WAEC SSCE - Further Mathematics - 2021 (Essay)
Question 1 Report
P and Q are two linear transformations in the X-Y plane defined by
P: (x, y) → (-3x + 6y, 4x + y) and
Q: (x, y) → (2x-3y, -4x - 6y).
(a) Write down the matrices of P and Q. (b) What is the image of (-2,-3) under the transformation Q?
(c) Obtain a single transformation representing the transformation Q followed by P.
(d) Find the image of (1,4) when transformed by Q followed by P.
(e) Find the image P\(^1\) of the point (-√2,2√2) under an anticlockwise rotation of 225° about the origin.
(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
Answer Details
(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
Question 2 Report
(a) The speed of a moving bus reduced from 45m/s to 5m/s with a uniform retardation of 10m/s\(^2\). Calculate the distance covered.
(b) A bucket full of water with mass 16kg is pulled out of a well with a light inextensible rope. Find its acceleration when the tension in the rope is 240N. [Take g= 10m/s\(^2\)]
(a) Uniform retardation, so use \(v^{2}=u^{2}-2as\) with \(u=45,\ v=5,\ a=10\):
\[5^{2}=45^{2}-2(10)s\Rightarrow 25=2025-20s\]
\[20s=2000\Rightarrow s=100\text{ m}\]
(b) The bucket is pulled upward, so applying Newton's second law upward with \(m=16\text{kg},\ T=240\text{N},\ g=10\text{m/s}^{2}\):
\[T-mg=ma\Rightarrow 240-16(10)=16a\]
\[80=16a\Rightarrow a=5\text{ m/s}^{2}\ (\text{upward})\]
Answer Details
(a) Uniform retardation, so use \(v^{2}=u^{2}-2as\) with \(u=45,\ v=5,\ a=10\):
\[5^{2}=45^{2}-2(10)s\Rightarrow 25=2025-20s\]
\[20s=2000\Rightarrow s=100\text{ m}\]
(b) The bucket is pulled upward, so applying Newton's second law upward with \(m=16\text{kg},\ T=240\text{N},\ g=10\text{m/s}^{2}\):
\[T-mg=ma\Rightarrow 240-16(10)=16a\]
\[80=16a\Rightarrow a=5\text{ m/s}^{2}\ (\text{upward})\]
Question 3 Report
Evaluate: \(^9{?}_1\) \(\frac{x(2x-3)}{?x}\) dx
9∫1 x(2x−3)√x dx = 2x2−3x)x1/2
= x −1/2 (2x2 - 3x)
= 2x3/2 - 3x1/2
= 2x3/23/2+1 - 3x1/21/2+1
= 2x3/25/2 - 3x1/23/2
= 9∫1 [4x3/25 - 6x1/23 ]
[ 4(9)3/25 - 6(9)1/23 ] - [ 4(9)1/25 - 6(1)1/23 ]
= [ 4(27)5 - 6(3)3 - [ 45 - 2 ]
= [ 1085- 2 ] - [ 65 ]
= 785 + 65= 845
164/5
Answer Details
9∫1 x(2x−3)√x dx = 2x2−3x)x1/2
= x −1/2 (2x2 - 3x)
= 2x3/2 - 3x1/2
= 2x3/23/2+1 - 3x1/21/2+1
= 2x3/25/2 - 3x1/23/2
= 9∫1 [4x3/25 - 6x1/23 ]
[ 4(9)3/25 - 6(9)1/23 ] - [ 4(9)1/25 - 6(1)1/23 ]
= [ 4(27)5 - 6(3)3 - [ 45 - 2 ]
= [ 1085- 2 ] - [ 65 ]
= 785 + 65= 845
164/5
Question 4 Report
Given that x = \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\) calculate, correct to the nearest degree, the angle between the vectors
Use \(\cos\theta=\dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|\,|\mathbf{y}|}\) with \(\mathbf{x}=\begin{pmatrix}-4\\3\end{pmatrix},\ \mathbf{y}=\begin{pmatrix}-9\\15\end{pmatrix}\).
Dot product: \(\mathbf{x}\cdot\mathbf{y}=(-4)(-9)+(3)(15)=36+45=81\).
