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

Redeem

ribbons in bag

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Wed Jul 20, 2011 12:36 pm
Thanked: 2 times

ribbons in bag

by sumasajja » Wed Aug 03, 2011 7:23 am
there are only two kinds of ribbons in a bag of 8 m and 13 m
combined length of all the ribbons is 100
A)no of 8m ribbons
B) no of 13m ribbons

a) A is greater
b) B is greater
c) both A and B are equal
d) data insufficient
Top
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Sun Jun 26, 2011 5:45 pm
Thanked: 2 times

by edge » Wed Aug 03, 2011 7:41 am
Multiples of 13: 13, 26, 39, 52, 65, 78, 91
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96

The only pair of numbers that add up exactly to 100 are 52 (13 * 4) and 48 (8 * 6).

Therefore, A = 6 and B = 4 and A > B.
Top
Junior | Next Rank: 30 Posts
Posts: 29
Joined: Tue Jul 12, 2011 1:07 pm
Thanked: 8 times
Followed by:1 members

by beatthegmat.garry » Wed Aug 03, 2011 7:51 am
Lets assume n1 be the no. of 8m ribbons.
Lets assume n2 be the no. of 13m ribbons.

Since total length=100, we can write:
8*n1 +13*n2=100 .........(1)

Lets guess values of n1 and n2 such that eqn.(1) will be satisfied
You will only find just one combination which is n1=6, n2=4.

So I believe there are 6 ribbons of 8m and 4 ribbon of 13m.

Hence (A) is greater.
Top
User avatar
GMAT Instructor
Posts: 509
Joined: Wed Apr 21, 2010 1:08 pm
Location: Irvine, CA
Thanked: 199 times
Followed by:85 members
GMAT Score:750

by tpr-becky » Wed Aug 03, 2011 7:58 am
This is a GRE problem but the ideas are the same (question type isn't on the GMAT)

Another way to approach these is to methodically approach the different multiples:

the most 13m ribbons you could have is 7 - 7(13) is 91 100 - 91 = 9 (not divisible by 8)
if you have 6 then 13(6) = 78 100 - 78 = 22 (not divisible by 8)
if you have 5 13(5) = 65; 100 - 65 = 35 (not divisible by 8)
if you have 4 13(4)= 52; 100 - 52 = 48 (that is 6(8)) - so there are more 8's elminate B and C.

you can then try the rest but none are divisible by 8 so the answer is A.
Becky
Master GMAT Instructor
The Princeton Review
Irvine, CA
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 312
Joined: Tue Aug 02, 2011 3:16 pm
Location: New York City
Thanked: 130 times
Followed by:33 members
GMAT Score:780

by gmatboost » Wed Aug 03, 2011 8:29 am
Consider 8a + 13b = 100

8a is EVEN
100 is EVEN

So, 13b must also be EVEN, since E + E = E

That means b must be EVEN, since 13 is not.

So, you should only spend time testing even values of b (6, 4, 2, maybe 0)
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.

Also, check out the most useful GMAT Math blog on the internet here.
Top
Post new topic Post Reply
• Page 1 of 1