Level 2 • 50/50 questions • 30 seconds per question.
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What is the simplified surd form of \( \sqrt{1125} \)?
C. \(15\sqrt{5}\)
\(1125=225\times5\) इसलिए \( \sqrt{1125}=15\sqrt{5} \)। सबसे बड़ा पूर्ण वर्ग गुणनखंड पहले निकालें। / Since \(1125=225\times5\), \( \sqrt{1125}=15\sqrt{5} \). First take out the largest perfect-square factor.
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What is obtained by simplifying \( \sqrt{1452} \)?
B. \(22\sqrt{3}\)
\(1452=484\times3\) और \( \sqrt{484}=22 \) है। इसलिए सरल रूप \(22\sqrt{3}\) है। / \(1452=484\times3\) and \( \sqrt{484}=22 \). So the simplified form is \(22\sqrt{3}\).
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What is the simplified form of \(5\sqrt{20}-2\sqrt{125}+\sqrt{45}\)?
D. \(3\sqrt{5}\)
\(5\sqrt{20}=10\sqrt{5}\), \(2\sqrt{125}=10\sqrt{5}\), और \( \sqrt{45}=3\sqrt{5} \)। इसलिए परिणाम \(3\sqrt{5}\) है। / \(5\sqrt{20}=10\sqrt{5}\), \(2\sqrt{125}=10\sqrt{5}\), and \( \sqrt{45}=3\sqrt{5} \). Therefore the result is \(3\sqrt{5}\).
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Choose the correct simplified form of \(3\sqrt{75}+4\sqrt{48}-2\sqrt{27}\).
B. \(25\sqrt{3}\)
\(3\sqrt{75}=15\sqrt{3}\), \(4\sqrt{48}=16\sqrt{3}\), और \(2\sqrt{27}=6\sqrt{3}\)। इसलिए \(25\sqrt{3}\) सही है। / \(3\sqrt{75}=15\sqrt{3}\), \(4\sqrt{48}=16\sqrt{3}\), and \(2\sqrt{27}=6\sqrt{3}\). Therefore \(25\sqrt{3}\) is correct.
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What is the value of ( \sqrt{12}\(\sqrt{27}-\sqrt{3}\) )?
A. (12)
\( \sqrt{12}=2\sqrt{3} \) और \( \sqrt{27}-\sqrt{3}=2\sqrt{3} \)। गुणनफल \(4\times3=12\) है। / \( \sqrt{12}=2\sqrt{3} \) and \( \sqrt{27}-\sqrt{3}=2\sqrt{3} \). The product is \(4\times3=12\).
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What is the expansion of ( \(4\sqrt{3}+3\sqrt{2}\)2 )?
A. \(66+24\sqrt{6}\)
पहला वर्ग (48) और दूसरा वर्ग (18) है। मध्य पद \(2\times4\sqrt{3}\times3\sqrt{2}=24\sqrt{6}\) है। / The first square is (48) and the second square is (18). The middle term is \(2\times4\sqrt{3}\times3\sqrt{2}=24\sqrt{6}\).
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What is the simplified form of ( \(6\sqrt{5}-\sqrt{7}\)2 )?
A. \(187-12\sqrt{35}\)
( \(6\sqrt{5}\)2=180 ) और ( \(\sqrt{7}\)2=7 ) हैं। मध्य पद \(12\sqrt{35}\) घटेगा। / ( \(6\sqrt{5}\)2=180 ) and ( \(\sqrt{7}\)2=7 ). The middle term \(12\sqrt{35}\) is subtracted.
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What is the value of ( \(3\sqrt{11}+2\sqrt{6}\)\(3\sqrt{11}-2\sqrt{6}\) )?
A. (75)
यह अंतर के वर्ग का रूप है। ( \(3\sqrt{11}\)2-\(2\sqrt{6}\)2=99-24=75 )। / This is a difference of squares. ( \(3\sqrt{11}\)2-\(2\sqrt{6}\)2=99-24=75 ).
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What is the rationalised form of \( \frac{1}{4\sqrt{3}+3\sqrt{5}} \)?
A. \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \)
संयुग्मी से गुणा करने पर हर (48-45=3) आता है। इसलिए रूप \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \) है। / Multiplying by the conjugate gives denominator (48-45=3). So the form is \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \).
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What is the rationalised form of \( \frac{5}{\sqrt{13}-\sqrt{8}} \)?
A. \( \sqrt{13}+\sqrt{8} \)
संयुग्मी से गुणा करने पर हर (13-8=5) आता है। अंश में भी (5) है इसलिए उत्तर \( \sqrt{13}+\sqrt{8} \) है। / Multiplying by the conjugate gives denominator (13-8=5). The numerator also has (5), so the answer is \( \sqrt{13}+\sqrt{8} \).
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What is the value of \( \frac{4+\sqrt{7}}{4-\sqrt{7}}+\frac{4-\sqrt{7}}{4+\sqrt{7}} \)?
A. \( \frac{46}{9} \)
मान ( \frac{\(4+\sqrt{7}\)2+\(4-\sqrt{7}\)2}{16-7} ) होगा। अंश (46) और हर (9) है। / The value is ( \frac{\(4+\sqrt{7}\)2+\(4-\sqrt{7}\)2}{16-7} ). The numerator is (46) and the denominator is (9).
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What is the value of ( \left\(\sqrt{7}+\sqrt{2}\right\)2-\left\(\sqrt{7}-\sqrt{2}\right\)2 )?
A. \(4\sqrt{14}\)
सूत्र ( (a+b)2-(a-b)2=4ab ) लगाएँ। यहाँ \(4\sqrt{7}\sqrt{2}=4\sqrt{14}\) मिलेगा। / Use ( (a+b)2-(a-b)2=4ab ). Here \(4\sqrt{7}\sqrt{2}=4\sqrt{14}\).
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What is the value of \( \frac{\sqrt{20}+\sqrt{45}}{\sqrt{5}} \)?
A. (5)
\( \sqrt{20}=2\sqrt{5} \) और \( \sqrt{45}=3\sqrt{5} \)। कुल \(5\sqrt{5}\) को \( \sqrt{5} \) से भाग देने पर (5) मिलता है। / \( \sqrt{20}=2\sqrt{5} \) and \( \sqrt{45}=3\sqrt{5} \). Dividing \(5\sqrt{5}\) by \( \sqrt{5} \) gives (5).
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What is \( \sqrt{9+4\sqrt{5}} \) equal to?
A. \(2+\sqrt{5}\)
( \(2+\sqrt{5}\)2=4+4\sqrt{5}+5=9+4\sqrt{5} )। वर्ग करके उत्तर की जाँच करें। / ( \(2+\sqrt{5}\)2=4+4\sqrt{5}+5=9+4\sqrt{5} ). Check the answer by squaring.
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What is the positive simplified form of \( \sqrt{14-4\sqrt{10}} \)?
A. \( \sqrt{10}-2 \)
( \(\sqrt{10}-2\)2=10-4\sqrt{10}+4=14-4\sqrt{10} )। धनात्मक मुख्य वर्गमूल \( \sqrt{10}-2 \) है। / ( \(\sqrt{10}-2\)2=10-4\sqrt{10}+4=14-4\sqrt{10} ). The positive principal square root is \( \sqrt{10}-2 \).
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Which is greater between \( \sqrt{130} \) and (11.4)?
A. \( \sqrt{130} \)
\(11.4^2=129.96\) जो (130) से थोड़ा कम है। इसलिए \( \sqrt{130}>11.4 \) है। / \(11.4^2=129.96\), which is slightly less than (130). Therefore \( \sqrt{130}>11.4 \).
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Between which two consecutive integers does \( \sqrt{255} \) lie?
B. (15) और (16)(15) and (16)
\(15^2=225\) और \(16^2=256\) है। इसलिए \( \sqrt{255} \) (15) और (16) के बीच है। / \(15^2=225\) and \(16^2=256\). Therefore \( \sqrt{255} \) lies between (15) and (16).
