Question

After how many decimal places, the decimal expansion of the rational number \(\frac{25}{2^2\times 5}\) will terminate?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is B

The problem asks us to determine the number of decimal places after which the decimal expansion of the rational number \(\frac{25}{2^2\times 5}\) will terminate. 
A rational number will have a terminating decimal expansion if its denominator can be expressed in the form \(2^m \times 5^n\), where \(m\) and \(n\) are non-negative integers.

First, let's simplify the given expression:

\[\frac{25}{2^2 \times 5} = \frac{25}{4 \times 5} = \frac{25}{20}\]

Now, simplify \(\frac{25}{20}\):

\[\frac{25}{20} = \frac{5}{4}\]

The denominator \(4\) is \(2^2\). Since it is in the form \(2^m\) (and implicitly \(5^0\)), the decimal expansion will terminate. 
The number of decimal places is determined by the largest power among \(m\) and \(n\). Here, \(m = 2\) and \(n = 0\). 
Thus, there are 2 decimal places.

Therefore, the number of decimal places in the terminating decimal expansion of the rational number \(\frac{25}{2^2 \times 5}\) is 2.

Answer 2

For the decimal expansion of a rational number to terminate, the denominator after simplification must have only 2 and 5 as prime factors. We simplify the denominator: \[ 2^2 \times 5 = 4 \times 5 = 20 \] The number 20 has only the prime factors 2 and 5. 
Hence, the decimal expansion will terminate after 2 decimal places.