Mathematical Problem Solving Techniques

Which of the following is the smallest integer greater than √63?

• 5

• 6

• 7

• 8

• 9

Answer explanation

The square root of 63 is 7.937. Both 6 and 7 are less than that, therefore 8 is the smallest integer greater than 7.937.

Tags

CCSS.6.EE.B.5

CCSS.6.EE.B.8

CCSS.6.NS.C.7

A motorcycle traveling at 45 mph on the highway is leaking oil, which is leaking at a rate of 5 fluid ounces per minute. If the motorcycle holds 600 fluid ounces of oil, how many miles will the motorcycle travel before there is no more oil?

• 9 miles

• 45 miles

• 90 miles

• 120 miles

• 180 miles

Answer explanation

If the motorcycle is losing 5 ounces every minute, it is losing 300 ounces per hour since there are 60 minutes in an hour. (5 × 60 = 300). Since the motorcycle can hold 600 fluid ounces total it will take 2 hours for the motorcycle to run out of oil (600 ÷ 300 = 2). If the motorcycle is traveling at 45 mph, in two hours it will have traveled 90 miles total. (45 𝑚𝑖𝑙𝑒𝑠 × 2 ℎ𝑜𝑢𝑟𝑠 = 90 𝑚𝑖𝑙𝑒𝑠 𝑡𝑜𝑡𝑎𝑙)

Tags

CCSS.6.EE.B.6

CCSS.6.EE.B.7

CCSS.6.EE.C.9

CCSS.6.RP.A.2

CCSS.6.RP.A.3

Josie had exactly 65 wreaths to sell at the craft show. After day one of selling, she had exactly 57 wreaths left. After day two of selling, Josie had exactly 49 wreaths. After day three, she has 41 wreaths remaining. Assuming Josie will continue to sell her wreaths at this same daily rate, how many of these flowers will she have left after Day 6?

• 50

• 45

• 41

• 17

• 6

Answer explanation

Since Josie started with 65 wreaths, and had 57 left after day one, she sold 8 wreaths on day one (65 − 57 = 8). After day two, she sold 8 wreaths on day two (57 − 49 = 8). Since Josie had 41 wreaths left after day three, she sold 8 wreaths on day three (49 − 41 = 8). Continuing this pattern, of selling 8 wreaths per day, at the end of day four Josie would have 41-8= 33 wreaths. After day five, 33-8= 25 wreaths. Finally, at the end of day six, Josie would have 25-8= 17 wreaths. Thus, Josie would have 17 wreaths left after day six.

Tags

CCSS.6.EE.B.6

CCSS.6.EE.B.7

CCSS.6.RP.A.3

• 0

• 180

• 200

• 1800

• 2000

Tags

CCSS.6.EE.A.2

CCSS.6.EE.B.5

CCSS.6.EE.B.6

• 4

• 6

• 8

• 12

• 24

Answer explanation

To solve for 𝑥 we need to distribute to eliminate parentheses. After distributing, we have 3𝑥 − 24 + 𝑥 = 24. Next, we can combine the like terms on the left side of the equation. The 3𝑥 and 𝑥 can be combined to have 4𝑥 − 24 = 24. We use inverse operations to eliminate the − 24, so we add 24 to both sides which gives us 4𝑥 = 48. We use inverse operations again and divide both sides by 4 to get x by itself. This gives us 𝑥 = 12.

Tags

CCSS.6.EE.B.5

CCSS.6.EE.B.6

• 180

• 7.2

• 45

• 72

• 51.84

Tags

CCSS.6.EE.A.2

CCSS.6.EE.B.6

CCSS.6.EE.B.7

CCSS.6.RP.A.3

In the standard (x,y) coordinate plane what is the midpoint of the line segment that has endpoints (6,2) and (2, -2)?

• (4, 4)

• (8, 4)

• (4, 0)

• (-4, -4)

• (8, 0)

Tags

CCSS.HSG.GPE.B.6

• 54°

• 72°

• 76°

• 80°

• 104°

Answer explanation

Using the properties of parallelograms, we know that ∠𝐹 is equal to ∠𝐷, so ∠𝐷 = 50°. We must then know that the sum of the angles of a triangle always equals 180°. This means ∠𝐶 + ∠𝐷 + ∠𝐸 = 180°. We will then substitute the measure of our angles into this equation to get 54° + 50° + ∠𝐸 = 180°. At this point, we will solve our equation to find ∠𝐸 = 76°.

Tags

CCSS.HSG.CO.C.11

• 8

• 10

• 12

• 14

• 20

Tags

CCSS.6.G.A.3

CCSS.6.NS.C.8

• 60

• 72

• 180

• 225

Tags

CCSS.6.EE.A.2

CCSS.6.EE.B.5

CCSS.6.EE.B.6

CCSS.6.EE.C.9