Level 14 • 50/50 questions • 35 seconds per question.
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If \(x=\sqrt{7}+3\), what type of number is (x-3)?
B. अपरिमेयIrrational
\(x-3=\sqrt{7}\) है जो अपरिमेय है। पहले परिमेय पदों को अलग करें। / \(x-3=\sqrt{7}\), which is irrational. First separate the rational terms.
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Which is the simplified form of \(\sqrt{32}+\sqrt{18}\)?
B. \(7\sqrt{2}\)
\(\sqrt{32}=4\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए योग \(7\sqrt{2}\) है। पहले मूलों को समान रूप में बदलें। / \(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the sum is \(7\sqrt{2}\). First convert radicals into like form.
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What is the value of (\(\sqrt{11}\)2-4)?
A. (7)
(\(\sqrt{11}\)2=11) इसलिए मान (7) है। वर्ग करने पर वर्गमूल हट सकता है। / (\(\sqrt{11}\)2=11), so the value is (7). Squaring can remove the square root.
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What is obtained after simplifying \(\frac{1}{\sqrt{5}-2}\)?
C. \(\sqrt{5}+2\)
संयुग्मी से गुणा करने पर हर (5-4=1) बनता है। इसलिए उत्तर \(\sqrt{5}+2\) है। / Multiplying by the conjugate makes the denominator (5-4=1). So the answer is \(\sqrt{5}+2\).
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If \(a=2\sqrt{3}+1\) and \(b=2\sqrt{3}-1\), what is (a+b)?
B. \(4\sqrt{3}\)
स्थिर पद कट जाते हैं और \(2\sqrt{3}+2\sqrt{3}=4\sqrt{3}\) मिलता है। समान मूलों को जोड़ें। / The constant terms cancel and \(2\sqrt{3}+2\sqrt{3}=4\sqrt{3}\). Add like radicals.
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What is the value of (\(4+\sqrt{7}\)\(4-\sqrt{7}\))?
B. (9)
यह \(a^2-b^2\) है इसलिए मान (16-7=9) है। संयुग्मी गुणन से मूल हट जाता है। / This is \(a^2-b^2\), so the value is (16-7=9). Conjugate multiplication removes the radical.
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Which is the simplified form of \(\sqrt{75}-\sqrt{12}\)?
A. \(3\sqrt{3}\)
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\), इसलिए अंतर \(3\sqrt{3}\) है। पहले पूर्ण वर्ग गुणनखंड निकालें। / \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(3\sqrt{3}\). First take out perfect-square factors.
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What type of number is the decimal \(2.303003000300003\ldots\)?
B. अपरिमेयIrrational
इसमें निश्चित दोहराने वाला खंड नहीं है इसलिए यह अपरिमेय है। असांत अनावर्ती दशमलव अपरिमेय होता है। / It has no fixed repeating block, so it is irrational. A non-terminating non-repeating decimal is irrational.
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What is the value of \(\frac{\sqrt{75}}{\sqrt{3}}\)?
A. (5)
\(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\) है। भाग में मूलों को एक साथ लिखकर सरल करें। / \(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\). Combine radicals in division and simplify.
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Which is the simplified form of \(\sqrt{80}+\sqrt{125}\)?
A. \(9\sqrt{5}\)
\(\sqrt{80}=4\sqrt{5}\) और \(\sqrt{125}=5\sqrt{5}\), इसलिए योग \(9\sqrt{5}\) है। समान मूलों को जोड़ना चाहिए। / \(\sqrt{80}=4\sqrt{5}\) and \(\sqrt{125}=5\sqrt{5}\), so the sum is \(9\sqrt{5}\). Like radicals should be added.
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Which number lies between \(\sqrt{15}\) and \(\sqrt{20}\)?
C. \(\sqrt{17}\)
क्योंकि (15<17<20), इसलिए \(\sqrt{17}\) इनके बीच है। वर्गमूल की तुलना अंदर की संख्याओं से करें। / Since (15<17<20), \(\sqrt{17}\) lies between them. Compare square roots using the numbers inside.
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If \(y=\sqrt{13}\), what is the value of \(y^2+6\)?
A. (19)
\(y^2=13\) इसलिए \(y^2+6=19\) है। पहले दिए गए मूल का वर्ग करें। / \(y^2=13\), so \(y^2+6=19\). First square the given radical.
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Which option is an example of an irrational number?
C. \(0.4141141114\ldots\)
तीसरा दशमलव असांत और अनावर्ती है इसलिए अपरिमेय है। आवर्ती और सांत दशमलव परिमेय होते हैं। / The third decimal is non-terminating and non-repeating, so it is irrational. Repeating and terminating decimals are rational.
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Which is the simplified form of \(\sqrt{147}\)?
A. \(7\sqrt{3}\)
\(\sqrt{147}=\sqrt{49\times3}=7\sqrt{3}\) है। सबसे बड़ा पूर्ण वर्ग गुणनखंड चुनें। / \(\sqrt{147}=\sqrt{49\times3}=7\sqrt{3}\). Choose the largest perfect-square factor.
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What is the rationalised form of \(\frac{3}{\sqrt{7}}\)?
A. \(\frac{3\sqrt{7}}{7}\)
अंश और हर को \(\sqrt{7}\) से गुणा करने पर \(\frac{3\sqrt{7}}{7}\) मिलता है। परिमेयकरण में मान समान रहता है। / Multiplying numerator and denominator by \(\sqrt{7}\) gives \(\frac{3\sqrt{7}}{7}\). Rationalisation keeps the value same.
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What is the value of (\(\sqrt{8}+\sqrt{2}\)2)?
B. (18)
\(\sqrt{8}=2\sqrt{2}\), इसलिए (\(3\sqrt{2}\)2=18) है। पहले कोष्ठक को सरल करें। / \(\sqrt{8}=2\sqrt{2}\), so (\(3\sqrt{2}\)2=18). Simplify the bracket first.
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What is the irrational part in \(6-\sqrt{10}\)?
B. \(-\sqrt{10}\)
(6) परिमेय है और \(-\sqrt{10}\) अपरिमेय भाग है। मिश्रित रूप में मूल वाला पद पहचानें। / (6) is rational and \(-\sqrt{10}\) is the irrational part. Identify the radical term in mixed form.
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What is the simplified form of \(\sqrt{45}+\sqrt{125}-\sqrt{20}\)?
B. \(6\sqrt{5}\)
\(\sqrt{45}=3\sqrt{5}\), \(\sqrt{125}=5\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\), इसलिए उत्तर \(6\sqrt{5}\) है। समान मूलों के गुणांक जोड़ें और घटाएं। / \(\sqrt{45}=3\sqrt{5}\), \(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(6\sqrt{5}\). Add and subtract coefficients of like radicals.
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If \(\sqrt{m}=9\), what is the value of (m)?
C. (81)
दोनों पक्षों का वर्ग करने पर (m=81) मिलता है। पूर्ण वर्गों को याद रखना उपयोगी है। / Squaring both sides gives (m=81). Remembering perfect squares is useful.
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If (n) is a positive integer and (n) is a perfect square, what type of number will \(\sqrt{n}\) be?
B. परिमेयRational
पूर्ण वर्ग का वर्गमूल पूर्णांक होता है इसलिए परिमेय होता है। पहले यह देखें कि संख्या पूर्ण वर्ग है या नहीं। / The square root of a perfect square is an integer, so it is rational. First check whether the number is a perfect square.
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What is the value of \(\sqrt{18}\times\sqrt{32}\)?
A. (24)
\(\sqrt{18}\times\sqrt{32}=\sqrt{576}=24\) है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है। / \(\sqrt{18}\times\sqrt{32}=\sqrt{576}=24\). The product of two irrationals can be rational.
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Which statement is correct about \(\sqrt{5}+\sqrt{7}\)?
