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07 Oct 2020, 00:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
85% (01:25) correct
15%
(02:02)
wrong
based on 266
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What is the sum of the even integers from 100 to 300 inclusive?
A. 20,196
B. 20,198
C. 20,200
D. 20,202
E. 20,204
A. 20,196
B. 20,198
C. 20,200
D. 20,202
E. 20,204
07 Oct 2020, 10:34
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No Need to Solve this question, as it is a 10 sec question with logic[ after seeing the options].
Now to find the sum of even numbers between 100 & 300-->
This would be your series: 100, 102, 104, 108, 110,......296, 298, 300. So if you sum the units digit of every even number in this range, you can easily get to know that only option C satisfies this.
Eg: 102, 104, 106, 108--> corresponding sum of units digits is 20 i.e. ending with 0 and since the same units digits will be repeated across the series, your sum of the entire series should be something ending in zero, which is only a case in option C.
Hence C, IMO.
Now to find the sum of even numbers between 100 & 300-->
This would be your series: 100, 102, 104, 108, 110,......296, 298, 300. So if you sum the units digit of every even number in this range, you can easily get to know that only option C satisfies this.
Eg: 102, 104, 106, 108--> corresponding sum of units digits is 20 i.e. ending with 0 and since the same units digits will be repeated across the series, your sum of the entire series should be something ending in zero, which is only a case in option C.
Hence C, IMO.
General Discussion
number of even integers = ((300-100)/2) + 1 =100 +1 = 101
formula of sum =n/2(2*a+(n-1)d)
here a = 1st term and d =distance between two even integers=2
=101/2*(2*100+(101-1)*2)
=101/2(200+200)
=101*200
=20200
formula of sum =n/2(2*a+(n-1)d)
here a = 1st term and d =distance between two even integers=2
=101/2*(2*100+(101-1)*2)
=101/2(200+200)
=101*200
=20200
