The decimal expansion of \(\frac{43}{2^3\times5^5}\) will terminate after how many decimal places?
Answer and explanation
Correct answer: 5
The denominator is \(2^3\times5^5\), and 43 has no common factor with 2 or 5. Multiply the numerator and denominator by \(2^2\): \(\frac{43}{2^3\times5^5}=\frac{172}{10^5}\). Hence, the decimal expansion terminates after 5 decimal places. Option 8 is incorrect because the exponents are not added; the larger exponent determines the required number of places. Exam tip: for a reduced denominator of the form \(2^m5^n\), the decimal terminates in at most \(\max(m,n)\) places.
Frequently asked questions
What is the correct answer to this question?
5
Why is this the correct answer?
The denominator is \(2^3\times5^5\), and 43 has no common factor with 2 or 5. Multiply the numerator and denominator by \(2^2\): \(\frac{43}{2^3\times5^5}=\frac{172}{10^5}\). Hence, the decimal expansion terminates after 5 decimal places. Option 8 is incorrect because the exponents are not added; the larger exponent determines the required number of places. Exam tip: for a reduced denominator of the form \(2^m5^n\), the decimal terminates in at most \(\max(m,n)\) places.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Decimal representation.
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