WAEC SSCE - Further Mathematics - 2021 (Essay)
Tambaya 1 Rahoto
\(^{5y}{C}_2\) = 190, find the value of y
Use \({}^{n}C_{2}=\dfrac{n(n-1)}{2}\) with \(n=5y\):
\[{}^{5y}C_{2}=\frac{5y(5y-1)}{2}=190\]
So \(5y(5y-1)=380\). Let \(n=5y\):
\[n^{2}-n-380=0\Rightarrow n=\frac{1\pm\sqrt{1+1520}}{2}=\frac{1\pm39}{2}\]
Taking the positive root, \(n=20\). Since \(n=5y\):
\[5y=20\Rightarrow \boxed{y=4}\]
(The negative root \(n=-19\) is rejected.)
Bayanin Amsa
Use \({}^{n}C_{2}=\dfrac{n(n-1)}{2}\) with \(n=5y\):
\[{}^{5y}C_{2}=\frac{5y(5y-1)}{2}=190\]
So \(5y(5y-1)=380\). Let \(n=5y\):
\[n^{2}-n-380=0\Rightarrow n=\frac{1\pm\sqrt{1+1520}}{2}=\frac{1\pm39}{2}\]
Taking the positive root, \(n=20\). Since \(n=5y\):
\[5y=20\Rightarrow \boxed{y=4}\]
(The negative root \(n=-19\) is rejected.)
Tambaya 2 Rahoto
(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) Daily distances form an AP: first term \(a=500\text{m}\), common difference \(d=250\text{m}\).
(i) The daily run reaches \(15\text{km}=15000\text{m}\) when the \(n\)th term equals \(15000\):
\[a+(n-1)d=15000\Rightarrow 500+250(n-1)=15000\]
\[250(n-1)=14500\Rightarrow n-1=58\Rightarrow n=59\text{ days}\]
(ii) Total distance is the sum of 59 terms:
\[S_{59}=\frac{n}{2}(a+l)=\frac{59}{2}(500+15000)=\frac{59}{2}(15500)=456250\text{ m}=456.25\text{ km}\]
(b) GP: second term \(ar=-3\); sum to infinity \(\dfrac{a}{1-r}=\dfrac{25}{2}\).
From \(ar=-3\), \(a=-\dfrac{3}{r}\). Substitute:
\[\frac{-3/r}{1-r}=\frac{25}{2}\Rightarrow -6=25r(1-r)\Rightarrow 25r^{2}-25r-6=0\]
\[r=\frac{25\pm\sqrt{625+600}}{50}=\frac{25\pm35}{50}\Rightarrow r=\tfrac65\ \text{or}\ r=-\tfrac15\]
A sum to infinity requires \(|r|<1\), so \(\boxed{r=-\tfrac15}\) (reject \(\tfrac65\)).
Bayanin Amsa
(a) Daily distances form an AP: first term \(a=500\text{m}\), common difference \(d=250\text{m}\).
(i) The daily run reaches \(15\text{km}=15000\text{m}\) when the \(n\)th term equals \(15000\):
\[a+(n-1)d=15000\Rightarrow 500+250(n-1)=15000\]
\[250(n-1)=14500\Rightarrow n-1=58\Rightarrow n=59\text{ days}\]
(ii) Total distance is the sum of 59 terms:
\[S_{59}=\frac{n}{2}(a+l)=\frac{59}{2}(500+15000)=\frac{59}{2}(15500)=456250\text{ m}=456.25\text{ km}\]
(b) GP: second term \(ar=-3\); sum to infinity \(\dfrac{a}{1-r}=\dfrac{25}{2}\).
From \(ar=-3\), \(a=-\dfrac{3}{r}\). Substitute:
\[\frac{-3/r}{1-r}=\frac{25}{2}\Rightarrow -6=25r(1-r)\Rightarrow 25r^{2}-25r-6=0\]
\[r=\frac{25\pm\sqrt{625+600}}{50}=\frac{25\pm35}{50}\Rightarrow r=\tfrac65\ \text{or}\ r=-\tfrac15\]
A sum to infinity requires \(|r|<1\), so \(\boxed{r=-\tfrac15}\) (reject \(\tfrac65\)).
