General Mathematics/Mathematics (Core) Essay (Theory)

$ 11\% = \dfrac{11}{100} = 0.11 \\[5ex] 10\% = \dfrac{10}{100} = 0.1 \\[5ex] 140\% = \dfrac{140}{100} = 1.4 \\[5ex] \text{time} = 3\dfrac{1}{2} = \dfrac{7}{2} = 3.5\;years \\[5ex] $ Let the:
account that pay 11% simple interest annually = first account
amount invested in account that pay 11% simple interest annually = x

account that pay 10% simple interest annually = second account
amount invested in account that pay 10% simple interest annually = y

The amount invested at 11% is 140% of the amount invested at 10%.
$x = 1.4 * y$
$x = 1.4y$

$ \underline{\text{Simple Interest Formula}} \\[3ex] P..........r..........t..........I \\[3ex] principal * rate * time = interest \\[3ex] $

Account	Principal	Rate	Time	Interest
First	1.4y	0.11	3.5	0.539y
Second	y	0.1	3.5	0.35y
Total Interest = GH¢ 22580.60 =				0.889y

$ 0.889y = 22580.6 \\[3ex] y = \dfrac{22580.6}{0.889} \\[5ex] y = 25400 \\[3ex] 1.4y = 1.4(25400) = 35560 \\[3ex] $ The amount invested at 11% was GH¢ 35560.00

Farm
Irrigating a 500 m² farm
Each square metre (m²) of the farm requires 50 litres of water.
This means that the farm needs $500 * 50 = 25000$ litres

$ 1\text{ litre} = 1\;dm^3 \\[3ex] 25000\text{ litres} = 25000\;dm^3 \\[3ex] $ Convert to cubic meters so we can compare the same units with that of the hemisphere.

$ 25000\;dm * dm * dm * \dfrac{10^{-1}\;m}{1\;dm} * \dfrac{10^{-1}\;m}{1\;dm} * \dfrac{10^{-1}\;m}{1\;dm} \\[5ex] = 25\;m^3 \\[3ex] $ This means that the farm needs $25000\;dm^3$ of water to be irrigated.

Hemispherical Tank
A hemisphere is half of a sphere

$ \text{radius, } r = 2\;m \\[3ex] \text{Volume of a Sphere} = \dfrac{4}{3}\pi r^3 \\[5ex] \text{Volume of a Hemisphere} = \dfrac{1}{2} * \dfrac{4}{3}\pi r^3 \\[5ex] = \dfrac{2}{3}\pi r^3 \\[5ex] = \dfrac{2}{3} * \dfrac{22}{7} * 2^3 \\[5ex] = 16.76190476\;m^3 \\[3ex] $ Would a full tank of water would be sufficient to irrigate the farm?
Compare Volumes: Hemisphere versus Farm
16.76190476 m³ < 25 m³
A full tank of water would not be sufficient to irrigate the farm.

(a.) The diagram is drawn as shown:


Not Drawn to Scale

(b.)
Let the:
Distance between the two speed boats = |GH| = t
|TH| = f

$ (b.) \\[3ex] \text{Alternate angles are congruent} \\[3ex] \beta = 36^\circ \\[3ex] \theta = 48^\circ \\[5ex] 36^\circ + \psi = 48^\circ \quad\dots\text{diagram} \\[3ex] \psi = 48 - 36 \\[3ex] \psi = 12^\circ \\[5ex] \theta + \phi = 180^\circ \quad\dots\text{angles on a straight line} \\[3ex] 48^\circ + \phi = 180^\circ \\[3ex] \phi = 180 - 48 \\[3ex] \phi = 132^\circ \\[5ex] \triangle TFH \\[3ex] \dfrac{f}{\sin 90^\circ} = \dfrac{100}{\sin\beta} \quad\dots\text{Sine Law} \\[5ex] \dfrac{f}{1} = \dfrac{100}{\sin 36^\circ} \\[5ex] f = \dfrac{100}{\sin 36^\circ}...eqn.(1) \\[5ex] \triangle TGH \\[3ex] \dfrac{t}{\sin\psi} = \dfrac{f}{\sin\phi} \quad\dots\text{Sine Law} \\[5ex] \dfrac{t}{\sin 12^\circ} = \dfrac{f}{\sin 132^\circ} \\[5ex] t = \dfrac{f\sin 12^\circ}{\sin 132^\circ} \\[5ex] \text{From eqn. (1), susbtitute for } f \\[3ex] t = \dfrac{100}{\sin 36^\circ} * \dfrac{\sin 12^\circ}{\sin 132^\circ} \\[5ex] t = 47.59778762 \\[3ex] t \approx 48\;m \quad\dots\text{to the nearest metre} $

