Level
2

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5 Problem
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Questions. This is the very first one. Directions are to solve five problems, and enter in the answer in the box like
$a_1 a_2 a_3 a_4 a_5$
, where as the subscript number is the problem number. (eg. if final result is
$54321$
, then answer to first problem is 5, second is 4, etc.) So let's begin! Note that
*
not all answers are single digit answers
*
.

1) A runner needs to get coach some water. He runs north 6 yards across the field, and to the left 8 yards. How long is the distance in inches between his starting point and his ending point, in a diagonal, assuming that there are no obstacles.

2) Mathfighters
**
MathBawse1 and Negato
**
were watching the Super Bowl at Dave and Busters. They only make up 0.5% of the whole population there. How many people are at Dave and Busters, with the information given?

3) A Super Bowl commercial pops up in your television screen. Commercial breaks are only supposed to be $\sqrt{64} \times 15$ seconds long. This particular commercial, however, is Ford's famous #nearlydouble commercial, which purposefully airs twice, each in a different setting, thus taking up $\frac{160}{2.5}$ seconds of time. The portion of the commercial break the Ford commercial takes up can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are co-prime positive integers. What is the positive value of $a-b$ ?

4) Four family members get together and eat some pie to celebrate the Super Bowl. The dad eats $\frac{1}{4}$ of the pie, the daughter eats $\frac{1}{2}$ of what's remaining, and the brother gulps down $\frac{1}{3}$ of whats remaining. Lets say that the diameter of the original pie is 12 units. The area of the remaining pie can be shown as $a\pi$ . Find a.

5) A football stadium consists of 3 squares. Given that each square has a diagonal of 20 meters, what is the area of the football stadium in square meters?

Thanks for testing out this problem! If you want more or have any suggestions, please comment below! Thanks again and enjoy your Sunday!

The answer is 36040079600.

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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 $10 \times 36 = \boxed{360}$

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

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: $\dfrac{64}{120} = \dfrac{8}{15} \Longrightarrow |8-15| = \boxed{7}$

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 $\dfrac{1}{4} \times \pi \times \left(\dfrac{12}{2}\right)^{2} = \boxed{9} \pi$

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: $\boxed{36040079600}$