Tambaya 3 Rahoto
Evaluate: \(^9{?}_1\) \(\frac{x(2x-3)}{?x}\) dx
Simplify the integrand first. Since \(\sqrt{x}=x^{1/2}\):
\[\frac{x(2x-3)}{\sqrt{x}}=\frac{2x^{2}-3x}{x^{1/2}}=2x^{3/2}-3x^{1/2}\]
Integrate:
\[\int_{1}^{9}\left(2x^{3/2}-3x^{1/2}\right)dx=\left[\frac{4}{5}x^{5/2}-2x^{3/2}\right]_{1}^{9}\]
At \(x=9\): \(9^{5/2}=3^{5}=243\) and \(9^{3/2}=27\), giving \(\tfrac45(243)-2(27)=194.4-54=140.4\).
At \(x=1\): \(\tfrac45-2=-1.2\).
\[140.4-(-1.2)=141.6=\frac{708}{5}\]
Bayanin Amsa
Simplify the integrand first. Since \(\sqrt{x}=x^{1/2}\):
\[\frac{x(2x-3)}{\sqrt{x}}=\frac{2x^{2}-3x}{x^{1/2}}=2x^{3/2}-3x^{1/2}\]
Integrate:
\[\int_{1}^{9}\left(2x^{3/2}-3x^{1/2}\right)dx=\left[\frac{4}{5}x^{5/2}-2x^{3/2}\right]_{1}^{9}\]
At \(x=9\): \(9^{5/2}=3^{5}=243\) and \(9^{3/2}=27\), giving \(\tfrac45(243)-2(27)=194.4-54=140.4\).
At \(x=1\): \(\tfrac45-2=-1.2\).
\[140.4-(-1.2)=141.6=\frac{708}{5}\]
Tambaya 4 Rahoto
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 \(P(X\le3)=1-P(X\ge4)\).
\[P(4)=\binom{6}{4}\left(\tfrac14\right)^{4}\left(\tfrac34\right)^{2}=\frac{15\times9}{4096}=\frac{135}{4096}\]
\[P(5)=\binom{6}{5}\left(\tfrac14\right)^{5}\left(\tfrac34\right)=\frac{6\times3}{4096}=\frac{18}{4096}\]
\[P(6)=\left(\tfrac14\right)^{6}=\frac{1}{4096}\]
\[P(X\ge4)=\frac{135+18+1}{4096}=\frac{154}{4096}=\frac{77}{2048}\]
\[P(X\le3)=1-\frac{77}{2048}=\frac{1971}{2048}\approx0.9624\]
Bayanin Amsa
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 \(P(X\le3)=1-P(X\ge4)\).
\[P(4)=\binom{6}{4}\left(\tfrac14\right)^{4}\left(\tfrac34\right)^{2}=\frac{15\times9}{4096}=\frac{135}{4096}\]
\[P(5)=\binom{6}{5}\left(\tfrac14\right)^{5}\left(\tfrac34\right)=\frac{6\times3}{4096}=\frac{18}{4096}\]
\[P(6)=\left(\tfrac14\right)^{6}=\frac{1}{4096}\]
\[P(X\ge4)=\frac{135+18+1}{4096}=\frac{154}{4096}=\frac{77}{2048}\]
\[P(X\le3)=1-\frac{77}{2048}=\frac{1971}{2048}\approx0.9624\]
Tambaya 5 Rahoto
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 \(=\dfrac{6+2\sqrt3}{4}=\dfrac{3+\sqrt3}{2}\).
\[p=\frac{3+\sqrt3}{2}-\frac{\sqrt3}{2}=\frac{3}{2}\]
\[\boxed{p=\tfrac32}\]
Bayanin Amsa
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 \(=\dfrac{6+2\sqrt3}{4}=\dfrac{3+\sqrt3}{2}\).
\[p=\frac{3+\sqrt3}{2}-\frac{\sqrt3}{2}=\frac{3}{2}\]
\[\boxed{p=\tfrac32}\]
Tambaya 6 Rahoto
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 \(79^{\circ}\).
Bayanin Amsa
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 \(79^{\circ}\).
Tambaya 7 Rahoto
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 \(q=3-9=-6\).
\[\boxed{p=9,\ q=-6}\]
Bayanin Amsa
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 \(q=3-9=-6\).
\[\boxed{p=9,\ q=-6}\]
Tambaya 8 Rahoto
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 (cf against upper boundaries). \(N = 400\).
| Height | Freq | Upper boundary | Cumulative freq |
|---|---|---|---|
| 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 |
(a)(ii) Plot (109.5, 27), (119.5, 85), ... , (169.5, 400) and join with a smooth ogive.