Magnitudes: \(|\mathbf{x}|=\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5\); \(|\mathbf{y}|=\sqrt{(-9)^{2}+15^{2}}=\sqrt{306}\approx17.49\).
\[\cos\theta=\frac{81}{5\times17.49}=\frac{81}{87.46}=0.9261\]
\[\theta=\cos^{-1}(0.9261)\approx22^{\circ}\ \text{(nearest degree)}\]
Answer Details
Use \(\cos\theta=\dfrac{\mathbf{x}\cdot\mathbf{y}}{|\mathbf{x}|\,|\mathbf{y}|}\) with \(\mathbf{x}=\begin{pmatrix}-4\\3\end{pmatrix},\ \mathbf{y}=\begin{pmatrix}-9\\15\end{pmatrix}\).
Dot product: \(\mathbf{x}\cdot\mathbf{y}=(-4)(-9)+(3)(15)=36+45=81\).
Magnitudes: \(|\mathbf{x}|=\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5\); \(|\mathbf{y}|=\sqrt{(-9)^{2}+15^{2}}=\sqrt{306}\approx17.49\).
\[\cos\theta=\frac{81}{5\times17.49}=\frac{81}{87.46}=0.9261\]
\[\theta=\cos^{-1}(0.9261)\approx22^{\circ}\ \text{(nearest degree)}\]
Question 5 Report
(a) Find the equation of the normal to the curve y = (x\(^2\) - x + 1)(x - 2) at the point where the curve cuts the X - axis.
(b) The coordinates of the pints P, Q and R are (-1, 2), (5, 1) and (3, -4) respectively. Find the equation of the line joining Q and the midpoint of \(\overline{PR}\).
(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
Answer Details
(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
Question 6 Report
The polynomial f(x) =2x\(^3\) + px+ qx - 5 has (x-1) as a factor and a remainder of 27 when divided by (x + 2), where p and q are constants. Find the values of p and q.
Reading the polynomial as \(f(x)=2x^{3}+px^{2}+qx-5\) (the two linear terms in the printed stem are a typographical slip for a quadratic and a linear term).
Since \((x-1)\) is a factor, \(f(1)=0\):
\[2+p+q-5=0\Rightarrow p+q=3\quad(\text{i})\]
The remainder is \(27\) when divided by \((x+2)\), so \(f(-2)=27\):
\[2(-8)+p(4)+q(-2)-5=27\Rightarrow -16+4p-2q-5=27\]
\[4p-2q=48\Rightarrow 2p-q=24\quad(\text{ii})\]
Add (i) and (ii): \(3p=27\Rightarrow p=9\). Then \(q=3-9=-6\).
\[\boxed{p=9,\ q=-6}\]
Answer Details
Reading the polynomial as \(f(x)=2x^{3}+px^{2}+qx-5\) (the two linear terms in the printed stem are a typographical slip for a quadratic and a linear term).
Since \((x-1)\) is a factor, \(f(1)=0\):
\[2+p+q-5=0\Rightarrow p+q=3\quad(\text{i})\]
The remainder is \(27\) when divided by \((x+2)\), so \(f(-2)=27\):
\[2(-8)+p(4)+q(-2)-5=27\Rightarrow -16+4p-2q-5=27\]
\[4p-2q=48\Rightarrow 2p-q=24\quad(\text{ii})\]
Add (i) and (ii): \(3p=27\Rightarrow p=9\). Then \(q=3-9=-6\).
\[\boxed{p=9,\ q=-6}\]
Question 7 Report
Given that (p + 1/2√3)(1 - √3)\(^2\) = 3- √3,
find x the value of p.
Reading the expression as \(\left(p+\tfrac12\sqrt3\right)(1-\sqrt3)^{2}=3-\sqrt3\).
