Level 13 • 50/50 questions • 35 seconds per question.
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If \(x=2+\sqrt{3}\), what type of number is (x-2)?
B. अपरिमेयIrrational
\(x-2=\sqrt{3}\) है जो अपरिमेय है। प्रश्न में पहले समान परिमेय पद हटाएं। / \(x-2=\sqrt{3}\), which is irrational. First cancel matching rational terms in such questions.
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Which is the simplified form of \(\sqrt{18}+\sqrt{8}\)?
C. \(5\sqrt{2}\)
\(\sqrt{18}=3\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\), इसलिए योग \(5\sqrt{2}\) है। पहले मूलों को सरल करें। / \(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the sum is \(5\sqrt{2}\). Simplify radicals first.
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What type of number is the value of (\(\sqrt{7}\)2+3)?
A. परिमेयRational
(\(\sqrt{7}\)2+3=10) है जो परिमेय है। अपरिमेय का वर्ग परिमेय हो सकता है। / (\(\sqrt{7}\)2+3=10), which is rational. The square of an irrational can be rational.
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What is obtained after simplifying \(\frac{1}{2+\sqrt{3}}\)?
D. \(2-\sqrt{3}\)
हर का परिमेयकरण करने पर \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\) मिलता है। संयुग्मी से गुणा करना याद रखें। / Rationalising the denominator gives \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\). Remember to multiply by the conjugate.
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If \(a=\sqrt{5}+1\) and \(b=\sqrt{5}-1\), what is (a+b)?
B. \(2\sqrt{5}\)
\(a+b=2\sqrt{5}\) है जो अपरिमेय है। समान मूल वाले पदों को जोड़ें। / \(a+b=2\sqrt{5}\), which is irrational. Add like radical terms.
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Which is the simplified form of \(\sqrt{50}-\sqrt{2}\)?
C. \(4\sqrt{2}\)
\(\sqrt{50}=5\sqrt{2}\), इसलिए अंतर \(4\sqrt{2}\) है। समान मूलों को घटाना आसान तरीका है। / \(\sqrt{50}=5\sqrt{2}\), so the difference is \(4\sqrt{2}\). Subtracting like radicals is the easy method.
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What is the simplified form of \(\sqrt{3}+\sqrt{12}+\sqrt{27}\)?
D. \(6\sqrt{3}\)
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए कुल \(6\sqrt{3}\) है। सभी पदों को पहले समान मूल में बदलें। / \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the total is \(6\sqrt{3}\). First convert all terms into like radicals.
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What type of number does decimal \(1.02002000200002\ldots\) represent?
B. अपरिमेयIrrational
इसमें निश्चित दोहराने वाला खंड नहीं है इसलिए यह अपरिमेय है। अनावर्ती दशमलव को आवर्ती समझने से बचें। / It has no fixed repeating block, so it is irrational. Avoid treating a non-repeating decimal as repeating.
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What is the value of \(\frac{\sqrt{27}}{\sqrt{3}}\)?
A. (3)
\(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{9}=3\) है। भाग में अंदर की संख्याओं का अनुपात लें। / \(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{9}=3\). In division take the ratio of the numbers inside roots.
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Which is the simplified form of \(\sqrt{45}+\sqrt{20}\)?
D. \(5\sqrt{5}\)
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\), इसलिए योग \(5\sqrt{5}\) है। पहले पूर्ण वर्ग गुणनखंड निकालें। / \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the sum is \(5\sqrt{5}\). First take out perfect-square factors.
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Which number lies between \(\sqrt{2}\) and \(\sqrt{8}\)?
C. \(\sqrt{5}\)
क्योंकि (2<5<8), इसलिए \(\sqrt{5}\) इनके बीच है। वर्गमूल की तुलना अंदर की संख्याओं से करें। / Since (2<5<8), \(\sqrt{5}\) lies between them. Compare square roots using the numbers inside.
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If \(x=\sqrt{11}\), what is the value of \(x^2-1\)?
B. (10)
\(x^2=11\) इसलिए \(x^2-1=10\) है। पहले दिए गए मूल का वर्ग निकालें। / \(x^2=11\), so \(x^2-1=10\). First square the given radical.
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Which option is an irrational number?
A. \(0.101001000100001\ldots\)
पहला दशमलव असांत और अनावर्ती है इसलिए अपरिमेय है। आवर्ती दशमलव परिमेय होते हैं। / The first decimal is non-terminating and non-repeating, so it is irrational. Repeating decimals are rational.
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What is the rationalised form of \(\frac{2}{\sqrt{5}}\)?
A. \(\frac{2\sqrt{5}}{5}\)
हर को परिमेय बनाने के लिए \(\sqrt{5}\) से गुणा करें। परिणाम \(\frac{2\sqrt{5}}{5}\) है। / Multiply by \(\sqrt{5}\) to make the denominator rational. The result is \(\frac{2\sqrt{5}}{5}\).
