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

Redeem

tough Q

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 279
Joined: Fri Nov 05, 2010 5:43 pm
Thanked: 15 times
Followed by:1 members

tough Q

by mehrasa » Thu Sep 08, 2011 11:20 pm
There are 26 students who have read a total of 56 books among them. The only
books they have read, though, are Aye, Bee, Cod, and Dee. If 10 students have only
read Aye, and 8 students have read only Cod and Dee, what is the smallest number
of books any of the remaining students could have read?
2 4 1 5 6

is there anyone help me to solve this?

[spoiler]OA:A[/spoiler]
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1309
Joined: Mon Apr 04, 2011 5:34 am
Location: India
Thanked: 310 times
Followed by:123 members
GMAT Score:750

by cans » Fri Sep 09, 2011 12:06 am
a,b,c,d - books.
10 ppl only A. total 10 books read.
only c&d = 8. total 16 books read.
remaining students = 8. remaining books = 30.
a person can read max 4. thus 8 ppl can read max 32.
if 7 ppl read 4 max, 28 books read. Thus left =2.
IMO A
If my post helped you- let me know by pushing the thanks button ;)

Contact me about long distance tutoring!
[email protected]

Cans!!
Top
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Wed Jul 20, 2011 3:10 am
Thanked: 2 times

Venn diagram approach

by pavand » Sat Sep 10, 2011 12:57 pm
26 students have read 4 books in such a way that they make 56 combinations
18 students have made 2 combinations
So 8 students should read a minimum (smallest value) of x book/books to make another 54 combinations

Minimum of x so that 8Cx <= 54

8C1 = 8
8C2 = 28
8C3 = 56
8C4 = 70

The answer is x=1 (Option C)

(If question asked is maximum (highest value), answer will be 2)
Last edited by pavand on Sat Sep 10, 2011 11:21 pm, edited 3 times in total.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 279
Joined: Fri Nov 05, 2010 5:43 pm
Thanked: 15 times
Followed by:1 members

by mehrasa » Sat Sep 10, 2011 8:00 pm
Hey pavand

it is correct that there are 30 other books to be covered by 8 students.. but why you suppose that there are 5 books to be covered while we have only 4 books (a,b,c,d).. Also i could not underestand how you conclude the last sentence
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 143
Joined: Sat Nov 06, 2010 9:06 pm
Thanked: 10 times
Followed by:1 members

by edvhou812 » Sat Sep 10, 2011 8:28 pm
10 read only Aye (1 Book); 8 have read both Cod and Dee (2 Books). This means 18 of 26 students have read 26 books.
Students Remaining: 26-18=8
Books Remaining: 56-26=30

So we need to divy up 30 books, with four different books available to read, among 8 students and we want to find out who can read the least amount of books among those 8 students. Since we are looking for the smallest, we can pile on the reading among the other 7:
7*4=28
30-28=2
The least that one student can possibly read among the 8 remaining students is 2. Answer: A
I don't know what to say, really. Three minutes to the biggest battle of our professional lives. You find out life's this game of inches, so is football. Because in either game - life or football - the margin for error is so small. I mean, one half a step too late or too early and you don't quite make it. One half second too slow, too fast and you don't quite catch it. I'll tell you this, in any fight it's the guy whose willing to die whose gonna win that inch. That's football guys, that's all it is. Now, what are you gonna do?
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 425
Joined: Wed Dec 08, 2010 9:00 am
Thanked: 56 times
Followed by:7 members
GMAT Score:690

by LalaB » Mon Sep 19, 2011 9:06 pm
10 students have read 10 A books
8 students have read 8C and 8D books (total-16 books)

then 8 students (26(total) -10-8) need to read 30 books (56(total)-10-16)

to minimize one point, we need to maximize other ones .so
30 total books minus -28 books (if each of 7 students have read 4 books (total 28) ) = 2 books

answ is A
Top
Post new topic Post Reply
• Page 1 of 1