First expand the square:
\[(1-\sqrt3)^{2}=1-2\sqrt3+3=4-2\sqrt3\]
So:
\[\left(p+\tfrac{\sqrt3}{2}\right)(4-2\sqrt3)=3-\sqrt3\]
Note \(4-2\sqrt3=2(2-\sqrt3)\). Divide both sides:
\[p+\frac{\sqrt3}{2}=\frac{3-\sqrt3}{4-2\sqrt3}\]
Rationalise the right side by \(\times\dfrac{4+2\sqrt3}{4+2\sqrt3}\): denominator \(16-12=4\); numerator \((3-\sqrt3)(4+2\sqrt3)=12+2\sqrt3-6=6+2\sqrt3\). So the right side \(=\dfrac{6+2\sqrt3}{4}=\dfrac{3+\sqrt3}{2}\).
\[p=\frac{3+\sqrt3}{2}-\frac{\sqrt3}{2}=\frac{3}{2}\]
\[\boxed{p=\tfrac32}\]
Answer Details
Reading the expression as \(\left(p+\tfrac12\sqrt3\right)(1-\sqrt3)^{2}=3-\sqrt3\).
First expand the square:
\[(1-\sqrt3)^{2}=1-2\sqrt3+3=4-2\sqrt3\]
So:
\[\left(p+\tfrac{\sqrt3}{2}\right)(4-2\sqrt3)=3-\sqrt3\]
Note \(4-2\sqrt3=2(2-\sqrt3)\). Divide both sides:
\[p+\frac{\sqrt3}{2}=\frac{3-\sqrt3}{4-2\sqrt3}\]
Rationalise the right side by \(\times\dfrac{4+2\sqrt3}{4+2\sqrt3}\): denominator \(16-12=4\); numerator \((3-\sqrt3)(4+2\sqrt3)=12+2\sqrt3-6=6+2\sqrt3\). So the right side \(=\dfrac{6+2\sqrt3}{4}=\dfrac{3+\sqrt3}{2}\).
\[p=\frac{3+\sqrt3}{2}-\frac{\sqrt3}{2}=\frac{3}{2}\]
\[\boxed{p=\tfrac32}\]
Question 8 Report
A bag contains 24 mangoes out of which six are bad. If 6 mangoes are selected randomly from the bag with replacement, find the probability that not more than 3 are bad.
Because selection is with replacement, each draw is independent with a constant probability of a bad mango:
\[p=\frac{6}{24}=\frac14,\qquad q=1-p=\frac34\]
Let \(X\) be the number of bad mangoes in \(n=6\) draws; \(X\sim\text{Bin}(6,\tfrac14)\). We need \(P(X\le3)\), which is easiest via the complement \(P(X\le3)=1-P(X\ge4)\).
\[P(4)=\binom{6}{4}\left(\tfrac14\right)^{4}\left(\tfrac34\right)^{2}=\frac{15\times9}{4096}=\frac{135}{4096}\]
\[P(5)=\binom{6}{5}\left(\tfrac14\right)^{5}\left(\tfrac34\right)=\frac{6\times3}{4096}=\frac{18}{4096}\]
\[P(6)=\left(\tfrac14\right)^{6}=\frac{1}{4096}\]
\[P(X\ge4)=\frac{135+18+1}{4096}=\frac{154}{4096}=\frac{77}{2048}\]
\[P(X\le3)=1-\frac{77}{2048}=\frac{1971}{2048}\approx0.9624\]
Answer Details
Because selection is with replacement, each draw is independent with a constant probability of a bad mango:
\[p=\frac{6}{24}=\frac14,\qquad q=1-p=\frac34\]
Let \(X\) be the number of bad mangoes in \(n=6\) draws; \(X\sim\text{Bin}(6,\tfrac14)\). We need \(P(X\le3)\), which is easiest via the complement \(P(X\le3)=1-P(X\ge4)\).
\[P(4)=\binom{6}{4}\left(\tfrac14\right)^{4}\left(\tfrac34\right)^{2}=\frac{15\times9}{4096}=\frac{135}{4096}\]
\[P(5)=\binom{6}{5}\left(\tfrac14\right)^{5}\left(\tfrac34\right)=\frac{6\times3}{4096}=\frac{18}{4096}\]
\[P(6)=\left(\tfrac14\right)^{6}=\frac{1}{4096}\]
\[P(X\ge4)=\frac{135+18+1}{4096}=\frac{154}{4096}=\frac{77}{2048}\]
\[P(X\le3)=1-\frac{77}{2048}=\frac{1971}{2048}\approx0.9624\]
Question 9 Report
A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement.
Calculate the probability of picking:
(a) same color of bulbs; (6) different color of bulbs (c) at least one red bulb.
