NEW HERE? LEARN MORE
closeclose
Last visit was: 11 May 2024, 07:09 It is currently 11 May 2024, 07:09
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

There are 5 boxes of different weights...

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
avatar
aayushagrawal
Intern
Intern
Joined: 01 Nov 2015
Posts: 14
Own Kudos [?]: 108 [16]
Given Kudos: 28
Send PM
Reviews Badge
There are 5 boxes of different weights... [#permalink] New post   08 Jun 2016, 06:08
4
Kudos
12
Bookmarks
00:00
A
B
C
D
E

Difficulty:

45% (medium)

Question Stats:

71% (02:52) correct 29% (02:46) wrong based on 108 sessions
Hide Show timer Statistics



There are 5 boxes of different weights. They are weighed in pairs taken 2 at a time and the weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. What is the sum of the weights of the 5 boxes ?

A) 250
B) 289
C) 290
D) 295
E) 305
ShowHide Answer
Official Answer
B
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11236
Own Kudos [?]: 32429 [2]
Given Kudos: 301
Send PM
Reviews Badge Forum Quiz Mode
There are 5 boxes of different weights... [#permalink] New post   08 Jun 2016, 06:52
2
Kudos
Expert Reply
aayushagrawal wrote:



There are 5 boxes of different weights. They are weighed in pairs taken 2 at a time and the weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. What is the sum of the weights of the 5 boxes ?

A) 250
B) 289
C) 290
D) 295
E) 305


Hi,

since the weights given are of 2 boxes taken at one time..
so the SUM of all the weights measured taking two boxes at any time will be 4 times total weight of all five combined..

We choose 2 boxes out of 5 in 5C2 ways = 5!/3!2! = 10 ways....
How many ways 1 box say A will not be included = choosing 2 out of 4 boxes other than A = 4C2 = 4!/2!2! =6..
so NUMBER of times A has been counted is 10-6 = 4.....
similarily all other boxes also are counted 4 times...
eg say A,B,C,D,E - AB,AC,AD,AE has A in it

Now since these 10 weights of pair of boxes given has 4 times each box..
we divide the SUM of these 10 weights to get SUM of single boxes

\(\frac{110+112+113+114+115+116+117+118+120+121}{4}\)...
UNITS digit is 6 of the NUMERATOR..

so when you divide by 4, 0 and 5 cannot be the UNITS digit..
ONLY B has units digit other than 0 and 5..
ans B
User avatar
L
chetan2u
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11236
Own Kudos [?]: 32429 [1]
Given Kudos: 301
Send PM
Reviews Badge Forum Quiz Mode
Re: There are 5 boxes of different weights... [#permalink] New post   08 Jun 2016, 09:48
1
Kudos
Expert Reply
fantaisie wrote:
chetan2u wrote:
aayushagrawal wrote:



There are 5 boxes of different weights. They are weighed in pairs taken 2 at a time and the weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. What is the sum of the weights of the 5 boxes ?

A) 250
B) 289
C) 290
D) 295
E) 305


Hi,

since the weights given are of 2 boxes taken at one time..
so the SUM of all the weights measured taking two boxes at any time will be 4 times total weight of all five combined..
\(\frac{110+112+113+114+115+116+117+118+120+121}{4}\)... UNITS digit is 6 of the NUMERATOR..
so when you divide by 4, 0 and 5 cannot be the UNITS digit..
ONLY B has units digit other than 0 and 5..
ans B


Thank you for the explanation chetan2u! I was just looking to understand exactly, why is it that the number is 4 times the total weight of all five boxes combined. Would it be correct to say that every number is counted 5 times instead of one, which makes it 4 times too much. And since it applies to every single number, we divide by 4 to cancel repetition? Thank you!


