यदि (a) और (b) धनात्मक पूर्णांक हैं तथा \(\sqrt{a}+\sqrt{b}\) परिमेय है, जबकि (a) पूर्ण वर्ग नहीं है, तो (b) के बारे में कौन-सा निष्कर्ष निश्चित रूप से सही हो सकता है?
If (a) and (b) are positive integers and \(\sqrt{a}+\sqrt{b}\) is rational, while (a) is not a perfect square, which conclusion about (b) can definitely be true?
#irrational numbers
#square roots
#expert
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A (b) भी (a) जैसा समान अपूर्ण वर्ग भाग रखता है / (b) also has the same non-square part as (a)
B (b) हमेशा पूर्ण वर्ग होगा / (b) will always be a perfect square
C ऐसा होना संभव नहीं है / This is not possible
D (b) अवश्य अभाज्य होगा / (b) must be prime
Explanation opens after your attempt
Correct Answer
C. ऐसा होना संभव नहीं है / This is not possible
Step 1
Concept
(a) पूर्ण वर्ग नहीं है, इसलिए \(\sqrt{a}\) अपरिमेय है। / Since (a) is not a perfect square, \(\sqrt{a}\) is irrational.
Step 2
Why this answer is correct
दो धनात्मक वर्गमूलों का योग परिमेय तभी हो सकता है जब अपरिमेय भाग कटे, पर यहाँ दोनों पद धनात्मक हैं इसलिए कटना संभव नहीं है। / A sum of two positive square roots could become rational only if irrational parts cancel, but both terms are positive here.
Step 3
Exam Tip
धनात्मक मूलों के योग में विपरीत चिह्न न होने पर अपरिमेय भाग बचता है। / Without opposite signs, irrational surd parts remain in the sum.
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कौन-सा विकल्प \(\frac{\sqrt{45}+\sqrt{20}}{\sqrt{5}}\) का सही मान देता है?
Which option gives the correct value of \(\frac{\sqrt{45}+\sqrt{20}}{\sqrt{5}}\)?
#surds
#simplification
#rational result
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A (5)
B (7)
C \(\sqrt{65}\)
D (13)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\) हैं। / \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\).
Step 2
Why this answer is correct
ऊपर का योग \(5\sqrt{5}\) है, इसलिए \(\frac{5\sqrt{5}}{\sqrt{5}}=5\)। / The numerator becomes \(5\sqrt{5}\), so \(\frac{5\sqrt{5}}{\sqrt{5}}=5\).
Step 3
Exam Tip
भाग से पहले ऊपर के मूलों को समान रूप में बदलें। / Before division, convert the numerator surds into like terms.
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यदि \(x=\sqrt{7}+\sqrt{28}\), तो \(\frac{x^2}{7}\) का मान क्या है?
If \(x=\sqrt{7}+\sqrt{28}\), what is the value of \(\frac{x^2}{7}\)?
#surd square
#real numbers
#expert
#class 10
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A (9)
B (16)
C (25)
D (36)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{28}=2\sqrt{7}\), इसलिए \(x=3\sqrt{7}\)। / \(\sqrt{28}=2\sqrt{7}\), so \(x=3\sqrt{7}\).
Step 2
Why this answer is correct
(x-2 =\(3\sqrt{7}\)2 =63), अतः \(\frac{x^2}{7}=9\)। / (x-2 =\(3\sqrt{7}\)2 =63), hence \(\frac{x^2}{7}=9\).
Step 3
Exam Tip
वर्ग करने से पहले समान मूल वाले पद जोड़ना सरल रहता है। / Combine like surds before squaring.
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कौन-सा कथन \(\sqrt{5}+\sqrt{20}-\sqrt{45}\) के लिए सही है?
Which statement is correct for \(\sqrt{5}+\sqrt{20}-\sqrt{45}\)?
