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If x is equal to the sum of the integers from 30 to 50, incl
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19 Dec 2010, 16:14
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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 ? (A) 810 (B) 811 (C) 830 (D) 850 (E) 851
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Re: Number Properties Question
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19 Dec 2010, 16:31



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If x is equal to the sum of the integers from 30 to 50, incl
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Updated on: 23 Jul 2018, 00:14
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 = ????
Originally posted by nimc2012 on 08 Feb 2012, 15:31.
Last edited by Bunuel on 23 Jul 2018, 00:14, edited 4 times in total.
Edited the question



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Re: Consecutive Numbers (Number Properties)
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08 Feb 2012, 15:37
nimc2012 wrote: 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 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 \(y=\frac{5030}{2}+1=11\) (check this: totallybasic94862.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. Hope it's clear.
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Re: If x is equal to the sum of the integers from 30 to 50
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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
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11 Feb 2013, 20:09
Since the integers from 3050 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 3050 = y. Therefore, x+y = 851. Answer: E



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Re: If x is equal to the sum of the integers from 30 to 50, incl
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28 Aug 2014, 09:54
I will use AP: Sum = n/2[2a+(n1)d) = 21/2[2*30+20*1] = 840 = sum from 30 50 Number of even terms from 30 50 are 11. so total = 240+11 =251 Hope this helps
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Re: If x is equal to the sum of the integers from 30 to 50
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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 (5030+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 (((5030)/2)+1) with common difference 2) Y=11
X+Y = 840+11 = 851
E



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If x is equal to the sum of the integers from 30 to 50, incl
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16 Dec 2015, 03:36
21 integers from 30 to 50 inclusive
so 11th integer is mean, it is 40
21*40=840
from 21 integers 11 ones is even
840+11=851
E



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Re: If x is equal to the sum of the integers from 30 to 50, incl
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16 Dec 2015, 09:13
Bunuel wrote: MasterGMAT12 wrote: Please help with the approach for the below question:
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 The sum of the integers from 30 to 50 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 \(y=\frac{5030}{2}+1=11\) (check this: totallybasic94862.html); x+y=840+11=851. Answer: E. Hi Bunuel how do you calculate no. of terms(i.e 21). By counting or is there any method. Counting no. of terms in given range or no. of hours have always been a problem for me. i always count which consumes a lot of time.



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Re: If x is equal to the sum of the integers from 30 to 50, incl
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17 Dec 2015, 09:43
sharma123 wrote: Bunuel wrote: MasterGMAT12 wrote: Please help with the approach for the below question:
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 The sum of the integers from 30 to 50 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 \(y=\frac{5030}{2}+1=11\) (check this: totallybasic94862.html); x+y=840+11=851. Answer: E. Hi Bunuel how do you calculate no. of terms(i.e 21). By counting or is there any method. Counting no. of terms in given range or no. of hours have always been a problem for me. i always count which consumes a lot of time. How many integers are from x to y inclusive? y  x + 1. For example, how many integers are from 2 to 7 inclusive? 7  2 + 1 = 6. Namely, 2, 3, 4, 5, 6 and 7. How many integers are from x to y NOT inclusive? y  x  1. For example, how many integers are from 2 to 7 NOT inclusive? 7  2  1 = 4. Namely, 3, 4, 5, and 6.
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Re: If x is equal to the sum of the integers from 30 to 50, incl
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18 Mar 2017, 10:37
21*(30+50)/2=840
(5030)/2=10+1=11 Answer is 851 or E




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