Total bulbs:
n(Red)=5, n(B) = 7, n(G) = 4
(5+7+4) = 16
p(R)= 5/16, p(B) = 7/16, p(G) = 4/16
(a) p(All same colour of bulbs)
= p(RR) Or p(BB) or p(GG)
516
* 415
+ 716
* 615
+ 416
* 315
= 112 + 740 + 120
= 10+21+6120 = 37120
(b) All different colors = 1-p (AIl the same colour) =
1 - 37120 = 83120
(c) p(At least one red) = p(RR) + p(RB) + p(RG)
516 * 415 + 516 * 715 + 516 * 415
= 112 + 748 + 112
8+748 = 1548
= 516
Answer Details
Total bulbs:
n(Red)=5, n(B) = 7, n(G) = 4
(5+7+4) = 16
p(R)= 5/16, p(B) = 7/16, p(G) = 4/16
(a) p(All same colour of bulbs)
= p(RR) Or p(BB) or p(GG)
516
* 415
+ 716
* 615
+ 416
* 315
= 112 + 740 + 120
= 10+21+6120 = 37120
(b) All different colors = 1-p (AIl the same colour) =
1 - 37120 = 83120
(c) p(At least one red) = p(RR) + p(RB) + p(RG)
516 * 415 + 516 * 715 + 516 * 415
= 112 + 748 + 112
8+748 = 1548
= 516
Question 10 Report
(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres.
(i) How many days will it take the jogger to reach a distance of 15km in training?
(ii) Calculate the total distance he would have run in the training.
(b) The second term of a Geometric Progression (GP) is -3. If its sum to infinity is 25/2, find its common ratios.
(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
Answer Details
(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
Question 11 Report
\(^{5y}{C}_2\) = 190, find the value of y
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.
Answer Details
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.
Question 12 Report
The position vectors of P, Q and R with respect to the origin are (4i-5j), (i+3j) and (-5i+2j) respectively. If PQRM is a parallelogram, find:
(a) the coordinates of M;
(b) the acute angle between \(\overline{PM}\) and \(\overline{PQ}\), correct to the nearest degree.
Write the position vectors as coordinates: \(P(4,-5)\), \(Q(1,3)\), \(R(-5,2)\).
(a) Coordinates of M
In the parallelogram \(PQRM\) the vertices are taken in order \(P\to Q\to R\to M\), so the diagonals \(PR\) and \(QM\) bisect each other. Equating their midpoints:
\[\text{mid}(PR)=\left(\tfrac{4+(-5)}{2},\tfrac{-5+2}{2}\right)=\left(-\tfrac{1}{2},-\tfrac{3}{2}\right).\]
\[\text{mid}(QM)=\left(\tfrac{1+m_1}{2},\tfrac{3+m_2}{2}\right).\]
So \(1+m_1=-1\Rightarrow m_1=-2\) and \(3+m_2=-3\Rightarrow m_2=-6\).
\[\boxed{M(-2,-6)}.\]
(b) Acute angle between \(\overline{PM}\) and \(\overline{PQ}\)
\[\overline{PM}=M-P=(-2-4,\,-6-(-5))=(-6,-1),\]
\[\overline{PQ}=Q-P=(1-4,\,3-(-5))=(-3,8).\]
Using the scalar (dot) product,
\[\overline{PM}\cdot\overline{PQ}=(-6)(-3)+(-1)(8)=18-8=10,\]
\[|\overline{PM}|=\sqrt{(-6)^2+(-1)^2}=\sqrt{37},\qquad |\overline{PQ}|=\sqrt{(-3)^2+8^2}=\sqrt{73}.\]
\[\cos\theta=\frac{10}{\sqrt{37}\,\sqrt{73}}=\frac{10}{\sqrt{2701}}\approx 0.1924.\]
\[\theta=\cos^{-1}(0.1924)\approx 78.9^{\circ}\approx 79^{\circ}.\]
Since this value is already acute, the acute angle is \(79^{\circ}\).
Answer Details
Write the position vectors as coordinates: \(P(4,-5)\), \(Q(1,3)\), \(R(-5,2)\).
