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sum of the even integers from 40 to 60 inclusive [#permalink]
 03 Jul 2009, 20:07
Question Stats:
42% (00:00) correct
57% (00:00) wrong based on 1 sessions
if x is equal to the sum of the even integers from 40 to 60 inclusive and y is the number of even integers from 40 to 60 inclusive,what is the value of x+y? 550 551 560 561 572 Guys I applied the formula for "sum of consecutive evn nos." but i am going wrong somewhere.Pease help. y=11 x=sum of consecutive even integers=n(n+1) where n= 1st even+last even/2 -1 Therefore,here n=40+60/2-1=50-1=49 So,x=49 x 50 =2450 Hence, x+y=2450+11=2461??!??!
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countdown-beginshas-ended-85483-40.html#p649902
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
03 Jul 2009, 20:47
Step 1 (y) - No of even numbers from 40-60 = 11 Step 2 (x) - Sum of those 11 nos is given by the formula - (F+L)*N/2 F= 40 L = 60 N =11 Therefore ==> 550 x+y = 550+11 = 561 tejal777 wrote: if x is equal to the sum of the even integers from 40 to 60 inclusive and y is the number of even integers from 40 to 60 inclusive,what is the value of x+y?
550
551
560
561
572
Guys I applied the formula for "sum of consecutive evn nos." but i am going wrong somewhere.Pease help.
y=11
x=sum of consecutive even integers=n(n+1)
where n= 1st even+last even/2 -1
Therefore,here n=40+60/2-1=50-1=49
So,x=49 x 50 =2450
Hence, x+y=2450+11=2461??!??!
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
04 Jul 2009, 00:10
This is solved by using AP series.
40,42,44,46..........................60
This is a AP series with common difference of 2
Number of terms = A + (n-1)d
A = first term (40)
n = number of terms (need to be calculated)
d = common difference (2 in this case)
60 = 40 + (n-1)2
or n = 11
Sum of series = [2A + (n-1)d ] * n/2
Sum = 550
So ans = 550 + 11 = 561
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
07 Jul 2009, 03:57
thanks guys i got the answer but I still dont know where im going wrong!!am i following the formulae wrong??please refer to link below from where i got the formula: http://www.beatthegmat.com/formula-for- ... 17241.html
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
07 Jul 2009, 08:29
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
07 Jul 2009, 08:47
For me,
step1: find y 40, 42, 44, ... 60 therefore y=11
step2: since consecutive numbers/even/odd, find the mean (40+60)/2=50
step3: x = 50(11)=550
step4: 550+11=561
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
08 Jul 2009, 03:10
sum of even integers = even number ( x is even)
number of even integers =11 ( y is odd )
so x+y = odd
A, C, E out ( all even)
Left with B and D :551, 561
If you have ever added even numbers you see that the pattern is 0,2,4,6,8 and 2+4+6+8 =20
{ there are 11 integers 5 in the 40's , 5 in the 50's and 60 , so when u add the u get 200+250+20+20+60 = 550}
hence the sum is 550 ( x=550) or x+y cannot be 551 since x =550
hence answer is D
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
29 Mar 2011, 07:41
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Another approach ('sum of pairs'): Step 1: 11 numbers Step 2: 40+60 = 42+58 = 100 (total 5 pairs, with exception of number 55 that does not have a pair) Step 3: 500 + 55 (the middle number with no pair)+ 11 = 561 Advantages: you don't need to know formulas nor you can make mistake in formulas 
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
29 Mar 2011, 19:38
x = (60 + 40)/2 * y 60 = 40 + (y-1)*2 => y = 20/2 + 1 = 11 so 50 * 11 + 11 = 561 Answer - D
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
30 Mar 2011, 06:43
I used to take a long time solving these kinds of problem until I learned about this formula:
Average = Sum of integers / number of terms
Average = (first term + last term) / 2 ==> this works for both consecutive and even integers
SOLUTION:
x (SUM) = Average x number of terms
Average = 40 +60 /2 = 50
number of terms = ((60 -40)/2)+1 = 11
x (SUM) = 50 x 11 = 550
Therefore,
x + y = 550 + 11 = 561
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Director
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
30 Mar 2011, 18:34
x = 40+42+....60 = (mean).N = (mean)y
=> x+y = = (51).11 = 561
Answer D.
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Re: sum of the even integers from 40 to 60 inclusive [#permalink]
06 Nov 2011, 12:04
lawschoolsearcher wrote: I used to take a long time solving these kinds of problem until I learned about this formula:
Average = Sum of integers / number of terms
Average = (first term + last term) / 2 ==> this works for both consecutive and even integers
SOLUTION:
x (SUM) = Average x number of terms
Average = 40 +60 /2 = 50
number of terms = ((60 -40)/2)+1 = 11
x (SUM) = 50 x 11 = 550
Therefore,
x + y = 550 + 11 = 561 How do you find the number of terms in a sequence?
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Re: sum of the even integers from 40 to 60 inclusive
[#permalink]
06 Nov 2011, 12:04
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