C. यह अपरिमेय हैIt is irrational
\(\sqrt{5}\) और \(\sqrt{7}\) अलग अपरिमेय मूल हैं और उनका योग अपरिमेय है। अलग मूलों को सीधे जोड़कर \(\sqrt{12}\) नहीं बनाते। / \(\sqrt{5}\) and \(\sqrt{7}\) are different irrational radicals and their sum is irrational. Different radicals are not added directly as \(\sqrt{12}\).
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What is the simplified value of \(\frac{\sqrt{112}}{\sqrt{7}}\)?
B. (4)
\(\frac{\sqrt{112}}{\sqrt{7}}=\sqrt{16}=4\) है। मूलों के भाग में अंदर का भागफल लें। / \(\frac{\sqrt{112}}{\sqrt{7}}=\sqrt{16}=4\). In division of roots take the quotient inside.
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What is the result of \(8\sqrt{2}-5\sqrt{2}\)?
B. \(3\sqrt{2}\)
समान मूलों के गुणांक घटाने पर \(3\sqrt{2}\) मिलता है। यह अपरिमेय है क्योंकि \(\sqrt{2}\) बचता है। / Subtracting coefficients of like radicals gives \(3\sqrt{2}\). It is irrational because \(\sqrt{2}\) remains.
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What is the value of (\sqrt{2}\(\sqrt{8}+\sqrt{18}\))?
A. (10)
\(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए कोष्ठक \(5\sqrt{2}\) और गुणनफल (10) है। पहले कोष्ठक सरल करें। / \(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the bracket is \(5\sqrt{2}\) and the product is (10). Simplify the bracket first.
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Which statement is correct when comparing \(\sqrt{29}\) and \(\sqrt{31}\)?
C. \(\sqrt{29}<\sqrt{31}\)
क्योंकि (29<31), इसलिए \(\sqrt{29}<\sqrt{31}\)। धनात्मक मूलों में अंदर की संख्याओं की तुलना करें। / Since (29<31), \(\sqrt{29}<\sqrt{31}\). For positive roots compare the numbers inside.
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What is the simplified form of (\sqrt{3}\times\(2\sqrt{3}+5\))?
A. \(6+5\sqrt{3}\)
वितरण करने पर \(2\times3+5\sqrt{3}=6+5\sqrt{3}\) मिलता है। मूल वाले पद और परिमेय पद अलग रखें। / Distributing gives \(2\times3+5\sqrt{3}=6+5\sqrt{3}\). Keep radical and rational terms separate.
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Which is the rationalised form of \(\frac{2}{\sqrt{3}+1}\)?
B. \(\sqrt{3}-1\)
संयुग्मी से गुणा करने पर (\frac{2\(\sqrt{3}-1\)}{3-1}=\sqrt{3}-1) मिलता है। हर को परिमेय बनाएं। / Multiplying by the conjugate gives (\frac{2\(\sqrt{3}-1\)}{3-1}=\sqrt{3}-1). Make the denominator rational.
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What is the sum of \(8+\sqrt{17}\) and \(8-\sqrt{17}\)?
A. (16)
अपरिमेय पद कट जाते हैं और योग (16) है। संयुग्मी संख्याओं का योग परिमेय होता है। / The irrational terms cancel and the sum is (16). The sum of conjugate numbers is rational.
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What is the product of \(8+\sqrt{17}\) and \(8-\sqrt{17}\)?
B. (47)
गुणनफल (82-\(\sqrt{17}\)2=64-17=47) है। संयुग्मी गुणन में \(a^2-b^2\) लगाएं। / The product is (82-\(\sqrt{17}\)2=64-17=47). Use \(a^2-b^2\) in conjugate multiplication.
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If \(q=5-\sqrt{2}\), what type of number is (q)?
C. अपरिमेयIrrational
परिमेय (5) में से अपरिमेय \(\sqrt{2}\) घटाने पर अपरिमेय संख्या मिलती है। अपरिमेय भाग बचा रहता है। / Subtracting irrational \(\sqrt{2}\) from rational (5) gives an irrational number. The irrational part remains.
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To construct \(\sqrt{5}\) on the number line, which right triangle hypotenuse can be used?
B. भुजाएँ (1) और (2)Legs (1) and (2)
समकोण त्रिभुज में कर्ण \(\sqrt{1^2+2^2}=\sqrt{5}\) होगा। निर्माण में पाइथागोरस प्रमेय लगती है। / In a right triangle, the hypotenuse is \(\sqrt{1^2+2^2}=\sqrt{5}\). Pythagoras theorem is used in construction.
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What is the value of \(\sqrt{54}\div\sqrt{6}\)?
B. (3)
\(\sqrt{54}\div\sqrt{6}=\sqrt{9}=3\) है। मूलों के भाग में अंदर का भागफल लें। / \(\sqrt{54}\div\sqrt{6}=\sqrt{9}=3\). In division of roots take the quotient inside.
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If the area of a square is (50) square units, what will be the simplified form of its side?
C. \(5\sqrt{2}\)
भुजा \(\sqrt{50}=5\sqrt{2}\) होगी। वर्ग में भुजा क्षेत्रफल के वर्गमूल के बराबर होती है। / The side will be \(\sqrt{50}=5\sqrt{2}\). In a square, side equals the square root of area.
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What is the rationalised form of \(\frac{1}{4-\sqrt{7}}\)?
A. \(\frac{4+\sqrt{7}}{9}\)
संयुग्मी से गुणा करने पर हर (16-7=9) बनता है। इसलिए उत्तर \(\frac{4+\sqrt{7}}{9}\) है। / Multiplying by the conjugate makes the denominator (16-7=9). So the answer is \(\frac{4+\sqrt{7}}{9}\).
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Which statement is correct about \(\sqrt{2}+\sqrt{50}\) and \(6\sqrt{2}\)?
C. दोनों बराबर हैंBoth are equal
\(\sqrt{50}=5\sqrt{2}\), इसलिए \(\sqrt{2}+\sqrt{50}=6\sqrt{2}\) है। तुलना से पहले सरल करें। / \(\sqrt{50}=5\sqrt{2}\), so \(\sqrt{2}+\sqrt{50}=6\sqrt{2}\). Simplify before comparing.
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What is the simplified form of \(3\sqrt{7}+2\sqrt{28}\)?
B. \(7\sqrt{7}\)
\(\sqrt{28}=2\sqrt{7}\), इसलिए \(3\sqrt{7}+4\sqrt{7}=7\sqrt{7}\) है। गुणांक और मूल दोनों ध्यान से देखें। / \(\sqrt{28}=2\sqrt{7}\), so \(3\sqrt{7}+4\sqrt{7}=7\sqrt{7}\). Watch both coefficients and radicals carefully.
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Which option will have a non-terminating non-repeating decimal expansion?
C. \(\sqrt{37}\)
\(\sqrt{37}\) अपरिमेय है क्योंकि (37) पूर्ण वर्ग नहीं है। अपरिमेय का दशमलव असांत अनावर्ती होता है। / \(\sqrt{37}\) is irrational because (37) is not a perfect square. An irrational number has a non-terminating non-repeating decimal.
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If \(a=\sqrt{3}+2\) and \(b=\sqrt{3}-2\), what is (ab)?
A. (-1)
(ab=\(\sqrt{3}\)2-22=3-4=-1) है। संयुग्मी गुणनफल परिमेय हो सकता है। / (ab=\(\sqrt{3}\)2-22=3-4=-1). A conjugate product can be rational.
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Which is the simplified form of \(\sqrt{7}+\sqrt{112}\)?
B. \(5\sqrt{7}\)
\(\sqrt{112}=4\sqrt{7}\), इसलिए योग \(5\sqrt{7}\) है। समान मूल जोड़ने से पहले सरल करें। / \(\sqrt{112}=4\sqrt{7}\), so the sum is \(5\sqrt{7}\). Simplify before adding like radicals.
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What is the value of (\(3\sqrt{2}\)2)?
C. (18)
(\(3\sqrt{2}\)2=9\times2=18) है। गुणांक और मूल दोनों का वर्ग करें। / (\(3\sqrt{2}\)2=9\times2=18). Square both the coefficient and the radical.
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If \(u=\sqrt{23}+4\) and \(v=\sqrt{23}-4\), what is the value of (uv)?
A. (7)
संयुग्मी गुणन में (uv=\(\sqrt{23}\)2-42=7) होता है। ऐसे प्रश्नों में \(a^2-b^2\) लगाएं। / In conjugate multiplication (uv=\(\sqrt{23}\)2-42=7). Use \(a^2-b^2\) in such questions.
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What is the simplified value of \(\frac{\sqrt{150}+\sqrt{54}}{\sqrt{6}}\)?
B. (8)
\(\sqrt{150}=5\sqrt{6}\) और \(\sqrt{54}=3\sqrt{6}\), इसलिए भाग देने पर (8) मिलता है। पहले अंश को समान मूल में बदलें। / \(\sqrt{150}=5\sqrt{6}\) and \(\sqrt{54}=3\sqrt{6}\), so division gives (8). First convert the numerator into like radicals.
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If \(\sqrt{k}\) lies between (7) and (8), which value of (k) is possible?
C. (57)
क्योंकि (49<57<64), इसलिए \(7<\sqrt{57}<8\)। वर्गमूल की सीमा वर्गों से तय करें। / Since (49<57<64), \(7<\sqrt{57}<8\). Decide square-root bounds using squares.
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What is the simplified form of (\sqrt{10}\(3\sqrt{10}-2\sqrt{40}\))?
B. (-10)
\(\sqrt{40}=2\sqrt{10}\), इसलिए कोष्ठक \(-\sqrt{10}\) और गुणनफल (-10) है। पहले कोष्ठक को सरल करें। / \(\sqrt{40}=2\sqrt{10}\), so the bracket is \(-\sqrt{10}\) and the product is (-10). Simplify the bracket first.
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Which option is the correct rationalised form of \(\frac{6}{\sqrt{19}}\)?
C. \(\frac{6\sqrt{19}}{19}\)
अंश और हर को \(\sqrt{19}\) से गुणा करने पर \(\frac{6\sqrt{19}}{19}\) मिलता है। परिमेयकरण में हर से मूल हटता है। / Multiplying numerator and denominator by \(\sqrt{19}\) gives \(\frac{6\sqrt{19}}{19}\). Rationalisation removes the radical from the denominator.
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Which statement is correct about \(\sqrt{12}+\sqrt{108}\) and \(8\sqrt{3}\)?
C. दोनों बराबर हैंBoth are equal
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{108}=6\sqrt{3}\), इसलिए योग \(8\sqrt{3}\) है। तुलना से पहले सरल रूप निकालें। / \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{108}=6\sqrt{3}\), so the sum is \(8\sqrt{3}\). Find simplified forms before comparing.
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If the side of a square is \(2\sqrt{7}\) units, what will be its area?
C. (28) वर्ग इकाई(28) square units
क्षेत्रफल (\(2\sqrt{7}\)2=28) वर्ग इकाई है। अपरिमेय भुजा का क्षेत्रफल परिमेय हो सकता है। / The area is (\(2\sqrt{7}\)2=28) square units. A square with irrational side can have rational area.
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What type of number is the decimal \(5.12012001200012\ldots\)?
C. अपरिमेयIrrational
इसमें निश्चित दोहराने वाला खंड नहीं है इसलिए यह अपरिमेय है। असांत अनावर्ती दशमलव को पहचानना जरूरी है। / It has no fixed repeating block, so it is irrational. Identifying non-terminating non-repeating decimals is important.
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What is the rationalised form of \(\frac{1}{5+\sqrt{6}}\)?
A. \(\frac{5-\sqrt{6}}{19}\)
संयुग्मी से गुणा करने पर हर (25-6=19) बनता है। इसलिए परिमेयकृत रूप \(\frac{5-\sqrt{6}}{19}\) है। / Multiplying by the conjugate makes the denominator (25-6=19). So the rationalised form is \(\frac{5-\sqrt{6}}{19}\).
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