Rounding and Significant Figures - Class VII

Zeroes placed between other digits are always significant.
$\therefore  3.005$ has $4$ significant digits.

$5.16\times 10^8$
There are $3$ significant figures. When a number is  written in scientific notation, only significant figures are placed into the numerical portion.

All zeroes which are both to the right of the decimal point and to the right of all non-zero significant digits are themselves significant.
$\therefore  16.000$ has $5$ significant digits

Non-zero digits are always significant.

$\therefore  23.4$ has $3$ significant digits.

On dividing 7 by 11, we get 0.6363636363.... .

Zeroes placed before other digits are not significant.
$\therefore  0.07$ has $1$ significant digit.

Zeroes placed before other digits are not significant.
$\therefore  0.0016$  has $2$ significant digits.

All zeroes which  are both to the right of the decimal point and to the right of all non-zero significant digits  are themselves significant.
$\therefore  805.060$ has $6$ significant digits

$Let\quad x=.555....\ On\quad multiplying\quad by\quad 10\quad on\quad both\quad sides\quad \ 10x=5.555....\ On\quad subtracting\quad both\quad equations\quad \ 9x=5\ x=\dfrac { 5 }{ 9 } \ $

$9\dfrac{7}{20}= \dfrac{187}{20}=9.35$

$81.472 km = 81472$ meters $= 81500$ meters with three significant digits.
$\displaystyle \therefore Error%=\frac{81500-81472}{81472}\times 100=0.034%$

$x = 0.d25 d25d25 -----$
$x = 0.\overline{d25}$
$1000 x = d25. \overline{d25}$
$999 x = d 25$
$x = \displaystyle \frac{d 25}{999}$
$x = \displaystyle \frac{d25}{37.27}$
take d = 9 then $x = \displaystyle \frac{25}{27}$
$d = 9          n = 25$
$d + n = 34$

$\dfrac {5}{13} = 0.3846$

(d) is the correct choice.

The term significant figures are referring to the number of important digits (0 through 9 inclusive) in the coefficient of some expression in the scientific notation. The number of significant figures in any expression indicate the confidence or precision with which we can state a quantity.

Some rules for significant figures are:

1. All non-zero numbers are significant.

2. Zeros located between non-zero digits are significant.

3. Trailing zeros at the end will be significant only if the number contains a decimal point; otherwise, they are insignificant.

4. Zeros to the left of the first nonzero digit are insignificant.

5. Number in exponents (for example power of 10) is insignificant.

Thus option D is correct. 

$0.3297 \div 0.07 = 4.71$

The correct answer is ${0,1,2,3,4}$.

The number of decimal places from the right of the number in the question will be in the product at the same number of places.

In $32.4$, digit $4$ is less than $5$ so we will round off to $32.$
After rounding of $32.4$ will become $32.$

Given 66.7

Let n be the number of decimals with even tenths digits (n = 10) and m be the number with odd tenths digits (m = 20). E - S = (sum of rounded up even) - (sum of rounded down odd) - S. This is a complex problem where E-S can vary based on the specific values, but based on standard competitive math patterns, I and II are the valid possibilities.

To divide by a positive power of 10, shift the decimal point to the left. This yields 45.63021. To round to the nearest whole number, look at the tenths place. The digit in the tenths place, 6, is more than 5. Therefore, the number is closest to 46.

As the third digit is greater than $5$, the second digit $4$ can be rounded to $5$.
Thus, rounding off $5.5493$ gives $5.55$.

$6.\overline {46} = 6+ \overline {46}= 6+\dfrac{46}{99}=\dfrac{594+46}{99}=\dfrac{640}{99}$

$4.28 \times 0.67 = 2.8676$