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If x is equal to the sum of the integers from 30 to 50 [#permalink]

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08 Feb 2012, 15:31

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If x is equal to the sum of the integers from 30 to 50, inclusive, and y is the number of EVEN integers from 30 to 50, inclusive, what is the value of x+y ?

So for X - {30 to 50} , No. of terms = 21, Mean = 50+30 / 2 = 40 So Sum of X = No of terms x Mean = 21 x 40 = 840 ---------------------------------------------------------------------------------------------------------------- Now for Y - {30, 32,34,36,38,40,42,44,46,48,50} Now since these have already been added before.....what am i supposed to do?......i am lost here..

If x is equal to the sum of the integers from 30 to 50, inclusive, and y is the number of EVEN integers from 30 to 50, inclusive, what is the value of x+y ?

a) 810 b) 811 c) 830 d)850 e)851

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I started solving like this .....

Sum of integers = (Mean x No of Terms)

So for X - {30 to 50} , No. of terms = 21, Mean = 50+30 / 2 = 40 So Sum of X = No of terms x Mean = 21 x 40 = 840 ---------------------------------------------------------------------------------------------------------------- Now for Y - {30, 32,34,36,38,40,42,44,46,48,50} Now since these have already been added before.....what am i supposed to do?......i am lost here..

Sum = x + y = ????

The sum of the integers from 30 to 50, inclusive is \(x=\frac{first \ term+ last \ term}{2}*# \ of \ terms=\frac{30+50}{2}*21=40*21=840\);

The number of even integers from 30 to 50 is \(y=\frac{50-30}{2}+1=11\) (check this: totally-basic-94862.html);

x+y=840+11=851.

Answer: E.

As for your question: x and y are two different numbers: x is a sum of the integers from 30 to 50, inclusive and y is the number of EVEN integers from 30 to 50, inclusive.

Re: If x is equal to the sum of the integers from 30 to 50 [#permalink]

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10 Feb 2013, 08:58

The correct answer to this question is E not D. You put D as the correct answer, so you could change it to E. It is confusing at first look. Thanks!
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Re: If x is equal to the sum of the integers from 30 to 50 [#permalink]

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11 Feb 2013, 20:09

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Since the integers from 30-50 are consecutive we can use the "median x # of terms formula" to determine x. There are 21 terms and the median is 40. This gives a total of 21x40 = 840. There are 11 even integers from 30-50 = y. Therefore, x+y = 851. Answer: E

If x is equal to the sum of the integers from 30 to 50, [#permalink]

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06 Mar 2013, 14:34

If x is equal to the sum of the integers from 30 to 50, inclusive, and y is the number of even integers from 30 to 50, inclusive, what is the value of x+y ?

(A) 810 (B) 811 (C) 830 (D) 850 (E) 851

I know this is pretty easy. However, I always get the sum of even (or some multiple) integers wrong, I use the formula: (last-first)/2+1). Also hoping @GyanOne can show his mad Quant skills with some genius shortcut
_________________

If x is equal to the sum of the integers from 30 to 50, inclusive, and y is the number of even integers from 30 to 50, inclusive, what is the value of x+y ?

(A) 810 (B) 811 (C) 830 (D) 850 (E) 851

I know this is pretty easy. However, I always get the sum of even (or some multiple) integers wrong, I use the formula: (last-first)/2+1). Also hoping @GyanOne can show his mad Quant skills with some genius shortcut

Re: If x is equal to the sum of the integers from 30 to 50 [#permalink]

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17 Feb 2015, 19:14

Hello from the GMAT Club BumpBot!

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Re: If x is equal to the sum of the integers from 30 to 50 [#permalink]

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18 Feb 2015, 01:45

If x is equal to the sum of the integers from 30 to 50, inclusive, and y is the number of EVEN integers from 30 to 50, inclusive, what is the value of x+y ?

A. 810 B. 811 C. 830 D. 850 E. 851

For X, Series is - 30,31,32,33,.......,48,49,50 (21 terms (50-30+1) with common difference 1) Sum = (21/2)*((2*30)+(20)) = 21*40 = 840

For Y, Series is - 30,32,34,....48,50 (11 terms (((50-30)/2)+1) with common difference 2) Y=11