Level 1 • 50/50 questions • 35 seconds per question.
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What is the simplified surd form of \( \sqrt{125} \)?
A. \(5\sqrt{5}\)
\(125=25\times5\) इसलिए \( \sqrt{125}=5\sqrt{5} \)। सबसे बड़ा पूर्ण वर्ग गुणनखंड चुनें। / Since \(125=25\times5\), \( \sqrt{125}=5\sqrt{5} \). Choose the largest perfect-square factor.
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What is obtained by simplifying \( \sqrt{147} \)?
B. \(7\sqrt{3}\)
\(147=49\times3\) इसलिए \( \sqrt{147}=7\sqrt{3} \)। पूर्ण वर्ग को बाहर निकालना मुख्य कदम है। / Since \(147=49\times3\), \( \sqrt{147}=7\sqrt{3} \). Taking out the perfect square is the main step.
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Which is the simplified form of \( \sqrt{192} \)?
C. \(8\sqrt{3}\)
\(192=64\times3\) इसलिए \( \sqrt{192}=8\sqrt{3} \)। करणी में सबसे बड़ा पूर्ण वर्ग खोजें। / Since \(192=64\times3\), \( \sqrt{192}=8\sqrt{3} \). Find the largest perfect square inside the radical.
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What is the simplified form of \(3\sqrt{8}+2\sqrt{18}\)?
A. \(12\sqrt{2}\)
\( \sqrt{8}=2\sqrt{2} \) और \( \sqrt{18}=3\sqrt{2} \) हैं। इसलिए योग \(6\sqrt{2}+6\sqrt{2}=12\sqrt{2}\) है। / \( \sqrt{8}=2\sqrt{2} \) and \( \sqrt{18}=3\sqrt{2} \). So the sum is \(6\sqrt{2}+6\sqrt{2}=12\sqrt{2}\).
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What is the simplified form of \(5\sqrt{12}-\sqrt{75}\)?
A. \(5\sqrt{3}\)
\(5\sqrt{12}=10\sqrt{3}\) और \( \sqrt{75}=5\sqrt{3} \) है। अंतर \(5\sqrt{3}\) होगा। / \(5\sqrt{12}=10\sqrt{3}\) and \( \sqrt{75}=5\sqrt{3} \). The difference is \(5\sqrt{3}\).
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What is the value of ( \sqrt{3}\(2+\sqrt{3}\) )?
A. \(2\sqrt{3}+3\)
वितरण नियम से \( \sqrt{3}\times2+\sqrt{3}\times\sqrt{3}=2\sqrt{3}+3 \)। करणी गुणा में \(\sqrt{a}\sqrt{a}=a\) याद रखें। / By distributive law, \( \sqrt{3}\times2+\sqrt{3}\times\sqrt{3}=2\sqrt{3}+3 \). Remember \( \sqrt{a}\sqrt{a}=a \) in surd multiplication.
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What is the expansion of ( \(2+\sqrt{7}\)2 )?
A. \(11+4\sqrt{7}\)
( (a+b)2=a-2+2ab+b-2 ) लगाएँ। यहाँ \(4+4\sqrt{7}+7=11+4\sqrt{7}\) मिलता है। / Use ( (a+b)2=a-2+2ab+b-2 ). Here \(4+4\sqrt{7}+7=11+4\sqrt{7}\).
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What is the simplified form of ( \(5-\sqrt{6}\)2 )?
A. \(31-10\sqrt{6}\)
( (a-b)2=a-2-2ab+b-2 ) से \(25-10\sqrt{6}+6=31-10\sqrt{6}\) मिलता है। संकेतों पर विशेष ध्यान दें। / Using ( (a-b)2=a-2-2ab+b-2 ), we get \(25-10\sqrt{6}+6=31-10\sqrt{6}\). Pay special attention to signs.
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What is the value of ( \(4+\sqrt{11}\)\(4-\sqrt{11}\) )?
B. (5)
यह संयुग्मी गुणन है। (42-\(\sqrt{11}\)2=16-11=5) होगा। / This is conjugate multiplication. (42-\(\sqrt{11}\)2=16-11=5).
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What is the value of ( \(3+\sqrt{10}\)\(3-\sqrt{10}\) )?
