We are given a set of math problems to solve. 4(a): A motorist leaves Windhoek at 05:55 and the trip to Tsumeb takes 4 hours 10 minutes, correct to the nearest 10 minutes. We need to find the latest possible time of arrival in Tsumeb. 4(b): The average speed is 100 km/h, correct to the nearest 5 km/h. We need to 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. We need to find how many light years it is to the system of Krul, 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. 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

We are given a set of math problems to solve.

4(a): A motorist leaves Windhoek at 05:55 and the trip to Tsumeb takes 4 hours 10 minutes, correct to the nearest 10 minutes. We need to find the latest possible time of arrival in Tsumeb.

4(b): The average speed is 100 km/h, correct to the nearest 5 km/h. We need to 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. We need to find how many light years it is to the system of Krul, 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.

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

4(a):

The trip takes 4 hours 10 minutes, correct to the nearest 10 minutes. This means the trip could take up to 4 hours and 15 minutes.

Departure time is 05:

5

5. Latest possible arrival time is 05:55 + 4 hours 15 minutes = 10:

1

0.

4(b):

The speed is 100 km/h, correct to the nearest 5 km/h, so the maximum speed is 105 km/h.

The time is 4 hours 10 minutes, correct to the nearest 10 minutes. This means the maximum time is 4 hours and 15 minutes, which is 4.25 hours.

Distance = Speed × Time

Maximum distance = 105 km/h × 4.25 h = 446.25 km.

5:

A light year is the distance travelled by light in 365 days. The speed of light is $3 \times 10^8$ km/h.

Distance of 1 light year = speed × time

Time = 365 days = 365 × 24 hours = 8760 hours.

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

The distance to Krul is $7 \times 10^{23}$ km.

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

Correct to 2 significant figures: $2.7 \times 10^{11}$ light years.

6:

$3(2 \times 10^6 - 4 \times 10^5) = 3(2000000 - 400000) = 3(1600000) = 4800000 = 4.8 \times 10^6$.

7:

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

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

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

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

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

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

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

3. Final Answer

4(a): 10:10

4(b): 446.25 km

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

6: $4.8 \times 10^6$

7: $x = 0$, $y = -2$