A, B and C divide a certain sum of money among themselves. The average of the amount with them is Rs.4520. Share of A is \(10\frac{2}{3}%\) % more than share of B and \(33\frac{1}{3}%\) % less than share of C. What is the share of B (in Rs.)?
3600
This question involves dividing a sum of money among three individuals, A, B, and C, based on their average share and specific percentage relationships between their shares. We are given the average amount they have and how A's share relates to B's and C's shares. Our goal is to find the exact share of B.
We are given that the average amount of money with A, B, and C is Rs. 4520. Since there are three people, the total sum of money is the average amount multiplied by the number of people.
Total Sum = Average Amount \(\times\) Number of People
Total Sum = \(4520 \times 3\)
Total Sum = \(13560\) Rs.
So, the sum of the shares of A, B, and C is Rs. 13560.
\(A + B + C = 13560\)
We are given two percentage relationships:
Let's convert these percentages into fractions:
Now, we can write the relationships algebraically:
From the relationships above, we can express B and C in terms of A:
We know that \(A + B + C = 13560\). Substitute the expressions for B and C in terms of A into this equation:
\(A + \frac{75}{83}A + \frac{3}{2}A = 13560\)
To solve for A, find a common denominator for the fractions. The denominators are 1, 83, and 2. The least common multiple of 83 and 2 is \(83 \times 2 = 166\).
\(\frac{166}{166}A + \frac{75 \times 2}{83 \times 2}A + \frac{3 \times 83}{2 \times 83}A = 13560\)
\(\frac{166}{166}A + \frac{150}{166}A + \frac{249}{166}A = 13560\)
Combine the fractions on the left side:
\(\frac{166 + 150 + 249}{166}A = 13560\)
\(\frac{565}{166}A = 13560\)
Now, solve for A:
\(A = 13560 \times \frac{166}{565}\)
Let's simplify the fraction. Both 13560 and 565 are divisible by 5.
\(13560 \div 5 = 2712\)
\(565 \div 5 = 113\)
So, \(A = 2712 \times \frac{166}{113}\)
Now, check if 2712 is divisible by 113. \(2712 \div 113 = 24\).
So, \(A = 24 \times 166\)
\(A = 3984\)
A's share is Rs. 3984.
We have the relationship \(B = \frac{75}{83}A\). Substitute the value of A we just found:
\(B = \frac{75}{83} \times 3984\)
Check if 3984 is divisible by 83. \(3984 \div 83 = 48\).
So, \(B = 75 \times 48\)
\(B = 3600\)
B's share is Rs. 3600.
Let's find C's share: \(C = \frac{3}{2}A = \frac{3}{2} \times 3984 = 3 \times 1992 = 5976\).
Total sum: \(A + B + C = 3984 + 3600 + 5976 = 13560\). This matches the total sum calculated from the average.
Percentage checks:
All conditions are satisfied, confirming that B's share is Rs. 3600.
| Individual | Share (Rs.) |
|---|---|
| A | 3984 |
| B | 3600 |
| C | 5976 |
| Total | 13560 |
The share of B is Rs. 3600.
| Concept | Description | Formula Used |
|---|---|---|
| Average | Sum divided by the number of items. | Average = Total Sum / Number of Items |
| Total Sum from Average | The sum of amounts when the average is known. | Total Sum = Average \(\times\) Number of Items |
| Percentage Increase | A quantity is (percentage)% more than another. | New Value = Original Value \(\times\) \((1 + \text{percentage}/100)\) |
| Percentage Decrease | A quantity is (percentage)% less than another. | New Value = Original Value \(\times\) \((1 - \text{percentage}/100)\) |
| Solving Linear Equations | Finding the value of an unknown variable in an equation. | Requires algebraic manipulation. |
When dealing with percentage increases or decreases in algebraic equations, it's often helpful to represent the percentages as decimals or fractions. For example:
Using fractions can sometimes make calculations easier, especially when dealing with repeating decimals or fractions like \(10\frac{2}{3}\)% or \(33\frac{1}{3}\)%, as seen in this problem.
In this solution, we used the fractional approach: \(10\frac{2}{3}\)% more means multiplying by \(1 + \frac{8}{75} = \frac{83}{75}\), and \(33\frac{1}{3}\)% less means multiplying by \(1 - \frac{1}{3} = \frac{2}{3}\).
Setting up the initial equation relating A, B, and C using one variable (in this case, A) was crucial for solving the problem effectively.
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