B. ( -1 )
संयुग्मी गुणन से (32-\(\sqrt{10}\)2=9-10=-1) मिलता है। मध्य पद कट जाते हैं। / Conjugate multiplication gives (32-\(\sqrt{10}\)2=9-10=-1). The middle terms cancel.
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What is the rationalised form of \( \frac{4}{\sqrt{5}} \)?
A. \( \frac{4\sqrt{5}}{5} \)
अंश और हर को \( \sqrt{5} \) से गुणा करें। इससे हर (5) बनता है। / Multiply numerator and denominator by \( \sqrt{5} \). This makes the denominator (5).
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What is the rationalised form of \( \frac{7}{2\sqrt{3}} \)?
A. \( \frac{7\sqrt{3}}{6} \)
अंश और हर को \( \sqrt{3} \) से गुणा करने पर \( \frac{7\sqrt{3}}{6} \) मिलता है। हर में करणी हटनी चाहिए। / Multiplying numerator and denominator by \( \sqrt{3} \) gives \( \frac{7\sqrt{3}}{6} \). The radical should be removed from the denominator.
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What is the rationalised form of \( \frac{1}{\sqrt{6}+\sqrt{5}} \)?
A. \( \sqrt{6}-\sqrt{5} \)
संयुग्मी \( \sqrt{6}-\sqrt{5} \) से गुणा करें और हर (6-5=1) बनेगा। इसलिए उत्तर \( \sqrt{6}-\sqrt{5} \) है। / Multiply by the conjugate \( \sqrt{6}-\sqrt{5} \) and the denominator becomes (6-5=1). So the answer is \( \sqrt{6}-\sqrt{5} \).
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What is the rationalised form of \( \frac{1}{\sqrt{7}-\sqrt{3}} \)?
A. \( \frac{\sqrt{7}+\sqrt{3}}{4} \)
संयुग्मी \( \sqrt{7}+\sqrt{3} \) से गुणा करें। हर (7-3=4) मिलेगा। / Multiply by the conjugate \( \sqrt{7}+\sqrt{3} \). The denominator becomes (7-3=4).
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What is the simplified form of \( \sqrt{45}+\sqrt{80}-\sqrt{20} \)?
B. \(6\sqrt{5}\)
\( \sqrt{45}=3\sqrt{5} \), \( \sqrt{80}=4\sqrt{5} \), और \( \sqrt{20}=2\sqrt{5} \)। इसलिए \(3\sqrt{5}+4\sqrt{5}-2\sqrt{5}=5\sqrt{5}\) नहीं बल्कि \(5\sqrt{5}\) होगा। / \( \sqrt{45}=3\sqrt{5} \), \( \sqrt{80}=4\sqrt{5} \), and \( \sqrt{20}=2\sqrt{5} \). So the value is \(3\sqrt{5}+4\sqrt{5}-2\sqrt{5}=5\sqrt{5}\).
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Choose the correct simplified form of \( \sqrt{45}+\sqrt{80}-\sqrt{20} \).
B. \(5\sqrt{5}\)
तीनों करणी को \( \sqrt{5} \) के पदों में बदलें। \(3\sqrt{5}+4\sqrt{5}-2\sqrt{5}=5\sqrt{5}\) सही है। / Convert all surds into terms of \( \sqrt{5} \). \(3\sqrt{5}+4\sqrt{5}-2\sqrt{5}=5\sqrt{5}\) is correct.
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What is the simplified form of \( \sqrt{28}+2\sqrt{63} \)?
B. \(8\sqrt{7}\)
\( \sqrt{28}=2\sqrt{7} \) और \(2\sqrt{63}=6\sqrt{7}\) है। योग \(8\sqrt{7}\) होगा। / \( \sqrt{28}=2\sqrt{7} \) and \(2\sqrt{63}=6\sqrt{7}\). The sum is \(8\sqrt{7}\).
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What is the simplified form of \(3\sqrt{27}-2\sqrt{12}\)?
A. \(5\sqrt{3}\)
\(3\sqrt{27}=9\sqrt{3}\) और \(2\sqrt{12}=4\sqrt{3}\) है। अंतर \(5\sqrt{3}\) मिलता है। / \(3\sqrt{27}=9\sqrt{3}\) and \(2\sqrt{12}=4\sqrt{3}\). The difference is \(5\sqrt{3}\).
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What is the value of \( \sqrt{27}\times\sqrt{12} \)?
