Level 15 • 50/50 questions • 35 seconds per question.
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If \(x=5+\sqrt{2}\), what type of number is (x-5)?
C. अपरिमेयIrrational
\(x-5=\sqrt{2}\) है जो अपरिमेय है। पहले परिमेय पदों को हटाकर सरल करें। / \(x-5=\sqrt{2}\), which is irrational. First remove rational terms and simplify.
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If \(x=\sqrt{12}+2\sqrt{3}\), what will be the simplified form of (x)?
D. \(4\sqrt{3}\)
\(\sqrt{12}=2\sqrt{3}\) इसलिए कुल \(4\sqrt{3}\) मिलता है। समान मूलों को जोड़ने से पहले सरल करें। / \(\sqrt{12}=2\sqrt{3}\) so the total is \(4\sqrt{3}\). Simplify before adding like radicals.
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Which is the simplified form of \(\sqrt{48}+\sqrt{27}\)?
A. \(7\sqrt{3}\)
\(\sqrt{48}=4\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए योग \(7\sqrt{3}\) है। समान मूलों को जोड़ने से पहले सरल करें। / \(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the sum is \(7\sqrt{3}\). Simplify before adding like radicals.
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What is the value of (\(\sqrt{15}+\sqrt{6}\)\(\sqrt{15}-\sqrt{6}\))?
A. (9)
यह \(a^2-b^2\) रूप है इसलिए मान (15-6=9) है। संयुग्मी गुणन में मूल हट जाता है। / This is the \(a^2-b^2\) form so the value is (15-6=9). Conjugate multiplication removes radicals.
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What is the value of (\(\sqrt{19}\)2+8)?
B. (27)
(\(\sqrt{19}\)2=19), इसलिए मान (27) है। वर्ग करने पर वर्गमूल हट जाता है। / (\(\sqrt{19}\)2=19), so the value is (27). Squaring removes the square root.
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Which is the rationalised form of \(\frac{4}{\sqrt{3}+1}\)?
C. \(2\sqrt{3}-2\)
संयुग्मी से गुणा करने पर (\frac{4\(\sqrt{3}-1\)}{2}=2\sqrt{3}-2) मिलता है। हर को परिमेय बनाते समय पूरा भिन्न सरल करें। / Multiplying by the conjugate gives (\frac{4\(\sqrt{3}-1\)}{2}=2\sqrt{3}-2). Simplify the whole fraction after rationalising.
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What is obtained by rationalising \(\frac{1}{\sqrt{6}+1}\)?
C. \(\frac{\sqrt{6}-1}{5}\)
संयुग्मी से गुणा करने पर हर (6-1=5) बनता है। इसलिए परिमेयकृत रूप \(\frac{\sqrt{6}-1}{5}\) है। / Multiplying by the conjugate makes the denominator (6-1=5). So the rationalised form is \(\frac{\sqrt{6}-1}{5}\).
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If \(a=\sqrt{8}-\sqrt{2}\), what is (a) equal to?
B. \(\sqrt{2}\)
\(\sqrt{8}=2\sqrt{2}\) इसलिए \(2\sqrt{2}-\sqrt{2}=\sqrt{2}\) है। समान मूलों के गुणांक घटाएं। / \(\sqrt{8}=2\sqrt{2}\) so \(2\sqrt{2}-\sqrt{2}=\sqrt{2}\). Subtract coefficients of like radicals.
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If \(a=3\sqrt{2}+4\) and \(b=3\sqrt{2}-4\), what is (a+b)?
B. \(6\sqrt{2}\)
स्थिर पद कट जाते हैं और \(3\sqrt{2}+3\sqrt{2}=6\sqrt{2}\) मिलता है। समान मूलों को जोड़ें। / The constant terms cancel and \(3\sqrt{2}+3\sqrt{2}=6\sqrt{2}\). Add like radicals.
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What is the value of (\(5+\sqrt{3}\)\(5-\sqrt{3}\))?
