Ram, Shyam, Rohan, Reeta and Mukesh are five members of a family who are weighed consecutively and their average weight is calculated after each member is weighed. If the average weight increases by 2 kg each time, how much heavier is Mukesh than Ram?
16 kg
Understanding the Family Weight Problem
This problem involves calculating the individual weights of family members given that the average weight of the group increases by a fixed amount as each new member is added. We are given that Ram, Shyam, Rohan, Reeta, and Mukesh are weighed consecutively, and the average weight increases by 2 kg after each person is included in the calculation.
Step-by-Step Solution
Let's denote the weight of each person as follows:
- Ram: \(w_1\)
- Shyam: \(w_2\)
- Rohan: \(w_3\)
- Reeta: \(w_4\)
- Mukesh: \(w_5\)
Let \(A_n\) be the average weight after weighing \(n\) members, and \(S_n\) be the total weight after weighing \(n\) members. The total weight is the sum of the individual weights up to that member.
We know that the average is calculated as the total weight divided by the number of members:
$$A_n = \frac{S_n}{n}$$
The problem states that the average weight increases by 2 kg each time a new member is added. This means:
- \(A_2 = A_1 + 2\)
- \(A_3 = A_2 + 2 = A_1 + 4\)
- \(A_4 = A_3 + 2 = A_1 + 6\)
- \(A_5 = A_4 + 2 = A_1 + 8\)
Now, let's find the weight of each person using the total weight \(S_n\):
1. After weighing Ram:
- Number of members, \(n=1\).
- \(S_1 = w_1\).
- \(A_1 = \frac{S_1}{1} = w_1\).
2. After weighing Shyam:
- Number of members, \(n=2\).
- \(S_2 = w_1 + w_2\).
- \(A_2 = \frac{S_2}{2} = \frac{w_1 + w_2}{2}\).
- We know \(A_2 = A_1 + 2\). So, \(\frac{w_1 + w_2}{2} = w_1 + 2\).
- Multiplying by 2, \(w_1 + w_2 = 2w_1 + 4\).
- Thus, \(w_2 = 2w_1 + 4 - w_1 = w_1 + 4\).
- The total weight is \(S_2 = 2A_2 = 2(A_1+2) = 2A_1+4\).
3. After weighing Rohan:
- Number of members, \(n=3\).
- \(S_3 = w_1 + w_2 + w_3\).
- \(A_3 = \frac{S_3}{3}\).
- We know \(A_3 = A_2 + 2 = (A_1+2) + 2 = A_1 + 4\). So, \(S_3 = 3A_3 = 3(A_1 + 4) = 3A_1 + 12\).
- The weight of the third person is \(w_3 = S_3 - S_2 = (3A_1 + 12) - (2A_1 + 4) = A_1 + 8\). Since \(A_1 = w_1\), \(w_3 = w_1 + 8\).
4. After weighing Reeta:
- Number of members, \(n=4\).
- \(S_4 = w_1 + w_2 + w_3 + w_4\).
- \(A_4 = \frac{S_4}{4}\).
- We know \(A_4 = A_3 + 2 = (A_1+4) + 2 = A_1 + 6\). So, \(S_4 = 4A_4 = 4(A_1 + 6) = 4A_1 + 24\).
- The weight of the fourth person is \(w_4 = S_4 - S_3 = (4A_1 + 24) - (3A_1 + 12) = A_1 + 12\). Since \(A_1 = w_1\), \(w_4 = w_1 + 12\).
5. After weighing Mukesh:
- Number of members, \(n=5\).
- \(S_5 = w_1 + w_2 + w_3 + w_4 + w_5\).
- \(A_5 = \frac{S_5}{5}\).
- We know \(A_5 = A_4 + 2 = (A_1+6) + 2 = A_1 + 8\). So, \(S_5 = 5A_5 = 5(A_1 + 8) = 5A_1 + 40\).
- The weight of the fifth person is \(w_5 = S_5 - S_4 = (5A_1 + 40) - (4A_1 + 24) = A_1 + 16\). Since \(A_1 = w_1\), \(w_5 = w_1 + 16\).
