1. an inter - city shuttle travels between two cities, amanfro and kotoku, covering the first half of the at an average speed of 60 km/h and the second half at 90 km/h. the total distance between the two cities is 300 km.
(a) calculate, for the entire journey, the average speed of the shuttle.
(b) if the speed of the shuttle in the first half of the journey is reduced to 50 km/h, determine on the overall time.
2. a stock broker, kojo, invested some amount in two accounts which pay 11% and 10% simple i annually. the amount invested at 11% is 140% of the amount invested at 10%. after 3 1/2 years he earned a total of gh₵ 22,580.60 as interest. how much was invested at 11%?
3. a contractor builds a rainwater collection system in a shape of a hemispherical tank of radius 2 to irrigate a 500 m² farm. each square metre of the farm requires 50 litres of water. determine, a mathematical statement, if a full tank of water would be sufficient to irrigate the farm? take π = 22/7
4. from the top of a lighthouse 100 m high, the angles of depression of two fishing v at 48° and 36°. the foot of the lighthouse and the sea level are on the same horizon and the two vessels are on the same side of the lighthouse.
(a) illustrate the information in a diagram.
(b) find, correct to the nearest metre, the distance between the two vessels.
5. the mid - point of the line joining y(2a, 3) and z( - 4, b) is m(1, 2a - 1). find:
(a) the values of a and b;
(b) |yz|.

Question

Accepted Answer

### 1. (a)
# Explanation:
## Step1: Calculate time for each half - journey
The total distance $d = 300$ km, so the distance of each half - journey $d_1=d_2 = 150$ km.
The time for the first half - journey $t_1=\frac{d_1}{v_1}$, where $v_1 = 60$ km/h. So $t_1=\frac{150}{60}=2.5$ h.
The time for the second half - journey $t_2=\frac{d_2}{v_2}$, where $v_2 = 90$ km/h. So $t_2=\frac{150}{90}=\frac{5}{3}$ h.
## Step2: Calculate total time and average speed
The total time $t=t_1 + t_2=2.5+\frac{5}{3}=\frac{5}{2}+\frac{5}{3}=\frac{15 + 10}{6}=\frac{25}{6}$ h.
The average speed $v=\frac{d}{t}$, with $d = 300$ km and $t=\frac{25}{6}$ h. So $v=\frac{300}{\frac{25}{6}}=300	imes\frac{6}{25}=72$ km/h.
# Answer:
$72$ km/h

### 1. (b)
# Explanation:
## Step1: Calculate new time for first half - journey
The new speed for the first half - journey $v_{1new}=50$ km/h, and $d_1 = 150$ km.
The new time for the first half - journey $t_{1new}=\frac{d_1}{v_{1new}}=\frac{150}{50}=3$ h.
The time for the second half - journey remains the same $t_2=\frac{150}{90}=\frac{5}{3}$ h.
The new total time $t_{new}=t_{1new}+t_2=3+\frac{5}{3}=\frac{9 + 5}{3}=\frac{14}{3}$ h.
The original total time $t=\frac{25}{6}$ h.
The change in time $\Delta t=t_{new}-t=\frac{14}{3}-\frac{25}{6}=\frac{28 - 25}{6}=\frac{1}{2}$ h.
# Answer:
The overall time increases by $0.5$ h.

### 2.
# Explanation:
## Step1: Let the amount invested at 10% be $x$.
The amount invested at 11% is $1.4x$.
The simple - interest formula is $I = Prt$, where $P$ is the principal amount, $r$ is the interest rate, and $t$ is the time.
For the 10% account, $I_1=x	imes0.1	imes3.5 = 0.35x$.
For the 11% account, $I_2=1.4x	imes0.11	imes3.5=1.4x	imes0.385 = 0.539x$.
The total interest $I = I_1+I_2$. Given $I=22580.60$.
So $0.35x + 0.539x=22580.60$.
Combining like terms, $0.889x=22580.60$.
Solving for $x$, $x=\frac{22580.60}{0.889}=25400$.
The amount invested at 11% is $1.4x$, so $1.4	imes25400 = 35560$.
# Answer:
$35560$

