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Bunuel
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ritwikkg
A and B have 7 and 8 common. So one of them has to be the number common to all three sets.

1. 7 is not in C -> leaves us with 8, SUFFicient

2. 8 is in C -> confirm that it's 8, SUFFicient


Hence D.

In (1), what if C contains just one number of 9?
In (2), what if C contains 8 and 7?
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ritwikkg
A and B have 7 and 8 common. So one of them has to be the number common to all three sets.

1. 7 is not in C -> leaves us with 8, SUFFicient

2. 8 is in C -> confirm that it's 8, SUFFicient


Hence D.

In (1), what if C contains just one number of 9?
In (2), what if C contains 8 and 7?

whoops! got it now, thanks!
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zhanbo ritwikkg
Sorry but as part of statement 1, how is C going to contain 9? 9 isn't present in either of the sets A and B. And the stimulus states that x is a number common in all three sets so x could be either 7/8/both 7 and 8.
Since statement 1 says that 7 is not there in the set C, it must be 8. It cannot be 0/1/2 since those numbers are not present in sets A and B and thus we would be violating the condition of the stimulus. IMO answer will be A.
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zhanbo ritwikkg
Sorry but as part of statement 1, how is C going to contain 9? 9 isn't present in either of the sets A and B. And the stimulus states that x is a number common in all three sets so x could be either 7/8/both 7 and 8.
Since statement 1 says that 7 is not there in the set C, it must be 8. It cannot be 0/1/2 since those numbers are not present in sets A and B and thus we would be violating the condition of the stimulus. IMO answer will be A.

We are only told: Set C is another set.

It can contain anything from none (an empty set) to a single number of 9 to everything in the universe including you and me.
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zhanbo ritwikkg
Sorry but as part of statement 1, how is C going to contain 9? 9 isn't present in either of the sets A and B. And the stimulus states that x is a number common in all three sets so x could be either 7/8/both 7 and 8.
Since statement 1 says that 7 is not there in the set C, it must be 8. It cannot be 0/1/2 since those numbers are not present in sets A and B and thus we would be violating the condition of the stimulus. IMO answer will be A.

We are only told: Set C is another set.

It can contain anything from none (an empty set) to a single number of 9 to everything in the universe including you and me.

You are right, C is an another set,but it cannot be an empty set since we are told that x is common across all three sets.So if it was an empty set, it would not have any common factor, but it also has to have a common factor. And the only two common factors, it can have are 7 and 8

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zhanbo
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zhanbo ritwikkg
Sorry but as part of statement 1, how is C going to contain 9? 9 isn't present in either of the sets A and B. And the stimulus states that x is a number common in all three sets so x could be either 7/8/both 7 and 8.
Since statement 1 says that 7 is not there in the set C, it must be 8. It cannot be 0/1/2 since those numbers are not present in sets A and B and thus we would be violating the condition of the stimulus. IMO answer will be A.

We are only told: Set C is another set.

It can contain anything from none (an empty set) to a single number of 9 to everything in the universe including you and me.

You are right, C is an another set,but it cannot be an empty set since we are told that x is common across all three sets.So if it was an empty set, it would not have any common factor, but it also has to have a common factor. And the only two common factors, it can have are 7 and 8

Posted from my mobile device

You are right.

I misread the sentence. I though x represents the number of common elements across all three sets.

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Bunuel
\(A = {2, 4, 7, 8}\)

\(B = {3, 6, 7, 8}\)

\(A\) and \(B\) are two sets containing the elements shown above, and \(C\) is another set. If \(x\) is a number common in all the three sets, what is the value of \(x\) ?

(1) 7 is not in set \(C\).

(2) 8 is in set \(C\).
x is common in A, B and C. What is x?

Statement 1 - 7 is not in set C.

The only number other than 7 which is common in A and B is 8, so x needs to be 8.

Hence, this statement is sufficient

Statement 2 - 8 is in set C.

Beware of the in and not in trap here. When something is in a set, doesn't guarantee if anything else would or would not be in that set, which means 7 can be or cannot be in set C.

Let's say set C is {2, 4, 8}, then x = 8
but if set C is {2, 4, 7, 8}, then we cannot get a unique value for x, as x could be either 7 or 8.

Hence, this statement is not sufficient.

Answer: A
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Bunuel
\(A = {2, 4, 7, 8}\)

\(B = {3, 6, 7, 8}\)

\(A\) and \(B\) are two sets containing the elements shown above, and \(C\) is another set. If \(x\) is a number common in all the three sets, what is the value of \(x\) ?

(1) 7 is not in set \(C\).

(2) 8 is in set \(C\).


Are You Up For the Challenge: 700 Level Questions
Visualizing the problem can be useful, so lets to put the sets in columns:
A = {2, 4, 7, 8}
B = {3, 6, 7, 8}
C = {?, ?, ?, ?}

Now we can clearly see the only possible values of X are either 7 or 8 as no other number is common in both sets A and B.

(1) 7 is not in set C --> this leaves us with only one possible value: 8. Therefore its SUFFICIENT
(2) 8 in in set C --> this does not imply 7 is not in set C also, therefore we have two possible values and no sure answer. NOT SUFFICIENT.


Answer A.
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