Let S be the set of integers x such that

i. 100 ≤ x ≤ 200

..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep."> Let S be the set of integers x such that

i. 100 ≤ x ≤ 200

..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." /> Let S be the set of integers x such that

i. 100 ≤ x ≤ 200

..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />

Question 58

Logical Reasoning Number representations Hard

<p>Let S be the set of integers x such that</p> <p>i. 100&nbsp;&le; x&nbsp;&le; 200</p> <p>ii.&nbsp;x is odd and</p> <p>iii. x is divisible by 3 but not by 7</p> <p>How many elements does S&nbsp;contain?</p>

(A) 16
(B) 12
(C) 11
(D) 13
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Step-by-Step Solution

We need to find the number of integers 'x' that satisfy all three given conditions. We will apply the conditions sequentially to narrow down the set of numbers.

1. Numbers Divisible by 3 in the range [100, 200]

First, let's find all the numbers between 100 and 200 (inclusive) that are divisible by 3.

  • The first multiple of 3 in this range is 102 (3 × 34).
  • The last multiple of 3 in this range is 198 (3 × 66).
  • The number of terms in this arithmetic progression is ((Last - First) / Step) + 1 = ((198 - 102) / 3) + 1 = 33.

There are 33 numbers divisible by 3 in the range.

2. Filtering for Odd Numbers

The second condition is that the numbers must be odd. The sequence of multiples of 3 (102, 105, 108, ...) alternates between even and odd. Since the sequence starts and ends with an even number and has an odd number of terms (33), there will be 17 even numbers and 16 odd numbers.

So, there are 16 odd numbers divisible by 3 in the range.

3. Excluding Numbers Divisible by 7

The third condition is that the numbers must not be divisible by 7. We need to find the numbers from our list of 16 that are also divisible by 7 and remove them.

  • A number divisible by both 3 and 7 must be divisible by their least common multiple, which is 21.
  • We need to find the odd multiples of 21 in the range [100, 200].
  • Let's list them:
    • 21 × 5 = 105 (Odd)
    • 21 × 6 = 126 (Even - ignore)
    • 21 × 7 = 147 (Odd)
    • 21 × 8 = 168 (Even - ignore)
    • 21 × 9 = 189 (Odd)

There are 3 numbers (105, 147, and 189) that are odd, divisible by 3, and also divisible by 7. These must be excluded.

4. Final Calculation

To find the total number of elements in set S, we subtract the excluded numbers from our count of odd numbers divisible by 3.

Total Elements = (Number of odd multiples of 3) - (Number of odd multiples of 21)

Total Elements = 16 - 3 = 13

Conclusion

The set S contains 13 elements.

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