Which number is greater than \(\sqrt{10}\) and less than \(\sqrt{17}\)?
Answer and explanation
Correct answer: 4
Since \(3^2=9<10<16=4^2\), we have \(3<\sqrt{10}<4\). Also, \(4^2=16<17<25=5^2\), so \(4<\sqrt{17}<5\). Hence, \(4\) is greater than \(\sqrt{10}\) and less than \(\sqrt{17}\). Option \(3\) is less than \(\sqrt{10}\), while \(5\) and \(6\) are greater than \(\sqrt{17}\). Exam tip: locate a number between consecutive perfect squares to compare its square root.
Frequently asked questions
What is the correct answer to this question?
4
Why is this the correct answer?
Since \(3^2=9<10<16=4^2\), we have \(3<\sqrt{10}<4\). Also, \(4^2=16<17<25=5^2\), so \(4<\sqrt{17}<5\). Hence, \(4\) is greater than \(\sqrt{10}\) and less than \(\sqrt{17}\). Option \(3\) is less than \(\sqrt{10}\), while \(5\) and \(6\) are greater than \(\sqrt{17}\). Exam tip: locate a number between consecutive perfect squares to compare its square root.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Number Systems. Topic: Irrational numbers.
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