in Step 3. In this case, the correct option is 75N. Therefore, the normal reaction to the plane is 75 N.

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The first person has 8 choices for a seat, the second person has 7 choices remaining, and the third person has 6 choices. However, the order in which they choose their seats does not matter, so we need to divide by the number of ways that the three people can be arranged. This is given by 3 factorial (3!), which is 3 x 2 x 1 = 6. Therefore, the total number of ways that 8 persons can be seated on a bench if only three seats are available is: 8 x 7 x 6 / 3! = 336 So the correct answer is (C) 336.

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Let the common difference of the AP be denoted by d. Then, the first three terms of the AP are 4, 4 + d, and 4 + 2d, respectively. The sum of the first three terms of the AP is given as 18. Therefore, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying the above equation, we get: 3d + 12 = 18 3d = 6 d = 2 Hence, the common difference of the AP is 2. Therefore, the first three terms of the AP are: 4, 6, 8 The product of the first three terms is: 4 x 6 x 8 = 192 Therefore, the answer is 192.

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To solve the inequality x\(^2\) - 2x - 3 ≤ 0, we can use factoring or the quadratic formula. Factoring gives us (x - 3)(x + 1) ≤ 0, which means that the expression is less than or equal to zero when x is between or equal to -1 and 3, since the factors change sign at these values. Therefore, the correct answer is {x: -1 ≤ x ≤ 3}. Alternatively, we can use the quadratic formula to find the roots of the equation x\(^2\) - 2x - 3 = 0, which are x = -1 and x = 3. Since the quadratic function is a parabola that opens upward, it is negative in the interval between these two roots. Therefore, the expression x\(^2\) - 2x - 3 is less than or equal to zero when x is between or equal to -1 and 3.

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The binary operation * is defined as: P * q = (q^2 - p^2) / (2pq). To find 3 * 2, we need to substitute the values of p and q in the equation: 3 * 2 = (2^2 - 3^2) / (2 * 3 * 2) = (-5) / (12) = -5/12 So, the answer is -5/12.

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To use the binomial expansion of (1 + x)\(^6\), we substitute x = 1.998, which gives: (1 + 1.998)\(^6\) = 1 + 6(1.998) + 15(1.998)\(^2\) + 20(1.998)\(^3\) + 6(1.998)\(^5\) + (1.998)\(^6\) We are interested in finding the value of (1.998)\(^6\), which is the last term on the right-hand side of the equation. We can solve for it by subtracting the other terms from both sides of the equation: (1.998)\(^6\) = (1 + 1.998)\(^6\) - 1 - 6(1.998) - 15(1.998)\(^2\) - 20(1.998)\(^3\) - 6(1.998)\(^5\) Using a calculator, we can evaluate the right-hand side of the equation to get: (1.998)\(^6\) ≈ 63.167 Therefore, the correct answer is option B, 63.167, rounded to three decimal places. Explanation: The binomial expansion of (1 + x)\(^6\) is a formula that allows us to expand the expression into a sum of terms involving powers of x. By substituting x = 1.998, we can use the formula to find the value of (1.998)\(^6\). We then use a calculator to evaluate the expression to obtain the final answer.

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To find the modal class mark, we need to first identify the class with the highest frequency, which in this case is the class with marks 20-29 (16 students). The modal class mark is the midpoint of this class, which is calculated by adding the lower and upper limits of the class and dividing by 2: Modal class mark = (20 + 29) / 2 = 24.5 Therefore, the modal class mark is 24.5, which is option (C).

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To calculate 3M - 2N, we need to multiply each matrix by its scalar factor and then subtract the results. First, we have: 3M = 3 x \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) Next, we have: 2N = 2 x \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) Now we can subtract these two matrices to get: 3M - 2N = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) = \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\) Therefore, the correct answer is (B) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).

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To find the radius of a circle from its equation in general form, we need to rewrite the equation in the standard form of a circle, which is: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) Where (h, k) is the center of the circle, and r is its radius. To do this, we complete the square for both x and y terms in the given equation. We start by rearranging the terms as follows: 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0 2x\(^2\) - 4x + 2y\(^2\) - 6y = 2 Now, we need to add and subtract appropriate constants to complete the square for x and y terms separately. For the x terms, we take half of the coefficient of x (-4/2 = -2) and square it to get 4. So, we add and subtract 4 to the equation: 2x\(^2\) - 4x + 4 - 4 + 2y\(^2\) - 6y = 2 We can now group the first three terms and factor it as a perfect square: 2(x - 1)\(^2\) + 2y\(^2\) - 6y - 2 = 0 For the y terms, we take half of the coefficient of y (-6/2 = -3) and square it to get 9. So, we add and subtract 9 to the equation: 2(x - 1)\(^2\) + 2(y - 3)\(^2\) - 2 - 9 = 0 2(x - 1)\(^2\) + 2(y - 3)\(^2\) = 11 Now, we have the equation in the standard form of a circle, where the center is at (1, 3) and the radius is the square root of 11/2. We can simplify this expression to get: sqrt(11/2) = sqrt(11)/sqrt(2) = sqrt(2)*sqrt(11)/2 Therefore, the answer is option (C), 17/√2.