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Which is the expansion of (\(\sqrt{10}+\sqrt{2}\)2)?
B. \(12+4\sqrt{5}\)
विस्तार \(10+2+2\sqrt{20}=12+4\sqrt{5}\) है। मध्य पद को सरल करना न भूलें। / The expansion is \(10+2+2\sqrt{20}=12+4\sqrt{5}\). Do not forget to simplify the middle term.
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Which option has a rational value?
C. (\(\sqrt{13}\)2)
(\(\sqrt{13}\)2=13) है जो परिमेय है। वर्ग करने से मूल हट सकता है। / (\(\sqrt{13}\)2=13), which is rational. Squaring can remove the radical.
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What is the irrational part of \(4+\sqrt{6}\)?
D. \(\sqrt{6}\)
(4) परिमेय है और \(\sqrt{6}\) अपरिमेय है। मिश्रित रूप में मूल वाला पद अपरिमेय भाग होता है। / (4) is rational and \(\sqrt{6}\) is irrational. In a mixed form the radical term is the irrational part.
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What is the simplified form of \(\sqrt{28}+\sqrt{63}-\sqrt{7}\)?
C. \(4\sqrt{7}\)
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए \(2\sqrt{7}+3\sqrt{7}-\sqrt{7}=4\sqrt{7}\) है। समान मूलों को जोड़ें और घटाएं। / \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so \(2\sqrt{7}+3\sqrt{7}-\sqrt{7}=4\sqrt{7}\). Add and subtract like radicals.
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If \(\sqrt{m}=6\), what is the value of (m)?
A. (36)
दोनों पक्षों का वर्ग करने पर (m=36) मिलता है। पूर्ण वर्ग पहचानना आवश्यक है। / Squaring both sides gives (m=36). Identifying perfect squares is essential.
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If (n) is not a perfect square and (n>0), what type of number will \(\sqrt{n}\) generally be?
D. अपरिमेयIrrational
धनात्मक पूर्ण संख्या का वर्गमूल अपरिमेय होता है जब वह पूर्ण वर्ग नहीं होती। यही वर्गमूल पहचान का मुख्य नियम है। / The square root of a positive integer is irrational when it is not a perfect square. This is the main rule for identifying square roots.
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What is the value of \(\sqrt{12}\times\sqrt{27}\)?
B. (18)
\(\sqrt{12}\times\sqrt{27}=\sqrt{324}=18\) है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है। / \(\sqrt{12}\times\sqrt{27}=\sqrt{324}=18\). The product of two irrationals can be rational.
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Which statement is correct about \(\sqrt{2}+\sqrt{3}\)?
C. यह अपरिमेय हैIt is irrational
\(\sqrt{2}\) और \(\sqrt{3}\) अलग अपरिमेय मूल हैं और उनका योग अपरिमेय है। इसे \(\sqrt{5}\) नहीं लिखते। / \(\sqrt{2}\) and \(\sqrt{3}\) are different irrational radicals and their sum is irrational. It is not written as \(\sqrt{5}\).
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Which is the simplified form of \(\sqrt{200}\)?
A. \(10\sqrt{2}\)
\(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}\) है। बड़े पूर्ण वर्ग गुणनखंड से हल छोटा होता है। / \(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}\). Using a large perfect-square factor makes the solution shorter.
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What is the simplified value of \(\frac{\sqrt{80}}{\sqrt{5}}\)?
B. (4)
\(\frac{\sqrt{80}}{\sqrt{5}}=\sqrt{16}=4\) है। भाग में मूलों को मिलाकर सरल करें। / \(\frac{\sqrt{80}}{\sqrt{5}}=\sqrt{16}=4\). Combine roots in division and simplify.
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Which option is a non-terminating and repeating decimal?
A. \(0.123123123\ldots\)
(123) बार-बार दोहरता है इसलिए यह आवर्ती और परिमेय है। आवर्ती दशमलव अपरिमेय नहीं होता। / (123) repeats again and again, so it is repeating and rational. A repeating decimal is not irrational.
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What is the result of \(5\sqrt{3}-2\sqrt{3}\)?
C. \(3\sqrt{3}\)
समान मूलों के गुणांक घटाने पर \(3\sqrt{3}\) मिलता है। यह अपरिमेय है। / Subtracting coefficients of like radicals gives \(3\sqrt{3}\). It is irrational.
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What is the value of (\sqrt{3}\(\sqrt{3}+\sqrt{12}\))?
D. (9)
\(\sqrt{12}=2\sqrt{3}\), इसलिए अंदर \(3\sqrt{3}\) है और गुणनफल (9) है। पहले कोष्ठक को सरल करें। / \(\sqrt{12}=2\sqrt{3}\), so the bracket is \(3\sqrt{3}\) and the product is (9). Simplify the bracket first.
