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If x is equal to the sum of the integers from 30 to 50, incl

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

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New post 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  [#permalink]

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New post 19 Dec 2010, 16:31
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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 \(y=\frac{50-30}{2}+1=11\) (check this: totally-basic-94862.html);

x+y=840+11=851.

Answer: E.


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

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New post Updated on: 23 Jul 2018, 00:14
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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 = ????

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)  [#permalink]

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New post 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 \(y=\frac{50-30}{2}+1=11\) (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.

Hope it's clear.
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Re: If x is equal to the sum of the integers from 30 to 50  [#permalink]

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

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New post 28 Aug 2014, 09:54
I will use AP:

Sum = n/2[2a+(n-1)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  [#permalink]

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New post 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

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  [#permalink]

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New post 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

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

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New post 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 \(y=\frac{50-30}{2}+1=11\) (check this: totally-basic-94862.html);

x+y=840+11=851.

Answer: E.


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Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.

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  [#permalink]

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New post 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 \(y=\frac{50-30}{2}+1=11\) (check this: totally-basic-94862.html);

x+y=840+11=851.

Answer: E.


!
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.

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  [#permalink]

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New post 18 Mar 2017, 10:37
21*(30+50)/2=840

(50-30)/2=10+1=11
Answer is 851 or E
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Re: If x is equal to the sum of the integers from 30 to 50, incl   [#permalink] 18 Mar 2017, 10:37
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