So the weights of the five members are:
- Ram (\(w_1\)): \(w_1\)
- Shyam (\(w_2\)): \(w_1 + 4\)
- Rohan (\(w_3\)): \(w_1 + 8\)
- Reeta (\(w_4\)): \(w_1 + 12\)
- Mukesh (\(w_5\)): \(w_1 + 16\)
The question asks how much heavier Mukesh is than Ram. This is the difference between Mukesh's weight and Ram's weight.
Difference = \(w_5 - w_1 = (w_1 + 16) - w_1 = 16\) kg.
Summary of Weights
| Member | Weight | Weight relative to Ram (\(w_1\)) |
|---|---|---|
| Ram | \(w_1\) | \(w_1\) |
| Shyam | \(w_2\) | \(w_1 + 4\) |
| Rohan | \(w_3\) | \(w_1 + 8\) |
| Reeta | \(w_4\) | \(w_1 + 12\) |
| Mukesh | \(w_5\) | \(w_1 + 16\) |
The difference in weight between Mukesh and Ram is 16 kg.
Final Answer on Weight Difference
Mukesh is 16 kg heavier than Ram.
Revision Table: Average Weight Concepts
| Concept | Formula | Description |
|---|---|---|
| Average | \(\text{Average} = \frac{\text{Sum of values}}{\text{Number of values}}\) | A measure of the central tendency of a set of numbers. |
| Total Sum | \(\text{Sum} = \text{Average} \times \text{Number of values}\) | The sum of all values in a set can be found if the average and number of values are known. |
| Effect of Adding a New Value on Average | If a new value is added, the new average changes based on whether the new value is greater than, equal to, or less than the previous average. | If the new value is greater than the previous average, the average increases. If it's less, the average decreases. If it's equal, the average stays the same. In this problem, the new person's weight must be such that it raises the average by exactly 2 kg. |
Additional Information: Arithmetic Progression of Weights
Notice that the weights of the members form an arithmetic progression: \(w_1, w_1+4, w_1+8, w_1+12, w_1+16\). The common difference is 4 kg.
If the average increases by a constant amount \(c\) each time a new member is added, the weights of the members will form an arithmetic progression. For \(n\) members, if the average increases by \(c\) after each addition (starting from the second member), the difference between consecutive members' weights will be \(2c\) (for \(w_2 - w_1\)) and then \(nc - (n-1)c = c\) adjusted by previous terms. A simpler way to see the pattern in this specific case (average increases by 2 kg) is:
- \(w_2 - w_1 = 4\)
- \(w_3 - w_2 = (w_1+8) - (w_1+4) = 4\)
- \(w_4 - w_3 = (w_1+12) - (w_1+8) = 4\)
- \(w_5 - w_4 = (w_1+16) - (w_1+12) = 4\)
Each new person added weighs 4 kg more than the previous person. This happens because the increase in average is constant (2 kg). The difference between the \(n\)-th person's weight and the \((n-1)\)-th person's weight is given by \(n \times \text{New Average} - (n-1) \times \text{Previous Average}\).
- \(w_2 = 2A_2 - 1A_1 = 2(A_1+2) - A_1 = 2A_1+4 - A_1 = A_1+4\)
- \(w_3 = 3A_3 - 2A_2 = 3(A_2+2) - 2A_2 = A_2+6 = (A_1+2)+6 = A_1+8\)
- \(w_4 = 4A_4 - 3A_3 = 4(A_3+2) - 3A_3 = A_3+8 = (A_1+4)+8 = A_1+12\)
- \(w_5 = 5A_5 - 4A_4 = 5(A_4+2) - 4A_4 = A_4+10 = (A_1+6)+10 = A_1+16\)
Since \(A_1 = w_1\), we get the same weights as before: \(w_1, w_1+4, w_1+8, w_1+12, w_1+16\).
The difference between the first and last person's weight is \((n-1) \times \text{common difference of weights}\). Here, \(n=5\) and the common difference is 4 kg. So, \((5-1) \times 4 = 4 \times 4 = 16\) kg.
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