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

Redeem

how to sovle plz help

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 96
Joined: Fri Apr 23, 2010 1:14 am
Thanked: 1 times
Followed by:1 members

how to sovle plz help

by quantskillsgmat » Thu Jan 05, 2012 10:10 pm
Q A shipping clerk has five boxes of different but unknown weights each weighting less than 100kgs. The clerk weights the box in pairs.The weight obtained are 110,112,113,114,115,116,117,118,120,and 121 kgs. what is the weight in kgs, of the heaviest box.
a)60kg
b)62kg
c)64kg
d)none.
Top
Source: — Problem Solving |
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pm
Thanked: 158 times
Followed by:21 members

by pemdas » Fri Jan 06, 2012 1:57 am
[spoiler]@quantskillsgmat, THIS IS NOT GMAT QUESTION, I AM ANSWERING WITH A SOLUTION PURELY FOR MY INTEREST TO THE PROBLEM SOLVING AREA [/spoiler]

there are ten diff. weights, and it's only possible if all diff.boxes are set in comb. 5C2=10
we need to solve 10 equations with 5 variables to obtain the answer!
let's see, boxes are ABCDE and pairs are A+B,A+C,A+D,B+C,B+D,D+C,A+E,B+E,C+E,D+E
each box is less than 100 kg.
let's sum up all pairs and get 4A+4B+4C+4D=Sum(110,112,113,114,115,116,117,118,120,and 121 kgs.)
ar.sequence (112...118)+110+120+121=(112+118)*7/2 +110+120+121=1156
4A+4B+4C+4D+4E=1156, A+B+C+D+E=289
remember again each box<100 and we have diff.boxes
choice a)60 +59+58+57+56=290 (greater than 289) Could be
choice b)62 +61+60+59+58=300 Could be
choice c)64 +63+62+61+60=310 Could be

choice d) n/a

So all answer choices Could be, and as we see the heaviest is difficult to assign.
Option d none suggests, as with 60,62 and 64 kg-s. the heaviest weights, the other weights could be (289-60)/4, (289-62)/4, (289-64)/4

d

it took me 4 min. to elaborate idea, 30 min to find my calc. under desk and numerous revisions for accuracy ...
quantskillsgmat wrote:Q A shipping clerk has five boxes of different but unknown weights each weighting less than 100kgs. The clerk weights the box in pairs.The weight obtained are 110,112,113,114,115,116,117,118,120,and 121 kgs. what is the weight in kgs, of the heaviest box.
a)60kg
b)62kg
c)64kg
d)none.
Success doesn't come overnight!
Top
Senior | Next Rank: 100 Posts
Posts: 96
Joined: Fri Apr 23, 2010 1:14 am
Thanked: 1 times
Followed by:1 members

by quantskillsgmat » Fri Jan 06, 2012 3:24 am
thanx for explaination . I followed same approach but answer of this question is 62kg
Top
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

by shankar.ashwin » Fri Jan 06, 2012 3:27 am
Let the 5 boxes be a,b,c,d, and e

Let (a,b) be the 2 boxes with minimum weight and (d,e) be the ones with max weight.

GIven, a+b = 110 and d+e = 121

'c' is the box which has the middle weight.

We know 4(a+b+c+d+e) = 1156 -> a+b+c+d+e = 289

Therefore, c = 289 - (a+b) - (d+e) = 58

Since we have 10 different pairs, we can say no 2 weights are the same.

a+b = 110

Now look at the next minimum weight which is 112, 2 kgs more than a+b, therefore b must be 2kgs less than c,

b = 56, therefore a = 54

Now (a,b,c) = (54,56,58)

Also given d+e = 121

Possibilities are (60,61) or (59,62) - Since both weights should be > 58 (c)

Now if we consider (60,61) as a possibility glancing through the options we can find 61+58(c) = 119 is not there in the mentioned set, so eliminate that possibility.

Hence (d,e) can only be (59,62) and the maximum weight is 62- Check if the numbers satisfies all possibilities - it does.

Again, no way it could be a GMAT problem.
Last edited by shankar.ashwin on Fri Jan 06, 2012 4:40 am, edited 1 time in total.
Top
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pm
Thanked: 158 times
Followed by:21 members

by pemdas » Fri Jan 06, 2012 3:47 am
quantskillsgmat wrote:thanx for explaination . I followed same approach but answer of this question is 62kg
in such case, we can only assign the heaviest weight to the pairs with the heaviest box, say it's box A and we solve
Q A shipping clerk has five boxes of different but unknown weights each weighting less than 100kgs. The clerk weights the box in pairs.The weight obtained are 110,112,113,114,115,116,117,118,120,and 121 kgs. what is the weight in kgs, of the heaviest box.
A+B,A+C,A+D,B+C,B+D,D+C,A+E,B+E,C+E,D+E

A+B=121,A+C=120,A+D=118,B+C=117,A+E=116 <-> 5A+B+C+D+E (total 592)
sum up remaining pairs of boxes such as below
B+D,D+C,B+E,C+E,D+E <-> 2B+2C+2D+2E=564 (obtained from 1156-592=564)

B+C+D+E=282

Now subtract, from 5A+B+C+D+E=592 the value B+C+D+E=282 to get 310 and divide by 5. We obtain 310/5=62 which is correct answer according to you

I tried to be consistent, though agree with Shankar it's not GMAT q.
Success doesn't come overnight!
Top
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pm
Thanked: 158 times
Followed by:21 members

by pemdas » Fri Jan 06, 2012 4:24 am
4(a+b+c+d) = 1156 is not accurate, neither is a+b+c+d = 289 <- We have five boxes which pairs total 1156 and you accounted for four boxes only

We don't need to deduce no 2 weights are the same, it's given in the question (boxes with unknown but diff. weights)

Shankar, there can be million possibilities of boxes weighing decimal points of grams
it's arduous to estimate/assign precise weights to the boxes. We can only estimate some minimum or maximum weight box here.
shankar.ashwin wrote:Let the 5 boxes be a,b,c,d, and e

Let (a,b) be the 2 boxes with minimum weight and (d,e) be the ones with max weight.

