In 2 hours, 1 student can paint 4 large signs or 12 small signs. Therefore, 3 students are required to paint the large signs ($\displaystyle 4\times 3=12$) and 5 students are required to paint the small signs ($\displaystyle 12\times 5=60$). In total, 8 students are required.

In the first three hours, he travels 180 miles.

$\displaystyle 3\times60=180$

In the next two hours, he travels 144 miles.

$\displaystyle 2\times 72=144$

for a total of 324 miles.

$\displaystyle 180+144=324$

Divide by the total number of hours to obtain the average traveling speed.

$\displaystyle \frac{324}{5}=64.8\ mph$

Tom runs a 100m race in a certain amount of time.  If John runs the same race, he takes 2 seconds longer.  If John ran at 8m/s, how fast did Tom run?

Let $\displaystyle x$ denote the amount of time that it took Tom to run the race.  Then it took John $\displaystyle x+2$ seconds to run the same race going 8m/s.  At 8m/s, it takes 12.5 seconds to finish a 100m race.  This means it took Tom 10.5 seconds to finish.  Running 100m in 10.5 seconds is the same as $\displaystyle 100/10.5 \approx 9.5m/s$

Jim arrived at the common destination in 70 minutes, or $\displaystyle \frac{7}{6}$ hours. His average speed was 54 miles per hour, so their workplace is 

$\displaystyle \frac{7}{6} \cdot 54 = 63$ miles away from Jim and Julia's home.

Julia traveled those 63 miles in 80 minutes, or $\displaystyle \frac{4}{3}$ hours, so her average speed was

$\displaystyle 63 \div \frac{4}{3} = \frac{63}{1} \cdot \frac{3} {4} = \frac{189}{4} = 47 \frac{1}{4}$ ,

or, rounded, 47 miles per hour.

Andy traveled for 50 minutes, or $\displaystyle \frac{5}{6}$ of one hour, at 45 miles per hour, so the office building is $\displaystyle \frac{5}{6} \times 45 = 37 \frac{1}{2}$ miles from Andy and Donna's house. Donna traveled this distance for 45 minutes, or $\displaystyle \frac{3}{4}$ of one hour, so her speed was:

$\displaystyle 37 \frac{1}{2} \div \frac{3}{4} = \frac{75}{2} \cdot \frac{4} {3} = 50$ miles per hour.

Let $\displaystyle r$ be the rate at which Kenny drove. Then Marie drove at a rate of $\displaystyle r - 6$ miles per hour. The two drove the same distance, so, since Kenny drove 70 minutes, or $\displaystyle \frac{7}{6}$ hours, and Marie drove for 80 minutes, or $\displaystyle \frac{8}{6}$ hours, we can use the formula $\displaystyle rt = d$ to set up the equation:

$\displaystyle \frac{7}{6} r = \frac{8}{6} (r - 6)$

$\displaystyle \frac{7}{6} r \cdot 6 = \frac{8}{6} (r - 6) \cdot 6$

$\displaystyle 7r = 8 (r - 6)$

$\displaystyle 7r = 8 r - 48$

$\displaystyle 7r - 7r + 48 = 8 r - 48 - 7r + 48$

$\displaystyle 48 = r$

Since Kenny traveled at 48 miles per hour for $\displaystyle \frac{7}{6}$ hours, the distance each drove is 

$\displaystyle 48 \cdot \frac{7}{6} = 56$ miles.

Jerry drove 60 miles per hour for 4 hours - that is, $\displaystyle 60 \times 4 = 240$ miles. 

He drove the remainder of the distance, or $\displaystyle 320-240 = 80$ miles over a period of $\displaystyle 6 \frac{2}{3} - 4 = 2\frac{2}{3}$ hours, so his average speed was 

$\displaystyle 80 \div 2 \frac{2}{3} = 80 \div \frac{8}{3} = 80 \times \frac{3} {8} = 30$ miles per hour.

If Sally drives q miles in 3 hours, her rate is 3/q miles per hour.  Plug this rate into the distance equation and solve for the time:

$\displaystyle \dpi{100} \small Distance = rate\times time$

$\displaystyle \dpi{100} \small r=\frac{q}{3}\times t$

$\displaystyle \dpi{100} \small t=\frac{3r}{q}$

We need to convert hours into minutes and multiply this by the 10 minute time interval:

$\displaystyle \small \frac{12\ miles}{1\ hour}x\frac{1\ hour}{60\ min}x\frac{10\ min}{1}=\frac{120\ miles}{60}=2\ miles$

If the driver needs to drive two laps, each one mile long, at an average rate of 60 miles per hour. To find the average speed, we need to add the speed for each lap together then divide by the number of laps. The equation would be as follows:

$\displaystyle \frac{X+Y}{2}=60$

In our case we know lap one was driven at $\displaystyle 40$ miles per hour. We substitute this value in for $\displaystyle X$ and solve for $\displaystyle Y$.

$\displaystyle \frac{40+Y}{2}=60$

$\displaystyle 40+Y=60(2)$

$\displaystyle 40+Y=120$

$\displaystyle Y=120-40$

$\displaystyle Y=80$

Thus to average $\displaystyle 60$ miles per hour for two laps with lap one being $\displaystyle 40$ miles per hour, lap two would have to have a rate of  $\displaystyle 80$ miles per hour.