Numerical Ability Reviewer - Part 21

Practice 20 Numerical Ability questions for the Civil Service Exam (CSE), each with the correct answer and an explanation.

20 questions80% passing markFree

Part 21 of our Numerical Ability reviewer has 20 practice questions in the style of the Civil Service Exam. Answer them in order, read the explanation after each one, and check your score against the 80% passing mark at the end. Questions and choices are reshuffled every time you retake the part.

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Sample questions from Part 21

Here are a few questions from this part. Take the full test to answer all 20 and get your score.

Sample question 1

An LGU employee processed the following numbers of documents per day over 9 days: 46, 23, 24, 35, 57, 50, 49, 16, 51. After removing the lowest value and the two highest values, what is the median of the remaining values?

  • 40.5
  • 39.5
  • 37.83
  • 46
Answer: 40.5
Sorted: 16, 23, 24, 35, 46, 49, 50, 51, 57. After removal: 23, 24, 35, 46, 49, 50 (6 values). The median is the average of the 3rd and 4th values = 40.5.

Sample question 2

The numbers of trees planted by 7 barangay volunteers are: 10, 24, 7, 17, 23, 5, 19. If every value is multiplied by 2, what is the new mean?

  • 15
  • 29
  • 30
  • 17
Answer: 30
Original mean = 105 ÷ 7 = 15. Multiplying every value by 2 multiplies the mean by 2: 15 × 2 = 30.

Sample question 3

The table shows the scores of 15 examinees in a mock Civil Service test.

ScoreNumber of examinees
601
757
856
901
What is the mean score?

  • 80
  • 77.5
  • 77
  • 79
Answer: 79
Multiply each score by its frequency and add: 60×1 + 75×7 + 85×6 + 90×1 = 1185. Mean = 1185 ÷ 15 = 79. (The simple average of the scores, 77.5, ignores the frequencies.)

Sample question 4

The mean score of 6 examinees is 85. When the score of one examinee, 55, is removed, what is the mean of the remaining 5 scores?

  • 92
  • 90
  • 91
  • 85
Answer: 91
Total of all 6 scores = 85 × 6 = 510. After removing 55: 510 − 55 = 455. New mean = 455 ÷ 5 = 91.

Sample question 5

The recorded mean score of 30 students was 67. Later, two scores were found to be wrongly entered: 61 should have been 78, and 89 should have been 66. What is the corrected mean?

  • 67.57
  • 66.8
  • 61
  • 67
Answer: 66.8
Net change in the total = (78 − 61) + (66 − 89) = -6. Corrected total = 2010 + -6 = 2004. Corrected mean = 2004 ÷ 30 = 66.8.

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