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# A game is played with a six sided, regularly numbered fair die. A play

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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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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 even-number 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 odd-number 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 sub-total of -0.9.
In total, for this scenario, he will get 1.1-0.9 = 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 max-min 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]
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]
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. https://www.gmatquantum.com/og13/218-pro ... 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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Really strange question. Can someone confirm that this is indeed a GMATPrep question?
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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
Hello,

Nice to improve your reading skills and comprehension skills. I spent 1 min on reading this question and got answrer in 10 secs.

Condition says the following

1.Ifnumber on the roll of the dice * 0.1$$\geq{current score}$$, then increase the score of player by .1

2. If number on the roll of the dice * 0.1 < current score, then decrease the score of player by .1

3. If number on the roll of the dice * 0.1 is even and is < current score, then no change in score of player.

So lets assume that say initially the score of the player is .1*3= .3
$$S_0$$=.3

On the first throw of the dice we get 2. so this is .1*2 = .2 and is even , So we have condition 3 here hence the score remains unchanged.
So $$S_1$$=.3

Now second throw we have 1, so it will be 0.1*1 = .1 and we see condition 2 coming into effect. So score will decrease by .1 and will be .2
$$S_2$$=.2

Now third throw we have 1, so it will be 0.1*1 = .1 and we see condition 2 coming into effect. So score will decrease by .1 and will be .1
$$S_3$$=.1

Now forth throw we have 1, so it will be 0.1*1 = .1 and we see condition 1 coming into effect. So score will increase by .1 and will be .2
$$S_4$$=.2

Clearly lowest possible is 0.1

Now assume that during some pint in the game the players score was .6, and on the throw of the dice he gets 6 so 6*.1\geq{.6}, so new score will be .7
Now on the next throw what ever number comes on the throw of the dice the score will either decrease or remain unchanged. . Why so? Because for a number to increase the throw of the dice *.1 should be $$\geq{current score}$$ . but this is possible only if we get 7 on throw of the dice. and getting 7 is not possible ( fair dice) except even the score will decrease

So maximum score possible is .7

So Max score is .7 and min is .1 so only option D
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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
chetan2u wrote:
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! This solution looks good, but it appears you did not make any use of the following information:

55% of the die rolls in a particular game are even

Is this information not needed at all or did you just not use it?
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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
Hey guys

It seems that I've misread the information: ..."A player starts with a number equal to 0.1n, where n is an integer between 1 and 6 inclusive..." I assumed that n was a tenths digit. Is it usually written the way it's been written in the question? Or if it doesn't explicitly mention that n is some sort of digit, I need to assume that in that cases it's multiplication what's given?

Thanks

chetan2u Bunuel
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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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Re: A game is played with a six sided, regularly numbered fair die. A play [#permalink]
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