The problem consists of three parts: (4) A motorist leaves Windhoek at 05:55 and travels to Tsumeb. The trip takes 4 hours 10 minutes, correct to the nearest 10 minutes. The average speed is 100 km/h, correct to the nearest 5 km/h. (a) Find the latest possible time of arrival in Tsumeb. (b) Find the maximum possible distance between Windhoek and Tsumeb. (5) A light year is the distance travelled by light in 365 days. The speed of light is $3 \times 10^8$ km/h. The distance to the star system Krul is $7 \times 10^{23}$ km. Find how many light years it is to the system of Krul, giving the answer in standard form correct to 2 significant figures. (6) Work out $3(2 \times 10^6 - 4 \times 10^5)$, giving the answer in standard form. (7) Given that $2 \times 10^3 + 3 \times 10^2 + 4 \times 10^x + 6 \times 10^y = 2304.06$, where $x$ and $y$ are integers, find the value of $x$ and $y$.

1. Problem Description

The problem consists of three parts:

(4) A motorist leaves Windhoek at 05:55 and travels to Tsumeb. The trip takes 4 hours 10 minutes, correct to the nearest 10 minutes. The average speed is 100 km/h, correct to the nearest 5 km/h.

(a) Find the latest possible time of arrival in Tsumeb.

(b) Find the maximum possible distance between Windhoek and Tsumeb.

(5) A light year is the distance travelled by light in 365 days. The speed of light is $3 \times 10^8$ km/h. The distance to the star system Krul is $7 \times 10^{23}$ km. Find how many light years it is to the system of Krul, giving the answer in standard form correct to 2 significant figures.

(6) Work out $3(2 \times 10^6 - 4 \times 10^5)$, giving the answer in standard form.

(7) Given that $2 \times 10^3 + 3 \times 10^2 + 4 \times 10^x + 6 \times 10^y = 2304.06$, where $x$ and $y$ are integers, find the value of $x$ and $y$.

2. Solution Steps

(4a) The trip takes 4 hours 10 minutes, correct to the nearest 10 minutes. Therefore, the longest possible time is 4 hours 15 minutes. The departure time is 05:

5

5. Arrival time = 05:55 + 4 hours 15 minutes = 10:

1

0.

(4b) The speed is 100 km/h, correct to the nearest 5 km/h. Therefore the maximum possible speed is 105 km/h. The time is 4 hours 10 minutes, correct to the nearest 10 minutes. So the maximum time is 4 hours 15 minutes, which is $4.25$ hours.

Distance = Speed $\times$ Time

Distance = $105 \times 4.25 = 446.25$ km.

(5) Speed of light = $3 \times 10^8$ km/h.

1 year = 365 days = $365 \times 24$ hours = 8760 hours.

1 light year = Speed of light $\times$ Time

1 light year = $3 \times 10^8 \times 8760 = 26280 \times 10^8 = 2.628 \times 10^{12}$ km.

Distance to Krul = $7 \times 10^{23}$ km.

Number of light years = $\frac{7 \times 10^{23}}{2.628 \times 10^{12}} = \frac{7}{2.628} \times 10^{11} = 2.663 \times 10^{11}$ light years.

To 2 significant figures, this is $2.7 \times 10^{11}$.

(6) $3(2 \times 10^6 - 4 \times 10^5) = 3(20 \times 10^5 - 4 \times 10^5) = 3(16 \times 10^5) = 48 \times 10^5 = 4.8 \times 10^6$.

(7) $2 \times 10^3 + 3 \times 10^2 + 4 \times 10^x + 6 \times 10^y = 2304.06$

$2000 + 300 + 4 \times 10^x + 6 \times 10^y = 2304.06$

$2300 + 4 \times 10^x + 6 \times 10^y = 2304.06$

$4 \times 10^x + 6 \times 10^y = 4.06$

$4 \times 10^x + 6 \times 10^y = 4 + 0.06$

$4 \times 10^x + 6 \times 10^y = 4 + 6 \times 10^{-2}$

$4 \times 10^x = 4$ implies $10^x = 1$, so $x = 0$.

$6 \times 10^y = 6 \times 10^{-2}$, so $y = -2$.

3. Final Answer

(4a) 10:10

(4b) 446.25 km

(5) $2.7 \times 10^{11}$ light years

(6) $4.8 \times 10^6$

(7) $x=0$, $y=-2$