(a) Coordinates of M
In the parallelogram \(PQRM\) the vertices are taken in order \(P\to Q\to R\to M\), so the diagonals \(PR\) and \(QM\) bisect each other. Equating their midpoints:
\[\text{mid}(PR)=\left(\tfrac{4+(-5)}{2},\tfrac{-5+2}{2}\right)=\left(-\tfrac{1}{2},-\tfrac{3}{2}\right).\]
\[\text{mid}(QM)=\left(\tfrac{1+m_1}{2},\tfrac{3+m_2}{2}\right).\]
So \(1+m_1=-1\Rightarrow m_1=-2\) and \(3+m_2=-3\Rightarrow m_2=-6\).
\[\boxed{M(-2,-6)}.\]
(b) Acute angle between \(\overline{PM}\) and \(\overline{PQ}\)
\[\overline{PM}=M-P=(-2-4,\,-6-(-5))=(-6,-1),\]
\[\overline{PQ}=Q-P=(1-4,\,3-(-5))=(-3,8).\]
Using the scalar (dot) product,
\[\overline{PM}\cdot\overline{PQ}=(-6)(-3)+(-1)(8)=18-8=10,\]
\[|\overline{PM}|=\sqrt{(-6)^2+(-1)^2}=\sqrt{37},\qquad |\overline{PQ}|=\sqrt{(-3)^2+8^2}=\sqrt{73}.\]
\[\cos\theta=\frac{10}{\sqrt{37}\,\sqrt{73}}=\frac{10}{\sqrt{2701}}\approx 0.1924.\]
\[\theta=\cos^{-1}(0.1924)\approx 78.9^{\circ}\approx 79^{\circ}.\]
Since this value is already acute, the acute angle is \(79^{\circ}\).
Question 13 Report
(a) A girl threw a stone horizontally with a velocity of 30m/s from the top of a cliff 50m high. How far from the foot of the cliff does the stone strike the ground? [Take g= 10m/s\(^2\)
(b) A body A, of mass 2kg is held in equilibrium by means of two strings AP and AR. AP is inclined at 56° to the upward vertical and AR is horizontal.
Find the tensions T\(_1\), and T\(_2\), in the strings [Take g= 10ms\(^2\)]
(a) Horizontal projectile from a cliff
Horizontally the stone travels at a constant \(30\,\text{m/s}\); vertically it starts with zero vertical velocity and falls under gravity. First find the time of flight from the vertical motion, using \(h=\tfrac{1}{2}gt^{2}\):
\[50=\tfrac{1}{2}(10)t^{2}\;\Rightarrow\; 50=5t^{2}\;\Rightarrow\; t^{2}=10\;\Rightarrow\; t=\sqrt{10}\approx 3.16\,\text{s}.\]
The horizontal distance (range) is
\[R=\text{(horizontal speed)}\times t=30\sqrt{10}\approx 94.9\,\text{m}.\]
The stone lands about \(94.9\,\text{m}\) from the foot of the cliff.
(b) Body in equilibrium on two strings
The weight of the body is \(W=mg=2\times 10=20\,\text{N}\), acting vertically downward. String \(AP\) (tension \(T_1\)) makes \(56^{\circ}\) with the upward vertical, and string \(AR\) (tension \(T_2\)) is horizontal. Resolve the forces at \(A\).
Vertical equilibrium: only \(T_1\) has a vertical component, and it supports the weight:
\[T_1\cos 56^{\circ}=20\;\Rightarrow\; T_1=\frac{20}{\cos 56^{\circ}}=\frac{20}{0.5592}\approx 35.8\,\text{N}.\]
Horizontal equilibrium: the horizontal component of \(T_1\) is balanced by \(T_2\):
\[T_2=T_1\sin 56^{\circ}=20\tan 56^{\circ}=20(1.4826)\approx 29.7\,\text{N}.\]
Hence \(T_1\approx 35.8\,\text{N}\) and \(T_2\approx 29.7\,\text{N}\).
Answer Details
(a) Horizontal projectile from a cliff
Horizontally the stone travels at a constant \(30\,\text{m/s}\); vertically it starts with zero vertical velocity and falls under gravity. First find the time of flight from the vertical motion, using \(h=\tfrac{1}{2}gt^{2}\):
\[50=\tfrac{1}{2}(10)t^{2}\;\Rightarrow\; 50=5t^{2}\;\Rightarrow\; t^{2}=10\;\Rightarrow\; t=\sqrt{10}\approx 3.16\,\text{s}.\]
The horizontal distance (range) is
\[R=\text{(horizontal speed)}\times t=30\sqrt{10}\approx 94.9\,\text{m}.\]
The stone lands about \(94.9\,\text{m}\) from the foot of the cliff.
