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A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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28 Dec 2015, 23:45
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A game is played with a six sided, regularly numbered fair die. A player starts with a number equal to 0.1n, where n is an integer between 1 and 6 inclusive. On each of 20 subsequent rolls, if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1; if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1; if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected. If 55% of the die rolls in a particular game are even, which of the following is a possible final value of that game i. 0.8 ii. 0.5 iii. 0.1 A. i only B. i and ii only C. i and iii only D. ii and iii only E. i, ii and iii
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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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28 Dec 2015, 23:58
ppb1487 wrote: A game is played with a six sided, regularly numbered fair die. A player starts with a number equal to 0.1n, where n is an integer between 1 and 6 inclusive. On each of 20 subsequent rolls, if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1; if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1; if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected. If 55% of the die rolls in a particular game are even, which of the following is a possible final value of that game
i. 0.8 ii. 0.5 iii. 0.1
A. i only B. i and ii only C. i and iii only D. ii and iii only E. i, ii and iii Hi, the Q may look very lengthy and full of values.. But here we can eaily find a solution if we understand the concepts.. three scenarios.. 1) if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1 2) if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1 3) if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected... then there are some %s given... What can we finally deduct.. THE MAX VALUE... YES, the max value can be 0.7... reason: after a person has got 0.6, there is only one value, which can increase 0.6 and that is by throwing 6.. this will take the value to 0.7 but cannot increase it further as there is no 7 on dice..so 0.8 is out.. only D is without (i)ans D Hope it was clear
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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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02 Nov 2017, 04:16
chetan2u wrote: ppb1487 wrote: A game is played with a six sided, regularly numbered fair die. A player starts with a number equal to 0.1n, where n is an integer between 1 and 6 inclusive. On each of 20 subsequent rolls, if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1; if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1; if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected. If 55% of the die rolls in a particular game are even, which of the following is a possible final value of that game
i. 0.8 ii. 0.5 iii. 0.1
A. i only B. i and ii only C. i and iii only D. ii and iii only E. i, ii and iii Hi, the Q may look very lengthy and full of values.. But here we can eaily find a solution if we understand the concepts.. three scenarios.. 1) if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1 2) if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1 3) if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected... then there are some %s given... What can we finally deduct.. THE MAX VALUE... YES, the max value can be 0.7... reason: after a person has got 0.6, there is only one value, which can increase 0.6 and that is by throwing 6.. this will take the value to 0.7 but cannot increase it further as there is no 7 on dice..so 0.8 is out.. only D is without (i)ans D Hope it was clear Hi chetan2u, I still don't get it why we could eliminate 0.8 and keep the other 2 options. Below are my reasons: 1. For 0.8: let's say that person starts with 0.6 (the highest point). He then throws the dice 20 times. As 55% of the rolls will be even (given as a condition), 11 of his rolls will be even and the remaining 9 will be odd. For the evennumber rolls, it's possible that all these 11 even rolls, he will get a 6 for each roll (yeah, this is quite extreme but still a possibility). In this case, he will get 1.1 points for all these 11 rolls. For the oddnumber rolls, clearly all of these rolls will be less than 6, so he will get 0.1 for each of these rolls, making the the subtotal of 0.9. In total, for this scenario, he will get 1.10.9 = 0.2 for 20 rolls, making his final score to be 0.2 + 0.6 = 0.8! How can we eliminate this value? 2. For the other two values (0.5 & 0.1), how can we be sure that there will be a combination of increases/decreases after 20 rolls to arrive at those points? Within a few minutes, I feel puzzled to verify these two values. Just because these two values are within the maxmin range of the possible scores, would it be safe to conclude that these two are two 'possible' values? We have some conditions that we must abide by, so I think making such a conclusion would be much of a guess. Thanks.



