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01 Oct 2021, 07:45
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
89% (01:11) correct
11%
(01:21)
wrong
based on 38
sessions
History
Date
Time
Result
Not Attempted Yet
If A = {products of 2 and an odd number} and B = {even multiples of 3}, which of the following numbers is in the intersection of A and B?
I. 6
II. 18
III. 30
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
I. 6
II. 18
III. 30
A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
04 Oct 2021, 00:11
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(I). 6: 6/3 = 2. Will be in both A and B.
(II). 18: 18/3 = 6. Will be in both A and B.
(III). 30: 30/3 = 10. Will be in both A and B.
Answer E
(II). 18: 18/3 = 6. Will be in both A and B.
(III). 30: 30/3 = 10. Will be in both A and B.
Answer E
04 Oct 2021, 01:02
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Bunuel
Solution:
Set A = {products of 2 and an odd number}. This means this set contains numbers like:
\(2\times 1=2\)
\(2\times 3=6\)
\(2\times 5=10\) and so on.
Set B = {even multiples of 3}. This means this set contains numbers like:
\(3\times 2=6\)
\(3\times 4=12\)
\(3\times 6=18\) and so on.
Now let us look into the numbers given to us:
1. 6: This will be in both sets. We can see above it is the \(2^{nd}\) element of set A and \(1^{st}\) of set B.
2. 18: This will also be in both sets. in set A there will be an element \(2\times 9=18\) (\(5^{th}\) element) and we see it is \(3^{rd}\) element of set B.
3. 30: This will also be in both sets. in set A there will be an element \(2\times 15=30\) (\(8^{th}\) element) and in set B there will be an element \(3\times 10=30\) (\(5^{th}\) element).
We see that all 3 numbers are in both sets.
Hence the right answer is Option E.
