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# Three dice, each with faces numbered from 1 through 6, were

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Three dice, each with faces numbered from 1 through 6, were [#permalink]

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06 Jul 2013, 02:39
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Three dice, each with faces numbered from 1 through 6, were tossed onto a game board. If one of the dice turned up 4, what is the sum of the numbers that turned up on all three dice?

1) The sum of the two of the numbers that turned up was 10
2) The sum of two of the numbers that turned up was 11
[Reveal] Spoiler: OA

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Last edited by fozzzy on 06 Jul 2013, 02:47, edited 1 time in total.

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Re: Three dice, each with faces numbered from 1 through 6, were [#permalink]

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06 Jul 2013, 02:45
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Three dice, each with faces numbered from 1 through 6, were tossed onto a game board. If one of the dice turned up 4, what is the sum of the numbers that turned up on all three dice?

(1) The sum of the two of the numbers that turned up was 10. The sum can be 4 + (5 + 5) = 14 or (4 + 6) + any > 10. Not sufficient.

(2) The sum of two of the numbers that turned up was 11 --> x+y=11 --> the first dice turned up 4, so it can be neither x nor y (because the other one will be 7, so more than 6). Thus the sum is 4 + (x + y) =15. Sufficient.

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Re: Three dice, each with faces numbered from 1 through 6, were [#permalink]

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08 Jul 2013, 20:39
This was a trickey one...

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Re: Three dice, each with faces numbered from 1 through 6, were [#permalink]

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08 Jul 2013, 21:30
Let there be three numbers D1,D2 and D3
D1=4
There are two possibilties for statement one...
(5,5,4) and (4,6,anything)
If D1=4 and D2=6 then D3 can be anything and we cannot find the sum.
Hence statement 1 is insufficient.

For statement 2...there is only one combination for making 11 i.e. (4,5,6)
Here we have two combinations
1. D1=4 D2=5 D3=6
2 D1=4 D2=6 D3=5
but since in both cases the sum is 15...hence statement 2 is sufficient.

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Re: Three dice, each with faces numbered from 1 through 6, were [#permalink]

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09 Jul 2013, 03:02
AMITAGARWAL2 wrote:
This was a trickey one...

The question conditions you to think of one case and ignores the case as provided by bunuel I fell for the trap answer choice.
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Re: Three dice, each with faces numbered from 1 through 6, were [#permalink]

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04 Mar 2014, 04:02
From question: One Dice was 4.

(1) Two of the dice add up to 10. Hence it could be the original 4 + 6 of the second dice and any number on the third dice. IS.
(2) Original: 4. Thus it is impossible for this Dice to be one of the dice which give us the sum 11. hence we have first dice 4 and the other 2 dice 5 and 6. Suff. B.

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Re: Three dice, each with faces numbered from 1 through 6, were [#permalink]

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04 Mar 2014, 05:35
Pretty straightforward.
From S1:We could have diff combinations like 4,6,2 or 4,6,3 that'd give us a different answer for sum of all three.Not suff.

From S2:If 2 numbers add up to be 11 the dice with 4 on top wont be there in these 2.So it's only possible that sum of nos. on other 2 dice is 11 in which case total of all three is 15.Suff.

Option B.

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Re: Three dice, each with faces numbered from 1 through 6, were [#permalink]

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18 May 2014, 09:50
When testing (1), you have two possible answers: you rolled the identified 4 and a 6 and an unknown 3rd die or you rolled the identified 4 and a 5 and a 5. (1) is insufficient to answer to sum of all 3 dice.

When testing (2), you only have one possible answer: the maximum value of each die is 6 so the sum of the two numbers that equal 11 cannot include the identified 4 die because there are no 7 value dice. So the only combination of two dice that sums 4 is 5 and 6. Because we don't care which die is 5 or 6, only in the summed value, we can answer the original question with only statement (2) thus the answer would be B.

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Re: Three dice, each with faces numbered from 1 through 6, were [#permalink]

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01 Jun 2015, 08:01
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Re: Three dice, each with faces numbered from 1 through 6, were   [#permalink] 01 Jun 2015, 08:01
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# Three dice, each with faces numbered from 1 through 6, were

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