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

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Veritas Prep

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

OA A.

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Hi All,

We're told that Set X consists of the first 100 positive EVEN integers and 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, we're asked for the sum of all unique terms in the new set. This question involves 'bunching' and requires some Arithmetic.

To start, we can find the sums of Set X and Set Y without too much trouble:

Set X includes the EVEN integers from 2 to 200, inclusive (2, 4, 6, ...... 198, 200). We can 'pair' terms into pairs of 202 (re: 2 + 200, 4 + 198, etc.). This will gives us 50 pairs of 202, which totals (50)(202) = 10,100

Set Y includes the multiples of 5 from 5 to 500, inclusive (5, 10, 15, ..... 495, 500). We can 'pair' terms into pairs of 505 (re: 5 + 500, 10 + 495, etc.). This will gives us 50 pairs of 505, which totals (50)(505) = 25,250

When combining these two sets, there WILL be duplicates though - all of the multiples of 10 from 10 to 200, inclusive will be in BOTH sets. We're told NOT to count the duplicate entries though, so we would have to subtract one of each of those terms from the total. Again, we can 'bunch' the numbers into 'pairs' of 210 (10 + 200, 20 + 190, etc.). This will gives us 10 pairs of 210, which totals (10)(210) = 2100.

Total = 10,100 + 25, 250 - 2100 = 33, 250

Final Answer: A

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

[Jeff@TargetTestPrep]

Veritas Prep

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

[swerve]

Sum of first 100 positive even integers: 2 + 4 + 6 + 8 + . . . + 200 = [100 (2+200)]/2 = 10100
Sum of first 100 positive multiples 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. Option A. Regards!