Topic Quiz • 150 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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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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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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