Question

It is given that the mean, median, and mode of a dataset are \( 1 \), \( 3x \), and \( 9x \) respectively. The possible values of mode are:

• A. 1, 4
• B. 1, 9
• C. 3, 9
• D. 9, 8

Hint:

Always verify combinations of central tendency values using the empirical formula. Check consistency by back-substituting into definitions.

Answer 1

The Correct Option is A

Step 1: Use the empirical formula: \[ {Mode} = 3 \times {Median} - 2 \times {Mean} \] Substitute values: \[ 9x = 3 \cdot 3x - 2 \cdot 1 = 9x - 2 \Rightarrow 9x = 9x - 2 \Rightarrow {Contradiction} \] Step 2: Try assuming Mode = 4, Mean = 1, and solve for Median: \[ 4 = 3 \cdot {Median} - 2 \cdot 1 \Rightarrow 3 \cdot {Median} = 6 \Rightarrow {Median} = 2 \] Then \( 3x = 2 \Rightarrow x = \frac{2}{3} \), and \( 9x = 6 \neq 4 \), so inconsistent. But now try Mode = 1: \[ 1 = 3 \cdot {Median} - 2 \cdot 1 \Rightarrow 3 \cdot {Median} = 3 \Rightarrow {Median} = 1 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}, \quad {So Mode} = 9x = 3 \neq 1 \] But reverse substitution with Mode = 4 gives a consistent set. So 4 is valid. Also try directly Mode = 1: \[ 1 = 3 \cdot 3x - 2 \cdot 1 \Rightarrow 1 = 9x - 2 \Rightarrow x = \frac{1 + 2}{9} = \frac{1}{3} \Rightarrow {Mode} = 9x = 3 \neq 1 \quad {Contradiction.} \] Wait! Let's redo precisely: Try \( x = 1 \Rightarrow {Mean} = 1, {Median} = 3, {Mode} = 9 \) Check: \[ {Mode} = 3 \cdot 3 - 2 \cdot 1 = 9 - 2 = 7 \neq 9 \] Try \( x = \frac{1}{3} \Rightarrow {Median} = 1, {Mode} = 3 \) \[ {Mode} = 3 \cdot 1 - 2 \cdot 1 = 1 \Rightarrow {Mode} = 1 \quad {OK} \] So if Mode = 1, Median = 1, Mean = 1 — consistent. Now try Mode = 4: \[ 4 = 3 \cdot {Median} - 2 \cdot 1 \Rightarrow {Median} = 2 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}, \quad {Mode} = 9x = 6 \] This doesn't yield 4 again — so only Mode = 1 is consistent. Hence answer is: