Actual value of a quantity is $1.354675$ and the numerical approximate is $1.354595$ then what is the number of significant digits? By using absolute error i.e. $\vert\text{true value} - \text{approx.} \vert < 0.5 \cdot 10 ^{-t}$, where $t = \text{number of significant digits in decimal place i.e. solution is accurate upto tTH decimal place}$.
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An approximation $\hat{x}$ to $x$ has $t$ significant digits if
$$\vert \hat{x} - x \vert <0.5 \cdot 10^{-t}.$$
In your case,
$$\vert \hat{x} - x \vert = \vert 1.354595 - 1.354675 \vert = 0.00008 = 0.8 \cdot 10^{-4} .$$
Notice that
$$ 0.5 \cdot 10^{-4} \lt 0.8 \cdot 10^{-4} \lt 0.5 \cdot 10^{-3}. $$
So the approximation has $t=3$ significant digits.
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$t=3$ counts only the significant fractional digits. Cf. Significant digits. – ccorn Jul 29 '21 at 22:45
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Yes the answer should be t=4 atleast and atmost 5 – Upstart Apr 12 '22 at 18:29