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If s is the sum of odd integers from 40 to 60, inclusive, and t is the

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If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]

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If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?

A. 480
B. 490
C. 980
D. 990
E. 995

Source: Ready4GMAT mobile app
[Reveal] Spoiler: OA

Kudos [?]: 7 [0], given: 84

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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]

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New post 12 Oct 2017, 11:39
ladyrenee95 wrote:
If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?

A. 480
B. 490
C. 980
D. 990
E. 995

Source: Ready4GMAT mobile app


Similar question: https://gmatclub.com/forum/if-x-is-equa ... 97544.html
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]

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New post 12 Oct 2017, 11:48
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There are total 10 odd numbers between 40 and 60.
hence, t=10

Now, the given sequence is 41,43,45,......,59
common difference is 2.
hence,
S= (41+59)*(10/2)
S=500.
S-t= 500-10= 490.

Hence, Answer is option B.

Kudos if it helps.

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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]

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Are these formulas correct? I am having trouble with odd numbers for some reason.

N for odd numbers = (Last odd - First odd)/2)) +1

Sum of odd numbers = (First odd + Last odd)*N))/2

1. N = (59-41)/2))+1 = 10

2. Sum of odds = (41+59)*10))/2 = 500

3. 500 - 10 = 490 B

Also, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?

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If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]

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New post 12 Oct 2017, 12:36
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ladyrenee95 wrote:
If s is the sum of odd integers from 40 to 60, inclusive, and t is the number of odd integers from 40 to 60, inclusive, what is s-t?

A. 480
B. 490
C. 980
D. 990
E. 995

Source: Ready4GMAT mobile app

The way I solve sums of arithmetic series, I need the number of terms no matter what. Start there.

Set t

Number of terms: number of odd integers from 40 to 60. First term is 41, last is 59.

Number of terms

\(\frac{LastTerm-FirstTerm}{interval} + 1\)

The "interval" for consecutive even and odds is always 2

\(\frac{59-41}{2} = \frac{18}{2} = (9 + 1)=\) 10

Set s

Sum of arithmetic series
(Average)(Number of terms) =

\(\frac{FirstTerm+LastTerm}{2}*(10)\) =

\(\frac{100}{2}(10) = (50)(10) = 500\)

s - t: 500 - 10 = 490

Answer B
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If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]

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ladyrenee95 wrote:
Are these formulas correct? I am having trouble with odd numbers for some reason.

N for odd numbers = (Last odd - First odd)/2)) +1

Sum of odd numbers = (First odd + Last odd)*N))/2

1. N = (59-41)/2))+1 = 10

2. Sum of odds = (41+59)*10))/2 = 500

3. 500 - 10 = 490 B

Also, on part 1 of my calculations, do you add one because it is inclusive and do not add one if it is not inclusive?

ladyrenee95 , sorry, I posted before I saw your questions.

Yes, your method is correct.

I think the reason you might be having trouble with odd numbers is that for consecutive odd integers, the interval (between the numbers), is two. Two is even; that could confuse. Or you might be forgetting to start with the first ODD term (e.g. here, first term is 41, not 40).

Bottom line: The "interval" between 1 and 3 is two. The interval between 2 and 4 is two. For consecutive odds and evens, in the formula, you divide by 2.

Take smaller numbers. Say, sum of odd integers from 0 to 6. So 1 + 3 + 5 = 9. There are three terms (1, 5, 9) whose sum is 9.

Number of terms in "odd integers from 0 to 6":

\(\frac{(Last - First)}{2}+ 1\)

\(\frac{(5-1)}{2} + 1 = (\frac{4}{2}+ 1) = (2 + 1) = 3\) Correct

SUM: (Average)(# of terms)

\(\frac{(First + Last)}{2}(3)\)

\(\frac{(1 + 5)}{2} (3) = (3)(3) = 9\)
Bingo. Dividing by 2 for consecutive odd integers works.

When subtracting, you add one because subtraction doesn't include the first number. If in doubt: use small numbers that replicate your situation.

For example, # of terms from 1 to 4? The numerals are 1, 2, 3, 4. There are 4 terms. BUT (4 - 1) = 3.
We need one more. 3 + 1 = 4.

There's a mnemonic: "add one before you're done." It almost always applies. If in doubt, replicate your situation with small numbers. Rare in sequence sums but possible: If you see the word "exclusive" or the phrase "exclusive of," that is when you really need to check.

I have seen your other answers using this method. The ones I've seen are correct! (The one about -190 to 195? Correct, but long. -190 to +190 = 0; you are left with only 5 numbers to sum. STILL - correct.)

Here is a fantastically written, comprehensive post on all kinds of sequences by benjiboo , scroll down for arithmetic sequence

And here is a thread on sequences. VeritasPrepKarishma 's posts are great

Hope this barrage of information helps. :-)
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Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the [#permalink]

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New post 13 Oct 2017, 10:23
N for odd numbers = (Last odd - First odd)/2)) +1

Sum of odd numbers = (First odd + Last odd)*N))/2

1. N = (59-41)/2))+1 = 10

2. Sum of odds = (41+59)*10))/2 = 500

3. 500 - 10 = 490 B

Kudos [?]: 8 [0], given: 194

Re: If s is the sum of odd integers from 40 to 60, inclusive, and t is the   [#permalink] 13 Oct 2017, 10:23
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