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Which is greater between \( -\sqrt{80} \) and ( -9 )?
A. \( -\sqrt{80} \)
\( \sqrt{80}\approx8.94 \) है इसलिए \( -\sqrt{80}\approx-8.94 \)। यह ( -9 ) से बड़ा है क्योंकि शून्य के अधिक निकट है। / \( \sqrt{80}\approx8.94 \), so \( -\sqrt{80}\approx-8.94 \). It is greater than ( -9 ) because it is closer to zero.
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What will be the decimal expansion of the simplified form of \( \frac{96}{384} \)?
A. समाप्त दशमलवTerminating decimal
\( \frac{96}{384}=\frac{1}{4} \) और \(4=2^2\) है। हर में केवल (2) और (5) होने पर दशमलव समाप्त होता है। / \( \frac{96}{384}=\frac{1}{4} \) and \(4=2^2\). A decimal terminates when the denominator has only (2) and (5) as prime factors.
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What will be the decimal expansion of the simplified form of \( \frac{35}{154} \)?
B. असमाप्त आवर्ती दशमलवNon-terminating repeating decimal
\( \frac{35}{154}=\frac{5}{22} \) है और हर में (11) है। इसलिए दशमलव असमाप्त आवर्ती होगा। / \( \frac{35}{154}=\frac{5}{22} \), and the denominator has (11). Therefore the decimal is non-terminating repeating.
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What will be the decimal expansion of the simplified form of \( \frac{132}{360} \)?
B. असमाप्त आवर्ती दशमलवNon-terminating repeating decimal
\( \frac{132}{360}=\frac{11}{30} \) है और हर में (3) भी है। इसलिए यह असमाप्त आवर्ती दशमलव देगा। / \( \frac{132}{360}=\frac{11}{30} \), and the denominator also has (3). Therefore it gives a non-terminating repeating decimal.
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What will be the decimal expansion of the simplified form of \( \frac{126}{224} \)?
A. समाप्त दशमलवTerminating decimal
\( \frac{126}{224}=\frac{9}{16} \) है और \(16=2^4\)। इसलिए दशमलव समाप्त होगा। / \( \frac{126}{224}=\frac{9}{16} \), and \(16=2^4\). Therefore the decimal will terminate.
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If (0.12112211122211112222...) is non-terminating and non-repeating, what type of number is it?
B. अपरिमेय वास्तविक संख्याIrrational real number
असमाप्त और अनावर्ती दशमलव अपरिमेय होता है। स्थिर आवृत्ति न होने पर इसे परिमेय न मानें। / A non-terminating and non-repeating decimal is irrational. Do not treat it as rational when there is no fixed repetition.
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How will (7.818181...) be classified as a real number?
B. परिमेय वास्तविक संख्याRational real number
(81) बार-बार आ रहा है इसलिए दशमलव आवर्ती है। हर आवर्ती दशमलव परिमेय वास्तविक संख्या होता है। / The block (81) repeats, so the decimal is repeating. Every repeating decimal is a rational real number.
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What is the simplified form of \( \sqrt{5}+\sqrt{20}+\sqrt{45} \)?
A. \(6\sqrt{5}\)
\( \sqrt{20}=2\sqrt{5} \) और \( \sqrt{45}=3\sqrt{5} \)। कुल (1+2+3=6) से \(6\sqrt{5}\) मिलता है। / \( \sqrt{20}=2\sqrt{5} \) and \( \sqrt{45}=3\sqrt{5} \). The total coefficient is (1+2+3=6), so the value is \(6\sqrt{5}\).
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What is the simplified form of \( \sqrt{7}+\sqrt{28}+\sqrt{63}+\sqrt{112} \)?
A. \(10\sqrt{7}\)
ये क्रमशः \( \sqrt{7},2\sqrt{7},3\sqrt{7},4\sqrt{7} \) हैं। कुल \(10\sqrt{7}\) होगा। / These are \( \sqrt{7},2\sqrt{7},3\sqrt{7},4\sqrt{7} \). The total is \(10\sqrt{7}\).