B. (22)
यह \(a^2-b^2\) है इसलिए मान (25-3=22) है। संयुग्मी गुणन में मूल हट जाता है। / This is \(a^2-b^2\), so the value is (25-3=22). Conjugate multiplication removes the radical.
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Which is irrational between \(0.414141\ldots\) and \(0.4141141114\ldots\)?
A. \(0.4141141114\ldots\)
\(0.4141141114\ldots\) में निश्चित दोहराव नहीं है इसलिए यह अपरिमेय है। आवर्ती दशमलव परिमेय होता है। / \(0.4141141114\ldots\) has no fixed repetition so it is irrational. A repeating decimal is rational.
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Which is the simplified form of \(\sqrt{98}-\sqrt{50}\)?
C. \(2\sqrt{2}\)
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए अंतर \(2\sqrt{2}\) है। पहले पूर्ण वर्ग गुणनखंड निकालें। / \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the difference is \(2\sqrt{2}\). First take out perfect-square factors.
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What is the simplified form of \(\sqrt{7}+\sqrt{28}+\sqrt{63}\)?
C. \(6\sqrt{7}\)
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए कुल \(6\sqrt{7}\) है। सभी पदों को समान मूल में बदलें। / \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the total is \(6\sqrt{7}\). Convert all terms to like radicals.
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What type of number is the decimal \(0.404004000400004\ldots\)?
D. अपरिमेयIrrational
इस दशमलव में निश्चित दोहराने वाला खंड नहीं है इसलिए यह अपरिमेय है। असांत अनावर्ती दशमलव को पहचानें। / This decimal has no fixed repeating block, so it is irrational. Identify non-terminating non-repeating decimals.
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What is the simplified value of \(\frac{\sqrt{98}-\sqrt{18}}{\sqrt{2}}\)?
B. (4)
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\) इसलिए भाग देने पर (4) मिलता है। पहले अंश को समान मूल में बदलें। / \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\) so division gives (4). First convert the numerator into like radicals.
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What is the value of \(\frac{\sqrt{108}}{\sqrt{3}}\)?
A. (6)
\(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{36}=6\) है। भाग में मूलों को एक साथ लिखकर सरल करें। / \(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{36}=6\). Combine radicals in division and simplify.
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Which is the expansion of (\(2+\sqrt{5}\)2)?
A. \(9+4\sqrt{5}\)
(\(2+\sqrt{5}\)2=4+4\sqrt{5}+5=9+4\sqrt{5}) है। वर्ग करते समय मध्य पद न भूलें। / (\(2+\sqrt{5}\)2=4+4\sqrt{5}+5=9+4\sqrt{5}). Do not forget the middle term while squaring.
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Which is the simplified form of \(\sqrt{162}+\sqrt{72}\)?
B. \(15\sqrt{2}\)
\(\sqrt{162}=9\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\), इसलिए योग \(15\sqrt{2}\) है। समान मूलों के गुणांक जोड़ें। / \(\sqrt{162}=9\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so the sum is \(15\sqrt{2}\). Add coefficients of like radicals.
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If \(\sqrt{k}\) lies between (8) and (9), which value of (k) is possible?
C. (73)
क्योंकि (64<73<81) इसलिए \(8<\sqrt{73}<9\)। वर्गमूल की सीमा वर्गों से तय करें। / Since (64<73<81), \(8<\sqrt{73}<9\). Decide square-root bounds using squares.
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Which number lies between \(\sqrt{21}\) and \(\sqrt{30}\)?
C. \(\sqrt{25}\)
क्योंकि (21<25<30), इसलिए \(\sqrt{25}\) इनके बीच है। वर्गमूलों की तुलना अंदर की संख्याओं से करें। / Since (21<25<30), \(\sqrt{25}\) lies between them. Compare square roots using the numbers inside.
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If \(z=\sqrt{17}\), what is the value of \(z^2-9\)?
