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If a equals the sum of the even integers from 2 to 20, inclusive, and
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Updated on: 04 Nov 2014, 04:44
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If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?
Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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04 Apr 2007, 20:36
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The sum is 10.
The fastest way is to notice that since we're taking a-b, all we are going to end up with is addings ten '1' since each even number is one larger than the odd.
Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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12 Jan 2016, 07:43
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faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?
A. 1 B. 10 C. 19 D. 20 E. 21
Hi, Another way.. Without getting into the sum etc, the answer can be found easily... 2-20 contains 10 integers and similarily 1-19 also has 10 integers.. smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers.. so a-b=1*10=10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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07 Feb 2016, 03:18
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chetan2u wrote:
faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?
A. 1 B. 10 C. 19 D. 20 E. 21
Hi, Another way.. Without getting into the sum etc, the answer can be found easily... 2-20 contains 10 integers and similarily 1-19 also has 10 integers.. smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers.. so a-b=1*10=10
how to count number of odd integers from 1 to n inclusive? Please advise
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If a equals the sum of the even integers from 2 to 20, inclusive, and
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07 Feb 2016, 03:29
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smartguy595 wrote:
chetan2u wrote:
faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?
A. 1 B. 10 C. 19 D. 20 E. 21
Hi, Another way.. Without getting into the sum etc, the answer can be found easily... 2-20 contains 10 integers and similarily 1-19 also has 10 integers.. smallest integer of a is greater by 1 than the corresponding integer of b.. and these is same for all 10 integers.. so a-b=1*10=10
how to count number of odd integers from 1 to n inclusive? Please advise
Hi, If n is odd, add 1to it and divide by 2... If n is even, straight way divide by 2... In this example, n is 19.. Add one,so 20.. 20/2=10..
smartguy595, at times it might be that you are given some integer in between to another integer ahead.. example, it may not be 1 to n, but n to x.. few points to note in that scenario..
1) both x and n are even total integers will be (x-n)+1... even will be 1 more than odd.. even= (x-n)/2+1 and odd=(x-n)/2..
2) x is even and n is odd.. total integers = x-n +1.. even will be equal to odd.. even= odd={(x-n)+1}/2 ..
3)x is odd and n is even.. total integers = x-n +1.. even will be equal to odd.. even= odd={(x-n)+1}/2 ..
4) both x and n are odd total integers will be (x-n)+1... odd will be 1 more than even.. odd= (x-n)/2+1 and even=(x-n)/2..
Your query on 1 to 19 will fall under 3) and 4) above.. hope this helps..
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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26 Mar 2016, 18:21
Here's a visual that should help.
Attachments
Screen Shot 2016-03-26 at 6.18.49 PM.png [ 103.38 KiB | Viewed 18017 times ]
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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29 Aug 2016, 00:33
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faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?
A. 1 B. 10 C. 19 D. 20 E. 21
Solution :
a = Sum of all even integers between 2 to 20 Theory : Sum of n even integers = n(n+1) Therefore a = n^2+n
b = sum of all odd integers between 1 to 19 Theory : Sum of n odd integers = n^2
Therefore a-b = n^2+n-n^2 = n
Here n = no of even integers between 2 to 20 = no of odd integers between 1 to 19 = 10
Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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29 Aug 2016, 01:26
faifai0714 wrote:
If a equals the sum of the even integers from 2 to 20, inclusive, and b equals the sum of the odd integers from 1 to 19, inclusive, what is the value of a - b ?
A. 1 B. 10 C. 19 D. 20 E. 21
Option B sum of even integers = 10*22/2= 110 sum of odd integers = 10*20/2 =100 a - b= 110-100 =10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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02 Sep 2017, 05:14
We can list numbers as 2 4 6 .....20 and odd numbers 1 3 5 ....19 There are a total of 10 even 10 odd numbers so if we subtract each corresponding odd number from even numbers we will get 1 so total will be 10
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Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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19 Mar 2018, 11:21
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Hi All,
Sequence questions are almost always based on a pattern. If you can spot the pattern, then you can avoid a lot of heavy calculations and deduce the correct answer in a much faster way. Here, if we compare the first term of A and the first term of B, we have…
2-1 = 1
Next, let's compare the second term of A and second term of B…
4-3 = 1
This pattern will continue all the way up to the 10th term of A and tenth term of B…
Re: If a equals the sum of the even integers from 2 to 20, inclusive, and
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07 Apr 2019, 06:04
no. of even digits between 2 and 20 = 10 no. of odd digits between 1 and 19 = 10 upon close inspection you can see the difference between each successive even and odd digit is 1 i.e 2-1,4-3 and so on. so the difference between a and b is 1*10=10 i.e B
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74% (01:27) correct 