$ \text{Given two points with coordinates say } (x_1, y_1) \text{ and } (x_2, y_2) \\[4ex] (a.) \\[3ex] \text{Midpoint} = M(1, 2r - 1) \\[3ex] x = 1 \\[3ex] y = 2r - 1 \\[5ex] \text{Midpoint of a line segment} = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) \\[5ex] \text{For } C(2r, 3) \text{ and } D(-4, s) \\[3ex] x_1 = 2r \hspace{3em} x_2 = -4 \\[3ex] y_1 = 3 \hspace{3.5em} y_2 = s \\[3ex] \implies \\[3ex] \text{x-coordinate of the Midpoint: } \dfrac{2r + (-4)}{2} = 1 \\[5ex] 2r - 4 = 1(2) \\[3ex] 2r = 2 + 4 \\[3ex] r = \dfrac{6}{2} \\[3ex] r = 3 \\[5ex] \text{y-coordinate of the Midpoint: } \dfrac{3 + s}{2} = 2r - 1 \\[5ex] \dfrac{3 + s}{2} = 2(3) - 1 \\[5ex] \dfrac{3 + s}{2} = 6 - 1 \\[5ex] 3 + s = 5(2) \\[3ex] s = 10 - 3 \\[3ex] s = 7 \\[5ex] \implies \\[3ex] C(2r, 3) \hspace{4.8em} D(-4, s) \\[3ex] \\[3ex] C[2(3), 3] \hspace{4em} D(-4, 7) \\[3ex] C(6, 3) \hspace{4.8em} D(-4, 7) \\[3ex] x_1 = 6 \hspace{4.8em} x_2 = -4 \\[3ex] y_1 = 3 \hspace{4.8em} y_2 = 7 \\[5ex] (b.) \\[3ex] \text{distance between two points} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[4ex] = \sqrt{(-4 - 6)^2 + (7 - 3)^2} \\[4ex] = \sqrt{(-10)^2 + 4^2} \\[4ex] = \sqrt{100 + 16} \\[3ex] = \sqrt{116} \\[3ex] = \sqrt{4 * 29} \\[3ex] = \sqrt{4} * \sqrt{29} \\[3ex] = 2\sqrt{29}\;units $

Cost of Building a 3-Bedroom Apartment = $64000
Made up of:
Labour Cost, Materials Cost, and Cntractor's Charges
in the ratio of:
12 : 15 : 5

$ \text{Sum of ratios} = 12 + 15 + 5 = 32 \\[3ex] \underline{\text{Before Inflation}} \\[3ex] \text{Labour Cost} = \dfrac{12}{32} * 64000 = \$24000 \\[5ex] \text{Materials Cost} = \dfrac{15}{32} * 64000 = \$30000 \\[5ex] \text{Contractor's Charges} = \dfrac{5}{32} * 64000 = \$10000 \\[5ex] \underline{\text{Due to Inflation}} \\[3ex] \underline{\text{Labour Cost}} \\[3ex] \text{increase by } p\% \\[3ex] = \dfrac{p}{100} * 24000 \\[5ex] = \$240p \\[3ex] \text{New Labour Cost} = \$24000 + \$240p \\[5ex] \underline{\text{Materials Cost}} \\[3ex] \text{increase by } 2p\% \\[3ex] = \dfrac{2p}{100} * 30000 \\[5ex] = \$600p \\[3ex] \text{New Materials Cost} = \$30000 + \$600p \\[5ex] \text{New Contractor's Charges} = \$10000 \quad\dots\text{remained the same} \\[3ex] $ After the increase, the cost of labour was two-thirds the cost of materials ⇒