A. (18)
\( \sqrt{27}\times\sqrt{12}=\sqrt{324}=18 \)। गुणा में करणी के अंदर की संख्याएँ गुणा करें। / \( \sqrt{27}\times\sqrt{12}=\sqrt{324}=18 \). In multiplication, multiply the numbers inside the radicals.
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What is the value of \( \sqrt{90}\div\sqrt{10} \)?
B. (3)
\( \frac{\sqrt{90}}{\sqrt{10}}=\sqrt{9}=3 \)। करणी भाग में अंदर की संख्याओं का भाग लें। / \( \frac{\sqrt{90}}{\sqrt{10}}=\sqrt{9}=3 \). In radical division, divide the radicands.
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What is the value of \( \frac{\sqrt{72}}{\sqrt{2}}+\sqrt{9} \)?
A. (9)
\( \frac{\sqrt{72}}{\sqrt{2}}=\sqrt{36}=6 \) और \( \sqrt{9}=3 \)। कुल मान (9) है। / \( \frac{\sqrt{72}}{\sqrt{2}}=\sqrt{36}=6 \) and \( \sqrt{9}=3 \). The total value is (9).
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What is the value of \( \sqrt{0.81}+\sqrt{0.04} \)?
B. (1.1)
\( \sqrt{0.81}=0.9 \) और \( \sqrt{0.04}=0.2 \) है। योग (1.1) होगा। / \( \sqrt{0.81}=0.9 \) and \( \sqrt{0.04}=0.2 \). The sum is (1.1).
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What is the value of \( \sqrt{2.56}-\sqrt{0.36} \)?
A. (1)
\( \sqrt{2.56}=1.6 \) और \( \sqrt{0.36}=0.6 \) है। अंतर (1) है। / \( \sqrt{2.56}=1.6 \) and \( \sqrt{0.36}=0.6 \). The difference is (1).
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What is the value of \( \sqrt{\frac{9}{16}}+\sqrt{\frac{25}{36}} \)?
A. \( \frac{19}{12} \)
पहला पद \( \frac{3}{4} \) और दूसरा \( \frac{5}{6} \) है। योग \( \frac{9}{12}+\frac{10}{12}=\frac{19}{12} \) मिलता है। / The first term is \( \frac{3}{4} \) and the second is \( \frac{5}{6} \). The sum is \( \frac{9}{12}+\frac{10}{12}=\frac{19}{12} \).
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What is the value of \( \sqrt{\frac{49}{100}}-\sqrt{\frac{4}{25}} \)?
B. \( \frac{3}{10} \)
\( \sqrt{\frac{49}{100}}=\frac{7}{10} \) और \( \sqrt{\frac{4}{25}}=\frac{2}{5} \) है। अंतर \( \frac{3}{10} \) है। / \( \sqrt{\frac{49}{100}}=\frac{7}{10} \) and \( \sqrt{\frac{4}{25}}=\frac{2}{5} \). The difference is \( \frac{3}{10} \).
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Which option is an irrational number between (2) and (3)?
B. \( \sqrt{8} \)
\(2^2<8<3^2\) इसलिए \( \sqrt{8} \) (2) और (3) के बीच है। (8) पूर्ण वर्ग नहीं है इसलिए यह अपरिमेय है। / Since \(2^2<8<3^2\), \( \sqrt{8} \) lies between (2) and (3). Since (8) is not a perfect square, it is irrational.
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Which is greater between \( \sqrt{18} \) and \( \frac{9}{2} \)?
B. \( \frac{9}{2} \)
\( \sqrt{18}\approx4.24 \) और \( \frac{9}{2}=4.5 \) है। इसलिए \( \frac{9}{2} \) बड़ा है। / \( \sqrt{18}\approx4.24 \) and \( \frac{9}{2}=4.5 \). Therefore \( \frac{9}{2} \) is greater.
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Between which two integers does \( \sqrt{26} \) lie?
B. (5) और (6)(5) and (6)
\(5^2=25\) और \(6^2=36\) है इसलिए \( \sqrt{26} \) (5) और (6) के बीच है। निकट पूर्ण वर्गों की तुलना करें। / \(5^2=25\) and \(6^2=36\), so \( \sqrt{26} \) lies between (5) and (6). Compare nearby perfect squares.
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Between which two integers will \( \sqrt{53} \) lie?