Hi,

It is 4 times the SUM of all five boxes because-

We choose 2 boxes out of 5 in 5C2 ways = 5!/3!2! = 10 ways....
How many ways 1 box say A will not be included = choosing 2 out of 4 boxes other than A = 4C2 = 4!/2!2! =6..
so NUMBER of times A has been counted is 10-6 = 4.....
similarily all other boxes also are counted 4 times...
eg say A,B,C,D,E - AB,AC,AD,AE has A in it

Now since these 10 weights of pair of boxes given has 4 times each box..
we divide the SUM of these 10 weights to get SUM of single boxes
User avatar
L
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18841
Own Kudos [?]: 22191 [1]
Given Kudos: 285
Location: United States (CA)
Send PM
Re: There are 5 boxes of different weights... [#permalink] New post   09 Oct 2017, 17:25
1
Kudos
Expert Reply
aayushagrawal wrote:



There are 5 boxes of different weights. They are weighed in pairs taken 2 at a time and the weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. What is the sum of the weights of the 5 boxes ?

A) 250
B) 289
C) 290
D) 295
E) 305


When the boxes are weighted 2 at a time, the weight of each box is counted 4 times. Thus, the sum 110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121 = 1,156 must be 4 times the sum of the weights of the 5 boxes. Therefore, the sum of the weights of the 5 boxes is 1,156/4 = 289.

Note: If you can’t see why the weight of each box is counted 4 times, we can label the boxes as A, B, C, D, and E. Thus, when we pair them up, we have:

AB
AC
AD
AE
BC
BD
BE
CD
CE
DE

We see that each of the letters appears 4 times in the list.

Answer: B
avatar
mukundsureka
Intern
Intern
Joined: 08 Jun 2016
Posts: 1
Own Kudos [?]: 1 [1]
Given Kudos: 2
Send PM
Re: There are 5 boxes of different weights... [#permalink] New post   08 Jun 2016, 06:30
1
Bookmarks
Let the 5 boxes be a, b, c, d and e. The ten weights correspond to all the possible combinations of 2 weights out of 5.
On adding all the given weights, each boxes weight gets added 4 times:
a+b, a+c, a+d, a+e.
Similarly for b:
b+c, b+d, b+e and a+b (already included above)
Similarly for c, d and e.
Therefore, 110+112+113+114 .....+120+121 = 4* (a+b+c+d+e)
or, 4* (a+b+c+d+e) = 1156
or, (a+b+c+d+e) = 1156/4 = 289
User avatar
fantaisie
Manager
Manager
Joined: 18 May 2016
Posts: 51
Own Kudos [?]: 106 [0]
Given Kudos: 105
Concentration: Finance, International Business
GMAT 1: 720 Q49 V39
GPA: 3.7
WE:Analyst (Investment Banking)
Send PM
There are 5 boxes of different weights... [#permalink] New post   08 Jun 2016, 09:40
chetan2u wrote:
aayushagrawal wrote:



There are 5 boxes of different weights. They are weighed in pairs taken 2 at a time and the weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. What is the sum of the weights of the 5 boxes ?

A) 250
B) 289
C) 290
D) 295
E) 305


Hi,

since the weights given are of 2 boxes taken at one time..
so the SUM of all the weights measured taking two boxes at any time will be 4 times total weight of all five combined..
\(\frac{110+112+113+114+115+116+117+118+120+121}{4}\)... UNITS digit is 6 of the NUMERATOR..
so when you divide by 4, 0 and 5 cannot be the UNITS digit..
ONLY B has units digit other than 0 and 5..
ans B


Thank you for the explanation chetan2u! I was just looking to understand exactly, why is it that the number is 4 times the total weight of all five boxes combined. Would it be correct to say that every number is counted 5 times instead of one, which makes it 4 times too much. And since it applies to every single number, we divide by 4 to cancel repetition? Thank you!
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32933
Own Kudos [?]: 828 [0]
Given Kudos: 0
Send PM
Re: There are 5 boxes of different weights... [#permalink] New post   15 Mar 2022, 17:48
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
gmatclubot
Re: There are 5 boxes of different weights... [#permalink]
15 Mar 2022, 17:48
Moderators:
Bunuel
Math Expert
93159 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions