Fourteen boys went to collect berries and returned with a total of 80

Question

Fourteen boys went to collect berries and returned with a total of 80 berries among themselves. If every boy collected at least one berry each, then what is the minimum number of pairs of boys that must have collected the same number of berries?

A. 0
B. 1
C. 2
D. 3
E. 4

Are You Up For the Challenge: 700 Level Questions

A. 0
B. 1
C. 2
D. 3
E. 4

Are You Up For the Challenge: 700 Level Questions

RamnathWizako · Answer

It is given that fourteen boys collected a total of 80 berries and each boy collected at least 1 berry.

Objective:  To find the minimum number of pairs of boys that collected the same number of berries.

If the number of boys collecting distinct number of berries is maximum then the pairs of boys collecting the same number of berries will be minimum.

If all 14 boys collected distinct number of berries, then the minimum number of berries collected by them will be:
1 + 2 + 3 + 4 + ........ + 13 + 14 =  \frac{(14∗15)}{2}  = 105

But, we have a total of only 80 berries. We have to bring down 105 to 80 with the least number of pairs of boys collecting the same number of berries.

If 1 + 2 + 3 + 4 + ........ + 13 + 14 has to be reduced accordingly, then the highest distinct number of berries should be removed and the lowest distinct number of berries collected should be duplicated.

1 2 3 4 5 6 7 8 9 10 11 12 13
1

i.e., 2 students collect a berry each and the rest of the 12 students collect 2, 3, 4, ....., 13 berries respectively.
Then, the total number of berries would be 105-14+1= 92. (which is still greater than 80!)

Repeating the same operation once again, i.e., removing the greatest distinct number and adding the least distinct number such that the sum is minimized:

1 2 3 4 5 6 7 8 9 10 11 12
1 2

i.e., 2 students collect a berry each, another set of 2 students collect 2 berries each, and the rest of the 10 students collect 3, 4, ....., 12 berries respectively.

Then, the total number of berries would be 92-13+2=81 (which is 1 greater than 80!!)

We want our total to be 80. We still need to bring down the total by 1.

In order to bring down the total by 1, the only way out is to remove 12 and add 11.

1 2 3 4 5 6 7 8 9 10 11
1 2 11

i.e., 2 students collect a berry each, another set of 2 students collect 2 berries each, another set of 2 students collect 11 berries each and the rest of the 8 students collect 3, 4, ....., 10 berries respectively.
Now, the total number of berries would be 81-12+11= 80. (Voilà!)

Thus, the minimum number of boys collecting the same number of berries will be  6  i.e., 3 pair of students will be collecting 1, 2 & 11 berries respectively.

The answer is  Choice (D)

Regor60 · Answer

Since each boy is allocated at least one item there remain after the initial distribution:

80-14 = 66

To minimize the number of pairs receiving the same number we need to maximize the number of distinct distributions of items.

This can be accomplished by distributing the 66 so that each successive person receives one more than the prior person, thus preserving the distinction requirement, or:

1+2+3... etcetera for a total of 66.

The sum of such an arithmetic series is:

n(n+1)/2 = 66

n(n+1) = 132

By observation 11*12 works, for:

n = 11

So 11 of 14 boys each receive a distinct number of items, leaving:

14-11 = 3 boys each receiving one item.

The number of pairs of boys that can be extracted from 3 is:

3!/2!1! = 3 pairs

RamnathWizako · Answer

Hello Regor60,
Thanks for pointing it out. 
There is no error in the question.
The answer is modified now.

Cheers.

raghav20 · Answer

Each boy has at least one berry, so if we allocate each boy 1 berry after which we have 66 remaining 
To have each boy have a unique number of berries with minimum berries we start allocating additional berries with the increase of one i.e. 1,2,3,4,5....... and on reaching 11, the total number of berries is exhausted, and we are left with 3 boys having 1 berry each which is option D

mimundertaker · Answer

The way I did this question was

We'll try to minimise number of boys with same number of berries i.e. maximise number of boys with different numbers of berries. So, if say boy 1 has 1 berry, let's give boy 2 - 2 berries, let's give boy 3 - berries and so on. When you finish this process with all 14 boys, you see that the sum of the berries used now is 105 i.e. 25 more than the number of berries we have which is 80.

Now, if we go back and take away the number of berries boy 13 and boy 14 had, which is 13 and 14 respectively, we're at 78 berries which is well within the upper limit of 80 that we have in this question. To distribute the remaining 2 berries (since every boy got at least 1 berry each), I gave boy 13 and boy 14 a berry each.

This means we have 3 people with 1 berry each (boy 1, boy 13 and boy 14) - that gives us 3 such pairs (boy1-boy13; boy13-boy14 and boy14-boy1).
Thus, answer - 3 pairs minimum.

Fourteen boys went to collect berries and returned with a total of 80

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Fourteen boys went to collect berries and returned with a total of 80

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