C. (7) और (8)(7) and (8)
\(7^2=49\) और \(8^2=64\) है इसलिए \( \sqrt{53} \) (7) और (8) के बीच है। वर्ग करके सीमा जाँचें। / \(7^2=49\) and \(8^2=64\), so \( \sqrt{53} \) lies between (7) and (8). Check limits by squaring.
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Which is greater between \( -\sqrt{20} \) and ( -4.4 )?
B. ( -4.4 )
\( \sqrt{20}\approx4.47 \) इसलिए \( -\sqrt{20}\approx-4.47 \)। ( -4.4 ) शून्य के अधिक निकट है इसलिए बड़ी है। / \( \sqrt{20}\approx4.47 \), so \( -\sqrt{20}\approx-4.47 \). ( -4.4 ) is closer to zero, so it is greater.
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C. \( \frac{5}{0} \) वास्तविक संख्या है\( \frac{5}{0} \) is a real number
\( \frac{5}{0} \) परिभाषित नहीं है क्योंकि हर शून्य है। ऐसी संख्या वास्तविक नहीं मानी जाती। / \( \frac{5}{0} \) is undefined because the denominator is zero. Such a number is not considered real.
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If (0.12122122212222...) is non-repeating, what type of number is it?
B. अपरिमेय वास्तविक संख्याIrrational real number
असमाप्त और अनावर्ती दशमलव अपरिमेय होता है। अनावर्ती पैटर्न पहचानना जरूरी है। / A non-terminating and non-repeating decimal is irrational. Identifying the non-repeating pattern is important.
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What type of real number is (1.232323...)?
B. परिमेय वास्तविक संख्याRational real number
(23) बार-बार आ रहा है इसलिए दशमलव आवर्ती है। आवर्ती दशमलव परिमेय होते हैं। / The block (23) repeats, so the decimal is repeating. Repeating decimals are rational.
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If (0.040040004...) has no fixed repetition, what is it?
C. अपरिमेय वास्तविक संख्याIrrational real number
स्थिर आवृत्ति न होने पर असमाप्त दशमलव अपरिमेय होता है। केवल आवर्ती दशमलव परिमेय होते हैं। / A non-terminating decimal without fixed repetition is irrational. Only repeating decimals are rational.
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What is the correct statement about the decimal form of \( \frac{7}{11} \)?
B. यह आवर्ती दशमलव हैIt is repeating
हर (11) में केवल (2) और (5) के गुणनखंड नहीं हैं। इसलिए दशमलव आवर्ती होगा। / The denominator (11) does not have only factors (2) and (5). Therefore the decimal will repeat.
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What is the correct statement about the decimal form of \( \frac{13}{40} \)?
A. यह समाप्त दशमलव हैIt is terminating
\(40=2^3\times5\) है इसलिए हर में केवल (2) और (5) हैं। ऐसा भिन्न समाप्त दशमलव देता है। / \(40=2^3\times5\), so the denominator has only (2) and (5). Such a fraction gives a terminating decimal.
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What is the correct statement about the decimal form of \( \frac{3}{28} \)?
B. यह असमाप्त आवर्ती दशमलव हैIt is non-terminating repeating decimal
\(28=2^2\times7\) में (7) गुणनखंड है। इसलिए दशमलव समाप्त नहीं होगा लेकिन आवर्ती होगा। / \(28=2^2\times7\) has factor (7). So the decimal will not terminate but will repeat.
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What type will \( \frac{17}{125} \) be in decimal form?
A. समाप्त दशमलवTerminating decimal
\(125=5^3\) है इसलिए हर में केवल (5) का गुणनखंड है। ऐसा भिन्न समाप्त दशमलव देता है। / \(125=5^3\), so the denominator has only factor (5). Such a fraction gives a terminating decimal.
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What will be the decimal expansion of \( \frac{19}{30} \)?
B. असमाप्त आवर्तीNon-terminating repeating
\(30=2\times3\times5\) में (3) भी है। इसलिए यह असमाप्त आवर्ती दशमलव होगा। / \(30=2\times3\times5\) also has (3). Therefore it will be a non-terminating repeating decimal.
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What type of number is \( \sqrt{2}+\sqrt{3} \)?
B. अपरिमेय वास्तविक संख्याIrrational real number
दो अलग-अलग अपरिमेय करणी पदों का योग यहाँ अपरिमेय है। यह वास्तविक संख्या रेखा पर स्थित है। / The sum of these two different irrational surds is irrational. It lies on the real number line.
