how to sovle plz help

[quantskillsgmat]

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.

[pemdas]

[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 ...

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.

[quantskillsgmat]

thanx for explaination . I followed same approach but answer of this question is 62kg

[shankar.ashwin]

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.

[pemdas]

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.

[pemdas]

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.

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.

[shankar.ashwin]

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.

[ronnie1985]

With the give information, the heaviest box cannot be ascertained. There must be another clue.

[Readerinline]

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!

[GMATGuruNY]

The intent of the problem seems to be as follows:

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.