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Sets X and Y consist of the same number of positive integers. The inte
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16 Oct 2023, 09:33
16 Oct 2023, 09:33
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Sets X and Y consist of the same number of positive integers. The integers in X are consecutive even integers, and the integers in Y are consecutive odd integers, such that the sum of the integers in X is greater than the sum of the integers in Y.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Median of the integers in X |
Average (arithmetic mean) of the integers in Y |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: Sets X and Y consist of the same number of positive integers. The inte
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08 Dec 2023, 15:47
08 Dec 2023, 15:47
Expert Reply
Re: Sets X and Y consist of the same number of positive integers. The inte
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11 Dec 2023, 05:53
11 Dec 2023, 05:53
1
The answer is A.
Let the number of integers = n
Let the smallest even integer = 2a
Let the smallest odd integer = 2b + 1
Set X --> 2a, 2a + 2, 2a + 2*2, ..., 2a + 2(n-1)
Set Y --> 2b + 1, (2b + 1) + 2, ..., (2b + 1) + 2(n-1)
For A.P., mean = median and mean = (1st number + last number)/2
Hence, Qty A = [2a + 2a + 2(n-1)]/2 = 2a + n - 1
& Qty B = [2b + 1 + (2b + 1) + 2(n-1)]/2 = 2b + 1 + n - 1 = 2b + n
Sum = Average * n
For Set X, sum = n(2a + n - 1)
For Set Y, sum = n(2b + n)
We know that n(2a + n - 1) > n(2b + n)
--> 2a + n - 1 > 2b + n
Hence, quantity A is greater.
Let the number of integers = n
Let the smallest even integer = 2a
Let the smallest odd integer = 2b + 1
Set X --> 2a, 2a + 2, 2a + 2*2, ..., 2a + 2(n-1)
Set Y --> 2b + 1, (2b + 1) + 2, ..., (2b + 1) + 2(n-1)
For A.P., mean = median and mean = (1st number + last number)/2
Hence, Qty A = [2a + 2a + 2(n-1)]/2 = 2a + n - 1
& Qty B = [2b + 1 + (2b + 1) + 2(n-1)]/2 = 2b + 1 + n - 1 = 2b + n
Sum = Average * n
For Set X, sum = n(2a + n - 1)
For Set Y, sum = n(2b + n)
We know that n(2a + n - 1) > n(2b + n)
--> 2a + n - 1 > 2b + n
Hence, quantity A is greater.
