6b. For a musical show, $x$ children were present. There were $60$ more adults than children. A adult paid D5 and a child paid D2. If a total of D1280 was collected , calculate the :
i. value of $x$
ii. ratio of the number of children to the number of adults.
iii. average amount paid per person
iv. percentage profit if the organizers spent D720 on the show.
7a. A woman looking out from a window at a building from a height of 30m, observed that the angle of depression of a top of a flag pole was $44^0$. If the foot of the pole was 25m from the foot of the building , and on the same horizontal ground, find, correct to the nearest whole number , the:
i. angle of depression of the foot of the pole from the woman;
ii. height of the flag pole.
In the diagram above, $O$ is the center of the circle of, $\angle OQR=32^o$ and $\angle TPR=15^o$. Calculate
i. $\angle QPR$
ii. $\angle TQO$
8. The marks scored by 50 students in a Geography examination are as follows:
60 40 53 37 62 44 39 45 48 39 46 40 61 23 59 65 74 40 68 50 71 26 38 46 51 54 67 73 56 43 69 32 58 67 51 59 52 48 60 47 58 47 59 51 50 51 36 70 40 42
8a. Using class intervals 21-30, 31-40,… prepare a frequency distribution table
8b. Calculate the mean mark of the distribution
8c. what is the percentage of the students scored more 60%
9a. Simplify $\frac{x+2}{x-2}-\frac{x+3}{x-1}$
9b. The graph of the equation $y=Ax^{2}+Bx+C$ passes through the points (0,0),(1,4) and (2,10). Find the
i. value of C
ii. values of A and B
iii. Co-ordinates of the other point where the graph cut the x-axis.
Answers
6a. Evaluate $\log\left( \frac{35 \times 48}{40 \div 56} \right)$
We are given:$\log 5 = 0.6990$
$\log 7 = 0.8451$
$\log 8 = 0.9031$
First simplify the expression:
$\frac{35 \times 48}{40 \div 56} = \frac{35 \times 48}{\frac{40}{56}} = \frac{1680}{\frac{40}{56}} = 1680 \times \frac{56}{40} = \frac{1680 \times 56}{40} = 2352$
So, we are to find:
$\log(2352) \approx \boxed{3.3714}$
6b. Musical Show Problem
Let number of children = $x$Then, number of adults = $x + 60$
Each child pays D2, each adult pays D5.
Total collected = D1280
i. Value of $x$:
$2x + 5(x + 60) = 1280$
$2x + 5x + 300 = 1280$
$7x = 980$
$x = 140$
Answer: 140 children
ii. Ratio of number of children to adults:
Children : Adults = $140 : (140 + 60) = 140 : 200 = 7 : 10$
Answer: 7:10
iii. Average amount paid per person:
Total people = \$140 + 200 = 340$
Average = $\frac{1280}{340} = \boxed{3.76}$
iv. Percentage profit:
Profit = $1280 - 720 = 560$
% Profit = $\frac{560}{720} \times 100 = 77.78%$
Answer: 77.78%
7a.
Given:Height of window from ground = 30m
Distance from base of building to pole = 25m
Angle of depression to top of pole = $44^0$
Let height of flag pole be $h$
We use tan formula:
$\tan(44^\circ) = \frac{30 - h}{25}$
$30 - h = 25 \tan(44^0) = 25 \times 0.9657 = 24.14$
$h = 30 - 24.14 = \boxed{6 \text{ m}}$
ii. Angle of depression to foot of pole:
$\tan(\theta) = \frac{30}{25} = 1.2$
$\theta = \tan^{-1}(1.2) = \boxed{50^0}$
In the diagram above, $O$ is the center of the circle of, $\angle OQR=32^o$ and $\angle TPR=15^o$. Calculate i. $\angle QPR$ ii. $\angle TQO$
7b.
We are given a circle with:$\angle OQR = 32^0$ (at center)
$\angle TPR = 15^0$
i. Find $\angle QPR$:
$\angle QOR = 32^0$, so arc $\widehat{QR}$ = $32^0$
Angle at circumference = $\frac{32}{2} = 16^0$
So, $\angle QPR = 16^0$
ii. Find $\angle TQO$:
In triangle $TPR$, $\angle TPR = 15^0$, and we found $\angle QPR = 16^0$
So, $\angle TPQ = 1^0$ difference between the two
Now, using triangle or further info, $\angle TQO$ cannot be fully determined without more data.
But if $TQ$ is a radius or triangle is right-angled, more info is needed.
8. Geography Scores
8a. Frequency Table
| Class Interval | Frequency |
|---|---|
| 21 - 30 | 3 |
| 31 - 40 | 8 |
| 41 - 50 | 14 |
| 51 - 60 | 13 |
| 61 - 70 | 7 |
| 71 - 80 | 5 |
8b. Mean Mark
Use midpoints:Mean = $\frac{\Sigma fx}{\Sigma f}$
| Class | Midpoint | f | fx |
|---|---|---|---|
| 21-30 | 25.5 | 3 | 76.5 |
| 31-40 | 35.5 | 8 | 284 |
| 41-50 | 45.5 | 14 | 637 |
| 51-60 | 55.5 | 13 | 721.5 |
| 61-70 | 65.5 | 7 | 458.5 |
| 71-80 | 75.5 | 5 | 377.5 |
$\text{Mean} = \frac{2555}{50} = 51.1$
8c. Percentage Scored > 60
Scores above 60: in classes 61-70 and 71-80 → $7 + 5 = 12$Percentage = $\frac{12}{50} \times 100 = 24%$
9a. Simplify Expression
$\frac{x+2}{x-2} - \frac{x+3}{x-1}$Find LCM: $(x - 2)(x - 1)$
Numerator: $(x + 2)(x - 1) - (x + 3)(x - 2)$
$= [x^{2} + x - 2] - [x^{2} - x - 6] = x^{2} + x - 2 - x^{2} + x + 6 = 2x + 4$
So, final answer: $\frac{2x + 4}{(x - 2)(x - 1)}$
9b. Quadratic Equation
Given points: (0,0), (1,4), (2,10)Equation: $y = Ax^2 + Bx + C$
Substitute:
(0,0): $0 = A(0)^2 + B(0) + C \Rightarrow C = 0$
(1,4): $4 = A(1)^2 + B(1) = A + B$
(2,10): $10 = 4A + 2B$
Now solve:
$A + B = 4$ → (i)
$4A + 2B = 10$ → (ii)
Multiply (i) by 2: $2A + 2B = 8$
Subtract: $2A = 2 \Rightarrow A = 1$
Then $B = 3$
Answer:
$A = 1$, $B = 3$, $C = 0$
Equation: $y = x^2 + 3x$
To find x-intercepts: Set $y = 0$
$x^2 + 3x = 0 \Rightarrow x(x + 3) = 0$
So, $x = 0$ or $x = -3$
Other x-intercept: (-3, 0)
