Level 1 • 50/50 questions • 30 seconds per question.
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What is the simplified surd form of \( \sqrt{675} \)?
A. \(15\sqrt{3}\)
\(675=225\times3\) इसलिए \( \sqrt{675}=15\sqrt{3} \)। सबसे बड़ा पूर्ण वर्ग गुणनखंड पहले खोजें। / Since \(675=225\times3\), \( \sqrt{675}=15\sqrt{3} \). First find the largest perfect-square factor.
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What is obtained by simplifying \( \sqrt{980} \)?
A. \(14\sqrt{5}\)
\(980=196\times5\) इसलिए \( \sqrt{980}=14\sqrt{5} \)। करणी में पूर्ण वर्ग को बाहर निकालें। / Since \(980=196\times5\), \( \sqrt{980}=14\sqrt{5} \). Take the perfect square outside the radical.
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What is the simplified form of \(2\sqrt{147}-3\sqrt{75}+\sqrt{300}\)?
B. \(8\sqrt{3}\)
\(2\sqrt{147}=14\sqrt{3}\), \(3\sqrt{75}=15\sqrt{3}\), और \( \sqrt{300}=10\sqrt{3} \)। कुल \(9\sqrt{3}\) नहीं बल्कि (14-15+10=9) से \(9\sqrt{3}\) होना चाहिए। / \(2\sqrt{147}=14\sqrt{3}\), \(3\sqrt{75}=15\sqrt{3}\), and \( \sqrt{300}=10\sqrt{3} \). The coefficient is (14-15+10=9), so the value is \(9\sqrt{3}\).
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Choose the correct simplified form of \(4\sqrt{72}+2\sqrt{200}-5\sqrt{50}\).
A. \(14\sqrt{2}\)
\(4\sqrt{72}=24\sqrt{2}\), \(2\sqrt{200}=20\sqrt{2}\), और \(5\sqrt{50}=25\sqrt{2}\)। इसलिए परिणाम \(19\sqrt{2}\) होना चाहिए। / \(4\sqrt{72}=24\sqrt{2}\), \(2\sqrt{200}=20\sqrt{2}\), and \(5\sqrt{50}=25\sqrt{2}\). Therefore the result should be \(19\sqrt{2}\).
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What is the value of ( \sqrt{18}\(2\sqrt{8}-\sqrt{50}\) )?
B. ( -6 )
\(2\sqrt{8}=4\sqrt{2}\) और \( \sqrt{50}=5\sqrt{2} \), इसलिए कोष्ठक \( -\sqrt{2} \) है। ( \sqrt{18}\times\(-\sqrt{2}\)=-\sqrt{36}=-6 )। / \(2\sqrt{8}=4\sqrt{2}\) and \( \sqrt{50}=5\sqrt{2} \), so the bracket is \( -\sqrt{2} \). ( \sqrt{18}\times\(-\sqrt{2}\)=-\sqrt{36}=-6 ).
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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 rationalised form of \( \frac{1}{3\sqrt{2}+2\sqrt{5}} \)?
B. \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \)
संयुग्मी से गुणा करने पर हर (18-20=-2) मिलता है। इसलिए रूप \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \) लिखा जा सकता है। / Multiplying by the conjugate gives denominator (18-20=-2). So the form can be written as \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \).
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What is the rationalised form of \( \frac{2}{\sqrt{7}-\sqrt{5}} \)?
A. \( \sqrt{7}+\sqrt{5} \)
संयुग्मी \( \sqrt{7}+\sqrt{5} \) से गुणा करने पर हर (7-5=2) आता है। अंश में भी (2) है इसलिए उत्तर \( \sqrt{7}+\sqrt{5} \) है। / Multiplying by the conjugate \( \sqrt{7}+\sqrt{5} \) gives denominator (7-5=2). The numerator also has (2), so the answer is \( \sqrt{7}+\sqrt{5} \).
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What is the value of \( \frac{3+\sqrt{2}}{3-\sqrt{2}}+\frac{3-\sqrt{2}}{3+\sqrt{2}} \)?
A. \( \frac{22}{7} \)
मान ( \frac{\(3+\sqrt{2}\)2+\(3-\sqrt{2}\)2}{9-2} ) होगा। अंश (22) और हर (7) है। / The value is ( \frac{\(3+\sqrt{2}\)2+\(3-\sqrt{2}\)2}{9-2} ). The numerator is (22) and the denominator is (7).
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What is the value of ( \left\(\sqrt{5}+\sqrt{3}\right\)2-\left\(\sqrt{5}-\sqrt{3}\right\)2 )?
A. \(4\sqrt{15}\)
सूत्र ( (a+b)2-(a-b)2=4ab ) लगाएँ। यहाँ \(4\sqrt{5}\sqrt{3}=4\sqrt{15}\) मिलेगा। / Use ( (a+b)2-(a-b)2=4ab ). Here \(4\sqrt{5}\sqrt{3}=4\sqrt{15}\).
