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If Q is an odd number and the median of Q consecutive intege
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19 Aug 2020, 12:04
19 Aug 2020, 12:04
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Question Stats:
80% (01:46) correct
20% (02:00) wrong based on 50 sessions
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If Q is an odd number and the median of Q consecutive integers is 120, what is the largest of these integers?
(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2
(A) (Q - 1)/2 + 120
(B) Q/2 + 119
(C) Q/2 + 120
(D) (Q + 119)/2
(E) (Q + 120)/2
Re: If Q is an odd number and the median of Q consecutive intege
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19 Aug 2020, 12:04
19 Aug 2020, 12:04
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Re: If Q is an odd number and the median of Q consecutive intege
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19 Aug 2020, 12:37
19 Aug 2020, 12:37
2
I took a simple value for Q to see what happened.
1st try: Q=3. The median is 120 and given that it's a set of an odd number of elements, we know that 120 is the second highest number. Given that the set is made of consecutive numbers, and that there are three numbers overall, we know that the highest number will be 121.
Looking at the answers, we see that we can write 121 = 120 + 1 = 120 + (3-1)/2 and we rule out the others.
2nd try: to check whether it's correct. Q=5. The median is still 120 and this will be the third highest number. Since it's a set of consecutive numbers, and since there are 5 numbers overall in this set, we know the highest number will be 122 = 120 + 2 = 120 + (5-1)/2
So Answer A works.
1st try: Q=3. The median is 120 and given that it's a set of an odd number of elements, we know that 120 is the second highest number. Given that the set is made of consecutive numbers, and that there are three numbers overall, we know that the highest number will be 121.
Looking at the answers, we see that we can write 121 = 120 + 1 = 120 + (3-1)/2 and we rule out the others.
2nd try: to check whether it's correct. Q=5. The median is still 120 and this will be the third highest number. Since it's a set of consecutive numbers, and since there are 5 numbers overall in this set, we know the highest number will be 122 = 120 + 2 = 120 + (5-1)/2
So Answer A works.
