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List L contains the 98 repeating decimals

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Manager
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Joined: 01 Sep 2016
Posts: 208

Kudos [?]: 184 [0], given: 33

GMAT 1: 690 Q49 V35
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List L contains the 98 repeating decimals [#permalink]

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New post 20 Sep 2017, 01:02
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Difficulty:

  95% (hard)

Question Stats:

32% (01:44) correct 68% (01:22) wrong based on 25 sessions

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List L contains the 98 repeating decimals 0.010101..., 0.020202..., ..., 0.989898....  If each value in list L is expressed as a fraction in lowest terms, what is the sum of the denominators of all 98 fractions?

(A) 5,096
(B) 5,608
(C) 6,770
(D) 6,894
(E) 9,702
[Reveal] Spoiler: OA

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Kudos [?]: 184 [0], given: 33

Director
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Joined: 25 Feb 2013
Posts: 542

Kudos [?]: 247 [0], given: 33

Location: India
Schools: Mannheim"19 (S)
GPA: 3.82
GMAT ToolKit User Reviews Badge
List L contains the 98 repeating decimals [#permalink]

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New post 20 Sep 2017, 07:33
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bkpolymers1617 wrote:
List L contains the 98 repeating decimals 0.010101..., 0.020202..., ..., 0.989898....  If each value in list L is expressed as a fraction in lowest terms, what is the sum of the denominators of all 98 fractions?

(A) 5,096
(B) 5,608
(C) 6,770
(D) 6,894
(E) 9,702


This is one hell of a question, that requires lot of calculations :sad:
A repeating decimals can be converted into fractions as -
let x=0.01010101...., multiply it by 100 to get

100x=1.0101010...., now subtract both the equation to get

99x=1, or x=\(\frac{1}{99}\)

So our question is to find the sum of denominators of the fractions: \(\frac{1}{99}+\frac{2}{99}+\frac{3}{99}+......\frac{98}{99}\)

Notice that the question asks "lowest terms" so fractions that are factors of 99 will get reduced. For eg. \(\frac{3}{99}=\frac{1}{33}\) and so on.

so from 1 to 98 numbers that are multiples of 11 are 8 (you can use AP formula to get it, 1st term 11 last term 88 and common difference 11)

From 1 to 98 numbers that are multipples of 9 are 10, and

From 1 to 98 numbers that are multiples of 3 are 32. But here, 33 & 66 are already included in muliples of 11 and all multiples of 9 are multiples of 3 so remove them. The final count is 32-(2+10) = 20

Now remaining number from 1 to 98 are = 98-(8+10+20)=60

Now SUM
multiples of 11 except 33 & 66 will have denominator 9 and 33 & 66 will have denominator 3. So sum of all denominators = 2*3+9*6 = 60

multiples of 9 will have denominator 11. So sum of denominators = 10*11 =110

multiples of 3 will have denominator 33. So sum of denominators = 33*20 = 660

Remaining fractions will have denominator 99. So sum of denominators = 99*60 = 5940

Sum of all = 5940+660+110+60 = 6770

Option C

Kudos [?]: 247 [0], given: 33

List L contains the 98 repeating decimals   [#permalink] 20 Sep 2017, 07:33
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List L contains the 98 repeating decimals

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