(b) Body in equilibrium on two strings
The weight of the body is \(W=mg=2\times 10=20\,\text{N}\), acting vertically downward. String \(AP\) (tension \(T_1\)) makes \(56^{\circ}\) with the upward vertical, and string \(AR\) (tension \(T_2\)) is horizontal. Resolve the forces at \(A\).
Vertical equilibrium: only \(T_1\) has a vertical component, and it supports the weight:
\[T_1\cos 56^{\circ}=20\;\Rightarrow\; T_1=\frac{20}{\cos 56^{\circ}}=\frac{20}{0.5592}\approx 35.8\,\text{N}.\]
Horizontal equilibrium: the horizontal component of \(T_1\) is balanced by \(T_2\):
\[T_2=T_1\sin 56^{\circ}=20\tan 56^{\circ}=20(1.4826)\approx 29.7\,\text{N}.\]
Hence \(T_1\approx 35.8\,\text{N}\) and \(T_2\approx 29.7\,\text{N}\).
Question 14 Report
The table shows the distribution of monthly income (in thousands of naira) of workers in a factory
| Monthly Income (N'1000) | 135-139 | 140-149 | 150-154 | 155-164 | 165-169 |
| Number of workers | 20 | 42 | 28 | 38 | 22 |
(a) Draw a histogram for the distribution.
(b) Use your graph to estimate the mode of the distribution.
(a) Histogram
Since the class intervals have unequal widths, the heights of the rectangles are the frequency densities:
\[\text{Frequency density}=\frac{\text{frequency}}{\text{class width}}.\]
| Monthly income (₦'000) | Class boundaries (₦'000) | Frequency | Class width | Frequency density |
|---|---|---|---|---|
| 135–139 | 134.5–139.5 | 20 | 5 | 4.0 |
| 140–149 | 139.5–149.5 | 42 | 10 | 4.2 |
| 150–154 | 149.5–154.5 | 28 | 5 | 5.6 |
| 155–164 | 154.5–164.5 | 38 | 10 | 3.8 |
| 165–169 | 164.5–169.5 | 22 | 5 | 4.4 |
Plot the class boundaries on the horizontal axis and frequency density on the vertical axis. The resulting histogram is:
(b) Estimated mode
The modal class is \(149.5\text{–}154.5\), since it has the greatest frequency density, \(5.6\).
Using the standard intersecting-diagonals construction on the modal rectangle gives
\[\begin{aligned} \text{Mode} &=149.5+\frac{5.6-4.2}{(5.6-4.2)+(5.6-3.8)}\times 5\\ &=149.5+\frac{1.4}{3.2}\times5\\ &=151.6875\approx151.7. \end{aligned}\]
Therefore, the estimated modal monthly income is \(151.7\) thousand naira, that is, approximately ₦151,700.
Answer Details
(a) Histogram
Since the class intervals have unequal widths, the heights of the rectangles are the frequency densities:
\[\text{Frequency density}=\frac{\text{frequency}}{\text{class width}}.\]
| Monthly income (₦'000) | Class boundaries (₦'000) | Frequency | Class width | Frequency density |
|---|---|---|---|---|
| 135–139 | 134.5–139.5 | 20 | 5 | 4.0 |
| 140–149 | 139.5–149.5 | 42 | 10 | 4.2 |
| 150–154 | 149.5–154.5 | 28 | 5 | 5.6 |
| 155–164 | 154.5–164.5 | 38 | 10 | 3.8 |
| 165–169 | 164.5–169.5 | 22 | 5 | 4.4 |
Plot the class boundaries on the horizontal axis and frequency density on the vertical axis. The resulting histogram is:
(b) Estimated mode
The modal class is \(149.5\text{–}154.5\), since it has the greatest frequency density, \(5.6\).
