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 \(\sqrt{37}+1\).
Frequently asked questions
What is the correct answer to this question?
7 and 8
Why is this the correct answer?
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 \(\sqrt{37}+1\).
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.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