(b)(i) Median at position \( \tfrac{400}{2} = 200\), in class 120 - 129 (\(L=119.5, \text{cf}=85, f=130\)):
\[ \text{Median} = 119.5 + \frac{200 - 85}{130}\times 10 = 119.5 + 8.85 \approx \mathbf{128.3 \text{ cm}} \](b)(ii) Interquartile range. \(Q_1\) at 100th (class 120 - 129):
\[ Q_1 = 119.5 + \frac{100 - 85}{130}\times 10 = 119.5 + 1.15 = 120.65 \]\(Q_3\) at 300th (class 130 - 139, \(L=129.5, \text{cf}=215, f=105\)):
\[ Q_3 = 129.5 + \frac{300 - 215}{105}\times 10 = 129.5 + 8.10 = 137.60 \] \[ \text{IQR} = Q_3 - Q_1 = 137.60 - 120.65 \approx \mathbf{17.0 \text{ cm}} \](b)(iii) Percentage with height at most 130 cm. Read cf at 130 (class 130 - 139):
\[ \text{cf}(130) = 215 + \frac{130 - 129.5}{10}\times 105 = 215 + 5.25 = 220.25 \] \[ \text{Percentage} = \frac{220.25}{400}\times 100 \approx \mathbf{55\%} \]Bayanin Amsa
(a)(i) Cumulative frequency table (cf against upper boundaries). \(N = 400\).
| Height | Freq | Upper boundary | Cumulative freq |
|---|---|---|---|
| 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 |
(a)(ii) Plot (109.5, 27), (119.5, 85), ... , (169.5, 400) and join with a smooth ogive.
(b)(i) Median at position \( \tfrac{400}{2} = 200\), in class 120 - 129 (\(L=119.5, \text{cf}=85, f=130\)):
\[ \text{Median} = 119.5 + \frac{200 - 85}{130}\times 10 = 119.5 + 8.85 \approx \mathbf{128.3 \text{ cm}} \](b)(ii) Interquartile range. \(Q_1\) at 100th (class 120 - 129):
\[ Q_1 = 119.5 + \frac{100 - 85}{130}\times 10 = 119.5 + 1.15 = 120.65 \]\(Q_3\) at 300th (class 130 - 139, \(L=129.5, \text{cf}=215, f=105\)):
\[ Q_3 = 129.5 + \frac{300 - 215}{105}\times 10 = 129.5 + 8.10 = 137.60 \] \[ \text{IQR} = Q_3 - Q_1 = 137.60 - 120.65 \approx \mathbf{17.0 \text{ cm}} \](b)(iii) Percentage with height at most 130 cm. Read cf at 130 (class 130 - 139):
\[ \text{cf}(130) = 215 + \frac{130 - 129.5}{10}\times 105 = 215 + 5.25 = 220.25 \] \[ \text{Percentage} = \frac{220.25}{400}\times 100 \approx \mathbf{55\%} \]Tambaya 9 Rahoto
(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 \(m=16\text{kg},\ T=240\text{N},\ g=10\text{m/s}^{2}\):
\[T-mg=ma\Rightarrow 240-16(10)=16a\]
\[80=16a\Rightarrow a=5\text{ m/s}^{2}\ (\text{upward})\]
Bayanin Amsa
(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 \(m=16\text{kg},\ T=240\text{N},\ g=10\text{m/s}^{2}\):
\[T-mg=ma\Rightarrow 240-16(10)=16a\]
\[80=16a\Rightarrow a=5\text{ m/s}^{2}\ (\text{upward})\]
Tambaya 10 Rahoto
(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) The curve \(y=(x^{2}-x+1)(x-2)\) cuts the x-axis where \(y=0\). Since \(x^{2}-x+1=0\) has discriminant \(1-4<0\) (no real roots), the only x-intercept is \(x-2=0\), i.e. the point \((2,0)\).
Expand: \(y=x^{3}-3x^{2}+3x-2\), so \(\dfrac{dy}{dx}=3x^{2}-6x+3\). At \(x=2\): \(3(4)-12+3=3\). This is the tangent gradient, so the normal gradient is \(-\tfrac13\).
Normal at \((2,0)\): \(y-0=-\tfrac13(x-2)\), i.e.
\[x+3y-2=0\]
(b) \(P(-1,2),\ Q(5,1),\ R(3,-4)\). Midpoint of \(\overline{PR}=\left(\tfrac{-1+3}{2},\tfrac{2-4}{2}\right)=(1,-1)\).