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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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02 Nov 2017, 05:18
romeotc wrote: chetan2u wrote: ppb1487 wrote: A game is played with a six sided, regularly numbered fair die. A player starts with a number equal to 0.1n, where n is an integer between 1 and 6 inclusive. On each of 20 subsequent rolls, if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1; if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1; if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected. If 55% of the die rolls in a particular game are even, which of the following is a possible final value of that game
i. 0.8 ii. 0.5 iii. 0.1
A. i only B. i and ii only C. i and iii only D. ii and iii only E. i, ii and iii Hi, the Q may look very lengthy and full of values.. But here we can eaily find a solution if we understand the concepts.. three scenarios.. 1) if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1 2) if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1 3) if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected... then there are some %s given... What can we finally deduct.. THE MAX VALUE... YES, the max value can be 0.7... reason: after a person has got 0.6, there is only one value, which can increase 0.6 and that is by throwing 6.. this will take the value to 0.7 but cannot increase it further as there is no 7 on dice..so 0.8 is out.. only D is without (i)ans D Hope it was clear Hi chetan2u, I still don't get it why we could eliminate 0.8 and keep the other 2 options. Below are my reasons: 1. For 0.8: let's say that person starts with 0.6 (the highest point). He then throws the dice 20 times. As 55% of the rolls will be even (given as a condition), 11 of his rolls will be even and the remaining 9 will be odd. For the evennumber rolls, it's possible that all these 11 even rolls, he will get a 6 for each roll (yeah, this is quite extreme but still a possibility). In this case, he will get 1.1 points for all these 11 rolls. For the oddnumber rolls, clearly all of these rolls will be less than 6, so he will get 0.1 for each of these rolls, making the the subtotal of 0.9. In total, for this scenario, he will get 1.10.9 = 0.2 for 20 rolls, making his final score to be 0.2 + 0.6 = 0.8! How can we eliminate this value? 2. For the other two values (0.5 & 0.1), how can we be sure that there will be a combination of increases/decreases after 20 rolls to arrive at those points? Within a few minutes, I feel puzzled to verify these two values. Just because these two values are within the maxmin range of the possible scores, would it be safe to conclude that these two are two 'possible' values? We have some conditions that we must abide by, so I think making such a conclusion would be much of a guess. Thanks. hi.. the scenario you have said is possible but the counting is not happening that way... the counting is happening.. 1) if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1 2) if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1 3) if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected... everything is happening on the players CURRENT NUMBER and what is the players current number initially, say 0.1*n, 0.1*6=0.6... what has he to put to increase this number.. 1) if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1 what can be the max rolled number, 6 so it becomes 0.1*6=o.6 this is EQUAL to the players current number 0.6, so players CURRENT number becomes 0.7 But after this to increase it, the number rolled has to be greater than or equal to current number 0.7 this is possible ONLY if number thrown is 7 or above, only then 0.7 or 0.8 possible but the numbers on dice is only till 6, so you will never be able to increase beyond 0.7 hope it helps.. you have to know CURRENT number here
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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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05 Dec 2017, 07:12
chetan2u wrote: ppb1487 wrote: A game is played with a six sided, regularly numbered fair die. A player starts with a number equal to 0.1n, where n is an integer between 1 and 6 inclusive. On each of 20 subsequent rolls, if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1; if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1; if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected. If 55% of the die rolls in a particular game are even, which of the following is a possible final value of that game
i. 0.8 ii. 0.5 iii. 0.1
A. i only B. i and ii only C. i and iii only D. ii and iii only E. i, ii and iii Hi, the Q may look very lengthy and full of values.. But here we can eaily find a solution if we understand the concepts.. three scenarios.. 1) if the number rolled times 0.1 is greater than or equal to the players current number, the players current number is incremented by 0.1 2) if the number rolled times 0.1 is less than the player's current number and is odd, the players number is decremented bu 0.1 3) if the number rolled times 0.1 is less than the player's current number and is even, the players number is unaffected... then there are some %s given... What can we finally deduct.. THE MAX VALUE... YES, the max value can be 0.7... reason: after a person has got 0.6, there is only one value, which can increase 0.6 and that is by throwing 6.. this will take the value to 0.7 but cannot increase it further as there is no 7 on dice..so 0.8 is out.. only D is without (i)ans D Hope it was clear Hi. During your GMAT test, did you get questions of this level? This is a really hard question.
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A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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17 Dec 2017, 07:18
I did this problem like this, 20*55/100=11 even and 9 odd. A dice has 6 possibilities. 1,2,3,4,5,6. Odd: 1,3,5 Even: 2,4,6 to make it easy, lets assume that for every 20 selections, we only get the same odd number and the same even number. hence, 1*9=9*.1=.9 2*11=22*.1=2.2 3*9=27*.1=2.7 4*11=44*.1=4.4 5*9=45*.1=4.5 6*11=66*.1=6.6 Using this, since all the odd's are less than the number selected, they are reduced by .1 and since all the even's are more than the number selected, they are increased by .1 we get, .8,2.6,4.5(odd) 2.3,4.5,6.7(even) Adding all possibilities, we only get a difference of .1,.3,.5,.7 or .9 hence it removes .8 as a possibility. ans: D you can use this link as a reference. http://www.gmatquantum.com/og13/218pro ... ition.html



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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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21 Feb 2018, 22:38
Really strange question. Can someone confirm that this is indeed a GMATPrep question?




Re: A game is played with a six sided, regularly numbered fair die. A play
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21 Feb 2018, 22:38