B. (8)
\(z^2=17\), इसलिए \(z^2-9=8\) है। पहले दिए गए मूल का वर्ग करें। / \(z^2=17\), so \(z^2-9=8\). First square the given radical.
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Which is the simplified form of \(\sqrt{243}\)?
C. \(9\sqrt{3}\)
\(\sqrt{243}=\sqrt{81\times3}=9\sqrt{3}\) है। सबसे बड़ा पूर्ण वर्ग गुणनखंड चुनें। / \(\sqrt{243}=\sqrt{81\times3}=9\sqrt{3}\). Choose the largest perfect-square factor.
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What is the rationalised form of \(\frac{7}{\sqrt{3}}\)?
C. \(\frac{7\sqrt{3}}{3}\)
अंश और हर को \(\sqrt{3}\) से गुणा करने पर \(\frac{7\sqrt{3}}{3}\) मिलता है। परिमेयकरण में हर से मूल हटता है। / Multiplying numerator and denominator by \(\sqrt{3}\) gives \(\frac{7\sqrt{3}}{3}\). Rationalisation removes the radical from the denominator.
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What is the value of (\(\sqrt{12}+\sqrt{3}\)2)?
B. (27)
\(\sqrt{12}=2\sqrt{3}\), इसलिए कोष्ठक \(3\sqrt{3}\) है और वर्ग (27) है। पहले कोष्ठक सरल करें। / \(\sqrt{12}=2\sqrt{3}\), so the bracket is \(3\sqrt{3}\) and its square is (27). Simplify the bracket first.
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What is the irrational part in \(9+\sqrt{14}\)?
B. \(\sqrt{14}\)
(9) परिमेय है और \(\sqrt{14}\) अपरिमेय भाग है। मिश्रित रूप में मूल वाला पद पहचानें। / (9) is rational and \(\sqrt{14}\) is the irrational part. Identify the radical term in a mixed form.
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What is the simplified form of \(\sqrt{147}+\sqrt{75}-\sqrt{27}\)?
A. \(9\sqrt{3}\)
\(\sqrt{147}=7\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए उत्तर \(9\sqrt{3}\) है। समान मूलों के गुणांक जोड़ें और घटाएं। / \(\sqrt{147}=7\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the answer is \(9\sqrt{3}\). Add and subtract coefficients of like radicals.
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If \(\sqrt{m}=11\), what is the value of (m)?
B. (121)
दोनों पक्षों का वर्ग करने पर (m=121) मिलता है। पूर्ण वर्गों को याद रखना उपयोगी है। / Squaring both sides gives (m=121). Remembering perfect squares is useful.
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C. अपरिमेयIrrational
पूर्ण वर्ग न होने पर धनात्मक पूर्ण संख्या का वर्गमूल अपरिमेय होता है। पहले पूर्ण वर्ग की जाँच करें। / The square root of a positive integer is irrational when it is not a perfect square. Check for a perfect square first.
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What is the value of \(\sqrt{24}\times\sqrt{54}\)?
A. (36)
\(\sqrt{24}\times\sqrt{54}=\sqrt{1296}=36\) है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है। / \(\sqrt{24}\times\sqrt{54}=\sqrt{1296}=36\). The product of two irrationals can be rational.
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Which statement is correct about \(\sqrt{13}+\sqrt{17}\)?
C. यह अपरिमेय हैIt is irrational
\(\sqrt{13}\) और \(\sqrt{17}\) अलग अपरिमेय मूल हैं और उनका योग अपरिमेय है। अलग मूलों को सीधे नहीं जोड़ा जाता। / \(\sqrt{13}\) and \(\sqrt{17}\) are different irrational radicals and their sum is irrational. Different radicals are not added directly.
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Which is the simplified form of \(\sqrt{392}\)?
A. \(14\sqrt{2}\)
\(\sqrt{392}=\sqrt{196\times2}=14\sqrt{2}\) है। बड़े पूर्ण वर्ग गुणनखंड से हल सरल होता है। / \(\sqrt{392}=\sqrt{196\times2}=14\sqrt{2}\). A large perfect-square factor makes the solution easier.
