Evaluate 11001two + 1111two

• $1 + 1 = 10_{\text{two}}$ (write 0, carry 1)
• $0 + 1 + 1 (\text{carry}) = 10_{\text{two}}$ (write 0, carry 1)
• $0 + 1 + 1 (\text{carry}) = 10_{\text{two}}$ (write 0, carry 1)
• $1 + 1 + 1 (\text{carry}) = 11_{\text{two}}$ (write 1, carry 1)
• $1 + 0 + 1 (\text{carry}) = 10_{\text{two}}$ (write 0, carry 1)
• The final carry is 1.

$$\begin{array}{c} \quad 11001_{\text{two}} \\ + \quad 01111_{\text{two}} \\ \hline 101000_{\text{two}} \end{array}$$

The correct option is B.
The sum is $\boxed{101000_{\text{two}}}$.

16. An aircraft flies from Q ($25^\circ$ N, $38^\circ$ E) to R ($72^\circ$ S, $38^\circ$ E). Calculate the distance QR along the line of longitude, correct to the nearest kilometre. (Take R = 6,400 km, $\pi = \frac{22}{7}$)

Since the points are in different hemispheres (North and South), add their latitudes.

$$\theta = 25^\circ + 72^\circ = 97^\circ$$

$$D = 6400 \text{ km} \times 97 \times \frac{22}{7 \times 180}$$ $$D = 6400 \times \frac{97 \times 22}{1260}$$ $$D = \frac{6400 \times 2134}{1260}$$ $$D = \frac{13657600}{1260}$$ $$D \approx 10839.365 \text{ km}$$

$$D \approx 10839 \text{ km}$$

The correct option is A.
The distance QR is $\boxed{10839 \text{ km}}$.

17. Find the gradient of the curve $y = 2x^2 - 3x + 4$ at the point $x=2$.

$$y = 2x^2 - 3x + 4$$ $$\frac{dy}{dx} = \frac{d}{dx}(2x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(4)$$ $$\frac{dy}{dx} = 4x - 3$$

$$\text{Gradient at } x=2 = 4(2) - 3$$ $$= 8 - 3$$ $$= 5$$

The correct option is B.
The gradient of the curve at $x=2$ is $\boxed{5}$.

18. Evaluate $\log_8 16$.

Assumption: The question "Evaluate log8 log16" is interpreted as $\log_{16} 8$ due to the available options. If it were $\log_8 16$, the answer would be $\frac{4}{3}$, which is not an option. If it were $\log_8(\log_{10} 16)$, the answer would not be a simple fraction.

$$(2^4)^k = 2^3$$ $$2^{4k} = 2^3$$

$$4k = 3$$ $$k = \frac{3}{4}$$

The correct option is A.
The value is $\boxed{\frac{3}{4}}$.