A shipping clerk has five boxes, each weighing less than 100 kg.

Question

A shipping clerk has five boxes, each weighing less than 100 kg. The clerk weighs the boxes in pairs, obtaining the following weights: 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121 kg. What are the weights of the lightest and heaviest boxes?

A. 52 and 66
B. 54 and 64
C. 54 and 62
D. 56 and 60
E. 58 and 62

A. 52 and 66
B. 54 and 64
C. 54 and 62
D. 56 and 60
E. 58 and 62

sherlocked221B · Answer

A shipping clerk has five boxes, each weighing less than 100 kg. The clerk weighs the boxes in pairs, obtaining the following weights: 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121 kg. What are the weights of the lightest and heaviest boxes?

A. 52 and 66
B. 54 and 64
C. 54 and 62
D. 56 and 60
E. 58 and 62

A. 52 and 66
B. 54 and 64
C. 54 and 62
D. 56 and 60
E. 58 and 62

Bunuel   KarishmaB   Sajjad1994   GMATinsight

Could you please share the solution for this?
Can we expect this question in the exam? Seems very lengthy!

My approach to the question:

1. Rule out D & E because based on the options -

In D, the greatest weight is 60, which is not possible since the greatest combination of 121 requires the other weight to be 61.   Rule out D  
In E, the smallest weight is 58, which is not possible since the least combination of 110 requires the other weight to be 52.   Rule out C

2. Now, trial and error for the remaining options -

Option A:  Lowest 52 and Highest 66
-> For greatest combination of 121, the other number apart from 66 will be 55.

Taking 55 as another value, and summing up with lowest 52 - the sum is 107, which is not in the list.   Rule out A

Option B:  Lowest 54 and Highest 64
-> For greatest combination of 121, the other number apart from 64 will be 57.

Taking 57 as another value, and summing up with lowest 54 - the sum is 111, which is not in the list.   Rule out B

After ruling out everything, C is the option.

To verify:

Option C:  Lowest 54 and Highest 62
-> For greatest combination of 121, the other number apart from 62 will be 59.
-> For least combination of 110, the other number apart from 54 is 56.

We have 4 weights already, taking 58 as other weight -> with trail and error to ensure the unit digit is in the given list, we end up with all the combinations from the list.

KarishmaB · Answer

There is no ad hoc hit and trial involved hence I would say it is fair play.

All sums are distinct which means weights of all boxes are distinct.

The greatest sum (121 so average of two greatest numbers is 60.5) will be obtained by adding the two greatest terms. Hence 60 cannot be the greatest number. Ignore (D)
The smallest sum will be obtained by adding two smallest terms (110 so average of two smallest numbers is 55). Hence 56 and 58 cannot be the smallest numbers. Ignore (D) and (E).

The greatest term cannot be 66 because to get 121, you would need 55 to be the second largest number. But then if 52 is the smallest, the 5 numbers must be 52, 53, 54, 55, 66. We don't have a sum of 105. Ignore (A)

This means the smallest number is 54 for sure. So next number must be 56 to get 110. Next number must be 58 to get 112 (we cannot have another 56). Next number must be 59 to get 113. And the greatest would be 62 to get a sum of 121 when we add 59 and 62. 
Numbers are 54, 56, 58, 59, 62

Answer (C)

LamboWalker · Answer

Firstly, amazing question!

Now, there are 5 boxes. Total ways of selecting 2 out 5 = 5c2 = 10 ways. Therefore, we are given 10 values corresponding to the total weight of each pair.

The only way I think to solve this is POE.

A - If there is a 52, then there would have to be a 58 in order to get the lowest value of 110. If there would have to be a 58, then there would have to be a value of 66 + 58 = 124. Since this is not a value present in the question, we eliminate A.

B - If there is a 64, there would have to be a 57. If there was a 57, then 57 + 54 i.e. 111 must have to be among the 10 values. Since that is not the case, eliminate B.

C - Since similar trials as above are not leading to a negative outcome, keep this for now.