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We can use the formula for combinations to solve for y:

^5yC₂ = ^(5y)! / _(2!(5y-2)!) = 190

Expanding the factorials:

^(5y)(5y-1) / 2 = 190

Solving for y:

(5y)(5y-1) = 380

25y^2 - 5y = 380

25y^2 - 5y - 380 = 0

Using the quadratic formula:

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-(−5) ± √((-5)^2 - 4(25)(-380)))/(2(25))

y = (5 ± √(25 + 90000))/(50)

y = (5 ± √(90025))/(50)

Taking the positive root:

y = (5 + 300)/(50)

y = 305/50

y = 6.1

So the value of y is approximately 6.1.

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x =  (−43) $\left(\begin{array}{c}-4\\ 3\end{array}\right)$ and y= (−915) $\left(\begin{array}{c}-9\\ 15\end{array}\right)$

Changing x and y to the form xi+yj

we have:

x= -4i + 3j and y = -9i - 15j

using:

cosØ = xy|x|ly| $\frac{𝑥𝑦}{|𝑥|𝑙𝑦|}$

where Ø is the angle between x and y

xy = (-4i+3j) (-9i - 15j)

-36 + 60 x 0 - 27 x 0- 45 = -81

|x| = √(42 ${}^2$ +32 ${}^2$) =√(16 +9) = √25 = 5

|y| = √(-9)2 ${}^2$ + (-15)2 ${}^2$ = √(81 +225) = √306

cosØ =−815√306 $\frac{-81}{5\surd 306}$

= −9√306170 $\frac{-9\surd 306}{170}$

17.49∗9170 $\frac{17.49\ast 9}{170}$

Ø = cos−1 ${}^{-1}$(-0.1029);

Ø = 95.91°

Ø = 96°

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The probability of selecting a bad mango from the bag is 6/24 = 1/4. This means that the probability of selecting a good mango is 3/4.

To find the probability that not more than 3 mangoes are bad, we need to consider four cases: selecting 0 bad mangoes, selecting 1 bad mango, selecting 2 bad mangoes, or selecting 3 bad mangoes.

Case 1: Selecting 0 bad mangoes

The probability of selecting a good mango on the first draw is 3/4. This probability remains the same for each of the 6 mangoes that are selected. Therefore, the probability of selecting 0 bad mangoes is (3/4)^6 = 0.1771.

Case 2: Selecting 1 bad mango

There are 6 ways to select 1 bad mango out of 6 mangoes, and 5 ways to select 5 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 1 bad mango is (6/24) * (18/24)^5 * 6 = 0.3934.

Case 3: Selecting 2 bad mangoes

There are 15 ways to select 2 bad mangoes out of 6 mangoes, and 4 ways to select 4 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 2 bad mangoes is (15/24)^2 * (9/24)^4 * 15 = 0.3147.

Case 4: Selecting 3 bad mangoes

There are 20 ways to select 3 bad mangoes out of 6 mangoes, and 3 ways to select 3 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 3 bad mangoes is (6/24)^3 * (18/24)^3 * 20 = 0.0983.

The total probability of not selecting more than 3 bad mangoes is the sum of the probabilities of these four cases: 0.1771 + 0.3934 + 0.3147 + 0.0983 = 0.9835.

Therefore, the probability of not more than 3 bad mangoes being selected is 0.9835 or approximately 98.35%.

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(a)

(i) To reach a distance of 15km, the jogger needs to cover 15000 meters.

Let's call the number of days it takes to reach this distance "n."

On the first day, he runs 500 meters. On the second day, he runs 500 + 250 = 750 meters. On the third day, he runs 750 + 250 = 1000 meters.

In general, on the nth day, he runs 500 + 250(n-1) meters.

So we can set up an equation:

500 + 750 + 1000 + ... + (500 + 250(n-1)) = 15000

Simplifying, we get:

250n^2 + 250n - 15000 = 0

Dividing both sides by 250:

n^2 + n - 60 = 0

This equation can be factored as:

(n + 6)(n - 10) = 0

Since we're looking for a positive value for n, the answer is n = 10.

So it will take the jogger 10 days to reach a distance of 15km in training.

(ii) We can use the formula for the sum of a geometric series:

S = a(1 - r^n)/(1 - r)

where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.

