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# Three fair dice were thrown simultaneously. One of the

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Manager
Joined: 28 Feb 2003
Posts: 100
Followers: 1

Kudos [?]: 3 [0], given: 0

Three fair dice were thrown simultaneously. One of the [#permalink]  19 May 2003, 22:35
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Three fair dice were thrown simultaneously. One of the number was 4. Find the sum of all the three numbers.

1) the sum of two of the numbers was 10
2) the sum of two of the numbers was 11
Manager
Joined: 28 Feb 2003
Posts: 100
Followers: 1

Kudos [?]: 3 [0], given: 0

B it is

was it easy to spot the trick?
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 78 [0], given: 0

About 2 minutes. You were right by naming the second set as the second one. If it were the first, it would be easier to crack.

Other people are invited to discuss. The problem is interesting!
Manager
Joined: 12 Mar 2003
Posts: 59
Followers: 1

Kudos [?]: 1 [0], given: 0

What's the trick ?

only 2 numbers on a dice can add to 11 ?
Manager
Joined: 12 Mar 2003
Posts: 59
Followers: 1

Kudos [?]: 1 [0], given: 0

for 2) 4 cannot be one of the 2 dice as 6 is max. therefore sum of three is 15 ?
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 78 [0], given: 0

Yes, the trick is that 6 is a maximal number.
SVP
Joined: 14 Dec 2004
Posts: 1707
Followers: 1

Kudos [?]: 52 [0], given: 0

Fortunately, I had a dice in mind while thinking on this.

"B" it is!
Manager
Joined: 12 Feb 2006
Posts: 71
Followers: 1

Kudos [?]: 2 [0], given: 0

i think i'm missing the trick

if one of the numbers is 4..

1) sum is 10 ( can't it be two 5's??
2) sum is 11 ( 6 & 5)

Can someone explain this
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