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# Set A contains all of the integers from 100 to 200, inclusive. Set B

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Set A contains all of the integers from 100 to 200, inclusive. Set B  [#permalink]

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28 Aug 2017, 23:35
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35% (medium)

Question Stats:

74% (02:01) correct 26% (02:19) wrong based on 102 sessions

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Set A contains all of the integers from 100 to 200, inclusive. Set B contains all of the integers from 25 to 125, inclusive. What is the sum of set A minus the sum of set B?

A. 7500
B. 7575
C. 8050
D. 9875
E. 12925

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Re: Set A contains all of the integers from 100 to 200, inclusive. Set B  [#permalink]

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29 Aug 2017, 01:49
Bunuel wrote:
Set A contains all of the integers from 100 to 200, inclusive. Set B contains all of the integers from 25 to 125, inclusive. What is the sum of set A minus the sum of set B?

A. 7500
B. 7575
C. 8050
D. 9875
E. 12925

Sum of Set A minus the sum of set B = Sum of integers from 126 to 200 - Sum of integers from 25 to 99 (Since integers 100 to 125 are common in both sets , we exclude them )
= 75/2 * (126+200) - 75/2 *(25+99)
= 75/2 ( 326 - 124)
= 75 * 202/2
= 75*101
= 7575

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Set A contains all of the integers from 100 to 200, inclusive. Set B  [#permalink]

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29 Aug 2017, 08:52
Bunuel wrote:
Set A contains all of the integers from 100 to 200, inclusive. Set B contains all of the integers from 25 to 125, inclusive. What is the sum of set A minus the sum of set B?

A. 7500
B. 7575
C. 8050
D. 9875
E. 12925

We can exclude common terms (in set A and set B), but I think the arithmetic is easier if we do not do so.
Both sets have the same number of terms. (The difference between largest term and smallest term in each set is the same). Leave the sums of the separate sets UNmultiplied.

(1) Find Sum of Set A (consecutive integers)

Sum = (Average) * (Number of terms)

Average = $$\frac{FirstTerm+LastTerm}{2}=\frac{100+200}{2}$$

Number of terms =
(Last - First) + 1 (add one before you're done):
(200 - 100) + 1 = (100 + 1) = 101 terms

SUM of Set A
$$\frac{200+100}{2}$$ * 101 = (150)(101)

(2) Sum of Set B

Average = $$\frac{FirstTerm+LastTerm}{2}=\frac{25+125}{2}$$

Number of terms:
(125 - 25) + 1 = (100 + 1) = 101 terms

SUM of Set B
$$\frac{125+25}{2}$$ * 101 = (75)(101)

3) (Sum of A) - (Sum of B) =

(150)(101) - (75)(101)
Factor out 101
101(150 - 75) =
101(75) =
(100 + 1)(75)

7500 + 75 = 7575

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Re: Set A contains all of the integers from 100 to 200, inclusive. Set B  [#permalink]

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01 Sep 2017, 12:02
Bunuel wrote:
Set A contains all of the integers from 100 to 200, inclusive. Set B contains all of the integers from 25 to 125, inclusive. What is the sum of set A minus the sum of set B?

A. 7500
B. 7575
C. 8050
D. 9875
E. 12925

Since we see there is overlap between sets, we can just determine the sum of the integers from 25 to 99 inclusive and 126 to 200 inclusive. Remember average = (first number + last number)/2.

For set B (excluding 100 to 125 inclusive):

sum = avg x quantity

sum = (25 + 99)/2 x (99 - 25 + 1)

sum = 62 x 75

For set A (excluding 100 to 125 inclusive):

sum = avg x quantity

sum = (126 + 200)/2 x (200 - 126 + 1)

sum = 163 x 75

Thus, A - B = 163 x 75 - 62 x 75 = 75(163 - 62) = 75(101) = 7,575.

Alternate solution:

Even though some of the numbers in B overlap with those in A, we can just find the sum of all the numbers in B and subtract that from the sum of all the numbers in A.

We use the formulas sum = average x quantity and average = (first number + last number)/2.

For set B:

sum = avg x quantity

sum = (25 + 125)/2 x (125 - 25 + 1)

sum = 75 x 101

For set A:

sum = avg x quantity

sum = (100 + 200)/2 x (200 - 100 + 1)

sum = 150 x 101

Thus, A - B = 150 x 101 - 75 x 101 = 101(150 - 75) = 101(75) = 7,575.

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Set A contains all of the integers from 100 to 200, inclusive. Set B  [#permalink]

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09 Mar 2019, 14:21
100 to 200 => 100 & 101 to 200 (100 numbers) ==> total 101 numbers

Sum of numbers from 101 to 200 can be obtained by taking an average of all the terms and then multiplying it by 100.
If you try to use the "mirror approach" to get the average, you will notice 150.5 divides the set of 100 numbers in the middle. So, the total sum is-

$$(150.5 * 100) + 100 = 15150$$

25 to 125 => 25 & 26 to 125 (100 numbers) ==> total 101 numbers

Apply the same approach here. Get the average of numbers from 26 to 125 by mirror approach and multiple by 100. Don't forget to add 25 to this number.

$$(75.5 * 100) + 25 = 7575$$

Finally, $$15150 - 7575 = 7575$$

Took me 1:48 minutes to get to the answer.
Set A contains all of the integers from 100 to 200, inclusive. Set B   [#permalink] 09 Mar 2019, 14:21
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