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Quantitative Aptitude

GATE CSE 2013 · Q62

MCQ2 marksEasy

Time

Out of all the 2-digit integers between 1 and 100, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7?

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Solution

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Official answer: D) 77/90

Quantitative Aptitude
  1. Total number of 2-digit integers between 1 and 100: The 2-digit integers range from 1010 to 9999 inclusive.

    Total outcomes, N=99−10+1=90\text{Total outcomes, } N = 99 - 10 + 1 = 90
  2. Number of 2-digit integers divisible by 7: The multiples of 77 in the range [10,99][10, 99] are:

    14,21,28,…,9814, 21, 28, \dots, 98

    Since 14=7×214 = 7 \times 2 and 98=7×1498 = 7 \times 14, the count is:

    n(divisible by 7)=14−2+1=13n(\text{divisible by } 7) = 14 - 2 + 1 = 13
  3. Number of 2-digit integers NOT divisible by 7:

    n(not divisible by 7)=90−13=77n(\text{not divisible by } 7) = 90 - 13 = 77
  4. Probability:

    P=7790P = \frac{77}{90}

Hence, the correct option is 77/90 (Option D).

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This Quantitative Aptitude previous year question is Q62 of the GATE CSE 2013 paper. Find more GATE CSE PYQs on Quantitative Aptitude in General Aptitude, or browse all GATE CSE PYQs.

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Quantitative Aptitude in GATE CSE

Quantitative Aptitude has 53 GATE CSE PYQs across 13 papers from 2011 to 2026. 27 of them are 2-mark MCQs like this one, and GATE CSE 2013 has 5.

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GateAI solution cardGATE CSE 2013 · Q62 · MCQ

Out of all the 2-digit integers between 1 and 100, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7?

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