1
Question
There are some boxes labelled 1, 2, 3, 4... and so on up to 2n, where n>6 such that, for k = 1, 2, 3, 4, .... 2n, there are exactly k boxes labelled k. The total number of boxes is N. Now, consider the following two cases for the number of footballs (fi) in the box labelled i:
Case 1: fi=i+1, if i is odd; = i + 2 if i is even.
Case 2: fi=i+2, if i is odd; = i + 1 if i is even.
The difference between the total number of footballs in the N boxes in Case 1 and Case 2 is:
There are some boxes labelled 1, 2, 3, 4... and so on up to 2n, where n>6 such that, for k = 1, 2, 3, 4, .... 2n, there are exactly k boxes labelled k. The total number of boxes is N. Now, consider the following two cases for the number of footballs (fi) in the box labelled i:
Case 1: fi=i+1, if i is odd; = i + 2 if i is even.
Case 2: fi=i+2, if i is odd; = i + 1 if i is even.
The difference between the total number of footballs in the N boxes in Case 1 and Case 2 is:
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Solution
The correct option is B
Method 1- Conventional Approach
Clearly, in both cases, fi=(i+1) or (i+1) + 1.
Therefore in both the cases, (i+1) is common.
As we need to find the difference in number of footballs, we can ignore (i+1).
Thus, effectively in each box, there will be 0 or 1 football.
We get:
Case 1: fi=0 if i is odd, = 1 if i is even.
Case 2: fi=1 if i is odd, = 0 if i is even.
Upon observation, we can conclude that the required difference is nothing but the difference in the number of boxes with odd numbered labels and the number of boxes with even numbered labels.
Also, number of boxes with label (x) is 1 more than number of boxes with label (x-1).
From 2n labels, we get n pairs of labels with consecutive numbers.
Therefore, required difference is n×1=n.
Or Case 1 - Case 2 = [{2+4+6+....+(2n)}-{1+3+5+....+(2n-1)}]
=2n(n+1)2−n2=n
Total number of boxes = 1+2+3+....+2n= n(2n+1) = N
Therefore, 2n2+n−N=0
n=−1±√1+8N4
Since n cannot be negative, n=(√8N+1)−14.
Therefore, answer option (b) is the correct answer choice.
Method 2- Assumption
This is a question based on Variables to Variables.
Assume n=1, hence the number of boxes labelled 1, will be 1 and the number of boxes labelled 2 will be 2. Total number of boxes (N) = 3.
Case 1
When i=1, f1=2 and when i=2, f2=4
Case 2
When i=1,f1=3 and when i=2, f2=3
Difference in total number of footballs in both cases = (2+4+4) - (3+3+3) = 1
Now, on substitution of N=3 in options, eliminate those options where you are not getting 1.
Only option (b) satisfies the required condition.
Method 1- Conventional Approach
Clearly, in both cases, fi=(i+1) or (i+1) + 1.
Therefore in both the cases, (i+1) is common.
As we need to find the difference in number of footballs, we can ignore (i+1).
Thus, effectively in each box, there will be 0 or 1 football.
We get:
Case 1: fi=0 if i is odd, = 1 if i is even.
Case 2: fi=1 if i is odd, = 0 if i is even.
Upon observation, we can conclude that the required difference is nothing but the difference in the number of boxes with odd numbered labels and the number of boxes with even numbered labels.
Also, number of boxes with label (x) is 1 more than number of boxes with label (x-1).
From 2n labels, we get n pairs of labels with consecutive numbers.
Therefore, required difference is n×1=n.
Or Case 1 - Case 2 = [{2+4+6+....+(2n)}-{1+3+5+....+(2n-1)}]
=2n(n+1)2−n2=n
Total number of boxes = 1+2+3+....+2n= n(2n+1) = N
Therefore, 2n2+n−N=0
n=−1±√1+8N4
Since n cannot be negative, n=(√8N+1)−14.
Therefore, answer option (b) is the correct answer choice.
Method 2- Assumption
This is a question based on Variables to Variables.
Assume n=1, hence the number of boxes labelled 1, will be 1 and the number of boxes labelled 2 will be 2. Total number of boxes (N) = 3.
Case 1
When i=1, f1=2 and when i=2, f2=4
Case 2
When i=1,f1=3 and when i=2, f2=3
Difference in total number of footballs in both cases = (2+4+4) - (3+3+3) = 1
Now, on substitution of N=3 in options, eliminate those options where you are not getting 1.
Only option (b) satisfies the required condition.