Step 1: Understanding the Question & Defining Variables:
The problem links the incomes, savings, and expenditures of three individuals through a series of equalities, ratios, and inequalities. We need to find the possible range for Ben's income.
Let the incomes of Adam, Ben, and Mary be \(I_A, I_B, I_M\).
Let their savings be \(S_A, S_B, S_M\).
Let their expenditures be \(E_A, E_B, E_M\).
The incomes are in the ratio 3:1:4. Let 'x' be the common ratio factor.
\[ I_A = 3x, \quad I_B = x, \quad I_M = 4x \]
Step 2: Key Formula or Approach:
The fundamental relationship for personal finance is:
\[ \text{Income} = \text{Expenditure} + \text{Savings} \implies \text{Expenditure} = \text{Income} - \text{Savings} \] Our strategy is to express all unknown quantities in terms of the variable 'x' and then use the given inequalities to establish a valid range for 'x'. This range will directly correspond to the range for Ben's income.
Step 3: Detailed Explanation:
First, use the given equalities and values:
We are given \(S_M = 50,000\).
We are also given \(S_A = E_B = S_M\).
Therefore, \(S_A = 50,000\) and \(E_B = 50,000\).
Next, express the expenditures of Adam and Mary in terms of 'x':
\[ E_A = I_A - S_A = 3x - 50,000 \] \[ E_M = I_M - S_M = 4x - 50,000 \] Now, we apply the two inequalities given in the problem:
Inequality 1: Mary's expenditure is less than thrice of Adam's expenditure.
\[ E_M<3 \times E_A \] \[ 4x - 50,000<3(3x - 50,000) \] \[ 4x - 50,000<9x - 150,000 \] \[ 150,000 - 50,000<9x - 4x \] \[ 100,000<5x \] \[ 20,000<x \] Inequality 2: Twice of Adam's expenditure is less than two times Ben's income.
\[ 2 \times E_A<2 \times I_B \] Dividing by 2, we get:
\[ E_A<I_B \] \[ 3x - 50,000<x \] \[ 2x<50,000 \] \[ x<25,000 \] Combining the results from both inequalities, we get the range for 'x':
\[ 20,000<x<25,000 \] Since Ben's income is \(I_B = x\), the range for Ben's income is:
\[ 20,000<I_B<25,000 \]
Step 4: Final Answer:
The possible range for Ben's income is between Rupees 20,000 and Rupees 25,000. This corresponds to option (B).
If \((2m+n) + (2n+m)=27\), find the maximum value of \((2m-3)\), assuming m and n are positive integers.