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John has more coins than the number of coins both Tom and Jane have

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John has more coins than the number of coins both Tom and Jane have  [#permalink]

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New post Updated on: 13 Jan 2019, 02:20
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A
B
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D
E

Difficulty:

  35% (medium)

Question Stats:

69% (01:18) correct 31% (01:04) wrong based on 48 sessions

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John has more coins than the number of coins both Tom and Jane have. How many more coins does John have?

1) John has 20% more coins than Tom
2) Tom has 20% more coins than Jane

Originally posted by philipssonicare on 13 Jan 2019, 00:01.
Last edited by philipssonicare on 13 Jan 2019, 02:20, edited 1 time in total.
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Re: John has more coins than the number of coins both Tom and Jane have  [#permalink]

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New post 13 Jan 2019, 03:06
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The answer is E.
This can be solved logically: the question asks about absolute numbers, but the statements only give us percentages - insufficient!
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Re: John has more coins than the number of coins both Tom and Jane have  [#permalink]

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New post 14 Jan 2019, 00:43
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philipssonicare wrote:
John has more coins than the number of coins both Tom and Jane have. How many more coins does John have?

1) John has 20% more coins than Tom
2) Tom has 20% more coins than Jane


Theoretically, the answer would be E because all the given info is ratios, no definite numerical values.

however, I find myself in front of an illogical finding upon combining the two statements:

John = \(\frac{6}{5}\) Tom = \((\frac{6}{5})^2\) Jane.

as John = Tom + Jane + More

Then More = John - \(\frac{5}{6}\) John - \(\frac{25}{36}\) John = \(-\frac{19}{36}\) John ---- > which is a negative value contradicting with the mentioned fact that John has more coins than both of Tom and Jane

Am I wrong?

(if so, changing the '20%'s to '200%'s would make it more plausible)
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Re: John has more coins than the number of coins both Tom and Jane have   [#permalink] 14 Jan 2019, 00:43
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