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What is the simplified value of \(\frac{\sqrt{180}}{\sqrt{5}}\)?
B. (6)
\(\frac{\sqrt{180}}{\sqrt{5}}=\sqrt{36}=6\) है। मूलों के भाग में अंदर का भागफल लें। / \(\frac{\sqrt{180}}{\sqrt{5}}=\sqrt{36}=6\). In division of roots take the quotient inside.
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What is the result of \(11\sqrt{3}-4\sqrt{3}\)?
B. \(7\sqrt{3}\)
समान मूलों के गुणांक घटाने पर \(7\sqrt{3}\) मिलता है। यह अपरिमेय है क्योंकि \(\sqrt{3}\) बचता है। / Subtracting coefficients of like radicals gives \(7\sqrt{3}\). It is irrational because \(\sqrt{3}\) remains.
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What is the value of (\sqrt{5}\(\sqrt{45}+\sqrt{80}\))?
A. (35)
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{80}=4\sqrt{5}\), इसलिए कोष्ठक \(7\sqrt{5}\) और गुणनफल (35) है। पहले कोष्ठक सरल करें। / \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\), so the bracket is \(7\sqrt{5}\) and the product is (35). Simplify the bracket first.
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Which statement is correct when comparing \(\sqrt{43}\) and \(\sqrt{47}\)?
C. \(\sqrt{43}<\sqrt{47}\)
क्योंकि (43<47), इसलिए \(\sqrt{43}<\sqrt{47}\)। धनात्मक मूलों में अंदर की संख्याओं की तुलना करें। / Since (43<47), \(\sqrt{43}<\sqrt{47}\). For positive roots compare the numbers inside.
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What is the simplified form of (\sqrt{7}\times\(2\sqrt{7}-3\))?
A. \(14-3\sqrt{7}\)
वितरण करने पर \(2\times7-3\sqrt{7}=14-3\sqrt{7}\) मिलता है। मूल और परिमेय पद अलग रखें। / Distributing gives \(2\times7-3\sqrt{7}=14-3\sqrt{7}\). Keep radical and rational terms separate.
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Which is the rationalised form of \(\frac{4}{\sqrt{5}+1}\)?
B. \(\sqrt{5}-1\)
संयुग्मी से गुणा करने पर (\frac{4\(\sqrt{5}-1\)}{5-1}=\sqrt{5}-1) मिलता है। हर को परिमेय बनाएं। / Multiplying by the conjugate gives (\frac{4\(\sqrt{5}-1\)}{5-1}=\sqrt{5}-1). Make the denominator rational.
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What is the sum of \(10+\sqrt{29}\) and \(10-\sqrt{29}\)?
A. (20)
अपरिमेय पद कट जाते हैं और योग (20) है। संयुग्मी संख्याओं का योग परिमेय होता है। / The irrational terms cancel and the sum is (20). The sum of conjugate numbers is rational.
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What is the product of \(10+\sqrt{29}\) and \(10-\sqrt{29}\)?
B. (71)
गुणनफल (102-\(\sqrt{29}\)2=100-29=71) है। संयुग्मी गुणन में \(a^2-b^2\) लगाएं। / The product is (102-\(\sqrt{29}\)2=100-29=71). Use \(a^2-b^2\) in conjugate multiplication.
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If \(q=7-\sqrt{11}\), what type of number is (q)?
C. अपरिमेयIrrational
परिमेय (7) में से अपरिमेय \(\sqrt{11}\) घटाने पर अपरिमेय संख्या मिलती है। अपरिमेय भाग बचा रहता है। / Subtracting irrational \(\sqrt{11}\) from rational (7) gives an irrational number. The irrational part remains.
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To construct \(\sqrt{10}\) on the number line, which right triangle hypotenuse can be used?