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A. \(2\sqrt{6}\)
समान हर (7-6=1) और अंश ( \sqrt{7}+\sqrt{6}-\(\sqrt{7}-\sqrt{6}\)=2\sqrt{6} ) है। / The common denominator is (7-6=1), and the numerator is ( \sqrt{7}+\sqrt{6}-\(\sqrt{7}-\sqrt{6}\)=2\sqrt{6} ).
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What is the value of ( \left\(\frac{1}{5-\sqrt{21}}\right\)+\left\(\frac{1}{5+\sqrt{21}}\right\) )?
A. \( \frac{5}{2} \)
समान हर (25-21=4) और अंश (10) है। इसलिए मान \( \frac{10}{4}=\frac{5}{2} \) है। / The common denominator is (25-21=4), and the numerator is (10). So the value is \( \frac{10}{4}=\frac{5}{2} \).
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What is the value of ( \left\(\sqrt{10}+\sqrt{3}\right\)\left\(\sqrt{10}-\sqrt{3}\right\) )?
A. (7)
यह संयुग्मी गुणन है। (10-3=7) मिलेगा। / This is conjugate multiplication. The value is (10-3=7).
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What is the value of \( \sqrt{20}+\sqrt{180}-\sqrt{80}-\sqrt{45} \)?
A. (0)
क्रमशः \(2\sqrt{5},6\sqrt{5},4\sqrt{5},3\sqrt{5}\) मिलते हैं। (2+6-4-3=1) से परिणाम \( \sqrt{5} \) होना चाहिए। / They become \(2\sqrt{5},6\sqrt{5},4\sqrt{5},3\sqrt{5}\). Since (2+6-4-3=1), the result should be \( \sqrt{5} \).
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If \(x=4-\sqrt{7}\), then \( \frac{1}{x} \) equals what?
A. \( \frac{4+\sqrt{7}}{9} \)
\( \frac{1}{4-\sqrt{7}} \) को संयुग्मी से गुणा करें। हर (16-7=9) और अंश \(4+\sqrt{7}\) होगा। / Multiply \( \frac{1}{4-\sqrt{7}} \) by the conjugate. The denominator is (16-7=9) and the numerator is \(4+\sqrt{7}\).
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If \(a=\sqrt{5}+\sqrt{2}\), what is the value of \(a^2+\frac{1}{a^2}\)?
A. (98)
\(a^2=7+2\sqrt{10}\) और \( \frac{1}{a^2}=7-2\sqrt{10} \) नहीं है क्योंकि (a\(\sqrt{5}-\sqrt{2}\)=3)। सही मान सीधे (a-2+\frac{1}{a-2}=\frac{\(a^2\)2+1}{a-2}) से कठिन है इसलिए विकल्प जाँचें। / \(a^2=7+2\sqrt{10}\) and \( \frac{1}{a^2} \) is not simply \(7-2\sqrt{10}\) because (a\(\sqrt{5}-\sqrt{2}\)=3). Check options carefully using reciprocal rules.
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If \(x=\sqrt{8}+3\), what is the value of \(x+\frac{1}{x-3}\)?
A. \(3+\frac{5\sqrt{2}}{2}\)
\(x-3=\sqrt{8}=2\sqrt{2}\) इसलिए \( \frac{1}{x-3}=\frac{\sqrt{2}}{4} \)। कुल \(3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}\) होना चाहिए। / \(x-3=\sqrt{8}=2\sqrt{2}\), so \( \frac{1}{x-3}=\frac{\sqrt{2}}{4} \). The total should be \(3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}\).
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What is the value of ( \sqrt{(a-b)2} ) in real numbers?
C. ( |a-b| )
मुख्य वर्गमूल हमेशा गैर-ऋणात्मक होता है। इसलिए ( \sqrt{(a-b)2}=|a-b| ) होता है। / The principal square root is always non-negative. Therefore ( \sqrt{(a-b)2}=|a-b| ).
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If (x=-9), what is the value of \( \sqrt{x^2}+3x \)?
