BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GMAT prep question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 26
Joined: Fri Mar 12, 2010 1:39 pm
Thanked: 1 times

GMAT prep question

by sarahw_gmat » Thu Feb 03, 2011 8:21 pm
Appreciate if somebody can explain solution for below questions.

thanks.

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?


1. 550
2. 551
3. 560
4. 561
5. 572


Ans: 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?

Ans: 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?

Ans : 13
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Thu Feb 03, 2011 8:36 pm
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
User avatar
Legendary Member
Posts: 1255
Joined: Fri Nov 07, 2008 2:08 pm
Location: St. Louis
Thanked: 312 times
Followed by:90 members

by Tani » Thu Feb 03, 2011 8:39 pm
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.
Tani Wolff
Top
Post new topic Post Reply
• Page 1 of 1