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There are n weights in a bag measuring 1kg, 2kg and so on till n kg. These weights are divided into 3 parts. The first part contains the weights 1kg, 4kg, 7kg, and so on. The second part contains the weights 2kg, 5kg, 8kg and so on. The third part contains the remaining weights. The average weights any two of the three parts is equal to the weight present in those parts but the average weight of the remaining one part is not equal to the weight present in that part. Which of the following can be a possible value of n?
- a)90
- b)93
- c)96
- d)88
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
There are n weights in a bag measuring 1kg, 2kg and so on till n kg. T...
We know that if in an AP the number of terms in a series is odd, the average of the terms of the series is equal to the middle term of the series. However if the number of terms in the series is even, the average of all the terms of the series is not equal to one of the terms of the series. Hence the three part contain terms 2x+1, 2x+1, 2x or 2x-1, 2x-1, 2x
Hence the total number of parts = 2x+1+2x+1+2x or 2x-1+2x-1+2x = 6x+2 or 6x-2
Among the options, the only number of the form 6x+2 or 6x-2 is 88. Hence 88 can be the required value of n.
Hence the total number of parts = 2x+1+2x+1+2x or 2x-1+2x-1+2x = 6x+2 or 6x-2
Among the options, the only number of the form 6x+2 or 6x-2 is 88. Hence 88 can be the required value of n.
Most Upvoted Answer
There are n weights in a bag measuring 1kg, 2kg and so on till n kg. T...
Given:
- n weights measuring 1kg, 2kg and so on till n kg
- The weights are divided into 3 parts:
- Part 1 contains weights 1kg, 4kg, 7kg, and so on.
- Part 2 contains weights 2kg, 5kg, 8kg, and so on.
- Part 3 contains the remaining weights.
- Average weights of any two of the three parts is equal to the weight present in those parts.
- Average weight of the remaining one part is not equal to the weight present in that part.
To find: Possible value of n
Solution:
Let's consider the weight in Part 1 as a, weight in Part 2 as b, and weight in Part 3 as c.
From the given condition, we can say that:
- Average weight of Part 1 and Part 2 = (a+b)/2
- Average weight of Part 2 and Part 3 = (b+c)/2
- Average weight of Part 1 and Part 3 ≠ (a+c)/2
We can simplify the above equations as:
- a + b = 2*(a+b)/2
- b + c = 2*(b+c)/2
- a + c ≠ 2*(a+c)/2
Simplifying further:
- a = b
- b = c
- a ≠ c
From the first two equations, we can say that a, b, and c are all equal to some multiple of 3. Let's assume that a = b = c = 3k.
Since there are n weights in total, we can say that 3k is the weight of (n/3) weights.
- Weight of Part 1 = 3k + 6k + 9k + ... = 3k*(1+2+3+...+(n/3))
- Weight of Part 2 = 3k + 6k + 9k + ... = 3k*(2+5+8+...+(n/3))
- Weight of Part 3 = n*(n+1)/2 - [(1+2+3+...+(n/3)) + (2+5+8+...+(n/3))]
Since a = b = c, we can say that the sum of weights in Part 1 and Part 2 is equal to twice the weight in either of them. Therefore,
- 3k*(1+2+3+...+(n/3)) = 2*3k*(2+5+8+...+(n/3))
- (1+2+3+...+(n/3)) = 2*(2+5+8+...+(n/3))
- (n/3)*(n/3 + 1)/2 = 2*(n/3)*(3/2 + (n/3-1)*3)
- n^2 + n - 36 = 0
- (n-5)(n+8) = 0
Therefore, n can be either 5 or -8. Since n has to be a positive integer, the possible value of n is 5.
However, we assumed earlier that a = b = c = 3k. For n = 5, this is not possible because the
- n weights measuring 1kg, 2kg and so on till n kg
- The weights are divided into 3 parts:
- Part 1 contains weights 1kg, 4kg, 7kg, and so on.
- Part 2 contains weights 2kg, 5kg, 8kg, and so on.
- Part 3 contains the remaining weights.
- Average weights of any two of the three parts is equal to the weight present in those parts.
- Average weight of the remaining one part is not equal to the weight present in that part.
To find: Possible value of n
Solution:
Let's consider the weight in Part 1 as a, weight in Part 2 as b, and weight in Part 3 as c.
From the given condition, we can say that:
- Average weight of Part 1 and Part 2 = (a+b)/2
- Average weight of Part 2 and Part 3 = (b+c)/2
- Average weight of Part 1 and Part 3 ≠ (a+c)/2
We can simplify the above equations as:
- a + b = 2*(a+b)/2
- b + c = 2*(b+c)/2
- a + c ≠ 2*(a+c)/2
Simplifying further:
- a = b
- b = c
- a ≠ c
From the first two equations, we can say that a, b, and c are all equal to some multiple of 3. Let's assume that a = b = c = 3k.
Since there are n weights in total, we can say that 3k is the weight of (n/3) weights.
- Weight of Part 1 = 3k + 6k + 9k + ... = 3k*(1+2+3+...+(n/3))
- Weight of Part 2 = 3k + 6k + 9k + ... = 3k*(2+5+8+...+(n/3))
- Weight of Part 3 = n*(n+1)/2 - [(1+2+3+...+(n/3)) + (2+5+8+...+(n/3))]
Since a = b = c, we can say that the sum of weights in Part 1 and Part 2 is equal to twice the weight in either of them. Therefore,
- 3k*(1+2+3+...+(n/3)) = 2*3k*(2+5+8+...+(n/3))
- (1+2+3+...+(n/3)) = 2*(2+5+8+...+(n/3))
- (n/3)*(n/3 + 1)/2 = 2*(n/3)*(3/2 + (n/3-1)*3)
- n^2 + n - 36 = 0
- (n-5)(n+8) = 0
Therefore, n can be either 5 or -8. Since n has to be a positive integer, the possible value of n is 5.
However, we assumed earlier that a = b = c = 3k. For n = 5, this is not possible because the
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There are n weights in a bag measuring 1kg, 2kg and so on till n kg. These weights are divided into 3 parts. The first part contains the weights 1kg, 4kg, 7kg, and so on. The second part contains the weights 2kg, 5kg, 8kg and so on. The third part contains the remaining weights. The average weights any two of the three parts is equal to the weight present in those parts but the average weight of the remaining one part is not equal to the weight present in that part. Which of the following can be a possible value of n?a)90b)93c)96d)88Correct answer is option 'D'. Can you explain this answer?
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