$ (a.) \\[3ex] (24000 + 240p) = \dfrac{2}{3} * (30000 + 600p) \\[5ex] 24000 + 240p = 20000 + 400p \\[3ex] 24000 - 20000 = 400p - 240p \\[3ex] 4000 = 160p \\[3ex] 160p = 4000 \\[3ex] p = \dfrac{4000}{160} \\[5ex] p = 25 \\[5ex] (b.) \\[3ex] \text{New Labour Cost} \\[3ex] = 24000 + 240(25) \\[3ex] = \$30000 \\[5ex] \text{New Materials Cost} \\[3ex] = 30000 + 600(25) \\[3ex] = \$45000 \\[5ex] \text{New Cost of Building the Apartment} \\[3ex] =\text{New Labour Cost} + \text{New Materials Cost} + \text{Contractor's Charges} \\[3ex] = 30000 + 45000 + 10000 \\[3ex] = \$85000 $

$ (a.) \\[3ex] \underline{\triangle QRS} \\[3ex] \text{Area} = 98\;m^2 \\[3ex] \perp\text{height} = 7\;m \\[3ex] \text{base} = |TR| + |TQ| \\[3ex] x + (2x + 3) \\[3ex] = x + 2x + 3 \\[3ex] = 3x + 3 \\[3ex] = 3(x + 1)\;m \\[3ex] \text{Area} = \dfrac{1}{2} * \text{base} * \perp\text{height} \\[5ex] 98 = \dfrac{1}{2} * 3(x + 1) * 7 \\[5ex] 98 * 2 * \dfrac{1}{3} * \dfrac{1}{7} = x + 1 \\[5ex] x + 1 = 9.\bar{3} \\[3ex] x = 9.\bar{3} - 1 \\[3ex] x = 8.\bar{3} \\[3ex] x \approx 8.3 \;m\quad\dots\text{to one decimal place} \\[5ex] (b.) \\[3ex] \underline{\triangle QTS} \\[3ex] \text{base} = 2x + 3 \\[3ex] = 2(8.\bar{3}) + 3 \\[3ex] = 19.\bar{6}\;m \\[3ex] \perp\text{height} = 7\;m \\[3ex] \text{Area} = \dfrac{1}{2} * 19.\bar{6} * 7 \\[5ex] = 68.8\bar{3}\;m^2 \\[5ex] \underline{\text{Pattern: } \triangle TRS} \\[3ex] \text{Area of the pattern} = \text{Area of } \triangle QRS - \text{Area of } \triangle QTS \\[3ex] = 98 - 68.8\bar{3} \\[3ex] = 29.1\bar{6} \\[3ex] \approx 29.2\;m^2 \\[5ex] (c.) \\[3ex] \underline{\triangle QTS} \\[3ex] \text{hyp} = |ST| \\[3ex] \text{leg} = \text{base} = 19.\bar{6}\;m \\[3ex] \text{leg} = \perp\text{height} = 7\;m \\[3ex] \text{hyp}^2 = \text{leg}^2 + \text{leg}^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] |ST|^2 = 19.\bar{6}^2 + 7^2 \\[3ex] |ST| = \sqrt{435.7777779} \\[3ex] |ST| = 20.87529109\;m \\[5ex] \underline{\triangle QRS} \\[3ex] \text{hyp} = |SR| \\[3ex] \text{leg} = \text{base} = 3(x + 1) \\[3ex] = 3(8.\bar{3} + 1) \\[3ex] = 28\;m \\[3ex] \text{leg} = \perp\text{height} = 7\;m \\[3ex] |SR|^2 = 28^2 + 7^2 \quad\dots\text{Pythagorean Theorem} \\[3ex] |SR| = \sqrt{833} \\[3ex] |SR| = 28.86173938\;m \\[5ex] \underline{\text{Pattern: } \triangle TRS} \\[3ex] \text{Perimeter} = |TR| + |SR| + |ST| \\[3ex] = 8.\bar{3} + 28.86173938 + 20.87529109 \\[3ex] = 58.0703638 \\[3ex] \approx 58.1\;m $

Let the number of days that Mr. Mensa planned to plough the land initially = p

Initially Planned (Ideality)
120 acres of farmland each day for p days = $120 * p = 120p$ acres

Reality
85 acres of farmland each day
Increase by 2 days = $p + 2$ days
85 acres of farmland each day for (p + 2) days = $85(p + 2)$ acres

Remaining
40 acres of farmland

The sum of the remaining acres to be ploughed and the number of acres already ploughed gives the total number of acres initially scheduled to be ploughed.
This implies that:

$ 40 + 85(p + 2) = 120p \\[3ex] 40 + 85p + 170 = 120p \\[3ex] 210 = 120p - 85p \\[3ex] 35p = 210 \\[3ex] p = \dfrac{210}{35} \\[5ex] p = 6 \\[5ex] 120p \\[3ex] = 120(6) \\[3ex] = 720\;acres \\[3ex] $ (a.)(i.) Mr. Mensa initially planned to plough the land for 14 days.
(ii.) The total area of the farmland is 720 acres.

