**Lower Elementary**

*Question:* A hummingbird can flap its wings 70 times per second. At that rate, how many times can the same hummingbird flap its wings in one and a half minutes?

*Answer:* 6300 times

*Note:* Since we are given the amount of wing flaps per second, to figure out how many times a hummingbird flaps its wings in a one and a half minutes we need to figure out how many seconds are in one and a half minutes. There are 60 seconds in 1 minute. That means there are 30 seconds in half a minute.

Since there are 70 flaps for each second to find the total number for one minutes multiply 70 flaps by 60 seconds.

70 x 60 = 4200

That means for half a minute (30 seconds) there are (half of 4200) 2100 flaps.

Adding these together we get the total: 4200 + 2100 = 6300 flaps.

We could also have changed the total minutes to seconds. Then multiplied by the number of flaps per second.

One minute has 60 seconds and a half minute has 30 seconds, so one and a half minutes has:

60 + 30 = 90 seconds

Now we multiply 90 seconds by the number of wing flaps per second, 70.

90 x 70 = 6300 wing flaps

**Upper Elementary**

*Question:* There are 45 people coming to your family reunion BBQ and hotdogs are being served. Hotdogs come in packs of 8 which cost $3.75 per pack and buns come in packs of 10 that cost $7.00 per pack. If you wanted to buy enough to for everyone to have 2 hotdogs, how much would it cost?

*Answer:* $115.00

*Note:* : Including yourself, there are 46 total people coming to the BBQ. Because you are buying enough hotdogs and buns to for each person to have 2 complete hotdogs (hotdog and bun) we need to buy 46 x 2 complete hotdogs or 92. Because hotdogs come in pack of 8, it is not possible to buy exactly 92. You would need to buy 12 packs totaling 96 hotdogs. Likewise, you would need to buy 10 packs of buns totaling 100 buns.

12 packs of hotdogs x $3.75 per pack = $45.00

10 packs of buns x $7.00 per pack = $70.00

Add them together to find the total cost.

$45.00 + $70.00 = $115.00

**Middle School:**

*Question:* A cylindrical tank is a quarter full of fuel. After adding 7 gallons were added to the tank, the tank 7/12 full. How much fuel can the tank hold?

*Answer:* 21 gallons

*Note:* : Let’s restate the problem as verbal math sentence to help solve this problem.

“1/4 of the tanks max capacity plus 7 gallons is equal to 7/12 the tanks max capacity.”

Translating this into a number sentence we where we let ** x** = the maximum capacity of the tank when full,

(1/4)

**+ 7 = (7/12)**

*x*

*x*Now solve for

**.**

*x*(1/4)

**+ 7 = (7/12)**

*x*

*x*7 = (7/12)

**– (1/4)**

*x***[Subtract (1/4)**

*x***from both sides.]**

*x*7 = (7/12)

**– (3/12)**

*x***[Find the Least Common Denominator of 12 and 4, 12.]**

*x*7 = (1/3)

**[Subtract 3/12 from 7/12 to get 4/12 = 1/3.]**

*x***= 21 [Divide both sides by 1/3 or multiply both sides by the reciprocal of 1/3 => 3.]**

*x***Algebra and Up**

*Question:* The area of this square **ABCD** is 36 units^{2}. What is the area of the largest square that intersects all four points **A**, **B**, **C**, and **D**?

*Answer:* 72 units^{2}

*Note:* The first step is to draw the largest square that goes through all points. We will use the diagonals of square **ABCD** for the height and width of our new square. This will maximize the area.

By inspection we can see that the larger square **EFGH** ends up being double the area of the smaller square.

**Explanation**

Because the height **BD** and line segment HE are parallel we can say that line **AB** is a transversal. When two parallel lines are cut be a transversal we know that the alternate interior angles are congruent. **AB** is congruent to **AB** because it is reflexive and angle **AOB** and **AEB** are congruent because they are right angles and all right angles are congruent. By *AAS* we can say that triangle **AOB** is congruent to triangle **AEB**. This can be proven for the three other outer triangles showing that we have doubled the area.

This problem can also be solved algebraically. Triangle **AEB** is a special 45-45-90 triangle. Because the hypotenuse is equal to 6 units and in every 45-45-90 triangle the hypotenuse is equal to ** x**√2, we set them equal and solve for

**which will be equal to the lengths of sides**

*x***AE**and

**EB**.

**√2 = 6**

*x***= 6/√2**

*x*Rationalize the denominator.

**= 3√2**

*x*Looking back at square

**EFGH**we see that 3√2 is equal to half a side. Doubling it then squaring it will give us the area of the whole square.

[2(3√2)]

^{2}= 72 units

^{2}