Line through \(Q(5,1)\) and \((1,-1)\): gradient \(=\dfrac{1-(-1)}{5-1}=\dfrac{2}{4}=\tfrac12\).
\[y-1=\tfrac12(x-5)\Rightarrow x-2y-3=0\]
Bayanin Amsa
(a) The curve \(y=(x^{2}-x+1)(x-2)\) cuts the x-axis where \(y=0\). Since \(x^{2}-x+1=0\) has discriminant \(1-4<0\) (no real roots), the only x-intercept is \(x-2=0\), i.e. the point \((2,0)\).
Expand: \(y=x^{3}-3x^{2}+3x-2\), so \(\dfrac{dy}{dx}=3x^{2}-6x+3\). At \(x=2\): \(3(4)-12+3=3\). This is the tangent gradient, so the normal gradient is \(-\tfrac13\).
Normal at \((2,0)\): \(y-0=-\tfrac13(x-2)\), i.e.
\[x+3y-2=0\]
(b) \(P(-1,2),\ Q(5,1),\ R(3,-4)\). Midpoint of \(\overline{PR}=\left(\tfrac{-1+3}{2},\tfrac{2-4}{2}\right)=(1,-1)\).
Line through \(Q(5,1)\) and \((1,-1)\): gradient \(=\dfrac{1-(-1)}{5-1}=\dfrac{2}{4}=\tfrac12\).
\[y-1=\tfrac12(x-5)\Rightarrow x-2y-3=0\]
Tambaya 11 Rahoto
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: \(|\mathbf{x}|=\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5\); \(|\mathbf{y}|=\sqrt{(-9)^{2}+15^{2}}=\sqrt{306}\approx17.49\).
\[\cos\theta=\frac{81}{5\times17.49}=\frac{81}{87.46}=0.9261\]
\[\theta=\cos^{-1}(0.9261)\approx22^{\circ}\ \text{(nearest degree)}\]
Bayanin Amsa
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: \(|\mathbf{x}|=\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5\); \(|\mathbf{y}|=\sqrt{(-9)^{2}+15^{2}}=\sqrt{306}\approx17.49\).
\[\cos\theta=\frac{81}{5\times17.49}=\frac{81}{87.46}=0.9261\]
\[\theta=\cos^{-1}(0.9261)\approx22^{\circ}\ \text{(nearest degree)}\]
Tambaya 12 Rahoto
(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 \(T_2\approx 29.7\,\text{N}\).
Bayanin Amsa
(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 \(T_2\approx 29.7\,\text{N}\).
Tambaya 13 Rahoto
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.
The box holds \(5\) red, \(7\) blue and \(4\) green bulbs, giving \(5+7+4=16\) bulbs. Two bulbs are drawn without replacement, so the number of equally likely selections is
\[\binom{16}{2}=\frac{16\times 15}{2}=120.\]
(a) Same colour
Count the same-colour pairs for each colour and add:
- Both red: \(\binom{5}{2}=10\)
- Both blue: \(\binom{7}{2}=21\)
- Both green: \(\binom{4}{2}=6\)
Favourable outcomes \(=10+21+6=37\), so
\[P(\text{same colour})=\frac{37}{120}.\]
(b) Different colours
Every draw is either the two bulbs matching or not, so "different colours" is the complement of "same colour":
\[P(\text{different})=1-\frac{37}{120}=\frac{83}{120}.\]
(c) At least one red
Use the complement "no red bulb". There are \(16-5=11\) non-red bulbs, giving \(\binom{11}{2}=\frac{11\times 10}{2}=55\) all-non-red pairs. Hence
\[P(\text{no red})=\frac{55}{120}=\frac{11}{24},\qquad P(\text{at least one red})=1-\frac{11}{24}=\frac{13}{24}.\]
Bayanin Amsa
The box holds \(5\) red, \(7\) blue and \(4\) green bulbs, giving \(5+7+4=16\) bulbs. Two bulbs are drawn without replacement, so the number of equally likely selections is
\[\binom{16}{2}=\frac{16\times 15}{2}=120.\]
(a) Same colour
Count the same-colour pairs for each colour and add:
- Both red: \(\binom{5}{2}=10\)
- Both blue: \(\binom{7}{2}=21\)
- Both green: \(\binom{4}{2}=6\)
Favourable outcomes \(=10+21+6=37\), so
\[P(\text{same colour})=\frac{37}{120}.\]
(b) Different colours
Every draw is either the two bulbs matching or not, so "different colours" is the complement of "same colour":
\[P(\text{different})=1-\frac{37}{120}=\frac{83}{120}.\]
(c) At least one red
Use the complement "no red bulb". There are \(16-5=11\) non-red bulbs, giving \(\binom{11}{2}=\frac{11\times 10}{2}=55\) all-non-red pairs. Hence
\[P(\text{no red})=\frac{55}{120}=\frac{11}{24},\qquad P(\text{at least one red})=1-\frac{11}{24}=\frac{13}{24}.\]
Tambaya 14 Rahoto
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) Reading the coefficients of \(x\) and \(y\) from each mapping:
\[P=\begin{pmatrix}-3&6\\4&1\end{pmatrix},\qquad Q=\begin{pmatrix}2&-3\\-4&-6\end{pmatrix}\]
(b) Image of \((-2,-3)\) under \(Q\): \((2(-2)-3(-3),\ -4(-2)-6(-3))=(-4+9,\ 8+18)=(5,26)\).
