Question

A number is written in scientific notation as $a\times {10}^b$, where a consists of $4$ digits without zero in it. When it is converted into standard form, the decimal point is moved to the right by $6$ places. If the total number of zero digits in the standard form is $p$, then which of the following statements is true?

Answer 1

Detailed step-by-step solution:
The number is written in scientific notation as $a\times {10}^b$.
$\therefore 1\le a<10$
Let us suppose $a=\mathrm{5.346.}(\because$ it consists of $4$ digits without zero in it)

$a\times {10}^b=5.346\times {10}^b$

It is given that the decimal point is shifted to the right by $6$ places.
This indicates the exponent is a positive integer, i.e., $b=6$.
$\to a\times {10}^b=5.346\times {10}^6$

$\to 5.346\times {10}^6=5,346,000$ (Converting to standard form)

$\to$ The number of zeros in the standard form will be $3$.

$\to p=3\text{and}b=6$

$\to$Option C is correct.