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Set X consists of the first 100 positive even integers. Set Y consists

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Set X consists of the first 100 positive even integers. Set Y consists  [#permalink]

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New post 06 Feb 2017, 13:16
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Set X consists of the first 100 positive even integers. Set Y consists of the first 100 positive multiples of 5. If the terms in both sets are combined to create a new set with no repeat elements, what is the sum of all unique terms in the new set?

a) 33250

b) 34350

c) 35750

d) 36300

e) 37500
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Re: Set X consists of the first 100 positive even integers. Set Y consists  [#permalink]

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New post 06 Feb 2017, 14:36
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Sum of first 100 positive even integers: 2 + 4 + 6 + 8 + ....+ 200 = [100 (2+200)]/2 = 10100

Sum of first 100 positive multiples of 5: 5 + 10 + 15 + ... + 500 = [100 (5+500)]/2 = 25250

Sum of multiples of 10 till 200: 10 + 20 + 30 + ... + 200 = [20 (10+200)]/2 = 2100

Therefore: Sum of all unique terms in the new set = 10100 + 25250 - 2100 = 33250 (A)
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Re: Set X consists of the first 100 positive even integers. Set Y consists  [#permalink]

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New post 12 Feb 2017, 07:32
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Matruco wrote:
Set X consists of the first 100 positive even integers. Set Y consists of the first 100 positive multiples of 5. If the terms in both sets are combined to create a new set with no repeat elements, what is the sum of all unique terms in the new set?

a) 33250

b) 34350

c) 35750

d) 36300

e) 37500


Official solution from Veritas Prep.

This problem heavily rewards you for knowing how to quickly calculate the sum of an evenly-spaced set. When you look at the two sets, you have:

\(Set X\): \(2, 4, 6, 8, 10….198, 200\)
\(Set Y\): \(5, 10, 15, 20…495, 500\)

And note that you do not need to perform a “count” to make sure you have exactly 100 terms in each set. Multiples of 2 are defined as 2 times an integer, so the first 100 terms will be 1 * 2, 2 * 2, 3 * 2… up to 100 * 2. Same for multiples of 5: 1 * 5, 2 * 5, 3 * 5…100 * 5. So you know that your ending number for each set will be 100 times the common multiple.
When you’re dealing with evenly-spaced set, the median equals the mean. So you can quickly calculate the average by averaging the first and last term. For Set X, that’s \(\frac{(2 + 200)}{2} = 101\), and for Set Y that’s \(\frac{(5 + 500)}{2} = 252.5\).

Now you’ll use the Average = Sum of Terms / Number of Terms rule to your advantage. The sum of the terms (what you want) can be calculated by multiplying the average times the number of terms. So for Set X that’s 100(101) = 10100. And for Set Y that’s 100(252.5) = 25250.

At this point you can sum those two sets to arrive at 35350. But wait – the question specified “unique terms” and “no repeat elements.” So here you will have to subtract out the numbers that are multiples of both 2 and 5. Which are those? The multiples of 10.

But be careful: Here you do not have 100 multiples of 10; you only have the multiples of 10 up to 200 (where the set of even numbers stops). So you’ll calculate the multiples of 10 between 10 and 200, inclusive. Again, take the average (105) and multiply by the number of terms (20) to get 2100. When you subtract that from 35350 you’ll have the correct answer, 33250.
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Re: Set X consists of the first 100 positive even integers. Set Y consists  [#permalink]

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New post 14 Feb 2017, 01:20
Matruco wrote:
Set X consists of the first 100 positive even integers. Set Y consists of the first 100 positive multiples of 5. If the terms in both sets are combined to create a new set with no repeat elements, what is the sum of all unique terms in the new set?

a) 33250

b) 34350

c) 35750

d) 36300

e) 37500


Similar question: https://gmatclub.com/forum/set-x-consis ... 86754.html
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Re: Set X consists of the first 100 positive even integers. Set Y consists  [#permalink]

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New post 18 Mar 2017, 13:48
find sum of 100 even intergers
find sum of 100 multiplies of 5
and then deduct the sum of multiplies of 10 from 10 to 200
10100+25250-2100=33250 (A)
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Re: Set X consists of the first 100 positive even integers. Set Y consists  [#permalink]

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New post 20 Apr 2018, 02:52
+1 for option A.

First find sum of all multiples of 2 = 10100
Next, find sum of all multiples of 5 = 25250

Sum of all common numbers = 2100

The final value = 35350-2100 = 33250.

Option A it is
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Re: Set X consists of the first 100 positive even integers. Set Y consists  [#permalink]

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New post 20 Aug 2018, 11:52
Matruco wrote:
Set X consists of the first 100 positive even integers. Set Y consists of the first 100 positive multiples of 5. If the terms in both sets are combined to create a new set with no repeat elements, what is the sum of all unique terms in the new set?

a) 33250

b) 34350

c) 35750

d) 36300

e) 37500


The sum of the numbers in X is (2 + 200)/2 x 100 = 202 x 50 = 10,100. Similarly, the sum of the numbers in Y is (5 + 500)/2 x 100 = 505 x 50 = 25,250. When the two sets of numbers are combined into one single set, if duplicate elements (i.e., the elements that are common in both sets) count as twice, then the sum of the new set would be 10,100 + 25,250 = 35,350. However, since the duplicate elements should only count once, not twice, we need to subtract, from 35,350, the sum of the duplicate elements. The duplicate elements are 10, 20, …, 200, with a sum of (10 + 200)/2 x 20 = 210 x 10 = 2,100. So the sum of the new set when the duplicate elements only count once is 35,350 - 2,100 = 33,250.

Answer: A
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Re: Set X consists of the first 100 positive even integers. Set Y consists &nbs [#permalink] 20 Aug 2018, 11:52
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