What type of decimal expansion will (\frac{2^5\cdot 17}{2^9\cdot 5^2\cdot 17^2}) have?
Answer and explanation
Correct answer: Non-terminating recurring
For a rational number, the decimal expansion terminates exactly when the denominator in lowest terms contains only the primes 2 and 5. Factors 2 and 5 can be combined to make a power of 10, so division ends. A remaining prime such as 17 prevents this and produces a repeating pattern of digits instead.
After cancellation, \(2^5\) removes part of \(2^9\), and one factor 17 removes part of \(17^2\). The denominator becomes \(2^4\cdot 5^2\cdot 17\). Because 17 remains, the denominator is not of the required form. Thus the decimal expansion is non-terminating recurring, so option B follows. Option A would be possible only if every factor other than 2 and 5 had cancelled.
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What is the correct answer to this question?
Non-terminating recurring
Why is this the correct answer?
For a rational number, the decimal expansion terminates exactly when the denominator in lowest terms contains only the primes 2 and 5. Factors 2 and 5 can be combined to make a power of 10, so division ends. A remaining prime such as 17 prevents this and produces a repeating pattern of digits instead.
After cancellation, \(2^5\) removes part of \(2^9\), and one factor 17 removes part of \(17^2\). The denominator becomes \(2^4\cdot 5^2\cdot 17\). Because 17 remains, the denominator is not of the required form. Thus the decimal expansion is non-terminating recurring, so option B follows. Option A would be possible only if every factor other than 2 and 5 had cancelled.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: Decimal expansion of rational numbers.
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