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

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

Bunuel · Accepted Answer

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.

nimc2012 · Answer

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 = ????

Bunuel · Answer

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.

Stiv · Answer

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!

sambam · Answer

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

navaljoshi · Answer

I will use AP: Sum = n/2 = 840 = sum from 30- 50 Number of even terms from 30 -50 are 11. so total = 240+11 =251 Hope this helps

dharam831 · Answer

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

Temurkhon · Answer

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

sharma123 · Answer

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.

Bunuel · Answer

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.

SeregaP · Answer

21*(30+50)/2=840

(50-30)/2=10+1=11
Answer is 851 or E

navya27 · Answer

you have add no. of terms that is 11 hence x+y = 840+11 = 851

you halfway approach was good though.

bumpbot · Answer

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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

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