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What is the value of \( \frac{\sqrt{12}+\sqrt{27}}{\sqrt{3}} \)?
A. (5)
\( \sqrt{12}=2\sqrt{3} \) और \( \sqrt{27}=3\sqrt{3} \)। कुल \(5\sqrt{3}\) को \( \sqrt{3} \) से भाग देने पर (5) मिलेगा। / \( \sqrt{12}=2\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \). Dividing \(5\sqrt{3}\) by \( \sqrt{3} \) gives (5).
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What is \( \sqrt{7+4\sqrt{3}} \) equal to?
A. \(2+\sqrt{3}\)
( \(2+\sqrt{3}\)2=4+4\sqrt{3}+3=7+4\sqrt{3} )। वर्ग करके जाँचने की आदत रखें। / ( \(2+\sqrt{3}\)2=4+4\sqrt{3}+3=7+4\sqrt{3} ). Build the habit of checking by squaring.
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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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Which is the correct ascending order?
B. \( \frac{11}{5}<\sqrt{5}<2.3 \)
\( \frac{11}{5}=2.2 \), \( \sqrt{5}\approx2.236 \), और (2.3) है। इसलिए आरोही क्रम \( \frac{11}{5}<\sqrt{5}<2.3 \) है। / \( \frac{11}{5}=2.2 \), \( \sqrt{5}\approx2.236 \), and then (2.3). So the ascending order is \( \frac{11}{5}<\sqrt{5}<2.3 \).
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Between which two consecutive integers does \( \sqrt{170} \) lie?
C. (13) और (14)(13) and (14)
\(13^2=169\) और \(14^2=196\) है। इसलिए \( \sqrt{170} \) (13) और (14) के बीच है। / \(13^2=169\) and \(14^2=196\). Therefore \( \sqrt{170} \) lies between (13) and (14).
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Which is greater between \( -\sqrt{30} \) and \( -\frac{11}{2} \)?
A. \( -\sqrt{30} \)
\( \sqrt{30}\approx5.477 \) और \( \frac{11}{2}=5.5 \) है। इसलिए \( -\sqrt{30} \) शून्य के अधिक निकट है और बड़ी है। / \( \sqrt{30}\approx5.477 \) and \( \frac{11}{2}=5.5 \). So \( -\sqrt{30} \) is closer to zero and is greater.
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What will be the decimal expansion of the simplified form of \( \frac{39}{312} \)?
A. समाप्त दशमलवTerminating decimal
\( \frac{39}{312}=\frac{1}{8} \) और \(8=2^3\) है। हर में केवल (2) और (5) होने पर दशमलव समाप्त होता है। / \( \frac{39}{312}=\frac{1}{8} \) and \(8=2^3\). 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 \( \frac{22}{77} \)?
B. असमाप्त आवर्ती दशमलवNon-terminating repeating decimal
\( \frac{22}{77}=\frac{2}{7} \) है और हर में (7) है। इसलिए दशमलव असमाप्त आवर्ती होगा। / \( \frac{22}{77}=\frac{2}{7} \), and the denominator has (7). Therefore the decimal is non-terminating repeating.
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What will be the decimal expansion of the simplified form of \( \frac{84}{150} \)?
A. समाप्त दशमलवTerminating decimal
\( \frac{84}{150}=\frac{14}{25} \) है और \(25=5^2\)। इसलिए यह समाप्त दशमलव देगा। / \( \frac{84}{150}=\frac{14}{25} \), and \(25=5^2\). Therefore it gives a terminating decimal.
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What will be the decimal expansion of the simplified form of \( \frac{105}{126} \)?
B. असमाप्त आवर्ती दशमलवNon-terminating repeating decimal
\( \frac{105}{126}=\frac{5}{6} \) है और हर में (3) है। इसलिए दशमलव असमाप्त आवर्ती होगा। / \( \frac{105}{126}=\frac{5}{6} \), and the denominator has (3). Therefore the decimal is non-terminating repeating.
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If (0.01001000100001...) 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। दशमलव में नियमित आवृत्ति है या नहीं यह पहचानें। / A non-terminating and non-repeating decimal is irrational. Check whether the decimal has a fixed repeating pattern.
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How will (5.232323...) be classified as a real number?
B. परिमेय वास्तविक संख्याRational real number
(23) बार-बार आ रहा है इसलिए दशमलव आवर्ती है। हर आवर्ती दशमलव परिमेय वास्तविक संख्या होता है। / The block (23) repeats, so the decimal is repeating. Every repeating decimal is a rational real number.
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What is the simplified form of \( \sqrt{2}+\sqrt{8}+\sqrt{18} \)?