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What type of number is \( \sqrt{5}\times\sqrt{20} \)?
A. परिमेय वास्तविक संख्याRational real number
\( \sqrt{5}\times\sqrt{20}=\sqrt{100}=10 \) है। इसलिए परिणाम परिमेय वास्तविक संख्या है। / \( \sqrt{5}\times\sqrt{20}=\sqrt{100}=10 \). Therefore the result is a rational real number.
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What type of number is \( \sqrt{7}\times\sqrt{14} \)?
B. अपरिमेय वास्तविक संख्याIrrational real number
\( \sqrt{7}\times\sqrt{14}=\sqrt{98}=7\sqrt{2} \) है। यह अपरिमेय वास्तविक संख्या है। / \( \sqrt{7}\times\sqrt{14}=\sqrt{98}=7\sqrt{2} \). This is an irrational real number.
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What is the product of \(2+\sqrt{13}\) and \(2-\sqrt{13}\)?
B. ( -9 )
संयुग्मी गुणन से (22-\(\sqrt{13}\)2=4-13=-9) मिलता है। इस रूप में अंतर के वर्ग का नियम लगाएँ। / Conjugate multiplication gives (22-\(\sqrt{13}\)2=4-13=-9). Use the difference of squares rule here.
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What is the value of \( \frac{2+\sqrt{3}}{2-\sqrt{3}} \) after rationalising?
A. \(7+4\sqrt{3}\)
हर के संयुग्मी \(2+\sqrt{3}\) से गुणा करें। अंश (\(2+\sqrt{3}\)2=7+4\sqrt{3}) और हर (1) होगा। / Multiply by the conjugate \(2+\sqrt{3}\). The numerator becomes (\(2+\sqrt{3}\)2=7+4\sqrt{3}) and the denominator becomes (1).
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What will be the denominator after rationalising \( \frac{\sqrt{5}-1}{\sqrt{5}+1} \)?
A. (4)
हर के संयुग्मी \( \sqrt{5}-1 \) से गुणा करने पर हर (5-1=4) होगा। संयुग्मी विधि में अंतर के वर्ग का प्रयोग करें। / Multiplying by the conjugate \( \sqrt{5}-1 \) makes the denominator (5-1=4). Use difference of squares in the conjugate method.
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What is the value of \( |-\sqrt{49}+3| \)?
A. (4)
\( -\sqrt{49}+3=-7+3=-4 \) है और ( |-4|=4 )। निरपेक्ष मान दूरी देता है। / \( -\sqrt{49}+3=-7+3=-4 \) and ( |-4|=4 ). Absolute value gives distance.
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What is the value of \( |\sqrt{16}-\sqrt{81}| \)?
A. (5)
\( \sqrt{16}=4 \) और \( \sqrt{81}=9 \) है। ( |4-9|=5 ) होगा। / \( \sqrt{16}=4 \) and \( \sqrt{81}=9 \). So ( |4-9|=5 ).
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What is the value of \( \sqrt{a^2} \) in real numbers?
C. ( |a| )
वास्तविक संख्याओं में \( \sqrt{a^2}=|a| \) होता है क्योंकि वर्गमूल अमुख्य रूप से गैर-ऋणात्मक होता है। संकेत का ध्यान रखें। / In real numbers, \( \sqrt{a^2}=|a| \) because the principal square root is non-negative. Pay attention to the sign.
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If (x=-5), what is the value of \( \sqrt{x^2} \)?
B. (5)
\( \sqrt{x^2}=|x| \) होता है। (x=-5) होने पर ( |-5|=5 ) मिलेगा। / \( \sqrt{x^2}=|x| \). When (x=-5), ( |-5|=5 ).
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What is the rationalised form of \( \frac{1}{\sqrt{3}+\sqrt{2}} \)?
C. \( \sqrt{3}-\sqrt{2} \)
संयुग्मी \( \sqrt{3}-\sqrt{2} \) से गुणा करने पर हर (3-2=1) बनता है। इसलिए परिमेय हर वाला रूप \( \sqrt{3}-\sqrt{2} \) है। / Multiplying by the conjugate \( \sqrt{3}-\sqrt{2} \) makes the denominator (3-2=1). So the rationalised form is \( \sqrt{3}-\sqrt{2} \).
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