sarahw_gmat wrote: Q. if x is equal to the sum of the even integers from 40 to 60, inclusive, and y is the number of even integers from 40 to 60, inclusive, what is the value of x+y?

A. 550
B. 551
C. 560
D. 561
E. 572

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

Since we're adding only the even integers, the distance between each successive pair is 2.
Thus, the number of even integers from 40 to 60 = (60-40)/2 + 1 = 11.

To determine the sum of evenly spaced integers:

Sum = (number of integers) * (average of biggest and smallest)

As shown above, number = 11.
Average of biggest and smallest = (60+40)/2 = 50.
Sum = (number of integers) * (average of biggest and smallest) = 11*50 = 550.

x+y = sum + number = 550 + 11 = 561.

The correct answer is D.

Q. if x is equal to the sum of the even integers from 40 to 60, inclusive, and y is the number of even integers from 40 to 60, inclusive, what is the value of x+y?
think of making pairs of the numbers you are adding:
40 + 60 = 100
42 + 58 = 100
44 + 56 = 100
46 + 54 = 100
48 + 52 = 100
+50 = 50
Total = 550
plus 11 terms = 561

Q. At a certain supplier, a machine of type A costs $20,000 and a machine of type B costs $50,000. Each machine can be purchased by making a 20% down payment and repaying the remainder of the cost and the finance charges over a period of time. If the finance charges are equal to 40% of the remainder of the cost, how much less would 2 machines of type A cost than 1 machine of type B under this arrangement?


1 type A machine:
down payment = 20% * $20,000 = $4,000
finance charges = 40% * $16,000 = $6,400
plus remaining cost = $20,000 - $4,000 = $16, 000
total for one type A machine = $26,400
Two type A machines = $52,800

1 type B machine:
down payment = 20% of $50,000 = $10,000
finance charges = 40% * $40,000 = $16,000
plus remaining cost = $40,000
Total = $66,000

$66,000 - $52,800 = $13,200


Q. In a class of 30 students, 2 students did not borrow any books from the library, 12 students each borrowed 1 book, 10 students each borrowed 2 books, and the rest of the students each borrowed at least 3 books. If the average number of books borrowed per student was 2, what is the maximum number of books that any single student could have borrowed?

2*0 = 0
12*1 = 12
10*2 = 20

Total 32 books and 24 students

total borrowed = 2*30 = 60
28 books left
6 students left
5 borrowed 3 each = 15
28 - 15 = 13 books left.
The sixth student borrowed 13.