A. ( -18 )
\( \sqrt{x^2}=|x|=9 \) और (3x=-27) है। इसलिए कुल (9-27=-18) मिलेगा। / \( \sqrt{x^2}=|x|=9 \) and (3x=-27). Therefore the total is (9-27=-18).
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What is the correct simplified form of \( |4-\sqrt{27}| \)?
A. \( \sqrt{27}-4 \)
\( \sqrt{27}>4 \) क्योंकि (27>16) है। इसलिए \(4-\sqrt{27}\) ऋणात्मक है और निरपेक्ष मान \( \sqrt{27}-4 \) होगा। / \( \sqrt{27}>4 \) because (27>16). So \(4-\sqrt{27}\) is negative and the absolute value is \( \sqrt{27}-4 \).
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What is the simplified form of \( |\sqrt{70}-9| \)?
A. \(9-\sqrt{70}\)
\( \sqrt{70}<9 \) क्योंकि (70<81) है। इसलिए निरपेक्ष मान \(9-\sqrt{70}\) होगा।
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What is the correct statement about \( \sqrt{32}+\sqrt{18} \) and \(7\sqrt{2}\)?
A. दोनों बराबर हैंBoth are equal
\( \sqrt{32}=4\sqrt{2} \) और \( \sqrt{18}=3\sqrt{2} \)। योग \(7\sqrt{2}\) है। / \( \sqrt{32}=4\sqrt{2} \) and \( \sqrt{18}=3\sqrt{2} \). Their sum is \(7\sqrt{2}\).
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Which relation is correct between \( \sqrt{75}+\sqrt{192} \) and \(13\sqrt{3}\)?
A. \( \sqrt{75}+\sqrt{192}=13\sqrt{3} \)
\( \sqrt{75}=5\sqrt{3} \) और \( \sqrt{192}=8\sqrt{3} \)। योग \(13\sqrt{3}\) है। / \( \sqrt{75}=5\sqrt{3} \) and \( \sqrt{192}=8\sqrt{3} \). Their sum is \(13\sqrt{3}\).
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What is the value of \( \frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}} \)?
A. \( \sqrt{3} \)
समान हर (5-3=2) और अंश \(2\sqrt{3}\) है। इसलिए मान \( \sqrt{3} \) है। / The common denominator is (5-3=2), and the numerator is \(2\sqrt{3}\). Therefore the value is \( \sqrt{3} \).
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What is the value of \( \frac{1}{\sqrt{13}+2}+\frac{1}{\sqrt{13}-2} \)?
A. \( \frac{2\sqrt{13}}{9} \)
समान हर (13-4=9) और अंश \(2\sqrt{13}\) है। इसलिए मान \( \frac{2\sqrt{13}}{9} \) है। / The common denominator is (13-4=9), and the numerator is \(2\sqrt{13}\). Therefore the value is \( \frac{2\sqrt{13}}{9} \).
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What is the simplified form of ( \(3+\sqrt{5}\)3 )?
A. \(72+32\sqrt{5}\)
पहले ( \(3+\sqrt{5}\)2=14+6\sqrt{5} ) निकालें। फिर \(3+\sqrt{5}\) से गुणा करने पर \(72+32\sqrt{5}\) मिलता है। / First find ( \(3+\sqrt{5}\)2=14+6\sqrt{5} ). Multiplying by \(3+\sqrt{5}\) gives \(72+32\sqrt{5}\).
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What is the simplified form of ( \(4-\sqrt{3}\)3 )?
A. \(100-51\sqrt{3}\)
पहले ( \(4-\sqrt{3}\)2=19-8\sqrt{3} ) निकालें। फिर \(4-\sqrt{3}\) से गुणा करने पर \(100-51\sqrt{3}\) मिलता है। / First find ( \(4-\sqrt{3}\)2=19-8\sqrt{3} ). Multiplying by \(4-\sqrt{3}\) gives \(100-51\sqrt{3}\).
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If \( \sqrt{n} \) lies between (18) and (19), which range is correct for (n)?
A. (324<n<361)
दोनों ओर वर्ग करने पर \(18^2<n<19^2\) मिलेगा। इसलिए (324<n<361) सही है। / Squaring both sides gives \(18^2<n<19^2\). Therefore (324<n<361) is correct.
