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# A bag contains at least 10 blue marbles and at least 10 red marbles,

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Manager
Joined: 20 Feb 2017
Posts: 164
Location: India
Concentration: Operations, Strategy
WE: Engineering (Other)
A bag contains at least 10 blue marbles and at least 10 red marbles,  [#permalink]

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14 Aug 2018, 20:23
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42% (02:25) correct 58% (02:14) wrong based on 30 sessions

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A bag contains at least 10 blue marbles and at least 10 red marbles, and contains marbles of no other color. Is the number of blue marbles in the bag greater than the number of red marbles?
1) If two marbles are chosen from the bag without replacement, the probability of picking two blue marbles is higher than the probability of picking one blue marble and one red marble.
2) If two marbles are chosen from the bag without replacement, the probability of picking one blue marble and one red marble is higher than the probability of picking two red marbles.

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Math Expert
Joined: 02 Aug 2009
Posts: 7764
A bag contains at least 10 blue marbles and at least 10 red marbles,  [#permalink]

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14 Aug 2018, 22:02
3
1
A bag contains at least 10 blue marbles and at least 10 red marbles, and contains marbles of no other color. Is the number of blue marbles in the bag greater than the number of red marbles?

Let the numbers be B and R..

1) If two marbles are chosen from the bag without replacement, the probability of picking two blue marbles is higher than the probability of picking one blue marble and one red marble.
You can straight way answer sufficient as blue will be more.
This means $$\frac{B}{B+R}*\frac{B-1}{B+R-1}>\frac{B}{B+R}*\frac{R}{B+R-1}*2$$..
Thus B-1>2R.....
Sufficient

2) If two marbles are chosen from the bag without replacement, the probability of picking one blue marble and one red marble is higher than the probability of picking two red marbles.
This means $$\frac{R}{B+R}*\frac{R-1}{B+R-1}<\frac{B}{B+R}*\frac{R}{B+R-1}*2$$..
Thus R-1<2B....
We cannot conclude anything about B and R from this
If R =12, and B=30.....yes
If R=20, and B =30.....no
Insufficient

A
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Re: A bag contains at least 10 blue marbles and at least 10 red marbles,  [#permalink]

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14 Aug 2018, 22:49
chetan2u wrote:
A bag contains at least 10 blue marbles and at least 10 red marbles, and contains marbles of no other color. Is the number of blue marbles in the bag greater than the number of red marbles?

Let the numbers be B and R..

1) If two marbles are chosen from the bag without replacement, the probability of picking two blue marbles is higher than the probability of picking one blue marble and one red marble.
You can straight way answer sufficient as blue will be more.
This means $$\frac{B}{B+R}*\frac{B-1}{B+R-1}>\frac{B}{B+R[/fraction]*[fraction]R/B+R-1}*2$$..
Thus B-1>2R.....
Sufficient

2) If two marbles are chosen from the bag without replacement, the probability of picking one blue marble and one red marble is higher than the probability of picking two red marbles.
This means $$\frac{R}{B+R}*\frac{R-1}{B+R-1}<\frac{B}{B+R}*\frac{R}{B+R-1}*2$$..
Thus R-1<2B....
We cannot conclude anything about B and R from this
If R =12, and B=30.....yes
If R=20, and B =30.....no
Insufficient

A

Sir, in the first statement calculation for B & R being picked.. why do you multiply by 2?

Can you please tell me the reason, i am getting confused.

Thanks.
Math Expert
Joined: 02 Aug 2009
Posts: 7764
Re: A bag contains at least 10 blue marbles and at least 10 red marbles,  [#permalink]

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14 Aug 2018, 22:57
duldtool wrote:
chetan2u wrote:
A bag contains at least 10 blue marbles and at least 10 red marbles, and contains marbles of no other color. Is the number of blue marbles in the bag greater than the number of red marbles?

Let the numbers be B and R..

1) If two marbles are chosen from the bag without replacement, the probability of picking two blue marbles is higher than the probability of picking one blue marble and one red marble.
You can straight way answer sufficient as blue will be more.
This means $$\frac{B}{B+R}*\frac{B-1}{B+R-1}>\frac{B}{B+R[/fraction]*[fraction]R/B+R-1}*2$$..
Thus B-1>2R.....
Sufficient

2) If two marbles are chosen from the bag without replacement, the probability of picking one blue marble and one red marble is higher than the probability of picking two red marbles.
This means $$\frac{R}{B+R}*\frac{R-1}{B+R-1}<\frac{B}{B+R}*\frac{R}{B+R-1}*2$$..
Thus R-1<2B....
We cannot conclude anything about B and R from this
If R =12, and B=30.....yes
If R=20, and B =30.....no
Insufficient

A

Sir, in the first statement calculation for B & R being picked.. why do you multiply by 2?

Can you please tell me the reason, i am getting confused.

Thanks.

Hi...

All blue marbles are identical and all red are identical..
So when I pick one blue after the other blue the order does not matter...
But if I pick - B and then R - or - R and then B, the order will be different
That is why we multiply by 2

Hope it helped
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Re: A bag contains at least 10 blue marbles and at least 10 red marbles,   [#permalink] 14 Aug 2018, 22:57
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