ATI TEAS 7
TEAS Test Math Prep
1. In Jim's school, there are 3 girls for every 2 boys. There are 650 students in total. Using this information, how many students are girls?
- A. 260
- B. 130
- C. 65
- D. 390
Correct answer: A
Rationale: To find the number of girls in Jim's school, we first establish the ratio of girls to boys as 3:2. This ratio implies that out of every 5 students (3 girls + 2 boys), 3 are girls and 2 are boys. Since there are a total of 650 students, we can divide them into 5 equal parts based on the ratio. Each part represents 650 divided by 5, which is 130. Therefore, there are 3 parts of girls in the school, totaling 3 multiplied by 130, which equals 390. Hence, there are 390 girls in Jim's school. Choice A, 260, is incorrect as it does not consider the correct ratio and calculation. Choice B, 130, is incorrect as it only represents one part of the total students, not the number of girls. Choice C, 65, is incorrect as it ignores the total number of students and the ratio provided.
2. Adrian measures the circumference of a circular picture frame with a radius of 3 inches. Which of the following is the best estimate for the circumference of the frame?
- A. 12 inches
- B. 16 inches
- C. 18 inches
- D. 24 inches
Correct answer: C
Rationale: To calculate the circumference of a circle, use the formula 2πr, where r is the radius. In this case, with a radius of 3 inches, the estimated circumference would be 2 x π x 3 = 6π ≈ 18.85 inches. Therefore, the best estimate for the circumference of the frame is 18 inches (Choice C). Choice A (12 inches) is too small as it corresponds to the diameter rather than the circumference. Choice B (16 inches) and Choice D (24 inches) are also incorrect as they do not reflect the accurate calculation based on the given radius.
3. Simplify the following expression: 5/9 × 15/36
- A. 5/36
- B. 8/27
- C. 10/17
- D. 15/27
Correct answer: A
Rationale: To simplify the given expression, multiply the numerators together and the denominators together. 5/9 × 15/36 = (5 × 15) / (9 × 36) = 75 / 324. Now, simplify the resulting fraction by finding the greatest common divisor (GCD) of 75 and 324, which is 3. Divide both the numerator and denominator by 3 to get the simplified fraction: 75 ÷ 3 / 324 ÷ 3 = 25 / 108. Therefore, the simplified form of 5/9 × 15/36 is 25/108, which is equivalent to 5/36. Choice A, 5/36, is the correct answer. Choice B, 8/27, is incorrect as it does not match the simplified form of the expression. Choice C, 10/17, is unrelated and does not result from the given multiplication. Choice D, 15/27, does not correspond to the simplification of the given expression.
4. What is the value of b in this equation? 5b - 4 = 2b + 17
- A. 13
- B. 24
- C. 7
- D. 21
Correct answer: C
Rationale: To find the value of b in the equation 5b - 4 = 2b + 17, you need to first simplify the equation. By subtracting 2b from both sides of the equation and adding 4 to both sides, you get 3b = 21. Then, dividing both sides of the equation by 3 gives you b = 7. Therefore, the value of b is 7, which corresponds to option C. Choice A (13) is incorrect as it does not match the correct calculation. Choice B (24) is incorrect as it is not the result of the correct algebraic manipulation. Choice D (21) is incorrect as it is not the value of b obtained after solving the equation step by step.
5. How will the number 89632 be written if rounded to the nearest hundred?
- A. 847.9
- B. 900
- C. 847.89
- D. 847.896
Correct answer: B
Rationale: Rounding the number 89632 to the nearest hundred means keeping only two digits before the decimal point. The digit in the hundredth place is the digit in the thousands place of the original number, which is 6. Since 6 is equal to or greater than 5, the digit in the hundredth place, which is 3, gets rounded up. Thus, the number 89632 rounded to the nearest hundred is 900. Choice A, 847.9, rounds the number to the nearest tenth, not hundredth. Choice C, 847.89, adds an extra decimal place which is not correct for rounding to the nearest hundred. Choice D, 847.896, adds more decimal places than necessary for rounding to the nearest hundred.
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