D - If the highest value itself is 60, then 121 cannot be the highest. Thus, eliminate D.

E - If the lowest value is 58, the minimum value would have to be a 116. However, since the minimum value is 110, this option is not possible. Eliminate.

Therefore, answer is C.

Arjunsaj · Answer

A shipping clerk has five boxes, each weighing less than 100 kg. The clerk weighs the boxes in pairs, obtaining the following weights: 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121 kg. What are the weights of the lightest and heaviest boxes?

A. 52 and 66
B. 54 and 64
C. 54 and 62
D. 56 and 60
E. 58 and 62

Rather than trial and error, we can find the solution using the formula itself, let the boxes be a,b,c,d,e with weights a<b<c<d<e.
we can see that a+b=110 and d+e= 121 => a+b+c+d+e= 231 -eq1
also 4*(a+b+c+d+e) = 110+112+113+114+115+116+117+118+120+121=1156 => a+b+c+d+e= 289 -eq2
from eq1 and eq2 we can get c = 58.
Now as a is the smallest weight and c is 3rd smallest weight their sum of weight will be 2nd smallest in the combination so a+c will be 112 implying a=54
Now as d is the largest weight and c is 3rd largest weight their sum of weight will be 2nd largest in the combination so e+c=120 implying e=62

wasario · Answer

Shouldn't equation 1 be a + b + d + e = 231? Not sure where you got the c from there. Great solution otherwise!

lordwells · Answer

Ok here is my try (it is for sure a little bit longer than required, but I tried to find out a method for working out not just the weights of the lightest and heaviest boxes, but those of all five boxes

1) first of all, all the pairs weight differently, that is to say the boxes have different weights one from the other. For the sake of simplicity, let's call them “a, b, c, d and e” where “a” is the lightest box and “e” is the heaviest;

2) the first two sums (that is the lowest ones) are necessarily and respectively given by

(a + b)

and

(a + c)

As a consequence 112 - 110 = 2 = c - b

Or c = b + 2;

3) Equivalently, the biggest and second-biggest sums are necessarily and respectively given by:

(d + e)

and

(c + e)

As a consequence 121 - 120 = 1 = d - c

Or d = c + 1

And, exploiting the result of point 2) above:

d = b + 3;

4) Now things are complicating a little bit, but we can work it out

In particular, the third and fourth lightest pairs of boxes are necessarily given by:

(a + d)

and

(b + c)

Unfortunately, in this case, we do not know whether is the lightest between the two (that is, the “necessarily and respectively” as per above is now just a “necessarily”)

But the info comes at hand nevertheless, since:

(Fourth lightest pair + Third lightest pair) -

(Second lightest pair) - (Lightest pair) = (114 + 113) - 112 - 110 = 5 = (b + c + a + d) - (a + c) - (a + b) = d - a

Or d = a + 5

And, exploiting the result of point 3) above:

a = b -

As per the result of point 4) above:

110 = a + b = 2*b - 2

b = 56;

6) Now it is pretty much trivial to work out the weights of all the other boxes which happen to be:

a = 54

b = 56

c = 58

d = 59

e = 62

Thus, the correct answer is C.

Posted from my mobile device

KunalShinde · Answer

I will share my approach.
We know that all the weights are distinct from their sums. Let's say a, b, c, d, e in descending order of their weights.

On the basis of 10 different combinations of weight we can conclude that there are 4 even number and 1 odd number.(With 2odds and 3evens total odd sums will be 6, likewise can be derived for other combinations)
4odds and 1even is not possible from the answer choices given.

Since the highest sum is odd then either 'a' or 'b' shall be odd. From the answer choices, 'a' must be even.
Hence we have, a:even, b:odd, c:even, d:even, e:even

Now, considering all the odd sums in descending order we can write
a+b=121
b+c=117
b+d=115
b+e=113

Therefore, there is a difference of '8' in 'a' and 'e'.

Only option choice that suffices is  option C

Posted from my mobile device

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A shipping clerk has five boxes, each weighing less than 100 kg.

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