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Which statement is correct when comparing \(\sqrt{17}\) and \(\sqrt{18}\)?
A. \(\sqrt{17}<\sqrt{18}\)
क्योंकि (17<18), इसलिए \(\sqrt{17}<\sqrt{18}\)। धनात्मक मूलों की तुलना अंदर की संख्या से करें। / Since (17<18), \(\sqrt{17}<\sqrt{18}\). Compare positive roots using the numbers inside.
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What is the simplified form of (\sqrt{2}\times\(3\sqrt{2}+1\))?
B. \(6+\sqrt{2}\)
वितरण करने पर \(3\times2+\sqrt{2}=6+\sqrt{2}\) मिलता है। मूल का गुणन ध्यान से करें। / Distributing gives \(3\times2+\sqrt{2}=6+\sqrt{2}\). Multiply radical terms carefully.
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Which is the rationalised form of \(\frac{3}{\sqrt{2}+1}\)?
C. \(3\sqrt{2}-3\)
हर के संयुग्मी \(\sqrt{2}-1\) से गुणा करने पर (3\(\sqrt{2}-1\)) मिलता है। इसलिए रूप \(3\sqrt{2}-3\) है। / Multiplying by the conjugate \(\sqrt{2}-1\) gives (3\(\sqrt{2}-1\)). So the form is \(3\sqrt{2}-3\).
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What is the sum of \(7+\sqrt{13}\) and \(7-\sqrt{13}\)?
A. (14)
अपरिमेय पद कट जाते हैं और योग (14) है। संयुग्मी संख्याओं का योग परिमेय हो सकता है। / The irrational terms cancel and the sum is (14). The sum of conjugate numbers can be rational.
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What is the product of \(7+\sqrt{13}\) and \(7-\sqrt{13}\)?
B. (36)
गुणनफल (72-\(\sqrt{13}\)2=49-13=36) है। संयुग्मी गुणन में \(a^2-b^2\) लगाएं। / The product is (72-\(\sqrt{13}\)2=49-13=36). Use \(a^2-b^2\) for conjugate multiplication.
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If \(p=4+\sqrt{5}\), what type of number is (p)?
D. अपरिमेयIrrational
परिमेय (4) में अपरिमेय \(\sqrt{5}\) जोड़ने से अपरिमेय संख्या मिलती है। अपरिमेय भाग बचा रहता है। / Adding irrational \(\sqrt{5}\) to rational (4) gives an irrational number. The irrational part remains.
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To construct \(\sqrt{2}\) on the number line, what hypotenuse length is used?
B. (1) और (1) भुजाओं वाला कर्णHypotenuse with legs (1) and (1)
समकोण त्रिभुज में (1) और (1) भुजाओं का कर्ण \(\sqrt{2}\) होता है। पाइथागोरस प्रमेय का उपयोग करें। / In a right triangle, legs (1) and (1) give hypotenuse \(\sqrt{2}\). Use Pythagoras theorem.
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What type of number is \(\sqrt{2}\times\sqrt{8}\)?
B. परिमेयRational
\(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\) है जो परिमेय है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है। / \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. The product of two irrationals can be rational.
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What is the value of \(\sqrt{20}\div\sqrt{2}\)?
B. \(\sqrt{10}\)
\(\sqrt{20}\div\sqrt{2}=\sqrt{10}\) है जो अपरिमेय है। भाग में मूलों को एक मूल में लिखें। / \(\sqrt{20}\div\sqrt{2}=\sqrt{10}\), which is irrational. Write the roots in division as one root.
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If the area of a square is (18) square units, what will be the simplified form of its side?
C. \(3\sqrt{2}\)
भुजा \(\sqrt{18}=3\sqrt{2}\) होगी। वर्ग में भुजा क्षेत्रफल के वर्गमूल के बराबर होती है। / The side will be \(\sqrt{18}=3\sqrt{2}\). In a square, side equals the square root of area.
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What is the rationalised form of \(\frac{1}{3-\sqrt{2}}\)?
A. \(\frac{3+\sqrt{2}}{7}\)
संयुग्मी से गुणा करने पर हर (9-2=7) बनता है। इसलिए रूप \(\frac{3+\sqrt{2}}{7}\) है। / Multiplying by the conjugate makes the denominator (9-2=7). So the form is \(\frac{3+\sqrt{2}}{7}\).
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Which statement is correct about \(\sqrt{3}+\sqrt{27}\) and \(\sqrt{48}\)?
C. दोनों बराबर हैंBoth are equal
\(\sqrt{3}+\sqrt{27}=4\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\) हैं। सरल रूप तुलना को आसान बनाता है। / \(\sqrt{3}+\sqrt{27}=4\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\). Simplified form makes comparison easy.