### 3.
# Explanation:
## Step1: Calculate volume of hemispherical tank
The volume of a hemispherical tank $V=\frac{2}{3}\pi r^{3}$, with $r = 2$ m and $\pi=\frac{22}{7}$.
$V=\frac{2}{3}	imes\frac{22}{7}	imes2^{3}=\frac{2}{3}	imes\frac{22}{7}	imes8=\frac{352}{21}	ext{ m}^3$.
Since $1	ext{ m}^3 = 1000$ L, $V=\frac{352}{21}	imes1000=\frac{352000}{21}	ext{ L}\approx16761.9$ L.
The farm requires $500	imes50 = 25000$ L of water.
The mathematical statement is: If $\frac{2}{3}	imes\frac{22}{7}	imes2^{3}	imes1000\geq500	imes50$, then a full tank is sufficient to irrigate the farm. Since $\frac{352000}{21}\approx16761.9<25000$, a full tank is not sufficient.
# Answer:
A full tank is not sufficient to irrigate the farm.

### 4. (a)
# Explanation:
Draw a vertical line to represent the lighthouse of height $h = 100$ m. Mark two points on the same side of the base of the lighthouse to represent the two fishing vessels. Draw lines from the top of the lighthouse to the two points. The angles of depression from the top of the lighthouse to the two vessels are $48^{\circ}$ and $36^{\circ}$ respectively.
# Answer:
Diagram should be drawn as described above.

### 4. (b)
# Explanation:
## Step1: Find distances from lighthouse to each vessel
Let the distance from the lighthouse to the nearer vessel be $x$ and to the farther vessel be $y$.
We know that $	an	heta=\frac{h}{d}$, where $h$ is the height of the lighthouse and $d$ is the horizontal distance from the lighthouse to the vessel.
For the angle of depression of $48^{\circ}$, $	an48^{\circ}=\frac{100}{x}$, so $x = \frac{100}{	an48^{\circ}}\approx100\div1.1106\approx90.04$ m.
For the angle of depression of $36^{\circ}$, $	an36^{\circ}=\frac{100}{y}$, so $y=\frac{100}{	an36^{\circ}}\approx100\div0.7265\approx137.64$ m.
The distance between the two vessels $d=y - x=137.64 - 90.04 = 47.6\approx48$ m.
# Answer:
$48$ m

### 5. (a)
# Explanation:
## Step1: Use mid - point formula
The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $M(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.
For points $Y(2a,3)$ and $Z(-4,b)$ with mid - point $M(1,2a - 1)$:
For the $x$ - coordinate: $\frac{2a-4}{2}=1$.
Cross - multiply: $2a-4 = 2$.
Add 4 to both sides: $2a=6$, so $a = 3$.
For the $y$ - coordinate: $\frac{3 + b}{2}=2a-1$. Substitute $a = 3$ into it, $\frac{3 + b}{2}=2	imes3-1=5$.
Cross - multiply: $3 + b = 10$, so $b = 7$.
# Answer:
$a = 3$, $b = 7$

### 5. (b)
# Explanation:
## Step1: Use distance formula
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}$.
Here, $x_1 = 2a=6$, $y_1 = 3$, $x_2=-4$, $y_2 = 7$.
$|YZ|=\sqrt{(-4 - 6)^{2}+(7 - 3)^{2}}=\sqrt{(-10)^{2}+4^{2}}=\sqrt{100 + 16}=\sqrt{116}=2\sqrt{29}\approx10.77\approx11$
# Answer:
$2\sqrt{29}\approx11$

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QUESTION IMAGE

Question

Explanation:

1. (a)

Step1: Calculate time for each half - journey

Step2: Calculate total time and average speed

Step1: Calculate new time for first half - journey

Step1: Let the amount invested at 10% be $x$.

Answer:

1. (b)