(b.) Based on the information:
Poki and Sofi are on: (a.) the same longitude
(b.) different latitudes
(c.) the same side (Northern Hemisphere) of the equator.

$ \text{Angular distance between them, } \theta \\[3ex] 6.8^\circ - 5^\circ = 1.8^\circ \\[3ex] \text{Distance between them, } L \\[3ex] L = \dfrac{\pi R \theta}{180} \\[5ex] = \dfrac{22}{7} * 6400 * 1.8 * \dfrac{1}{180} \\[5ex] = 201.1428571 \\[3ex] \approx 201\;km \quad\dots\text{to the nearest whole number} $

(a.) The diagram that represents the information is drawn as shown:

Not Drawn to Scale

$ \underline{\text{Diagram}} \\[3ex] |OD| = |OC| = k \quad\dots\text{midpoint, }O \\[3ex] |PD| = h \\[3ex] |PO| = d \\[3ex] |RO| = c \\[3ex] |RC| = 10m \\[3ex] |RP| = e \\[3ex] (b.) \\[3ex] \underline{\triangle ROC} \\[3ex] \tan 28^\circ = \dfrac{10}{k} \quad\dots\text{SOHCAHTOA} \\[5ex] k\tan 28 = 10 \\[3ex] k = \dfrac{10}{\tan 28} \\[5ex] \sin 28^\circ = \dfrac{10}{c} \quad\dots\text{SOHCAHTOA} \\[5ex] c\sin 28 = 10 \\[3ex] c = \dfrac{10}{\sin 28} \\[5ex] c = 21.30054468\;m \\[5ex] \underline{\triangle POD} \\[3ex] (i.) \\[3ex] \tan 70^\circ = \dfrac{h}{k} \quad\dots\text{SOHCAHTOA} \\[5ex] h = k\tan 70 \\[3ex] h = \dfrac{10}{\tan 28} * \tan 70 \\[5ex] h = 51.67253496 \\[3ex] h \approx 52\;m \\[5ex] \cos 70^\circ = \dfrac{k}{d} \quad\dots\text{SOHCAHTOA} \\[5ex] d\cos 70 = k \\[3ex] d = \dfrac{k}{\cos 70} \\[5ex] d = \dfrac{10}{\tan 28} * \dfrac{1}{\cos 70} \\[5ex] d = 54.98876315\;m \\[5ex] \underline{\triangle POR} \\[3ex] (ii.) \\[3ex] \angle POD + \angle POR + \angle ROC = 180^\circ \quad\dots\text{sum of angles on a straight line} \\[3ex] 70^\circ + \angle POR + 28^\circ = 180 \\[3ex] \angle POR = 180 - 70 - 28 \\[3ex] \angle POR = 82^\circ \\[5ex] e^2 = c^2 + d^2 - 2cd \cos\angle POR \quad\dots\text{Cosine Law} \\[3ex] e^2 = 21.30054468^2 + 54.98876315^2 - 2(21.30054468)(54.98876315) \cos 82^\circ \\[3ex] e^2 = 3151.452985 \\[3ex] e = \sqrt{3151.452985} \\[3ex] e = 56.13780353 \\[3ex] e \approx 56\;m $

(a.)(i.) The diagram that Illustrates the information is drawn as shown:

Not Drawn to Scale

(ii.) The diagram that Illustrates the updated information is drawn as shown:

Not Drawn to Scale

$ \text{Reflex } \angle AOB + \text{Acute } \angle AOB = 360^\circ \quad\dots\text{sum of angles at a point} \\[3ex] \text{Reflex } \angle AOB + 88 = 360 \\[3ex] \text{Reflex } \angle AOB = 360 - 88 \\[3ex] \text{Reflex } \angle AOB = 272^\circ \\[5ex] \text{Reflex } \angle AOB = 2 * \angle ACB \quad\dots\text{angle at centre is twice angle at circumference} \\[3ex] 2 * \angle ACB = 272 \\[3ex] \angle ACB = \dfrac{272}{2} \\[5ex] \angle ACB = 136^\circ \\[3ex] $ (b.)
A carpenter, Kwaw, is able to make 400 kitchen stools each day in 8 hours.
How many stools can Kwaw make in an hour?
This implies that the:
$ \text{Kwaw's rate of work} = \dfrac{400\;stools}{8\;hours} = 50\;stools/hour \\[5ex] $ Another carpenter, Danso, works at $\dfrac{3}{2}$ times Kwaw's rate when making same kind of kitchen stools.

