Level 1 • 9/50 questions • 30 seconds per question.
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What is the expansion of ( \(3\sqrt{2}+2\sqrt{3}\)2 )?
A. \(30+12\sqrt{6}\)
\(a^2+2ab+b^2\) लगाने पर \(18+12\sqrt{6}+12=30+12\sqrt{6}\) मिलता है। मध्य पद को न भूलें। / Using \(a^2+2ab+b^2\), we get \(18+12\sqrt{6}+12=30+12\sqrt{6}\). Do not forget the middle term.
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What is the simplified form of ( \(5\sqrt{3}-2\sqrt{5}\)2 )?
A. \(95-20\sqrt{15}\)
( \(5\sqrt{3}\)2=75 ) और ( \(2\sqrt{5}\)2=20 ) हैं। मध्य पद \(2\times5\sqrt{3}\times2\sqrt{5}=20\sqrt{15}\) घटेगा। / ( \(5\sqrt{3}\)2=75 ) and ( \(2\sqrt{5}\)2=20 ). The middle term \(2\times5\sqrt{3}\times2\sqrt{5}=20\sqrt{15}\) is subtracted.
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What is the value of ( \(2\sqrt{7}+\sqrt{5}\)\(2\sqrt{7}-\sqrt{5}\) )?
A. (23)
यह अंतर के वर्ग का रूप है। ( \(2\sqrt{7}\)2-\(\sqrt{5}\)2=28-5=23 )। / This is a difference of squares. ( \(2\sqrt{7}\)2-\(\sqrt{5}\)2=28-5=23 ).
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What is the positive simplified form of \( \sqrt{11-6\sqrt{2}} \)?
A. \(3-\sqrt{2}\)
( \(3-\sqrt{2}\)2=9-6\sqrt{2}+2=11-6\sqrt{2} )। धनात्मक मुख्य वर्गमूल चुनें। / ( \(3-\sqrt{2}\)2=9-6\sqrt{2}+2=11-6\sqrt{2} ). Choose the positive principal square root.
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Which is greater between \( \sqrt{50} \) and (7.08)?
B. (7.08)
\(7.08^2=50.1264\) जो (50) से बड़ा है। इसलिए \(7.08>\sqrt{50}\) है। / \(7.08^2=50.1264\), which is greater than (50). Therefore \(7.08>\sqrt{50}\).
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If \( \sqrt{n} \) lies between (12) and (13), which range is correct for (n)?
A. (144<n<169)
दोनों ओर वर्ग करने पर \(12^2<n<13^2\) मिलेगा। इसलिए (144<n<169) सही है। / Squaring both sides gives \(12^2<n<13^2\). Therefore (144<n<169) is correct.
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If \(15<\sqrt{m}<16\), what is the correct range for \(m\)?
A. \(225<m<256\)
चूँकि \(\sqrt{m}\) धनात्मक है, असमानता के दोनों पक्षों का वर्ग लेने पर दिशा नहीं बदलती। इसलिए \(15^2<m<16^2\) मिलता है, यानी \(225<m<256\)। विकल्प C (\(196<m<225\)) गलत है क्योंकि वह \(14<\sqrt{m}<15\) के वर्ग का परिणाम होगा; विकल्प B सिर्फ मूल असमानता को बिना वर्ग किए दे रहा है; विकल्प D \(16<\sqrt{m}<17\) के वर्ग का परिणाम है। परीक्षा-सूत्र: धनात्मक प्राथमिकमालों के लिए असमानता को वर्गित करते समय संकेत बदलने की चिंता नहीं करनी चाहिए — सीधे-साथी रूप में वर्ग कर दें और सीमाओं को संख्याओं में बदलकर लिखें। / Since \(\sqrt{m}\) is positive, squaring every part of the inequality preserves the order. Squaring gives \(15^2<m<16^2\), i.e. \(225<m<256\). Option C (\(196<m<225\)) would correspond to squaring \(14<\sqrt{m}<15\) and so is incorrect; option B simply restates the original range for \(\sqrt{m}\) rather than for \(m\); option D corresponds to squaring \(16<\sqrt{m}<17\). Exam tip: when all terms are nonnegative you may square an inequality termwise — then compute the numerical squares to get the correct interval for the variable inside the square.
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When is \( \sqrt{a}+\sqrt{b}=\sqrt{a+b} \) generally not correct?
A. जब (a) और (b) दोनों धनात्मक होंWhen (a) and (b) are both positive
सामान्यतः \( \sqrt{a}+\sqrt{b} \neq \sqrt{a+b} \) होता है। उदाहरण के लिए \( \sqrt{4}+\sqrt{9}=5 \) लेकिन \( \sqrt{13} \) नहीं। / Generally \( \sqrt{a}+\sqrt{b} \neq \sqrt{a+b} \). For example, \( \sqrt{4}+\sqrt{9}=5 \), not \( \sqrt{13} \).
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What is the value of \( \sqrt{a}\times\sqrt{a} \) where \(a\geq0\)?
A. (a)
एक ही गैर-ऋणात्मक संख्या के वर्गमूल को स्वयं से गुणा करने पर वही संख्या मिलती है। इसलिए मान (a) है। / Multiplying the square root of the same non-negative number by itself gives the number. Therefore the value is (a).
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