A positive integer, x, is divisible by 15, 25, and 30. Which of the following is the smallest possible value of x?
a) 75 b) 90 c) 120 d) 150 e) 450 From an old ACT Exam. SOLUTION The answer is choice d. The smallest number that is divisible by 15, 25, and 30 is the Least Common Multiple (LCM) of all the numbers. First factor all the given numbers. 15 = 3 * 5 25 = 5 * 5 30 = 2 * 3 * 5 Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs. 2: 1 occurrence 3: 1 occurrence 5: 2 occurrences LCM = 2 * 3 * 5 * 5 = 150 Therefore, the answer is choice d) 150.
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There are 25 buildings on Maple Street. Of these 25 buildings, 10 have fewer than 6 rooms, 10 have more than 7 rooms, and 4 have more than 8 rooms. What is the total number of buildings on Maple street that have 6,7, or 8 rooms?
a) 5 b) 9 c) 11 d) 14 e) 15 SOLUTION: The answer is choice c. GIVEN: There are 25 buildings. 10 buildings have less than 6 rooms. 10 buildings have more than 7 rooms. (EXTRA INFORMATION) 4 buildings have more than 8 rooms. ANSWER Therefore, 25 - 10 - 4 = 11 buildings that have 6, 7, or 8 rooms. |
Math Problem of the Day ANSWERS PART II
Here are the solutions to The Math Problems of The Day. The problems and solutions shown here come
from a variety of sources and I attempt to vary the difficulty level of
the problems from day to day. Archives
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