$ \text{Danso's rate of work} = \dfrac{3}{2} * \dfrac{50\;stools}{1\;hour} = 75\;stools/hour \\[5ex] \text{In 8 hours, Danso can make:} \\[3ex] 8\;hours * \dfrac{75\;stools}{1\;hour} = 600\;stools\quad\dots\text{Unity Fraction Method} \\[5ex] $ (ii.)
Kwaw can make 50 stools in 1 hour.
Danso can make 75 stools in 1 hour.
If both Kwaw and Danso decide to work together,
50 + 75 = 125
they can make 125 stools in 1 hour.
Determine the time required to make 2000 kitchen stools.

Proportional Reasoning Method

Number of kitchen stools	Time (hours)
125	1
2000	what

$ 125 * what = 1 * 2000 \\[3ex] what = \dfrac{2000}{125} \\[5ex] what = 16 \\[3ex] $ If they work together, they can make 2000 kitchen stools in 16 hours.

(a.) The diagram that Illustrates the information is drawn as shown:

Not Drawn to Scale

$ (b.) \\[3ex] a : b = 2 : 3 \quad\dots\text{Given} \\[3ex] \dfrac{a}{b} = \dfrac{2}{3} \\[5ex] a = \dfrac{2b}{3} ...eqn.(1) \\[5ex] (i.) \\[3ex] \text{Perimeter} = 2(2a + b) + (3a + 5) \quad\dots\text{Formula} \\[3ex] = 4a + 2b + 3a + 5 \\[3ex] = 7a + 2b + 5 \\[5ex] \text{Also}: \\[3ex] \text{Perimeter} = 7b + 3 \quad\dots\text{Given} \\[3ex] \implies \\[3ex] 7b + 3 = 7a + 2b + 5 \\[3ex] 7b - 2b - 7a = 5 - 3 \\[3ex] 5b - 7a = 2 \\[3ex] \text{From eqn.(1), substitute for } a \\[3ex] 5b - 7\left(\dfrac{2b}{3}\right) = 2 \\[5ex] 5b - \dfrac{14b}{3} = 2 \\[5ex] LCD = 3 \\[3ex] 3(5b) - 3\left(\dfrac{14b}{3}\right) = 3(2) \\[5ex] 15b - 14b = 6 \\[3ex] b = 6\;m \\[5ex] \text{Substitute for } b \text{ in eqn.(1)} \\[3ex] a = \dfrac{2(6)}{3} \\[5ex] a = 4\;m \\[5ex] (ii.) \\[3ex] \underline{\text{Sides of the Isosceles Triangle}} \\[3ex] 2a + b \\[3ex] = 2(4) + 6 \\[3ex] = 14\;m \\[3ex] 3a + 5 \\[3ex] = 3(4) + 5 \\[3ex] = 17\;m \\[3ex] $ The updated diagram is:

Not Drawn to Scale

$ \underline{\text{Sides}} \\[3ex] |DS| = |RS| = r = 14\;m \\[3ex] |DR| = e = 17\;m \\[5ex] \underline{\text{SemiPerimeter}} \\[3ex] s = \dfrac{2(14) + 17}{2} \\[5ex] s = 22.5\;m \\[5ex] \underline{\text{Difference between the semiperimeter and the sides}} \\[3ex] s - r = 22.5 - 14 = 8.5\;m\;\;(twice) \\[3ex] s - e = 22.5 - 17 = 5.5\;m \\[5ex] \underline{\text{Product of the semiperimeter and the differences}} \\[3ex] s(s - r)(s - r)(s - e) \\[3ex] = 22.5(8.5)^2(5.5) \\[3ex] = 8940.9375\;m^4 \\[3ex] \text{Area} = \sqrt{s(s - r)(s - r)(s - e)} \quad\dots\text{Heron's Formula} \\[3ex] = \sqrt{8940.9375} \\[3ex] = 94.55653071 \\[3ex] \approx 94.6\;m^2 \quad\dots\text{to one decimal place} $

