The problem consists of three independent questions: 4) A motorist leaves Windhoek at 05:55 on his way to Tsumeb. The trip takes him 4 hours 10 minutes, correct to the nearest 10 minutes. He drives at an average speed of 100 km/h, correct to the nearest 5 km/h. (a) Find his latest possible time of arrival in Tsumeb. (b) Find the maximum possible distance between Windhoek and Tsumeb. 5) A light year is the distance traveled 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. How many light years is it to the system of Krul? Give your answer in standard form correct to 2 significant figures. 6) Work out $3(2 \times 10^6 - 4 \times 10^5)$, giving your answer in standard form. 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 independent questions:

4) A motorist leaves Windhoek at 05:55 on his way to Tsumeb. The trip takes him 4 hours 10 minutes, correct to the nearest 10 minutes. He drives at an average speed of 100 km/h, correct to the nearest 5 km/h.

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

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

5) A light year is the distance traveled 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. How many light years is it to the system of Krul? Give your answer in standard form correct to 2 significant figures.

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

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

Question 4:

(a) The latest possible departure time is 05:55 + 00:05 = 06:

0

0. The longest possible journey time is 4 hours 20 minutes.

Therefore, the latest possible arrival time is 06:00 + 4 hours 20 minutes = 10:

2

0.

(b) The maximum possible speed is 105 km/h.

The maximum possible journey time is 4 hours 20 minutes = $4 + \frac{20}{60} = 4 + \frac{1}{3} = \frac{13}{3}$ hours.

Therefore, the maximum possible distance is $105 \times \frac{13}{3} = 35 \times 13 = 455$ km.

Question 5:

Distance traveled by light in 1 year = $3 \times 10^8 \text{ km/h} \times 365 \text{ days} \times 24 \text{ h/day} = 3 \times 10^8 \times 365 \times 24 = 2.628 \times 10^{12}$ km.

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

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

Question 6:

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

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

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

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

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

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

$4 \times 10^x + 6 \times 10^y = 4 \times 1 + 6 \times 0.01$

$4 \times 10^x + 6 \times 10^y = 4 \times 10^0 + 6 \times 10^{-2}$

So $x = 0$ and $y = -2$.

3. Final Answer

4) (a) 10:20

(b) 455 km

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

6) $4.8 \times 10^6$

$x = 0$ and $y = -2$