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What is the simplified form of \(2\sqrt{5}+3\sqrt{20}\)?
B. \(8\sqrt{5}\)
\(\sqrt{20}=2\sqrt{5}\), इसलिए \(2\sqrt{5}+6\sqrt{5}=8\sqrt{5}\) है। गुणांक और मूल दोनों ध्यान से देखें। / \(\sqrt{20}=2\sqrt{5}\), so \(2\sqrt{5}+6\sqrt{5}=8\sqrt{5}\). Watch both coefficients and radicals carefully.
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Which option will have a terminating decimal expansion?
D. \(\frac{7}{16}\)
\(\frac{7}{16}\) परिमेय है और हर \(2^4\) है, इसलिए दशमलव सांत होगा। हर के अभाज्य गुणनखंड देखें। / \(\frac{7}{16}\) is rational and the denominator is \(2^4\), so the decimal will terminate. Check prime factors of the denominator.
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If \(a=\sqrt{2}+1\) and \(b=\sqrt{2}-1\), what is (ab)?
A. (1)
(ab=\(\sqrt{2}\)2-12=1) है। संयुग्मी रूप का गुणनफल परिमेय देता है। / (ab=\(\sqrt{2}\)2-12=1). The product of conjugate forms gives a rational result.
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Which is the simplified form of \(\sqrt{2}+\sqrt{32}\)?
B. \(5\sqrt{2}\)
\(\sqrt{32}=4\sqrt{2}\), इसलिए योग \(5\sqrt{2}\) है। समान मूलों को जोड़ने से पहले सरल करें। / \(\sqrt{32}=4\sqrt{2}\), so the sum is \(5\sqrt{2}\). Simplify before adding like radicals.
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What is the value of (\(2\sqrt{3}\)2)?
C. (12)
(\(2\sqrt{3}\)2=4\times3=12) है। गुणांक और मूल दोनों का वर्ग करें। / (\(2\sqrt{3}\)2=4\times3=12). Square both the coefficient and the radical.
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Which option shows correct rationalisation of the denominator?
A. \(\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}\)
हर और अंश को \(\sqrt{7}\) से गुणा करने पर \(\frac{\sqrt{7}}{7}\) मिलता है। परिमेयकरण में मान नहीं बदलना चाहिए। / Multiplying numerator and denominator by \(\sqrt{7}\) gives \(\frac{\sqrt{7}}{7}\). Rationalisation should not change the value.
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If \(r=\sqrt{15}-\sqrt{6}\) and \(s=\sqrt{15}+\sqrt{6}\), what is the value of (rs)?
B. (9)
संयुग्मी गुणन में (rs=\(\sqrt{15}\)2-\(\sqrt{6}\)2=9) होता है। ऐसे प्रश्नों में \(a^2-b^2\) लगाएं। / In conjugate multiplication (rs=\(\sqrt{15}\)2-\(\sqrt{6}\)2=9). Use \(a^2-b^2\) in such questions.
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What is the simplified value of \(\frac{\sqrt{98}+\sqrt{50}}{\sqrt{2}}\)?
A. (12)
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए भाग देने पर (12) मिलता है। पहले अंश को सरल करें। / \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so division gives (12). Simplify the numerator first.
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If \(\sqrt{k}\) lies between (6) and (7), which value of (k) is possible?
C. (43)
क्योंकि (36<43<49), इसलिए \(6<\sqrt{43}<7\)। वर्गमूल की सीमा वर्गों से तय करें। / Since (36<43<49), \(6<\sqrt{43}<7\). Decide bounds of square roots using squares.
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What is the simplified form of (\sqrt{5}\(2\sqrt{5}-3\sqrt{20}\))?
A. (-20)
\(\sqrt{20}=2\sqrt{5}\), इसलिए अंदर \(2\sqrt{5}-6\sqrt{5}=-4\sqrt{5}\) और गुणनफल (-20) है। पहले कोष्ठक सरल करें। / \(\sqrt{20}=2\sqrt{5}\), so inside becomes \(2\sqrt{5}-6\sqrt{5}=-4\sqrt{5}\) and the product is (-20). Simplify the bracket first.
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Which option is an equal form of an irrational number with a rationalised denominator?
C. \(\frac{4}{\sqrt{11}}=\frac{4\sqrt{11}}{11}\)
अंश और हर को \(\sqrt{11}\) से गुणा करने पर \(\frac{4\sqrt{11}}{11}\) मिलता है। परिमेयकरण में बराबर मान रखना जरूरी है। / Multiplying numerator and denominator by \(\sqrt{11}\) gives \(\frac{4\sqrt{11}}{11}\). Rationalisation must keep the value equal.
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