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Between which two consecutive integers does \\(\sqrt{37}+1\\) lie on the number line?

Answer and explanation

Correct answer: 7 and 8

Since \(6^2=36<37<49=7^2\), we get \(6<\sqrt{37}<7\). Adding 1 to all parts gives \(7<\sqrt{37}+1<8\), so the expression lies between 7 and 8. Exam tip: To bound a square root, compare the number with the consecutive perfect squares around it. Option B is incorrect because it represents the interval for \(\sqrt{37}\), not for \(\sqrt{37}+1\).

Related tags

Number LineSquare RootsInteger BoundsRepresenting Real Numbers On The Number LinePolynomialsMathematicsClass 10 Mcq

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