Numerical Ability Reviewer - Part 19

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

20 questions80% passing markFree

Part 19 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 19

Sample questions from Part 19

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

Sample question 1

What is the smallest number that must be subtracted from 378 so that the result is divisible by 11?

  • 4
  • 6
  • 15
  • 7
Answer: 4
378 ÷ 11 gives a remainder of 4. Subtract the remainder 4 to get 374, a multiple of 11.

Sample question 2

When a positive integer n is divided by 36, the remainder is 17. What is the remainder when 2n + 1 is divided by 9?

  • 0
  • 7
  • 8
  • 11
Answer: 8
Since 9 divides 36, n ≡ 17 ≡ 8 (mod 9). Then 2n + 1 ≡ 2(8) + 1 = 17 ≡ 8 (mod 9).

Sample question 3

How many integers between 156 and 1456, inclusive, are divisible by both 6 and 18?

  • 72
  • 80
  • 73
  • 75
Answer: 72
Divisible by both means divisible by LCM(6, 18) = 18. Count = ⌊1456/18⌋ − ⌊155/18⌋ = 80 − 8 = 72.

Sample question 4

What is the sum of all the positive factors of 72?

  • 196
  • 195
  • 123
  • 201
Answer: 195
The factors are 1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 + 72 = 195.

Sample question 5

3 LGU service vehicles leave the motor pool together at 6:00 AM. They return to the motor pool every 21, 12, 40 minutes, respectively. At what time will they all be at the motor pool together again?

  • 10:00 AM
  • 7:24 AM
  • 8:00 PM
  • 7:39 PM
Answer: 8:00 PM
They coincide every LCM(21, 12, 40) = 840 minutes, i.e., 14 h 0 min after 6:00 AM, which is 8:00 PM.

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