while at the local ice skating rink cora went around the rink 27 times total she slipped and fell 20 of the 27 times she skated around the rink what a

ATI TEAS 7

TEAS Practice Math Test

1. While at the local ice skating rink, Cora went around the rink 27 times in total. She slipped and fell 20 of the 27 times she skated around the rink. What approximate percentage of the times around the rink did Cora not slip and fall?

A. 37%
B. 74%
C. 26%
D. 15%

Correct answer: C

Rationale: To find the approximate percentage of the times Cora did not slip and fall, subtract the times she fell (20) from the total times around the rink (27), which gives 7. Then, divide the number of times she did not slip and fall (7) by the total times around the rink (27) and multiply by 100 to get the percentage. So, 7 divided by 27 equals 0.259, which rounds to approximately 26%. Therefore, the correct answer is 26%. Choice A (37%) is incorrect because it does not reflect the calculation based on the given information. Choice B (74%) is incorrect as it is not the result of the correct calculation. Choice D (15%) is incorrect as it does not match the calculated percentage based on the scenario provided.

2. Half of a circular garden with a radius of 11.5 feet needs weeding. Find the area in square feet that needs weeding. Round to the nearest hundredth. Use 3.14 for π.

A. 2.4
B. 207.64
C. 15.1
D. 30.1

Correct answer: B

Rationale: The formula for the area of a full circle is calculated as Area = π × (radius²). When finding the area of half a circle, we multiply by 0.5. Thus, the formula becomes Area = 0.5 × π × (radius²). Given that the radius of the circular garden is 11.5 feet, the calculation using π = 3.14 is as follows: Area = 0.5 × 3.14 × (11.5²) = 0.5 × 3.14 × 132.25 = 0.5 × 415.27 = 207.64 square feet. Therefore, the correct answer is B. Choices A, C, and D are incorrect because they do not reflect the correct calculation for finding the area of half a circular garden with a radius of 11.5 feet.

3. What is the least common denominator of two fractions?

A. The smallest number that is a multiple of both denominators
B. The smallest number that both fractions can divide into evenly
C. The least common multiple of both denominators
D. The greatest common factor of both denominators

Correct answer: C

Rationale: The least common denominator of two fractions is the least common multiple of both denominators. This is because the least common denominator is the smallest number that both denominators can divide into evenly, ensuring that both fractions can be expressed with a common denominator. Choice A is incorrect as the least common denominator is a multiple of both denominators, not a number that multiplies into both. Choice B is incorrect because the common denominator needs to be a multiple of both denominators, not just a number they can divide into evenly. Choice D is incorrect as the greatest common factor is not used to find the least common denominator, but rather the least common multiple.

4. Which of the following is the decimal form of 87.5%?

A. 875
B. 8,750
C. 0.875
D. 8.75

Correct answer: C

Rationale: To convert a percentage to a decimal, you move the decimal point two places to the left. Therefore, 87.5% as a decimal is 0.875. Choice A (875) is incorrect as it represents the percentage without converting to a decimal. Choice B (8,750) is incorrect as it represents the percentage in whole numbers without decimal conversion. Choice D (8.75) is incorrect as it represents 875% instead of 87.5%.

5. A gift box has a length of 14 inches, a height of 8 inches, and a width of 6 inches. How many square inches of wrapping paper are needed to wrap the box?

A. 56
B. 244
C. 488
D. 672

Correct answer: C

Rationale: To find the surface area of a rectangular prism, you use the formula SA = 2lw + 2wh + 2hl, where l is the length, w is the width, and h is the height. Substituting the given dimensions, the calculation would be SA = 2(14)(6) + 2(6)(8) + 2(8)(14) = 168 + 96 + 224 = 488 square inches. Therefore, 488 square inches of wrapping paper are needed to wrap the box. Choice A (56), Choice B (244), and Choice D (672) are incorrect because they do not represent the correct surface area calculation for the given box dimensions.