5 Essential SAT Math Practice Questions | SAT Math Practice

Think you can handle five SAT math questions in a row without getting one wrong? Question 1 Suppose a, b, c, and d are real nonzero numbers. If d = a/c and b = c/a, which of the following is equivalent to d? So we have d = (a * b) / c. Given b = c/a, cross-multiplying gives a * b = c. Substituting a * b = c into our equation for d gives c / c, which simplifies to 1. Hence, the correct answer is Option A. Question 2 Kronhorst International has a marketing division with 23 employees and a sales division with 18 employees. 9 employees work in both divisions. How many employees work only in one department, and how many work only in marketing? We can solve this using a Venn diagram. Total Marketing: 23 Total Sales: 18 Both Departments: 9 To find those working only in marketing: 23 - 9 = 14. To find those working only in sales: 18 - 9 = 9. For employees working in only one department, we combine those only in marketing and those only in sales: 14 + 9 = 23. So the answers are 23 and 14. Correcting the answer key accordingly, the correct answer is Option C. Question 3 The probability of event A taking place is x/3, and event B is 2x/5. These events cannot take place at the same time. What is the probability of either event A or event B taking place? Since events A and B cannot occur simultaneously, they are mutually exclusive. Thus, P(A ∩ B) = 0. Using the addition rule for probability: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) P(A ∪ B) = x/3 + 2x/5 - 0 Finding a common denominator of 15: (5x + 6x) / 15 = 11x/15 Therefore, the correct answer is Option C. Question 4 Consider a data set with 10 values. Each value is 8 more than the previous value. If the largest value is 109, what is the median of this data set? We have 10 ordered values ending at 109. Moving backward by subtracting 8 each time gives us the full data set: 37, 45, 53, 61, 69, 77, 85, 93, 101, 109 To find the median (the middle value), we cancel terms equally from both ends until two middle terms remain: 69 and 77. Taking their average: (69 + 77) / 2 = 146 / 2 = 73 Hence, the correct answer is Option D. Question 5 What is the probability that a randomly generated three-digit integer will be less than 200 and divisible by both 6 and 9? First, find total sample space n(S) of all 3-digit integers (100 to 999): Using Arithmetic Progression: t_n = a + (n - 1) * d 999 = 100 + (n - 1) * 1 899 = n - 1 = n = 900 So, n(S) = 900. Next, consider 3-digit integers less than 200 (100 to 199) divisible by both 6 and 9. The LCM of 6 and 9 is 18, so these numbers must be multiples of 18: 108, 126, 144, 162, 180, 198 There are 6 favorable outcomes, so n(A) = 6. Probability = n(A) / n(S) = 6 / 900 = 1/150. The required probability is 1/150.