Question

Adam's savings are equal to Ben's expenditure, which in turn is equal to Mary's savings. If Mary's savings are Rupees 50,000 and the incomes of Adam, Ben, and Mary are in the ratio 3:1:4, Mary's expenditure is less than thrice of Adam's expenditure and twice of Adam's expenditure is less than two times Ben's income, then which of the following could be true about Ben's income (\(I_B\))?

• A. \(18000<I_B<22000\)
• B. \(20000<I_B<25000\)
• C. \(23000<I_B<28000\)
• D. Data Inconsistent

Hint:

In complex word problems with multiple variables, the key is to define a single variable (like a common ratio 'x') and express all other quantities in terms of it. This simplifies the problem into solving inequalities for that single variable.

Answer 1

The Correct Option is B

Approach: Put everything in terms of the single ratio unit \(x\), use Income \(=\) Expenditure \(+\) Savings to convert the word-statements into two clean inequalities in \(x\), and intersect them.

Step 1: Set variables. Incomes are in ratio \(3:1:4\), so \(I_A = 3x,\ I_B = x,\ I_M = 4x\).
Given \(S_M = 50{,}000\) and \(S_A = E_B = S_M\), we get \(S_A = 50{,}000\) and \(E_B = 50{,}000\).

Step 2: Expenditures. Using Expenditure \(=\) Income \(-\) Savings:
\[ E_A = 3x - 50{,}000, \qquad E_M = 4x - 50{,}000. \]
Step 3: Inequality 1 \(-\) Mary's expenditure \(<\) thrice Adam's.
\[ 4x - 50{,}000 < 3(3x - 50{,}000) \implies 4x - 50{,}000 < 9x - 150{,}000 \]
\[ \implies 100{,}000 < 5x \implies x > 20{,}000. \]
Step 4: Inequality 2 \(-\) twice Adam's expenditure \(<\) twice Ben's income.
\[ 2E_A < 2I_B \implies E_A < I_B \implies 3x - 50{,}000 < x \implies 2x < 50{,}000 \implies x < 25{,}000. \]
Step 5: Combine. \(20{,}000 < x < 25{,}000\), and since \(I_B = x\),
\[ 20{,}000 < I_B < 25{,}000. \]
Final Answer: Option (B), \(20{,}000 < I_B < 25{,}000\).

Answer 2

Approach: Normalize everything by 50,000 first (call it \(k = x/50{,}000\)), turning both conditions into simple fraction inequalities in \(k\) instead of working with large numbers throughout.

Step 1: Base facts. \(I_A = 3x,\ I_B = x,\ I_M = 4x\), and \(S_A = E_B = S_M = 50{,}000\), so \(E_A = 3x - 50{,}000\) and \(E_M = 4x - 50{,}000\).

Step 2: Substitute \(x = 50{,}000\,k\). Then \(E_A = 50{,}000(3k-1)\) and \(E_M = 50{,}000(4k-1)\).

Step 3: Condition 1 – \(E_M < 3E_A\). \[ 4k - 1 < 3(3k-1) \implies 4k - 1 < 9k - 3 \implies 2 < 5k \implies k > 0.4. \]

Step 4: Condition 2 – \(E_A < I_B\). \[ 50{,}000(3k-1) < 50{,}000\,k \implies 3k - 1 < k \implies 2k < 1 \implies k < 0.5. \]

Step 5: Convert back. \(0.4 < k < 0.5 \implies 20{,}000 < x < 25{,}000\), and since \(I_B = x\):
\[ 20{,}000 < I_B < 25{,}000. \]

Final Answer: Option (B), \(20{,}000<I_B<25{,}000\).