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# The sum of the odd/even integers

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The sum of the odd/even integers [#permalink]  20 Jan 2016, 18:36
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Question Stats:

65% (01:06) correct 34% (00:45) wrong based on 156 sessions
 Quantity A Quantity B The sum of the odd integers from 1 to 199 The sum of the even integers from 2 to 198

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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Question: 7
Page: 458
Difficulty: medium
[Reveal] Spoiler: OA

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Last edited by Carcass on 22 Jan 2016, 07:30, edited 1 time in total.
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Re: The sum of the odd/even integers [#permalink]  22 Jan 2016, 07:45
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Solution

The most straightforward way to solve this question, among the others, is to count the numbers in the sets

From quantity $$A$$ we do have that

$$199+1=\frac{200}{2}=100$$ So we have the precise middle of the set, the average

Then count the numbers in the set $$199-1=\frac{198}{2}=99+1=100$$ (we divide by two because we want only the odd numbers, one yes one no)

Multiply $$100*100=10000$$

Same for quantity $$B$$

$$198+2=\frac{200}{2}=100$$

$$198-2=\frac{196}{2}=98+1=99$$

$$99*100=9900$$

$$A > B$$

The answer is$$A$$

Note: if we were in GMAT Land the question would specify that the numbers at the extreme of the set are inclusive or not. Here, we assume that they are inclusive.
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Re: The sum of the odd/even integers [#permalink]  27 Apr 2016, 08:42
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Let’s fit the range in between any of them. The odd integers from 1 to 199 can be placed between the range of even integers from 2 to 198 or vice versa.

If the range swallows other range, the sum of the numbers of swallowing range must be greater than that of swallowed range. For example, 1, 2, 3,…………..8 or 2,3,,,,,,,,,,,,7.Sum of all numbers in the first range must be greater than that of 2nd range.
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Re: The sum of the odd/even integers [#permalink]  30 Jun 2017, 11:04
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sandy wrote:
 Quantity A Quantity B The sum of the odd integers from 1 to 199 The sum of the even integers from 2 to 198

We have:
Quantity A: 1 + 3 + 5 + ... + 197 + 199
Quantity B: 2 + 4 + 6 + ... + 196 + 198

Subtract (2 + 4 + 6 + ... + 196 + 198) from both quantities to get:
Quantity A: (1 + 3 + 5 + ... + 197 + 199) - (2 + 4 + 6 + ... + 196 + 198)
Quantity B: (2 + 4 + 6 + ... + 196 + 198) - (2 + 4 + 6 + ... + 196 + 198)

Simplify:
Quantity A: 199 - 198 + 197 - 196 + . . . . + 5 - 4 + 3 - 2 + 1
Quantity B: 0

Rewrite Quantity A as follows:
Quantity A: (199 - 198) + (197 - 196) + . . . . + (5 - 4) + (3 - 2) + 1
Quantity B: 0

Simplify:
Quantity A: 1 + 1 + 1 + 1 + ... + 1 + 1 + 1 + 1
Quantity B: 0

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Re: The sum of the odd/even integers [#permalink]  20 Sep 2017, 08:17
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Probably less fast than Carcass answer but we can also see that what we are asked is the sum of an arithmetic progression that is equal to $$S_n=\frac{n}{2}(f+l)$$ where n is the number of elements of our progression, f is the first element and l is the last one. Applying this rule to the column A we would get that n = 100 since 199/2=99.5 but the list terminates with an odd number. Thus the formula becomes $$\frac{100}{2}(1+199)=10,000$$. Using the same rationale for column B, we get that B = 9,900. Thus, A is larger!
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Re: The sum of the odd/even integers [#permalink]  18 May 2018, 08:51
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sandy wrote:
 Quantity A Quantity B The sum of the odd integers from 1 to 199 The sum of the even integers from 2 to 198

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Another approach is to compare the sums in parts

We have:
Quantity A: 1 + 3 + 5 + ... + 197 + 199
Quantity B: 2 + 4 + 6 + ... + 196 + 198

