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AMahmoodi
rephrasing the question: is R>=2?

st1. if 9 books either red or blue we don't know exactly how many are red. insuff.
st2. if 7 books are either blue or yellow it imply that 3 are red so it is sufficient.

B

how can you be sure that there are only 3 colours ?
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AMahmoodi
rephrasing the question: is R>=2?

st1. if 9 books either red or blue we don't know exactly how many are red. insuff.
st2. if 7 books are either blue or yellow it imply that 3 are red so it is sufficient.

B

how can you be sure that there are only 3 colours ?

hmm. you are absolutely right.

then It must be 1&2 combined. from 2 if we assume that there is just 1 yellow book then there should be 6 blue books to add up to 7. then if there are 6 blue books from st1 we can conclude that there are 3 red books. and then we can conclude that there is 3 or more red books. suff
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N is the number of Red books, so we are asked whether N is greater or equal to 2

Statement 1: Can be 8 red, 1 blue and 1 other; in this case the answer is yes. But can be also 1 red, 8 blue and 1 other; in this case the anwer is no. Insufficient
Statement 2: Provide no information about red books, so insufficient

Taking both Statements together the answer is no. From Statement 2 we know that maximum value of blue books is 6 and combining this with Statement 1 we can infer that the number of red books must be at least 3. Sufficient
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chetan2u
There are 10 books of different colours lying on a table. Is the probability of picking a red book more than \(\frac{1}{10}\)?

1) there are 9 books which are either Red or Blue.
2) there are 7 books which are either Blue or Yellow.

chetan2u please can you explain. I thought answer is E.
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chetan2u
There are 10 books of different colours lying on a table. Is the probability of picking a red book more than \(\frac{1}{10}\)?

1) there are 9 books which are either Red or Blue.
2) there are 7 books which are either Blue or Yellow.

(1) INSUFFICIENT. red books could be any number ranging from 1 to 8.
(2) INSUFFICIENT. red books could be any number ranging from 0 to 3.

(1)+(2) SUFFICIENT.
From (1),there could be only one book with different color from red and blue.
From (2),we have yellow.Hence,we have 1 yellow,6 blue and 3 red.
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How do we know that there are only 3 colors present in the population? If there was one other color wouldn't that make the result choice E?
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emockus
How do we know that there are only 3 colors present in the population? If there was one other color wouldn't that make the result choice E?

Hi,from statement 1 we know that the total number of red and blue is nine.
Statement 2 introduce another color-yellow.
So far,we have 10 books,nine is either red or blue.Hence,there is only a room left for yellow book.
If we add another color,then the total number of books must be greater than 10.
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statement 1: there are 9 either red or blue books. So to satisfy the condition if the table consists of 1 red book and 8 blue books.The probability of picking red book is 1/10
Then answer to the question is no.
.But if the number of red books become 2,3 or so on then the probability will be 1/5,3/10 ,so on. Then the answer to the question is yes.
Hence the statement 1 is insufficient.

statement 2: the statement doesn't say anything about the number of red books.

combining statement 1&2 there are 9 red or blue books and 7 blue or yellow books. Now the maximum number of blue books that can be on the table is 7 hence there will be minimum of 2 red books on the table. hence the probability of picking up red books will be minimum 2/10 = 1/5 >1/10.So sufficient.
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chetan2u
There are 10 books of different colours lying on a table. Is the probability of picking a red book more than \(\frac{1}{10}\)?

1) there are 9 books which are either Red or Blue.
2) there are 7 books which are either Blue or Yellow.


I dont understand what i am missing :S . If there are 10 books of different colours how can R+B=9, B+Y=7?
Shouldn't there be 10 different colours?
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chetan2u
There are 10 books of different colours lying on a table. Is the probability of picking a red book more than \(\frac{1}{10}\)?

1) there are 9 books which are either Red or Blue.
2) there are 7 books which are either Blue or Yellow.


I dont understand what i am missing :S . If there are 10 books of different colours how can R+B=9, B+Y=7?
Shouldn't there be 10 different colours?

Please understand that it is given that the number of books are 10 but they have different colours. What I construed is that those 10 might have 3,4,5 or more different colours. Details regarding their colours can be inferred from the statements provided.
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10 books of different color then statement says 9 books of red or blue
Question need rephrasing
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RS81
10 books of different color then statement says 9 books of red or blue
Question need rephrasing


Different colours could mean 2 colours or 3 colours etc.
if 9 are either B or R, that means the 10th book is of a different colour, neither red nor blue.

When you combine, you will get yellow as 1, 6 are blue and remaining 3 are red.
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chetan2u
There are 10 books of different colours lying on a table. Is the probability of picking a red book more than \(\frac{1}{10}\)?

1) there are 9 books which are either Red or Blue.
2) there are 7 books which are either Blue or Yellow.

From 1
R+B=9
From 2
B+Y=7
From 1 and 2
R-Y=2
Y=R-2
So total will be
R+9-R+R-2=10
R=3
chetan2u
I think B=7 is not possible as the option 2 says "there are 7 books which are either Blue or Yellow"
there must be some Yellow books. Correct me please
Though C is the answer
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chetan2u
There are 10 books of different colours lying on a table. Is the probability of picking a red book more than \(\frac{1}{10}\)?

1) there are 9 books which are either Red or Blue.
2) there are 7 books which are either Blue or Yellow.

From 1
R+B=9
From 2
B+Y=7
From 1 and 2
R-Y=2
Y=R-2
So total will be
R+9-R+R-2=10
R=3
chetan2u
I think B=7 is not possible as the option 2 says "there are 7 books which are either Blue or Yellow"
there must be some Yellow books. Correct me please
Though C is the answer


Yes, you are correct.
My post right above yours talks of final solution as R=3, B=6 and Y=1.

The initial solution was just to say that you don’t require to solve as we can be sure that probability >1/10
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How can we conclude that there are only red,blue and yellow books.

Posted from my mobile device
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Wording on this question is not good...The stem says 10 books are each of a different colour...not worth pursuing any further. Red simply cannot be greater than 1.
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chetan2u

Hi Chetan,

i love your questions but I have my doubts on this one:

(1) EITHER Red or Blue <-> If it is not red, it must be blue AND If it is not blue, it must be red. We can therefore assume that all books could be blue.

(2) Again, we can assume all books to be blue

(1) and (2): Only blue books
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