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Set A contains the consecutive integers ranging from x to y,
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10 Aug 2010, 13:54
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Set A contains the consecutive integers ranging from x to y, inclusive. If the number of integers in set A that are less than 75 is equal to the number of integers that are greater than 75, what is the value of 3x+ 3y? A. 225 B. 300 C. 372 D. 450 E. 528 Range x to y is a consecutive range of “75” integers before and after number 75, so 150 total integers (75 + 75)? This can’t be as simple as plugging in 75 for both x and y (answer D: 450)....that’s definitely a trap answer. But I don’t quite see where else to go.
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Re: range problem
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10 Aug 2010, 15:06
lifeisshort wrote: Set A contains the consecutive integers ranging from x to y, inclusive. If the number of integers in set A that are less than 75 is equal to the number of integers that are greater than 75, what is the value of 3x+ 3y?
A 225
B 300
C 372
D 450
E 528
Range x to y is a consecutive range of “75” integers before and after number 75, so 150 total integers (75 + 75)? This can’t be as simple as plugging in 75 for both x and y (answer D: 450)....that’s definitely a trap answer. But I don’t quite see where else to go. "The number of integers in set A that are less than 75 is equal to the number of integers that are greater than 75" means that if there are \(k\) integer less than 75 than there are also \(k\) integers more than 75. So if for example there are 2 integers less than 75 then the set is {73, 74, 75, 76, 77} and if there is 1 integer less than 75 then the set is {74, 75, 76}. Note that no matter how many integers are less and more than 75, the sum of the smallest (\(x\)) and the greatest (\(y\)) will always be 150: For set {73, 74, 75, 76, 77} > 73+77=150; For set {74, 75, 76} > 74+76=150; ... So 3x+3y=3(x+y)=3*150=450. Answer: D. Side notes: knowing that "the number of integers in set A that are less than 75 is equal to the number of integers that are greater than 75" we can say that 75 is a median value. Also consecutive integers is evenly spaced set and for every evenly spaced set median=mean=average of first and last terms=(x+y)/2, so 75=(x+y)/2 > x+y=150 > 3(x+y)=450. Hope it helps.
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Re: range problem
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10 Aug 2010, 19:39
given to us is that the series is consecutive. We need to find out 3(x+y)
x+y = sum of smallest and largest variable
now in a consecutive series the: mean = (smallest + largest )/ 2 now given to us that 75 is the middle value, i.e. median
for a consecutive series median = mean so median = 75 smallest + largest = mean * 2 = 2 * 75 = 150
smallest = x and largest = y x+y = 150 3(x+y) = 450 (answer)



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Re: integer problem
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17 Mar 2011, 22:42
3(x+y) = 6 times the mean = 6 * 75 = 450. Pls verify the reasoning.



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Re: integer problem
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17 Mar 2011, 23:22
AnkitK wrote: Set A contains the consecutive integers ranging from x to y,inclusive.If the number of integers in set A that are less than 75 is equal to the number of integers that are greater than 75 ,what is the value of 3x+3y?
A.225 B.300 C.372 D.450 E.528 I will breakdown what gmat1220 rightly solved. For any set of consecutive numbers, the average of the set will be the average of first and last number. Since, we know that the number of elements are evenly spaced and symmetrically distributed on both sides of 75, 75 must be the mean of the set. \(\frac{x+y}{2}=75\) \(x+y=75*2=150\) Multiply both sides by 3 \(3x+3y=3*150=450\) Ans: "D"
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Re: integer problem
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Updated on: 09 Apr 2011, 22:12
AnkitK wrote: Set A contains the consecutive integers ranging from x to y,inclusive.If the number of integers in set A that are less than 75 is equal to the number of integers that are greater than 75 ,what is the value of 3x+3y?
A.225 B.300 C.372 D.450 E.528 Let the integers in set A = {1,2,......74,75,76,......y} x=1 No. of integers less than 75 = 74 Similarly, no. of integers greater than 75 is also 74. Which means y = 149 (Using Last  First + 1 formula) so 3x + 3y = 3*1 + 3*149 = 3 + 447 = 450 Answer (D)
Originally posted by mniyer on 09 Apr 2011, 16:44.
Last edited by mniyer on 09 Apr 2011, 22:12, edited 1 time in total.



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Re: integer problem
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09 Apr 2011, 20:39
AnkitK wrote: Set A contains the consecutive integers ranging from x to y,inclusive.If the number of integers in set A that are less than 75 is equal to the number of integers that are greater than 75 ,what is the value of 3x+3y?
A.225 B.300 C.372 D.450 E.528 Something that comes to mind immediately when I read this question: There is one unique value of 3x + 3y. Number of integers in set A that are less than 75 is equal to the number of integers greater than 75  I just assume that there are 0 integers less than 75 and 0 are more than 75 such that x = y = 75 Then 3 (x+y) = 3*150 = 450 Such plugging numbers can be very useful in PS questions since you have a unique answer (though admittedly, I am not a big fan of plugging numbers since they can drive you crazy in DS questions but they can save you a lot of time in some PS questions)
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Re: integer problem
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18 May 2011, 03:48
first + last term/2 = average (x+y)/2 = 75 =150 3*(x+y) = 450
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Re: Set A contains the consecutive integers ranging from x to y,
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12 Aug 2013, 15:16
lifeisshort wrote: Set A contains the consecutive integers ranging from x to y, inclusive. If the number of integers in set A that are less than 75 is equal to the number of integers that are greater than 75, what is the value of 3x+ 3y?
A. 225 B. 300 C. 372 D. 450 E. 528
Range x to y is a consecutive range of “75” integers before and after number 75, so 150 total integers (75 + 75)? This can’t be as simple as plugging in 75 for both x and y (answer D: 450)....that’s definitely a trap answer. But I don’t quite see where else to go. .............. Imagine this set, A = {1,2,3,4,5} , 3 is the mean of the set. And 3 = (1+5)/2 less than 3 are two numbers and more than 3 are two numbers. so the set contains an odd number of integers. Now, A = {x,..................75,................y} same from example, 75 = (x+y)/2 or, x+y = 150 or, 3x+3y = 450
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Re: Set A contains the consecutive integers ranging from x to y,
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