Compare the 1st number in each quantity (1 and 2). At this point in the sums, Quantity B is 1 greater than Quantity A.
Compare the 2nd number in each quantity (3 and 4). At this point in the sums, Quantity B is 2 greater than Quantity A.
Compare the 3rd number in each quantity (5 and 6). At this point in the sums, Quantity B is 3 greater than Quantity A.
Compare the 4th number in each quantity (7 and 8). At this point in the sums, Quantity B is 4 greater than Quantity A.
.
.
.
Compare the 48th number in each quantity (195 and 196). At this point in the sums, Quantity B is 98 greater than Quantity A.
Compare the 49th number in each quantity (197 and 198). At this point in the sums, Quantity B is 99 greater than Quantity A.

At this point, Quantity B is 99 greater than Quantity A.
HOWEVER, we have now run out of numbers in Quantity B. Yet we still have the number 199 left to add to Quantity A.
When we add this last value (199) to Quantity A, we help overcome the lead that Quantity B previously had, making Quantity A the bigger quantity.

RELATED VIDEO (I cover the strategy of comparing in parts starting at 2:16)

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Re: The sum of the odd/even integers [#permalink]  18 May 2018, 09:08
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Just make a shorter range of numbers to do that easier, for example from 1 to 11 and from 2 to 10, the idea is the same, or even better. From 1 to 5 and from 2 to 4, even the last number 5 is higher than 4, and you have one more number.
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Re: The sum of the odd/even integers [#permalink]  30 Jul 2019, 20:01
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GreenlightTestPrep wrote:
sandy wrote:
 Quantity A Quantity B The sum of the odd integers from 1 to 199 The sum of the even integers from 2 to 198

We have:
Quantity A: 1 + 3 + 5 + ... + 197 + 199
Quantity B: 2 + 4 + 6 + ... + 196 + 198

Subtract (2 + 4 + 6 + ... + 196 + 198) from both quantities to get:
Quantity A: (1 + 3 + 5 + ... + 197 + 199) - (2 + 4 + 6 + ... + 196 + 198)
Quantity B: (2 + 4 + 6 + ... + 196 + 198) - (2 + 4 + 6 + ... + 196 + 198)

Simplify:
Quantity A: 199 - 198 + 197 - 196 + . . . . + 5 - 4 + 3 - 2 + 1
Quantity B: 0

Rewrite Quantity A as follows:
Quantity A: (199 - 198) + (197 - 196) + . . . . + (5 - 4) + (3 - 2) + 1
Quantity B: 0

Simplify:
Quantity A: 1 + 1 + 1 + 1 + ... + 1 + 1 + 1 + 1
Quantity B: 0

[Reveal] Spoiler:
A

How does the generalized formula n(n+1)/2 adjust to the sequence of the problem? Sum of odd integers from 1 to 199.
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Re: The sum of the odd/even integers [#permalink]  10 May 2020, 03:57
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There is a formula to calculate sum of consecutive integers.
Sum of consecutive int = num of int * average

for 1-199:
Num of int = (largest - smallest)/2 + 1
=(199-1)/2 + 1 = 198/2 + 1
= 99 + 1 = 100

Avg = (largest + smallest)/2
=(199+1)/2
=100

So sum = Num * avg
= 100 * 100 = 10000

for 2-198:
Num of int = (largest - smallest)/2 + 1
=(198-2)/2 + 1 = 196/2 + 1
= 98 + 1 = 99

Avg = (largest + smallest)/2
=(198+2)/2
=100

So sum = Num * avg
= 99 * 100 = 9900

So quantity A is greater
Experts please confirm if my solution is correct? Thank you
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Re: The sum of the odd/even integers [#permalink]  10 May 2020, 05:33
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That's a perfectly valid solution, @Farina
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Re: The sum of the odd/even integers [#permalink]  11 May 2020, 16:11
GreenlightTestPrep wrote:
That's a perfectly valid solution, @Farina

Thank you very much
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Re: The sum of the odd/even integers   [#permalink] 11 May 2020, 16:11
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