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If \(21<\sqrt{m}<22\), which range is correct for \(m\)?
A. 441<m<484
चूँकि \(\sqrt{m}\) धनात्मक है, दोनों तरफों का वर्ग लेने पर असमानता की दिशा अपरिवर्तित रहती है: \(21^2<m<22^2\)। अतः \(441<m<484\) सही है। निकटवर्ती भ्रामक विकल्पों में, \(21<m<22\) (विकल्प B) \(\sqrt{m}\) की सीमा है न कि \(m\) की; \(400<m<441\) (विकल्प C) और \(484<m<529\) (विकल्प D) क्रमशः \(20^2\)–\(21^2\) और \(22^2\)–\(23^2\) के अंतराल हैं, इसलिए वे गलत हैं। परीक्षा-सुझाव: जब सीमाएँ धनात्मक हों तो पक्षों का वर्ग लेते समय असमानता की दिशा नहीं बदलती — हमेशा वर्ग करके सीमा पर पहुँचें और सीमाएँ कड़ी/मुलायम (strict/non-strict) बनी रहती हैं। / Since \(\sqrt{m}\) is positive, squaring all parts preserves the inequality: \(21^2<m<22^2\). Thus \(441<m<484\) is correct. Closest distractors: option B (\(21<m<22\)) gives the range of \(\sqrt{m}\), not of \(m\); option C (\(400<m<441\)) equals \(20^2\)–\(21^2\) and option D (\(484<m<529\)) equals \(22^2\)–\(23^2\), so both are outside the required range. Exam tip: when squaring an inequality ensure the expressions are nonnegative — then you can square termwise without flipping the inequality sign; square the exact endpoints to get the correct interval.
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Which example shows that \( \sqrt{a+b}=\sqrt{a}+\sqrt{b} \) is false?
A. \( \sqrt{4+9}\neq\sqrt{4}+\sqrt{9} \)
\( \sqrt{13} \) और (2+3=5) बराबर नहीं हैं। इसलिए वर्गमूल को जोड़ के अंदर सीधे नहीं बाँटते। / \( \sqrt{13} \) and (2+3=5) are not equal. Therefore square root cannot be distributed directly over addition.
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What is the value of \( \sqrt{ab}\times\sqrt{ab} \) where \(ab\geq0\)?
A. (ab)
किसी गैर-ऋणात्मक संख्या के वर्गमूल को स्वयं से गुणा करने पर वही संख्या मिलती है। इसलिए मान (ab) है। / Multiplying the square root of a non-negative number by itself gives the same number. Therefore the value is (ab).
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What is the correct statement about \( \sqrt{36}+\sqrt{49} \) and \( \sqrt{85} \)?
A. \( \sqrt{36}+\sqrt{49}>\sqrt{85} \)
\( \sqrt{36}+\sqrt{49}=6+7=13 \) और \( \sqrt{85}<10 \) है। इसलिए पहला मान बड़ा है। / \( \sqrt{36}+\sqrt{49}=6+7=13 \), and \( \sqrt{85}<10 \). Therefore the first value is greater.
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Which product gives a rational number?
A. \( \sqrt{3}\times\sqrt{27} \)
\( \sqrt{3}\times\sqrt{27}=\sqrt{81}=9 \) परिमेय है। बाकी गुणनफल पूर्ण वर्ग के वर्गमूल में नहीं बदलते। / \( \sqrt{3}\times\sqrt{27}=\sqrt{81}=9 \), which is rational. The other products do not become square roots of perfect squares.
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What is the value of \( \frac{1}{3+\sqrt{8}}+\frac{1}{3-\sqrt{8}} \)?
A. (6)
समान हर (9-8=1) और अंश \(3-\sqrt{8}+3+\sqrt{8}=6\) है। इसलिए मान (6) है। / The common denominator is (9-8=1), and the numerator is \(3-\sqrt{8}+3+\sqrt{8}=6\). Therefore the value is (6).
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