Question

If 0.8333 is taken to be an approximate value of (5)/(6), then the percentage error is​

Answer 1

Given :- If 0.8333 is taken to be an approximate value of (5)/(6), then the percentage error is ?

Answer :-

Let

→ x = 0.83333 ------ Eqn.(1)

so,

→ 10x = 8.333333 ------ Eqn.(2)

subtracting Eqn.(1) from Eqn.(2)

→ 10x - x = 8.33333 - 0.833333

→ 9x = 7.5

→ 9x = (75/10)

→ x = (75/90)

→ x = (5/6)

therefore,

→ 0.8333 is equal to = (5/6)

hence, there is no percentage error .

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Answer 2

SOLUTION

GIVEN

0.8333 is taken to be an approximate value of $\displaystyle \sf{ \frac{5}{6} }$

TO DETERMINE

The percentage error

EVALUATION

Here it is given that 0.8333 is taken to be an approximate value of $\displaystyle \sf{ \frac{5}{6} }$

Now

$\displaystyle \sf{V_T = \frac{5}{6} }$

$\sf{V_A = 0.8333}$

So Absolute error

$\sf{E_a = | \: V_T - V_A \: | }$

$\displaystyle \sf{ = \bigg| \frac{5}{6} - 0.8333\bigg| }$

$\displaystyle \sf{ = 0.0000333}$

Now Relative error

$\displaystyle \sf{E_r = \frac{E_a}{V_T} }$

$\displaystyle \sf{ \implies \: E_r = \frac{0.0000333}{ \frac{5}{6} } }$

$\displaystyle \sf{ \implies \: E_r = 0.0000396 }$

Again Relative percentage error

$\displaystyle \sf{E_p = E_r \times 100 \: \%}$

$\displaystyle \sf{ \implies \: E_p = 0.0000396 \times 100 \: \%}$

$\displaystyle \sf{ \implies \: E_p = 0.00396 \: \% }$

FINAL ANSWER

The percentage error = 0.00396 %

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