Numerical Ability Reviewer - Part 18

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

20 questions80% passing markFree

Part 18 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.

Start Part 18

Sample questions from Part 18

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

Sample question 1

Which of the following numbers is divisible by 2?

  • 7,268
  • 7,267
  • 4,907
  • 7,269
Answer: 7,268
Rule: a number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). 7,268: last digit 8 is even; 7,269: last digit 9 is odd; 7,267: last digit 7 is odd; 4,907: last digit 7 is odd. The answer is 7,268.

Sample question 2

What is the greatest common factor (GCF) of 96 and 264?

  • 24
  • 1056
  • 20
  • 48
Answer: 24
96 = 24 × 4 and 264 = 24 × 11, and 4 and 11 share no factor. GCF = 24.

Sample question 3

What is the greatest digit that can replace ★ so that the number 3687★ is divisible by 15?

  • 6
  • 5
  • 0
  • 7
Answer: 0
Test each digit: valid digits are 0 (e.g., 36870 ÷ 15 = 2458). The greatest is 0.

Sample question 4

Which of the following numbers is divisible by 10?

  • 3,815
  • 5,659
  • 4,020
  • 6,968
Answer: 4,020
Rule: a number is divisible by 10 if its last digit is 0. 4,020: last digit 0 is 0; 3,815: last digit 5 is not 0; 5,659: last digit 9 is not 0; 6,968: last digit 8 is not 0. The answer is 4,020.

Sample question 5

What is the remainder when 2,892 is divided by 10?

  • 9
  • 2
  • 0
  • 8
Answer: 2
10 × 289 = 2,890, and 2,892 − 2,890 = 2. So the remainder is 2.

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