The sum of all proper fractions whose denominators are less than or equal to 100 is:

This question was previously asked in
Haryana CET Previous Year Paper (Held On: 6 Nov 2022 Shift 2)
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  1. 2124
  2. 2475
  3. 1925
  4. Not attempted

Answer (Detailed Solution Below)

Option 2 : 2475
Free
Haryana CET Full Test 1
100 Qs. 100 Marks 105 Mins

Detailed Solution

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Formula used :

(1 + 2 + 3 + 4 +     ......... + 99) = [ n ( n + 1 ) ] ÷ 2

Solution :

We can write all proper fractions whose denominators are less than or equal to 100 in given form :

⇒ 1/2 + (1/3 + 2/3) + (1/4 + 2/4 + 3/4) + ...  (1/100 + 2/100 + .. ..  99/100)

⇒ 1/2 + 3/3 + 6/4 + ..... + 4950/100

⇒ 1/2 + 1 + 3/2 + 2 + ........ + 99/2

We can also write this equation in following way :

⇒ 1/2 + 2/2 + 3/2 + ...... + 99/2

⇒ 1/2 ( 1 + 2 + 3 + ..... + 99 )

⇒  /2 [ ( 99 × 100) ÷ 2 ]

⇒ 99 × 25

⇒ 2475

Hence the correct answer is "2475".

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