If x is the sum of the first 10 positive multiples of 3 and

Max@Math Revolution wrote:[GMAT math practice question]

If x is the sum of the first 10 positive multiples of 3 and y is the sum of the first 10 positive multiples of 2, which of the following is a factor of x – y?

A. 2
B. 3
C. 5
D. 7
E. 9

For any EVENLY SPACED SET:
AVERAGE = (BIGGEST + SMALLEST)/2.
SUM = (COUNT)(AVERAGE).

The first 10 positive multiples of 3 are as follows:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
Thus:
Average = (biggest + smallest)/2 = (30+3)/2 = 16.5.
x = sum = (count)(average) = 10*16.5 = 165.

The first 10 positive multiples of 2 are as follows:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
Thus:
Average = (biggest + smallest)/2 = (20+2)/2 = 11.
y = sum = (count)(average) = 10*11 = 110.

Result:
x-y = 165 - 110 = 55 = 5*11.
Of the five answer choices, only C is a factor of x-y.

The correct answer is C.

=>
x – y = ( 3 + 6 + 9 + … + 30 ) - ( 2 + 4 + 6 + … + 20 )
= 3(1 + 2 + … + 10 ) – 2( 1 + 2 + 3 + … + 10 )
= ( 3 – 2 ) ( 1 + 2 + 3 + … + 10 )
= 55
= 5 * 11

Therefore, the answer is C.
Answer: C

Max@Math Revolution wrote:[GMAT math practice question]

If x is the sum of the first 10 positive multiples of 3 and y is the sum of the first 10 positive multiples of 2, which of the following is a factor of x – y?

A. 20
B. 35
C. 55
D. 75
E. 90

Recall that the sum = average x quantity. The first positive multiple of 3 is 3 and the last (10th) multiple of 3 is 30. Thus, the average is (30 + 3)/2 = 33/2, and the sum of the first 10 positive multiples of 3 is:

33/2 x 10 = 165

Similarly, the first positive multiple of 2 is 2 and the last (10th) multiple of 2 is 20. Thus the average is (20 + 2)/2 = 11, and the sum of the first 10 positive multiples of 2 is:

11 x 10 = 110

x - y = 165 - 110 = 55 = 5 x 11

We see 5 is a factor of 55.

Alternate Solution:

The first positive multiple of 3 is 3 and the last (10th) multiple of 3 is 30. Thus,

x = 3 + 6 + 9 + … + 30

Similarly, the first positive multiple of 2 is 2 and the last (10th) multiple of 3 is 20. Thus,

y = 2 + 4 + 6 + … + 20

Let’s subtract the second equation from the first:

x - y = (3 - 2) + (6 - 4) + (9 - 6) + … + (30 - 20)

x - y = 1 + 2 + 3 + … + 10

Recall that the formula for the sum of the first n consecutive positive integers is n(n + 1)/2, where n is the largest number in the list. Thus, we have:

x - y = (10 x 11)/2 = 55

We see 5 is a factor of 55.

Answer: C