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If x ? y is defined for all x and y as the sum of all integers from x

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If x ? y is defined for all x and y as the sum of all integers from x [#permalink]

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New post 05 Sep 2017, 01:10
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Question Stats:

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If x ? y is defined for all x and y as the sum of all integers from x to y, inclusive, what is the value of 5 ? 150?

A. 22,630
B. 22,475
C. 11,315
D. 11,238
E. 5,658
[Reveal] Spoiler: OA

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If x ? y is defined for all x and y as the sum of all integers from x [#permalink]

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New post 05 Sep 2017, 01:18
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x ? y = the sum of all integers from x to y, inclusive
Find : 5 ? 150

5 ? 150 = 5+6+7...+150

Sum of Continues series a+(a+1)+(a+2)+.... L = S = \(\frac{n}{2} *(a+L)\)
where n is number of terms, a is first term, L is last term

Here number of terms = (150-5)+1 = 146 ( Here +1 as 150 and 5 both are included)

Sum=\(\frac{146}{2}* (5+150) = \frac{146}{2} *(155)\)

= 11315

Answer: C

Last edited by Nikkb on 12 Nov 2017, 04:59, edited 1 time in total.

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If x ? y is defined for all x and y as the sum of all integers from x [#permalink]

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New post 05 Sep 2017, 12:25
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Bunuel wrote:
If x ? y is defined for all x and y as the sum of all integers from x to y, inclusive, what is the value of 5 ? 150?

A. 22,630
B. 22,475
C. 11,315
D. 11,238
E. 5,658

The ? strange symbol just gives the rule that, in this problem, the task is to find the sum of an arithmetic series of integers, from 5 to 150 inclusive.

Sum = (Average) * (Number of terms)

Average = \(\frac{First Term + Last Term}{2}\)

Average = \(\frac{150 + 5}{2}\) = \(\frac{155}{2}\) Leave it.*

Number of terms = (Last Term - First term + 1)

Number of terms = (150 - 5) + 1 = 146

Sum of series: \(\frac{155}{2}\) * (146) = 155 * 73

If you do the arithmetic fully, 155 * 73 = 11,315

ANSWER C

To simplify the arithmetic:

Three answers, A, D,and E, can be eliminated immediately because the units' digit of the answer must be five (3*5 = 15).

Between B and C: Answers B and C are far enough apart that you can round numbers to speed the arithmetic.

150 * 70 = 10,500. That's close to C. To get close to Answer B, the number of terms would have to double.

Either way:

Answer C

*If the average of consecutive integers is not an integer because the sum of the first and last terms is odd (odd/2 = not integer), the number of terms will be even, and the 2 in the denominator will be factored out. So write the two factors, average and number of terms, before calculating; that way it is clear that dividing by 2 will lead to an integer.

Last edited by genxer123 on 05 Sep 2017, 14:02, edited 1 time in total.

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Re: If x ? y is defined for all x and y as the sum of all integers from x [#permalink]

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New post 05 Sep 2017, 12:37
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Since x ? y is defined for all x and y as the sum of all integers from x to y, inclusive and
we have been asked to find the value of 5 ? 150, one easy way is to find the sum of the
first 150 numbers and subtract 1+2+3+4(10)

The sum of first n numbers is \(\frac{n(n+1)}{2} = \frac{150*151}{2} = 75*151 = 11325\)

Therefore 5 ? 150 = 11325 - 10 = 11315(Option C)
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Re: If x ? y is defined for all x and y as the sum of all integers from x [#permalink]

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New post 07 Sep 2017, 16:48
Bunuel wrote:
If x ? y is defined for all x and y as the sum of all integers from x to y, inclusive, what is the value of 5 ? 150?

A. 22,630
B. 22,475
C. 11,315
D. 11,238
E. 5,658


We can use the formula sum = (avg)(quantity).

Let’s first get the average of the integers from 5 to 150 inclusive:

average = (5 + 150)/2 = 155/2

quantity = 150 - 5 + 1 = 146

sum = 155/2 x 146 = 11,315

Answer: C
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Re: If x ? y is defined for all x and y as the sum of all integers from x   [#permalink] 07 Sep 2017, 16:48
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