Question

Average is 1/60 and observations is 1/x , 1/2x , 1/3x . Than find the value of x ?​

Answer 1

$\frac{ \frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} }{3} = \frac{1}{60} \\ \\ \frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} = \frac{3}{60} \\ \\\frac{2x + x}{2 {x}^{2} } + \frac{1}{3x} = \frac{1}{20} \\ \\ \frac{3x}{2 {x}^{2} } + \frac{1}{3x} = \frac{1}{20} \\ \\ \frac{3x \times 3x + 2 {x}^{2} }{2 {x}^{2} \times 3x} = \frac{1}{20} \\ \\ \frac{9 {x}^{2} + 2 {x}^{2} }{ {6x}^{3} } = \frac{1}{20} \\ \\ \frac{11 {x}^{2} }{6 {x}^{3} } = \frac{1 }{20} \\\\ 20 \times 11 {x}^{2} = 6 {x}^{3} \\ \\220 {x}^{2} = 6 {x}^{3} \\ \\ {x}^{2} = \frac{6 {x}^{3} }{220} \\ \\ {x}^{2} = \frac{3 {x}^{3} }{110} \\\\ 110 {x}^{2} = 3 {x}^{3 } \\ \\ 110 {x}^{2} - 3 {x}^{3} = 0 \\\\ {x}^{2} ( 110 - 3x) = 0 \\ \\ {x}^{2} = 0 \: or \: 110 - 3x = 0 \\\\ x = 0 \: or \: x = \frac{110}{3}$

As x is in denominator in eqns so it can't be zero here, value if x should be 110/3.

Answer. Value of x= 110/3

Answer 2

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