Which of the following comparisons between \(\sqrt{90}\) and \(\sqrt{99}\) is correct?
Answer and explanation
Correct answer: \(\sqrt{99}>\sqrt{90}\)
The square-root function is increasing for non‑negative real numbers: if \(a>b\ge0\) then \(\sqrt{a}>\sqrt{b}\). Since \(99>90\), we have \(\sqrt{99}>\sqrt{90}\). Numerically \(\sqrt{90}\approx9.4868\) and \(\sqrt{99}\approx9.9499\). Option B reverses the order, C is false because the radicands are different, and D is wrong because the comparison is definite. Exam tip: either compare the radicands directly or compute approximate square roots to decide quickly.
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What is the correct answer to this question?
\(\sqrt{99}>\sqrt{90}\)
Why is this the correct answer?
The square-root function is increasing for non‑negative real numbers: if \(a>b\ge0\) then \(\sqrt{a}>\sqrt{b}\). Since \(99>90\), we have \(\sqrt{99}>\sqrt{90}\). Numerically \(\sqrt{90}\approx9.4868\) and \(\sqrt{99}\approx9.9499\). Option B reverses the order, C is false because the radicands are different, and D is wrong because the comparison is definite. Exam tip: either compare the radicands directly or compute approximate square roots to decide quickly.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Irrational numbers and real numbers.
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