A. \(6\sqrt{2}\)
\( \sqrt{8}=2\sqrt{2} \) और \( \sqrt{18}=3\sqrt{2} \)। कुल (1+2+3=6) से \(6\sqrt{2}\) मिलता है। / \( \sqrt{8}=2\sqrt{2} \) and \( \sqrt{18}=3\sqrt{2} \). The total coefficient is (1+2+3=6), so the value is \(6\sqrt{2}\).
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What is the simplified form of \( \sqrt{3}+\sqrt{12}+\sqrt{27}+\sqrt{48} \)?
A. \(10\sqrt{3}\)
ये क्रमशः \( \sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3} \) हैं। कुल \(10\sqrt{3}\) होगा। / These are \( \sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3} \). The total is \(10\sqrt{3}\).
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What is the value of ( \left\(\frac{1}{\sqrt{5}-2}\right\)-\left\(\frac{1}{\sqrt{5}+2}\right\) )?
A. (4)
समान हर (5-4=1) होगा और अंश ( \sqrt{5}+2-\(\sqrt{5}-2\)=4 ) है। इसलिए मान (4) है। / The common denominator is (5-4=1), and the numerator is ( \sqrt{5}+2-\(\sqrt{5}-2\)=4 ). Hence the value is (4).
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What is the value of ( \left\(\frac{1}{3-\sqrt{7}}\right\)+\left\(\frac{1}{3+\sqrt{7}}\right\) )?
A. (3)
समान हर (9-7=2) और अंश \(3+\sqrt{7}+3-\sqrt{7}=6\) है। इसलिए मान \( \frac{6}{2}=3 \) है। / The common denominator is (9-7=2), and the numerator is \(3+\sqrt{7}+3-\sqrt{7}=6\). So the value is \( \frac{6}{2}=3 \).
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What is the value of ( \left\(\sqrt{6}+\sqrt{2}\right\)\left\(\sqrt{6}-\sqrt{2}\right\) )?
A. (4)
यह संयुग्मी गुणन है। (6-2=4) मिलेगा। / This is conjugate multiplication. The value is (6-2=4).
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What is the value of \( \sqrt{12}+\sqrt{75}-\sqrt{48}-\sqrt{27} \)?
A. (0)
क्रमशः \(2\sqrt{3},5\sqrt{3},4\sqrt{3},3\sqrt{3}\) मिलते हैं। (2+5-4-3=0) इसलिए मान (0) है। / They become \(2\sqrt{3},5\sqrt{3},4\sqrt{3},3\sqrt{3}\). Since (2+5-4-3=0), the value is (0).
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If \(x=3-\sqrt{5}\), then \( \frac{1}{x} \) equals what?
A. \( \frac{3+\sqrt{5}}{4} \)
\( \frac{1}{3-\sqrt{5}} \) को संयुग्मी से गुणा करें। हर (9-5=4) और अंश \(3+\sqrt{5}\) होगा। / Multiply \( \frac{1}{3-\sqrt{5}} \) by the conjugate. The denominator is (9-5=4) and the numerator is \(3+\sqrt{5}\).
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If \(a=\sqrt{3}+\sqrt{2}\), what is the value of \(a^2+\frac{1}{a^2}\)?
B. (98)
\(a^2=5+2\sqrt{6}\) और \( \frac{1}{a^2}=5-2\sqrt{6} \)। योग (10) होना चाहिए। / \(a^2=5+2\sqrt{6}\) and \( \frac{1}{a^2}=5-2\sqrt{6} \). The sum should be (10).
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If \(x=\sqrt{5}+2\), what is the value of \(x+\frac{1}{x}\)?
A. \(2\sqrt{5}\)
\( \frac{1}{\sqrt{5}+2}=\sqrt{5}-2 \) है। इसलिए \(x+\frac{1}{x}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)। / \( \frac{1}{\sqrt{5}+2}=\sqrt{5}-2 \). Hence \(x+\frac{1}{x}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\).
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If (x=-7), what is the value of \( \sqrt{x^2}-2x \)?
A. (21)
\( \sqrt{x^2}=|x|=7 \) और (-2x=14) है। इसलिए कुल (21) मिलेगा। / \( \sqrt{x^2}=|x|=7 \) and (-2x=14). Therefore the total is (21).
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What is the correct simplified form of \( |3-\sqrt{20}| \)?
A. \( \sqrt{20}-3 \)
\( \sqrt{20}>3 \) इसलिए \(3-\sqrt{20}\) ऋणात्मक है। निरपेक्ष मान इसे \( \sqrt{20}-3 \) बना देता है। / Since \( \sqrt{20}>3 \), \(3-\sqrt{20}\) is negative. Absolute value changes it to \( \sqrt{20}-3 \).
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What is the simplified form of \( |\sqrt{30}-6| \)?
