If \(x=\sqrt{a}\) and \(12.6<x<12.7\) on the number line, which value of \(a\) is possible?
Answer and explanation
Correct answer: \(159\)
Since \(x\) is positive, squaring the inequality gives \(12.6^2<a<12.7^2\). Therefore, \(158.76<a<161.29\). Among the given options, only \(159\) lies in this interval. The value \(158\) is too small, while \(162\) and \(163\) are too large. Exam tip: For \(x=\sqrt a\), square the bounds on \(x\) to find the required interval for \(a\).
Frequently asked questions
What is the correct answer to this question?
\(159\)
Why is this the correct answer?
Since \(x\) is positive, squaring the inequality gives \(12.6^2<a<12.7^2\). Therefore, \(158.76<a<161.29\). Among the given options, only \(159\) lies in this interval. The value \(158\) is too small, while \(162\) and \(163\) are too large. Exam tip: For \(x=\sqrt a\), square the bounds on \(x\) to find the required interval for \(a\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Representing real numbers on the number line.
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