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# A die with x sides has consecutive integers on its sides. If

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VP
Joined: 13 Jun 2004
Posts: 1119
Location: London, UK
Schools: Tuck'08
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A die with x sides has consecutive integers on its sides. If [#permalink]  06 Sep 2004, 18:06
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A die with x sides has consecutive integers on its sides. If the probability of NOT getting a 4 on either of two tosses is 36/49, how many sides does the die have?

(A) 4
(B) 5
(C) 7
(D) 8
(E) 13
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5068
Location: Singapore
Followers: 23

Kudos [?]: 194 [0], given: 0

Hi,

could you explain the logic behind this working:

(x-1)/x = sqrt 36/49
or x = 7
Director
Joined: 20 Jul 2004
Posts: 593
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Kudos [?]: 54 [0], given: 0

ywilfred wrote:
Hi,

could you explain the logic behind this working:

(x-1)/x = sqrt 36/49
or x = 7

for a die with x sides, to not get a particular number (say 4)
= 1 - [1(x-1) + (x-1)1 + 1)] / x^2
= (x-1)^2/x^2

=> (x-1)^2/x^2 = 36/49
and n = 7

But, a very quicker way, I saw "49" in the denominator and "two rolls" (which requires x^2 in denominator), so decided it should be 7.
Manager
Joined: 16 Jul 2003
Posts: 71
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7

let x=no of sides
P(no 4 on either toss) = P(no 4 on 1st toss) * P(no 4 on 2nd toss)
=[1-P(4 on 1st toss)] * [1-P(4 on 2nd toss)]
= [1- 1/x] * [1-1/x]

36/49 = [1-1/x]^2
=> x=7
VP
Joined: 13 Jun 2004
Posts: 1119
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 31 [0], given: 0

yep, no need for me to repeat what you've all said very well on previous posts...but on this one plug in the answers seems to be the fastest way.

I confirm OA is C
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