(c) "Q followed by P" means \(P\) acts second, so the single matrix is \(PQ\):
\[PQ=\begin{pmatrix}-3&6\\4&1\end{pmatrix}\begin{pmatrix}2&-3\\-4&-6\end{pmatrix}=\begin{pmatrix}-30&-27\\4&-18\end{pmatrix}\]
i.e. \((x,y)\to(-30x-27y,\ 4x-18y)\).
(d) Image of \((1,4)\): \((-30(1)-27(4),\ 4(1)-18(4))=(-30-108,\ 4-72)=(-138,-68)\).
(e) Anticlockwise rotation of \(225^{\circ}\): \(\begin{pmatrix}\cos225^{\circ}&-\sin225^{\circ}\\\sin225^{\circ}&\cos225^{\circ}\end{pmatrix}=\begin{pmatrix}-\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&-\frac{\sqrt2}{2}\end{pmatrix}\).
Apply to \((-\sqrt2,\ 2\sqrt2)\):
\(x'=-\tfrac{\sqrt2}{2}(-\sqrt2)+\tfrac{\sqrt2}{2}(2\sqrt2)=1+2=3\)
\(y'=-\tfrac{\sqrt2}{2}(-\sqrt2)-\tfrac{\sqrt2}{2}(2\sqrt2)=1-2=-1\)
So \(P'=(3,-1)\).
Bayanin Amsa
(a) Reading the coefficients of \(x\) and \(y\) from each mapping:
\[P=\begin{pmatrix}-3&6\\4&1\end{pmatrix},\qquad Q=\begin{pmatrix}2&-3\\-4&-6\end{pmatrix}\]
(b) Image of \((-2,-3)\) under \(Q\): \((2(-2)-3(-3),\ -4(-2)-6(-3))=(-4+9,\ 8+18)=(5,26)\).
(c) "Q followed by P" means \(P\) acts second, so the single matrix is \(PQ\):
\[PQ=\begin{pmatrix}-3&6\\4&1\end{pmatrix}\begin{pmatrix}2&-3\\-4&-6\end{pmatrix}=\begin{pmatrix}-30&-27\\4&-18\end{pmatrix}\]
i.e. \((x,y)\to(-30x-27y,\ 4x-18y)\).
(d) Image of \((1,4)\): \((-30(1)-27(4),\ 4(1)-18(4))=(-30-108,\ 4-72)=(-138,-68)\).
(e) Anticlockwise rotation of \(225^{\circ}\): \(\begin{pmatrix}\cos225^{\circ}&-\sin225^{\circ}\\\sin225^{\circ}&\cos225^{\circ}\end{pmatrix}=\begin{pmatrix}-\frac{\sqrt2}{2}&\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}&-\frac{\sqrt2}{2}\end{pmatrix}\).
Apply to \((-\sqrt2,\ 2\sqrt2)\):
\(x'=-\tfrac{\sqrt2}{2}(-\sqrt2)+\tfrac{\sqrt2}{2}(2\sqrt2)=1+2=3\)
\(y'=-\tfrac{\sqrt2}{2}(-\sqrt2)-\tfrac{\sqrt2}{2}(2\sqrt2)=1-2=-1\)
So \(P'=(3,-1)\).
Tambaya 15 Rahoto
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.
Bayanin Amsa
(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.