A. भुजाएँ (1) और (3)Legs (1) and (3)
समकोण त्रिभुज में कर्ण \(\sqrt{1^2+3^2}=\sqrt{10}\) होगा। निर्माण में पाइथागोरस प्रमेय लगती है। / In a right triangle, the hypotenuse is \(\sqrt{1^2+3^2}=\sqrt{10}\). Pythagoras theorem is used in construction.
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What type of number is \(\sqrt{6}\times\sqrt{24}\)?
B. परिमेयRational
\(\sqrt{6}\times\sqrt{24}=\sqrt{144}=12\) है जो परिमेय है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है। / \(\sqrt{6}\times\sqrt{24}=\sqrt{144}=12\), which is rational. The product of two irrationals can be rational.
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What is the value of \(\sqrt{96}\div\sqrt{6}\)?
B. (4)
\(\sqrt{96}\div\sqrt{6}=\sqrt{16}=4\) है। मूलों के भाग में अंदर का भागफल लें। / \(\sqrt{96}\div\sqrt{6}=\sqrt{16}=4\). In division of roots take the quotient inside.
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If the area of a square is (72) square units, what will be the simplified form of its side?
C. \(6\sqrt{2}\)
भुजा \(\sqrt{72}=6\sqrt{2}\) होगी। वर्ग में भुजा क्षेत्रफल के वर्गमूल के बराबर होती है। / The side will be \(\sqrt{72}=6\sqrt{2}\). In a square, side equals the square root of area.
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What is the rationalised form of \(\frac{1}{6-\sqrt{5}}\)?
A. \(\frac{6+\sqrt{5}}{31}\)
संयुग्मी से गुणा करने पर हर (36-5=31) बनता है। इसलिए उत्तर \(\frac{6+\sqrt{5}}{31}\) है। / Multiplying by the conjugate makes the denominator (36-5=31). So the answer is \(\frac{6+\sqrt{5}}{31}\).
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Which statement is correct about \(\sqrt{8}+\sqrt{200}\) and \(12\sqrt{2}\)?
C. दोनों बराबर हैंBoth are equal
\(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{200}=10\sqrt{2}\), इसलिए योग \(12\sqrt{2}\) है। तुलना से पहले सरल करें। / \(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{200}=10\sqrt{2}\), so the sum is \(12\sqrt{2}\). Simplify before comparing.
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What is the simplified form of \(4\sqrt{11}+3\sqrt{44}\)?
A. \(10\sqrt{11}\)
\(\sqrt{44}=2\sqrt{11}\), इसलिए \(4\sqrt{11}+6\sqrt{11}=10\sqrt{11}\) है। गुणांक और मूल दोनों ध्यान से देखें। / \(\sqrt{44}=2\sqrt{11}\), so \(4\sqrt{11}+6\sqrt{11}=10\sqrt{11}\). Watch both coefficients and radicals carefully.
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If \(a=\sqrt{5}+3\) and \(b=\sqrt{5}-3\), what is (ab)?
A. (-4)
(ab=\(\sqrt{5}\)2-32=5-9=-4) है। संयुग्मी गुणनफल परिमेय हो सकता है। / (ab=\(\sqrt{5}\)2-32=5-9=-4). A conjugate product can be rational.
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Which is the simplified form of \(\sqrt{13}+\sqrt{208}\)?
B. \(5\sqrt{13}\)
\(\sqrt{208}=4\sqrt{13}\), इसलिए योग \(5\sqrt{13}\) है। समान मूल जोड़ने से पहले सरल करें। / \(\sqrt{208}=4\sqrt{13}\), so the sum is \(5\sqrt{13}\). Simplify before adding like radicals.
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What is the value of (\(4\sqrt{3}\)2)?
C. (48)
(\(4\sqrt{3}\)2=16\times3=48) है। गुणांक और मूल दोनों का वर्ग करें। / (\(4\sqrt{3}\)2=16\times3=48). Square both the coefficient and the radical.
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