IPMAT Indore 2023 (MCQ) - If a three-digit number is chosen at random, what is the probability that it is divisible neither by 3 nor by 4? | PYQs + Solutions | AfterBoards
Skip to main contentSkip to question navigationSkip to solution
IPMAT Indore Free Mocks Topic Tests

IPMAT Indore 2023 (MCQ) PYQs

IPMAT Indore 2023

Number System
>
Divisibility Rules

Medium

If a three-digit number is chosen at random, what is the probability that it is divisible neither by 3 nor by 4?

Correct Option: 2
Total 3 digit numbers =999100+1=900= 999-100+1 = 900\newlineFormula used: [highestlowest+1][\text{highest}-\text{lowest}+1]
Every 3rd number is divisible by 3.\newlineProbability(number not divisible by 3)=23\text{Probability(number not divisible by 3)} = \frac23
Every 4th number is divisible by 4.\newlineProbability(number not divisible by 4)=34\text{Probability(number not divisible by 4)} = \frac34
Hence, 23×34=12\dfrac23 \times \dfrac34 = \boxed{\frac12}

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question