GIven, a+b = 110 and d+e = 121

'c' is the box which has the middle weight.

We know 4(a+b+c+d) = 1156 -> a+b+c+d = 289

Therefore, c = 289 - (a+b) - (d+e) = 58

Since we have 10 different pairs, we can say no 2 weights are the same.

a+b = 110

Now look at the next minimum weight which is 112, 2 kgs more than a+b, therefore b must be 2kgs less than c,

b = 56, therefore a = 54

Now (a,b,c) = (54,56,58)

Also given d+e = 121

Possibilities are (60,61) or (59,62) - Since both weights should be > 58 (c)

Now if we consider (60,61) as a possibility glancing through the options we can find 61+58(c) = 119 is not there in the mentioned set, so eliminate that possibility.

Hence (d,e) can only be (59,62) and the maximum weight is 62- Check if the numbers satisfies all possibilities - it does.

Again, no way it could be a GMAT problem.
Success doesn't come overnight!
Top
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 am
Thanked: 230 times
Followed by:21 members

by shankar.ashwin » Fri Jan 06, 2012 4:52 am
Ah, I missed 'e' there... I am not really sure if there are several possibilities in question for the given set of equations.

As you say, we have 10 set of linear equations here. Generally solving for a linear equation gives you only one distinct value for each variable.

Ofcourse we cannot directly write the equations here, but I believe there are enough hints to frame them.

Again, no point discussing problems such as these outside the scope of GMAT.

@ quantskillsgmat - Sorry to mention it again, refrain posting questions which are not from GMAT material. GRE or CAT or other exam questions could be posted in other forums where people would help. A couple of them are fine but most of your questions don't seem to be GMAT problems or at least post a disclaimer. I hope I am not asking too much here.
Top
User avatar
Legendary Member
Posts: 626
Joined: Fri Dec 23, 2011 2:50 am
Location: Ahmedabad
Thanked: 31 times
Followed by:10 members

by ronnie1985 » Mon Jan 09, 2012 1:49 am
With the give information, the heaviest box cannot be ascertained. There must be another clue.
Follow your passion, Success as perceived by others shall follow you
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Mon Jun 02, 2014 9:43 pm

by Readerinline » Mon Jun 02, 2014 9:50 pm
I found a pretty simple solution to the question.
Let the boxes be A, B, C, D, E..A being the lightest.
The only certainty is that
A+B is tge lightest pair.
A+C is tge second lightest pair
D+E is the heaviest pair.
C+E is the second heaviest.
Now we calculated C=58kgs by equating 4(A+B+C+D+E)=1156.
C+E = 120.
Thus E = 120-58 = 62 kgs.

Please respond if there is any unattended mistake in this solution!
Thank you!
Top
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 » Tue Jun 03, 2014 6:02 am
The intent of the problem seems to be as follows:
quantskillsgmat wrote:A shipping clerk has five boxes of different weights. Each of the weights is an integer value less than 100. The clerk weighs the boxes in pairs, yielding the following results: 110, 112, 113, 114, 115 ,116, 117, 118, 120, and 121 kgs. What is the weight, in kgs, of the heaviest box?

a)60kg
b)62kg
c)64kg
I've eliminated D, since the GMAT would not include none as an answer choice.
While clearly not a GMAT problem, the problem above does offer a useful take-away:
To solve, we can PLUG IN THE ANSWERS.

In ascending order, let the weights of the 5 boxes be a, b, c, d and e.
Since the combined weight of the 2 heaviest boxes = 121, we get:
d+e = 121.
Since the combined weight of the next greatest pair = 120, we get:
c+e = 120.
Since the combined weight of the 2 lightest boxes = 110, we get:
a+b = 110.

The answer choices represent possible values of e.

Answer choice C: e=64
Since e=64 and d+e = 121, d=57.
Since e=64 and c+e = 120, c=56.
Since a and b must be different integers less than 56, the greatest possible value of a+b = 54+55 = 109.
Doesn't work, since a+b = 110.
Eliminate C.

Answer choice A: e=60
Since e=60 and d+e = 121, d=61.
Not possible, since d must be less than e.
Eliminate A.

The correct answer is B.

Answer choice B: e=62
Since e=62 and d+e = 121, d=59.
Since e=62 and c+e = 120, c=58.
Since a and b must be different integers less than 56 and a+b = 110, two cases seem possible:
Case 1: a=54 and b=56
Case 2: a=53 and b=57.

In Case 2, b+e = 57+62 = 119, which is not included in the list of weights.
Thus, Case 2 is not viable.

Since only Case 1 works, the weights are as follows:
a=54, b=56, c=58, d=59, e=62.
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
Post new topic Post Reply
• Page 1 of 1