5 Yards, 5 Problems

1. 6 yards up, 8 yards left. Using the Pythagorean theorem, shortest distance between initial and ending points is $\sqrt{6^{2}+8^{2}} = 10$ yards. 1 yard is apparently 36 inches (thanks, Google!) and so, the final answer is $10 \times 36 = \boxed{360}$

2. If $0.5 \%$ of the crowd represents 2 people, $100 \%$ would be represented by $\dfrac{100 \% \times 2}{0.5 \%} = \boxed{400}$ people.

3. The total commercial time is $\sqrt{64} \times 15 = 120$ seconds, of which $\dfrac{160}{2.5} = 64$ seconds are taken up by the Ford commercial. Hence, the ratio of Ford airtime to total commercial time is: $\dfrac{64}{120} = \dfrac{8}{15} \Longrightarrow |8-15| = \boxed{7}$

4. The dad eats $\dfrac{1}{4}$ of the pie, leaving $1-\dfrac{1}{4} = \dfrac{3}{4}$ of the pie for his kids. The daughter gulps down $\dfrac{1}{2} \times \dfrac{3}{4} = \dfrac{3}{8}$ of the pie, leaving $\dfrac{3}{4} - \dfrac{3}{8} = \dfrac{3}{8}$ of the pie for her brother. The brother gobbles up $\dfrac{1}{3} \times \dfrac{3}{8} =\dfrac{1}{8}$ of the pie, leaving $\dfrac{3}{8} - \dfrac{1}{8} = \dfrac{1}{4}$ of the pie. If the diameter of the pie was 12 units, the remaining area is given by $\dfrac{1}{4} \times \pi \times \left(\dfrac{12}{2}\right)^{2} = \boxed{9} \pi$

5. Finally, if the diagonal of a square is 20 metres, its length is $\dfrac{20}{\sqrt{2}}$ , since the ratio of hypotenuse to "leg" of a right-angled isosceles triangle is $\sqrt{2}$ . This gives us a total area of $3 \times \left(\dfrac{20}{\sqrt{2}}\right)^{2} = \boxed{600}$ square metres.

Concatenating the answers, we get: $\boxed{36040079600}$

Wow! You came a long way solving 5 problems! Here's the way I did it if you got it wrong.

1) This is Pythagorean Theorem in place. The first thing that pops up to me is that the path is like a right triangle; the legs are 6 and 8, and the hypotenuse is unknown. Luckily this is an easy multiple of a Pythagorean pair (3,4,5). So multiply by 2 and you get (6,8,10), so our answer is 10 yards. BUT the catch here is inches . Recall that $36 \text{ inches} = 1 \text{ yard}$ . So 10 yards equates to $\boxed{360 \text{ inches}}$

2) To get the whole population, divide the 2 people by the percentage. In this case, $\frac{2}{0.005} = \boxed{400}$

3) The fraction is $\frac{\frac{160}{2.5}}{\sqrt{64} \times{} 15} \Rightarrow \frac{\frac{320}{5}}{8 \times 15} \Rightarrow \frac{64}{120} \Rightarrow \frac{8}{15}$ . Thus $|8-15| = \boxed{7}$

4) The area of the pie is $\pi r^2 \Rightarrow \pi 6^2 \Rightarrow 36\pi$ . Now our equation is $\frac{2}{3}(\frac{1}{2}(\frac{3}{4}36))$ is how we get the area remaining. So $\frac{2}{3}(\frac{1}{2}(27)) \Rightarrow \frac{2}{3}(\frac{27}{2}) \Rightarrow \frac{54}{6} = \boxed{9}$ when $\pi$ is stripped out.

5) One square in the stadium has an area of $\frac{Diagonal^2}{2} \Rightarrow \frac{20^2}{2} \Rightarrow \frac{400}{2} \Rightarrow 200$ meters squared. Multiply it by 3 and you get the area of the whole stadium, which is $\boxed{600 \text{ meters}}$

If you find this problem hard to handle, or would like to add something, let me know by commenting below! Thanks guys!

Let us take the answers as $a_1,a_2,a_3,a_4,a_5$ respectively for the 5 problems and the answer to be given is $a_1a_2a_3a_4a_5$ .

1) For the 1st problem, if we plot the distance travelled in the directions as specified, then we get a right angled triangle with the two sides being 6 yards and 8 yards and the required diagonal distance is the hypotenuse whose length is $=\sqrt{6^2+8^2}=\sqrt{100}=10\text{ yards }=(10\times 36)\text{ inches }=360\text{ inches }$ . So, $a_1=\boxed{360}$

2) For the 2nd problem, it is given that the 2 persons make $0.5$ % of the total no. of people (say $x$ ). Then we have,

$\frac{2}{x}\times 100 = 0.5 \implies x=\frac{200}{0.5} \implies x=\frac{2000}{5}=400$ . So, $a_2=\boxed{400}$

3) For the 3rd problem, the time taken by the Ford commercial $=\frac{160}{2.5}=\frac{1600}{25}=64$ sec. while total commercial time $=\sqrt{64}\times 15 = 8\times 15=120$ sec. So, portion of break taken by Ford commercial $=\frac{64}{120}=\frac{8}{15}$ . So, this is in $\frac{a}{b}$ form with $a=8,b=15$ . So, positive value of $a-b = |8-15|=|-7|=7$ . So, $a_3=\boxed{7}$

4) For the 4th problem, the pie has a diameter of 12 units, so the total area of pie at the beginning $=\pi (\frac{12}{2})^2=\pi \times 6^2 = 36 \pi$ . Now, the dad eats $\frac{1}{4}$ th of the pie $=\frac{1}{4}\times 36 \pi=9 \pi$ . So, remaining pie $=36 \pi-9 \pi = 27 \pi$ . Then, daughter eats $\frac{1}{2}$ of remaining $27 \pi$ area of pie, i.e., $\frac{1}{2}\times 27 \pi = \frac{27}{2} \pi$ . So, remaining pie $=27 \pi-\frac{27}{2} \pi = \frac{27}{2} \pi$ . Then, brother eats $\frac{1}{3}$ of whats remaining, i.e, $\frac{1}{3}\times \frac{27}{2} \pi = \frac{27}{6} \pi$ . Then, final remaining part of pie $=\frac{27}{2} \pi-\frac{27}{6} \pi= \frac{81-27}{6} \pi=\frac{54}{6} \pi=9 \pi$ . This is in form $a \pi$ with $a=9$ . So, $a_4=\boxed{9}$ .

5) For the 5th problem, let each square have side $a$ . Then, we know that diagonal of square $=\sqrt{2}a$ . Then, from the problem, we get---

$\sqrt{2}a=20 \implies a=10\sqrt{2}$

Then, area of each square $=a^2=(10\sqrt{2})^2=200$ . So, total area of 3 squares(area of stadium) $=3\times 200=600$ . So, $a_5=\boxed{600}$ .

So, the answer to be given $=a_1a_2a_3a_4a_5=\boxed{36040079600}$