Using the standard intersecting-diagonals construction on the modal rectangle gives
\[\begin{aligned} \text{Mode} &=149.5+\frac{5.6-4.2}{(5.6-4.2)+(5.6-3.8)}\times 5\\ &=149.5+\frac{1.4}{3.2}\times5\\ &=151.6875\approx151.7. \end{aligned}\]
Therefore, the estimated modal monthly income is \(151.7\) thousand naira, that is, approximately ₦151,700.
Question 15 Report
The table shows the frequency distribution of heights (in cm) of pupils in a certain school.
| Heights | 100-109 | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 |
| Frequency | 27 | 58 | 130 | 105 | 50 | 25 | 5 |
(a) (i) Construct a cumulative frequency table. (ii) Use the table to draw a cumulative frequency curve.
(b) Using the curve, estimate the: (i)median height; (ii) inter quartile range (iii) percentage of students whose heights are most 130cm.
(a) (i) Cumulative frequency table
| Height class (cm) | Frequency | Upper class boundary (cm) | Cumulative frequency |
|---|---|---|---|
| 100–109 | 27 | 109.5 | 27 |
| 110–119 | 58 | 119.5 | 85 |
| 120–129 | 130 | 129.5 | 215 |
| 130–139 | 105 | 139.5 | 320 |
| 140–149 | 50 | 149.5 | 370 |
| 150–159 | 25 | 159.5 | 395 |
| 160–169 | 5 | 169.5 | 400 |
The total number of pupils is \(N=400\). Include the starting point \((99.5,0)\).
(a) (ii) Less-than cumulative frequency curve (ogive)
Plot the upper class boundaries against their cumulative frequencies and join successive points with a smooth increasing curve.
(b) Estimates from the ogive
(i) Median height
\[\frac{N}{2}=\frac{400}{2}=200\]
At cumulative frequency 200, the corresponding height is approximately \(128.3\text{ cm}\).
Median height \(\approx 128.3\text{ cm}\).
(ii) Interquartile range
\[Q_1=\frac{N}{4}=100,\qquad Q_3=\frac{3N}{4}=300\]
From the curve, \(Q_1\approx120.7\text{ cm}\) and \(Q_3\approx137.6\text{ cm}\).
\[\text{Interquartile range}=Q_3-Q_1=137.6-120.7\approx16.9\text{ cm}.\]
(iii) Percentage of pupils whose heights are at most \(130\text{ cm}\)
At \(130\text{ cm}\), the cumulative frequency is approximately \(220\).
\[\text{Percentage}=\frac{220}{400}\times100\%\approx55\%.\]
Therefore, approximately \(55\%\) of the pupils have heights at most \(130\text{ cm}\).
Answer Details
(a) (i) Cumulative frequency table
| Height class (cm) | Frequency | Upper class boundary (cm) | Cumulative frequency |
|---|---|---|---|
| 100–109 | 27 | 109.5 | 27 |
| 110–119 | 58 | 119.5 | 85 |
| 120–129 | 130 | 129.5 | 215 |
| 130–139 | 105 | 139.5 | 320 |
| 140–149 | 50 | 149.5 | 370 |
| 150–159 | 25 | 159.5 | 395 |
| 160–169 | 5 | 169.5 | 400 |
The total number of pupils is \(N=400\). Include the starting point \((99.5,0)\).
(a) (ii) Less-than cumulative frequency curve (ogive)
Plot the upper class boundaries against their cumulative frequencies and join successive points with a smooth increasing curve.
(b) Estimates from the ogive
(i) Median height
\[\frac{N}{2}=\frac{400}{2}=200\]
At cumulative frequency 200, the corresponding height is approximately \(128.3\text{ cm}\).
Median height \(\approx 128.3\text{ cm}\).
(ii) Interquartile range
\[Q_1=\frac{N}{4}=100,\qquad Q_3=\frac{3N}{4}=300\]
From the curve, \(Q_1\approx120.7\text{ cm}\) and \(Q_3\approx137.6\text{ cm}\).
\[\text{Interquartile range}=Q_3-Q_1=137.6-120.7\approx16.9\text{ cm}.\]
(iii) Percentage of pupils whose heights are at most \(130\text{ cm}\)
At \(130\text{ cm}\), the cumulative frequency is approximately \(220\).
\[\text{Percentage}=\frac{220}{400}\times100\%\approx55\%.\]
Therefore, approximately \(55\%\) of the pupils have heights at most \(130\text{ cm}\).
