If a equals the sum of the even integers from 2 to 20, inclusive, and

Answer is 10

Yes! there is really a faster way to solve it.

Sum of consecutive odd or even integers = (no. of odd or even ints) * (first int + last int) / 2

Here A = sum of even ints from 2 to 20, inclusive
number of even ints = 10,
first int + last int = 2+20 = 22
A = 10*22 / 2 = 110

B = sum of odd ints from 1 to 19, inclusive
number of odd ints = 10,
first int + last int = 1+19 = 20
A = 10*20 / 2 = 100

A-B = 110 - 100 = 10

Hope it helps

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.

Like 2-1 = 1, then 4-3 = 1.... 20-19 = 1

That's just adding up a bunch of 1s.

A = 2+4+6 ... +20
B = 1+3+5 ...+19

We want A-B
essentially 2-1 + 4-3 + 6-5 + ... + 20-19

thus ... 1 + 1 + 1 .. + 1 (ten times)

Ans: 10 B

Good explanation ywilfred....

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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how to count number of odd integers from 1 to n inclusive? Please advise
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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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Here's a visual that should help.

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

Therefore, a-b = n = 10
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Option B
sum of even integers = 10*22/2= 110
sum of odd integers = 10*20/2 =100
a - b= 110-100 =10

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

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…

20-19 = 1

We have 10 of these identical results: 10(1) = 10

Final Answers:

GMAT assassins aren't born, they're made,
Rich

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

We can also use an extremely simple formula to find the sum of evenly spaced integers.

\(mean * (# of integers) = sum of integers\)

Simply put,

There are 10 terms from 2 to 20 inclusive.
The mean is 22/2 = 11.

a = 11*10 = 110

For the odd set,

There are 10 terms from 1 to 19 inclusive.
The mean is 20/2 = 10

b = 10*10 = 100

110-100 = 10

a = 2+4 + ..+ 20
b= 1+ 3+....+ 19

both have equal number of terms i.e. 10.
a-b = (2-1) + (4-2) + ...+ (20-19)
= 1+ 1 ....10 times = 10
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Solution:

We are given that 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, and we need to determine the value of a - b.

When comparing the two sets of numbers, we can see that each number from 2 to 20 (which we can call list a) is ONE GREATER than each number from 1 to 19 (which we can call list b). That is, we can pair a number from list a such that it is one more than a corresponding number from list b. Let’s explore this idea further, starting with the least numbers in each list and working up.

1st number in list a = 2
1st number in list b = 1

2nd number in list a = 4
2nd number in list b = 3

Last number in list a = 20
Last number in list b = 19

Again, notice that each number in list a is ONE MORE than the corresponding number in list b.

Since there are 10 numbers in each list, the sum of the numbers in list a must be 10 greater than the sum of the numbers in list b.

Answer: B
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