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Exercise 5.1 · Q2

Q.The number of integers from 100 to 500 that are divisible by 5 are:

(a) 80
(b) 81
(c) 75
(d) none of these
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✓ Free question

The multiples of 55 from 100100 to 500500 form an A.P.; the count is found with the standard "number of terms" formula.

For an A.P. with first term a1a_1, last term ana_n, common difference dd:

n=an−a1d+1n = \frac{a_n - a_1}{d} + 1

  1. Identify the A.P. of multiples of 55 in range: first term a1=100a_1 = 100 (itself divisible by 55), last term an=500a_n = 500 (also divisible by 55), common difference d=5d=5.
  2. Apply the term-count formula:

n=500−1005+1=4005+1n = \frac{500-100}{5}+1 = \frac{400}{5}+1

  1. Compute: 4005=80\dfrac{400}{5}=80, so n=80+1=81n = 80+1 = 81.
  2. Self-check: listing a few terms — 100,105,110,…,500100,105,110,\ldots,500 — the sequence has 8181 terms; equivalently, multiples of 55 up to 500500 number 500/5=100500/5=100, and up to 9595 (just below 100100) number 95/5=1995/5=19, so terms from 100100 to 500500 inclusive =100−19=81=100-19=81. ✓
  3. Verify against options: (a) 8080 ✗, (b) 8181 ✓, (c) 7575 ✗, (d) none of these ✗.
✓Final answer

There are 8181 integers from 100100 to 500500 divisible by 55 — option (b)

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