A. \(6-\sqrt{30}\)
\( \sqrt{30}<6 \) इसलिए \( \sqrt{30}-6 \) ऋणात्मक है। निरपेक्ष मान \(6-\sqrt{30}\) होगा। / Since \( \sqrt{30}<6 \), \( \sqrt{30}-6 \) is negative. The absolute value is \(6-\sqrt{30}\).
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What is the correct statement about \( \sqrt{27}+\sqrt{12} \) and \( \sqrt{75} \)?
A. दोनों बराबर हैंBoth are equal
\( \sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3} \)। \( \sqrt{75}=5\sqrt{3} \) इसलिए दोनों बराबर हैं। / \( \sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3} \). Also \( \sqrt{75}=5\sqrt{3} \), so both are equal.
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Which relation is correct between \( \sqrt{48}+\sqrt{108} \) and \(10\sqrt{3}\)?
A. \( \sqrt{48}+\sqrt{108}=10\sqrt{3} \)
\( \sqrt{48}=4\sqrt{3} \) और \( \sqrt{108}=6\sqrt{3} \)। योग \(10\sqrt{3}\) है। / \( \sqrt{48}=4\sqrt{3} \) and \( \sqrt{108}=6\sqrt{3} \). Their sum is \(10\sqrt{3}\).
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What is the value of \( \frac{1}{\sqrt{3}-\sqrt{2}}-\frac{1}{\sqrt{3}+\sqrt{2}} \)?
A. \(2\sqrt{2}\)
समान हर (3-2=1) और अंश ( \sqrt{3}+\sqrt{2}-\(\sqrt{3}-\sqrt{2}\)=2\sqrt{2} ) है। / The common denominator is (3-2=1), and the numerator is ( \sqrt{3}+\sqrt{2}-\(\sqrt{3}-\sqrt{2}\)=2\sqrt{2} ).
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What is the value of \( \frac{1}{\sqrt{11}+3}+\frac{1}{\sqrt{11}-3} \)?
A. \( \sqrt{11} \)
समान हर (11-9=2) और अंश \(2\sqrt{11}\) है। इसलिए मान \( \sqrt{11} \) है। / The common denominator is (11-9=2), and the numerator is \(2\sqrt{11}\). Therefore the value is \( \sqrt{11} \).
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What is the simplified form of ( \(2+\sqrt{3}\)3 )?
A. \(26+15\sqrt{3}\)
पहले ( \(2+\sqrt{3}\)2=7+4\sqrt{3} ) निकालें। फिर \(2+\sqrt{3}\) से गुणा करने पर \(26+15\sqrt{3}\) मिलता है। / First find ( \(2+\sqrt{3}\)2=7+4\sqrt{3} ). Multiplying by \(2+\sqrt{3}\) gives \(26+15\sqrt{3}\).
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What is the simplified form of ( \(3-\sqrt{2}\)3 )?
A. \(45-29\sqrt{2}\)
पहले ( \(3-\sqrt{2}\)2=11-6\sqrt{2} ) निकालें। फिर \(3-\sqrt{2}\) से गुणा करने पर \(45-29\sqrt{2}\) मिलता है। / First find ( \(3-\sqrt{2}\)2=11-6\sqrt{2} ). Multiplying by \(3-\sqrt{2}\) gives \(45-29\sqrt{2}\).
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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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What is the correct statement about \( \sqrt{16}+\sqrt{25} \) and \( \sqrt{41} \)?
A. \( \sqrt{16}+\sqrt{25}>\sqrt{41} \)
\( \sqrt{16}+\sqrt{25}=4+5=9 \) और \( \sqrt{41}<7 \) है। इसलिए पहला मान बड़ा है। / \( \sqrt{16}+\sqrt{25}=4+5=9 \), and \( \sqrt{41}<7 \). Therefore the first value is greater.
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Which product made from \( \sqrt{2} \), \( \sqrt{3} \), and \( \sqrt{5} \) is rational?
A. \( \sqrt{2}\times\sqrt{8} \)
\( \sqrt{2}\times\sqrt{8}=\sqrt{16}=4 \) परिमेय है। बाकी गुणनफल पूर्ण वर्ग के वर्गमूल में नहीं बदलते। / \( \sqrt{2}\times\sqrt{8}=\sqrt{16}=4 \), which is rational. The other products do not become square roots of perfect squares.
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What is the value of \( \frac{1}{2+\sqrt{3}}+\frac{1}{2-\sqrt{3}} \)?
A. (4)
समान हर (4-3=1) और अंश \(2-\sqrt{3}+2+\sqrt{3}=4\) है। इसलिए मान (4) है। / The common denominator is (4-3=1), and the numerator is \(2-\sqrt{3}+2+\sqrt{3}=4\). Therefore the value is (4).
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