Numerical Ability Reviewer - Part 20

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

20 questions80% passing markFree

Part 20 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 20

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

Sample question 1

The five-digit number 6A41B is divisible by 72. What is the value of A + B?

  • 6
  • 9
  • 7
  • 11
Answer: 7
72 = 8 × 9. For 8, the last three digits 41B must be divisible by 8: 416 = 8 × 52, so B = 6. For 9, the digit sum 6 + A + 4 + 1 + 6 = 17 + A must be a multiple of 9, so A = 1. Number: 61,416. A + B = 7.

Sample question 2

What digit should replace A so that 73A25 is divisible by 11?

  • 8
  • 4
  • 6
  • 2
Answer: 4
Subtract the sum of digits in even places from the sum in odd places (from the left): (7 + A + 5) − (3 + 2) = 7 + A, which must be a multiple of 11. So A = 4. Check: 73,425 = 11 × 6,675.

Sample question 3

What is the remainder when 2¹⁰⁰ is divided by 7?

  • 3
  • 4
  • 2
  • 1
Answer: 2
Powers of 2 mod 7 repeat in a cycle of 3: 2, 4, 1. Since 100 = 3 × 33 + 1, 2¹⁰⁰ ≡ 2¹ = 2.

Sample question 4

How many integers from 1 to 500 are divisible by 6 or by 15?

  • 100
  • 116
  • 83
  • 133
Answer: 100
Multiples of 6: ⌊500/6⌋ = 83. Multiples of 15: ⌊500/15⌋ = 33. Both (LCM 30): ⌊500/30⌋ = 16. Total = 83 + 33 − 16 = 100.

Sample question 5

How many integers from 1 to 1,000 are divisible by 3 but NOT by 5?

  • 266
  • 333
  • 200
  • 267
Answer: 267
Multiples of 3: 333. Multiples of both 3 and 5 (15): 66. Divisible by 3 but not 5: 333 − 66 = 267.

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