Two runners ride from A to B a distance of 55 km, and the first arrives 30 minutes before the second. They then ride from B to A, the first giving the second a start of 4 km and yet arrive 6 minutes before him. Find the speed of A.

Total volume = 2800 cubic feet

Area of ceiling = 2800/28 = 100sqfeet

Total time taken = $\frac{100}{0.125}$ = 800min

= 13hr 20min

Time taken to paint four walls= 40 – 13× $\frac{1}{3}$ = 26hr 40min

Volume of tube $=\frac{2}{3}\pi a^3+\pi a^{2}b$ $\frac{}{}{}^{}{}^{}$

When it is half-full $=\frac{1}{3}\pi a^3+\frac{1}{2}\pi a^{2}b$ $\frac{}{}{}^{}\frac{}{}{}^{}$

Volume of Cylinder

$=\frac{1}{3}\pi a^3+\frac{1}{2}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{1}{2}\pi a^{2}b-\frac{1}{3}\pi a^3$

Length in the cylinder

$=\left(\frac{1}{2}\pi a^{2}b-\frac{1}{3}\pi a^3\right)÷\pi a^2=\frac{1}{2}b-\frac{1}{3}a$

The height above the lowest point

$\frac{1}{2}b-\frac{1}{3}a+a=\frac{2}{3}a+\frac{1}{2}b$

When it is $\frac{1}{4}$ $\frac{}{}$ full, volume $=\frac{1}{6}\pi a^3+\frac{1}{4}{\pi }^{a}b$ $\frac{}{}{}^{}\frac{}{}{}^{}$

Volume of cylinder

$=\frac{1}{6}\pi a^3+\frac{1}{4}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{1}{4}\pi a^{2}b-\frac{1}{2}\pi a^3$

Length in the cylinder

$=\left(\frac{1}{4}\pi a^{2}b-\frac{1}{2}\pi a^3\right)÷\pi a^2=\frac{1}{4}b-\frac{1}{2}a$

The height above the lowest point

$=\frac{1}{4}b-\frac{1}{2}a+a=\frac{1}{2}a+\frac{1}{4}b$

When it is full, volume 

$=\frac{3}{4}\left(\frac{2}{3}\pi a^3+\pi a^{2}b\right)=\frac{1}{2}\pi a^3+\frac{3}{4}\pi a^{2}b$

Volume in cylinder

$=\frac{1}{2}\pi a^2+\frac{3}{4}\pi a^{2}b-\frac{2}{3}\pi a^3=\frac{3}{4}\pi a^{2}b-\frac{1}{6}\pi a^3$

Length in the cylinder

$=\left(\frac{3}{4}\pi a^{2}b-\frac{1}{6}\pi a^3\right)÷\pi a^2=\frac{3}{4}b-\frac{1}{6}a$

The height above the lowest point

$=\frac{3}{4}b-\frac{1}{6}a+a=\frac{5}{6}a+\frac{3}{4}b$

Capacity of the tube

$\pi {(1)}^2(7-1)+\frac{2}{3}\pi {(1)}^3=\frac{20}{3}\pi$

Since M is the midpoint of side PQ, the length of MQ is 2.

Hence, the area of? MQR = $\frac{1}{2}$ × 2 × 4 = 4.

Also area of? NSR = 4. Thus, the unshaded area of the figure = 4 + 4 = 8.

Hence, the area of quadrilateral PMRN

= Area of the square PQRS – The unshaded area of the figure

= 16 – 8 = 8