In this case, we know that the second term is -3, so a = 500 * (-3) = -1500.

We also know that the sum to infinity is 25/2, so S = 25/2.

Plugging in these values and solving for r:

25/2 = (-1500)(1 - r^10)/(1 - r)

r = 1/2 or -1 (discarded since a negative ratio would make the terms alternate in sign, and the second term is negative)

So the common ratio is 1/2.

The total distance the jogger would have run in the training is the sum of the terms of the geometric series:

S = a/(1 - r) = (-1500)/(1 - 1/2) = 3000 meters.

(b)

We know that the second term of the GP is -3, so we can write the first two terms as:

a, ar

where ar = -3.

We also know that the sum to infinity is 25/2, so we can use the formula:

S = a/(1 - r)

25/2 = a/(1 - r)

a = 25/2 - 25r/2

Substituting this value of a into the equation ar = -3:

(25/2 - 25r/2)r = -3

Simplifying:

25r^2 - 50r + 6 = 0

We can solve for r using the quadratic formula:

r = (50

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(a) The matrix representation of a linear transformation can be obtained by writing the image of the standard basis vectors (1,0) and (0,1). The matrix of P is given by: P = [[-3, 6], [4, 1]] Similarly, the matrix of Q is given by: Q = [[2, -3], [-4, -6]] (b) To find the image of a point (-2,-3) under the transformation Q, we first write the point as a column vector: (-2,-3) = [-2, -3]^T Next, we multiply the matrix Q with the vector: Q * [-2, -3]^T = [2, -3] * [-2, -3]^T = [-12, 18]^T So the image of (-2,-3) under the transformation Q is (-12, 18). (c) To find the single transformation representing the transformation Q followed by P, we multiply the matrices Q and P: Q * P = [[2, -3], [-4, -6]] * [[-3, 6], [4, 1]] = [[36, -18], [-24, -30]] So the single transformation representing the transformation Q followed by P is given by the matrix [[36, -18], [-24, -30]]. (d) To find the image of (1,4) when transformed by Q followed by P, we first write the point as a column vector: (1, 4) = [1, 4]^T Next, we multiply the matrix representing Q followed by P with the vector: [[36, -18], [-24, -30]] * [1, 4]^T = [36, -18] * 1 + [-24, -30] * 4 = [36, -18] + [-96, -120] = [-60, -138]^T So the image of (1,4) when transformed by Q followed by P is (-60, -138). (e) To find the image P^1 of the point (-√2,√2) under an anticlockwise rotation of 225° about the origin, we can first rotate the point by 225° in the counterclockwise direction and then apply the transformation P. The counterclockwise rotation of 225° can be represented by the matrix: R = [[cos(225), -sin(225)], [sin(225), cos(225)]] = [[-√2/2, √2/2], [-√2/2, -√2/2]] Next, we multiply the matrix R with the vector representing the point (-√2,√2): R * [-√2, √2]^T = [[-√2/2, √2/2], [-√2/2, -√2/2]] * [-√2, √2]^T = [-√2/2 * -√2 + √2/2 * √2, -√2/2 * -√2 - √2/2 * √2]^T = [√2, -√2]^T So the image of the point (-√2,√2) under the counterclockwise rotation of 225° is (√2, -√2). Finally, to find

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To solve this problem, we first need to understand what a parallelogram is. A parallelogram is a four-sided shape with opposite sides that are parallel to each other. This means that if we draw lines from opposite corners of the parallelogram, these lines will intersect at the midpoint of the two sides they are drawn from.

Now let's look at the given position vectors of P, Q, and R. We can use these to find the position vector of M, which is opposite to P in the parallelogram. To do this, we need to use the fact that the position vector of the midpoint of a line segment is the average of the position vectors of the two endpoints. So, the position vector of M is:

¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ =  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ - (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ = (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

MR¯¯¯¯¯¯¯¯¯ $\overline{𝑀𝑅}$ =  (−52) $\left(\begin{array}{c}-5\\ 2\end{array}\right)$ - (xy) $\left(\begin{array}{c}𝑥\\ 𝑦\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

 (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

⇒ -5 - x = -3; x = -2

⇒ 2 - y = 8; y = -6
The coordinate of M (-2,-6)

(b) PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$ = OM¯¯¯¯¯¯¯¯¯ $\overline{𝑂𝑀}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=   (−2−6) $\left(\begin{array}{c}-2\\ -6\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−6−1) $\left(\begin{array}{c}-6\\ -1\end{array}\right)$

    PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ = OQ¯¯¯¯¯¯¯¯ $\overline{𝑂𝑄}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

|PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$| = √(-62 ${}^2$ + [-12 ${}^2$]) = √37

|PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$| = √(-32 ${}^2$ + 82 ${}^2$) = √73

cosØ = PM.PQ|PM¯¯¯¯¯¯¯¯¯¯||PQ¯¯¯¯¯¯¯¯| $\frac{𝑃𝑀.𝑃𝑄}{|\overline{𝑃𝑀}||\overline{𝑃𝑄}|}$

cosØ = [−6i−j].[−3i+8j]√37.√73 $\frac{[-6𝑖-𝑗].[-3𝑖+8𝑗]}{\surd 37.\surd 73}$ = 10√2710 $\frac{10}{\surd 2710}$

⇒ Ø = cos−1 ${}^{-1}$ 10√2710 $\frac{10}{\surd 2710}$

: Ø = 78.91° ≈ 79°

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To find the values of p and q, we can use the Remainder Theorem and synthetic division. The Remainder Theorem states that if a polynomial f(x) is divided by x - a, then the remainder is equal to f(a).

So, in this case, when we divide f(x) by (x - 1), the remainder is 27, meaning that f(1) = 27. We can use this information to find one of the constants, p.

Next, we can use synthetic division to divide f(x) by (x + 2) and find the remainder. The process of synthetic division involves dividing each term of the polynomial by the leading coefficient of the divisor, and then using the coefficients to find the remainder.

In this case, the polynomial f(x) is divided by (x + 2), giving us:

2x^3 + px + qx - 5 
          ÷ x + 2
2          1    p/2    (q - 5)/2
-2x^2 + (p - 4)x + (q - 5)

Since the remainder is 27, we have:

-2x^2 + (p - 4)x + (q - 5) = 27

Solving for p and q, we have:

p - 4 = 0
p = 4
q - 5 = 27
q = 32

So, the values of p and q are p = 4 and q = 32.

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(a) To find the equation of the normal to the curve y = (x² - x + 1)(x - 2) at the point where the curve cuts the X-axis, we first need to find the x-coordinate of the point where the curve intersects the X-axis. This is where y = 0, so we can solve the equation (x² - x + 1)(x - 2) = 0 to find the roots of the equation. The roots are x = 1 ± i√3 and x = 2, but since we want the point where the curve intersects the X-axis, we take x = 2.

Next, we need to find the gradient of the curve at this point. We can do this by differentiating the equation y = (x² - x + 1)(x - 2) with respect to x, giving:

dy/dx = 3x² - 8x + 1

Substituting x = 2, we get:

dy/dx = 3(2)² - 8(2) + 1 = -7

Therefore, the gradient of the curve at the point where it intersects the X-axis is -7.

Since the normal to the curve is perpendicular to the tangent at the point of intersection, we know that the gradient of the normal is the negative reciprocal of the gradient of the tangent. So the gradient of the normal is 1/7.

Finally, we can find the equation of the normal using the point-slope form of the equation of a straight line:

y - 0 = (1/7)(x - 2)

Simplifying this equation gives:

y = (1/7)x - 2/7

So the equation of the normal to the curve y = (x² - x + 1)(x - 2) at the point where the curve cuts the X-axis is y = (1/7)x - 2/7.

(b) To find the equation of the line joining Q and the midpoint of PR, we first need to find the coordinates of the midpoint of PR. The midpoint of a line segment with endpoints (x₁, y₁) and (x₂, y₂) is ((x₁ + x₂)/2, (y₁ + y₂)/2). So the midpoint of PR is ((-1 + 3)/2, (2 - 4)/2), which simplifies to (1, -1).

Next, we need to find the gradient of the line joining Q and the midpoint of PR. We can use the gradient formula, which gives:

m = (y₂ - y₁)/(x₂ - x₁)

Substituting the coordinates of Q and the midpoint of PR gives:

m = (1 - (-1))/(5 - 1) = 1/2

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(a) The initial velocity of the bus is 45 m/s and the final velocity is 5 m/s, and the retardation is 10 m/s². We can use the equation of motion to find the distance covered. The equation is:

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered.

Substituting the given values, we get:

5² = 45² + 2(-10)s

25 = 2025 - 20s

20s = 2000

s = 100 meters

Therefore, the distance covered by the bus is 100 meters.

(b) The weight of the bucket is given by W = mg, where m is the mass of the bucket and g is the acceleration due to gravity. The tension in the rope is equal to the weight of the bucket, so we have:

T = W = mg

The net force acting on the bucket is given by F = T - mg, where F is the force accelerating the bucket.

Using Newton's second law of motion, F = ma, where a is the acceleration of the bucket. Substituting the values, we get:

ma = T - mg

a = (T - mg) / m

Substituting the given values, we get:

a = (240 - 16 x 10) / 16

a = 10 m/s²

Therefore, the acceleration of the bucket is 10 m/s².