(a.) The cumulative frequency table is shown:

Cumulative Frequency Table

Class Interval	Tally	Frequency	Upper Class Boundary	Cumulative Frequency
1 — 10	III	3	$ \dfrac{10 + 11}{2} = 10.5 $	3
11 — 20	IIII	5	$ \dfrac{20 + 21}{2} = 20.5 $	3 + 5 = 8
21 — 30	IIII II	7	$ \dfrac{30 + 31}{2} = 30.5 $	8 + 7 = 15
31 — 40	IIII I	6	$ \dfrac{40 + 41}{2} = 40.5 $	15 + 6 = 21
41 — 50	IIII IIII	10	$ \dfrac{50 + 51}{2} = 50.5 $	21 + 10 = 31
51 — 60	IIII II	7	$ \dfrac{60 + 61}{2} = 60.5 $	31 + 7 = 38
61 — 70	IIII	5	$ \dfrac{70 + 71}{2} = 70.5 $	38 + 5 = 43
71 — 80	IIII	4	$ \dfrac{80 + 81}{2} = 80.5 $	43 + 4 = 47
81 — 90	II	2	$ \dfrac{80 + 81}{2} = 80.5 $	47 + 2 = 49
91 — 100	I	1	$ \dfrac{90 + 91}{2} = 90.5 $	49 + 1 = 50
		$\Sigma Frequency = 50$

(b.) The coordinates to plot on the graph are:
(Upper Class Boundary, Cumulative Frequency)
(10.5, 3)
(20.5, 8)
(30.5, 15)
(40.5, 21)
(50.5, 31)
(60.5, 38)
(70.5, 43)
(80.5, 47)
(90.5, 49)
(100.5, 50)
The cumulative frequency curve for the distribution is drawn as shown:


$ \text{From the Ogive}: \\[3ex] (c.) \\[3ex] N = \Sigma Frequency = 50 \\[3ex] \text{Upper Quartile position} = \dfrac{3N}{4}th \\[5ex] = \dfrac{3 * 50}{4}th \\[5ex] = 37.5th \\[5ex] \text{Upper Quartile} = 37.5th\;\;value \\[3ex] = 59.5 \quad\dots\text{red line on the Ogive} \\[3ex] (d.) \\[3ex] \text{Passed} = 10\%\;\;of\;\;N \\[3ex] = \dfrac{10}{100} * 50 \\[5ex] = 5\text{ students} \\[3ex] \text{Failed} = 50 - 5 \\[3ex] = 45\text{ students} \\[3ex] \text{5 students passed implies that 45 students failed} \\[3ex] \text{Pass Mark} = 45th\;\;value \\[3ex] = 79.5 \quad\dots\text{black line on the Ogive} \\[3ex] $

$ \angle GUF = \angle HUE = 110^\circ \quad\dots\text{vertical angles} \\[3ex] \angle GFU = \angle GFH = 30^\circ \quad\dots\text{diagram} \\[3ex] \underline{\triangle GUF} \\[3ex] \angle UGF + \angle GUF + \angle GFU = 180^\circ \quad\dots\text{sum of angles in a triangle} \\[3ex] \angle UGF + 110^\circ + 30^\circ = 180^\circ \\[3ex] \angle UGF = 180 - 110 - 30 \\[3ex] \angle UGF = 40^\circ \\[3ex] \angle EGF = \angle UGF = 40^\circ \quad\dots\text{diagram} \\[5ex] (i.) \\[3ex] \angle EOF = 2 * \angle EGF \quad\dots\text{angle at centre = twice the angle at circumference} \\[3ex] = 2 * 40 \\[3ex] = 80^\circ \\[5ex] (ii.) \\[3ex] \angle UEH = 30^\circ \quad\dots\text{angles in the same segment} \\[3ex] $ Let the set of:
Arts students = A
Students who are good at French = F
Non-Arts students = N

The Venn Diagram illustration of the information is drawn as shown:

The Venn Diagram illustrating the conclusions is drawn as shown:

As seen from the diagram:
(α.) Dennis is a non-Arts student who is good at French.
Because of this possibility, the conclusion is not valid.

(β.) Rose is an Arts student who is not good at French.
Because of this possibility, the conclusion is not valid.