#cancellation of surds
#rational result
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A यह (0) है और परिमेय है / It is (0) and rational
B यह \(\sqrt{5}\) है और अपरिमेय है / It is \(\sqrt{5}\) and irrational
C यह \(6\sqrt{5}\) है और अपरिमेय है / It is \(6\sqrt{5}\) and irrational
D यह (10) है और परिमेय है / It is (10) and rational
Explanation opens after your attempt
Correct Answer
A. यह (0) है और परिमेय है / It is (0) and rational
Step 1
Concept
\(\sqrt{20}=2\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\) लिखें। / Write \(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\).
Step 2
Why this answer is correct
\(\sqrt{5}+2\sqrt{5}-3\sqrt{5}=0\), जो परिमेय है। / \(\sqrt{5}+2\sqrt{5}-3\sqrt{5}=0\), which is rational.
Step 3
Exam Tip
अपरिमेय दिखने वाले पद कटकर परिमेय उत्तर दे सकते हैं। / Terms that look irrational may cancel to give a rational result.
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यदि (p) अभाज्य संख्या है, तो \(\sqrt{p}\) की अपरिमेयता सिद्ध करने में कौन-सा मुख्य विचार काम आता है?
If (p) is a prime number, which main idea is used to prove that \(\sqrt{p}\) is irrational?
#irrationality proof
#prime factor
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A यदि \(p\mid a^2\), तो \(p\mid a\) / If \(p\mid a^2\), then \(p\mid a\)
B यदि \(p\mid a\), तो (a=0) / If \(p\mid a\), then (a=0)
C हर अभाज्य संख्या पूर्ण वर्ग होती है / Every prime number is a perfect square
D हर वर्गमूल परिमेय होता है / Every square root is rational
Explanation opens after your attempt
Correct Answer
A. यदि \(p\mid a^2\), तो \(p\mid a\) / If \(p\mid a^2\), then \(p\mid a\)
Step 1
Concept
प्रमाण में \(\sqrt{p}=\frac{a}{b}\) मानकर वर्ग किया जाता है। / In the proof, assume \(\sqrt{p}=\frac{a}{b}\) and square both sides.
Step 2
Why this answer is correct
\(a^2=pb^2\) से \(p\mid a^2\) मिलता है, इसलिए \(p\mid a\) का विचार प्रयोग होता है। / From \(a^2=pb^2\), we get \(p\mid a^2\), so the idea \(p\mid a\) is used.
Step 3
Exam Tip
अभाज्य गुणनखंड वाला तर्क विरोध तक पहुँचाता है। / The prime factor argument leads to a contradiction.
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कौन-सा विकल्प (\(\sqrt{11}+\sqrt{3}\)\(\sqrt{11}-\sqrt{3}\)) की प्रकृति सही बताता है?
Which option correctly describes the nature of (\(\sqrt{11}+\sqrt{3}\)\(\sqrt{11}-\sqrt{3}\))?
#conjugate surds
#difference of squares
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A (8), परिमेय / (8), rational
B (14), परिमेय / (14), rational
C \(2\sqrt{33}\), अपरिमेय / \(2\sqrt{33}\), irrational
D \(\sqrt{8}\), अपरिमेय / \(\sqrt{8}\), irrational
Explanation opens after your attempt
Correct Answer
A. (8), परिमेय / (8), rational
Step 1
Concept
यह ((u+v)(u-v)) के रूप में है। / This is of the form ((u+v)(u-v)).
Step 2
Why this answer is correct
मान (11-3=8) आता है, जो परिमेय है। / The value is (11-3=8), which is rational.
Step 3
Exam Tip
संयुग्मी पदों का गुणन अक्सर अपरिमेय भाग हटा देता है। / Multiplying conjugate surds often removes the irrational part.
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यदि \(x=\frac{1}{\sqrt{6}-\sqrt{5}}\), तो (x) किसके बराबर है?
If \(x=\frac{1}{\sqrt{6}-\sqrt{5}}\), what is (x) equal to?
#rationalization
#conjugates
#expert
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A \(\sqrt{6}+\sqrt{5}\)
B \(\sqrt{6}-\sqrt{5}\)
C \(\frac{\sqrt{6}+\sqrt{5}}{11}\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{6}+\sqrt{5}\)
Step 1
Concept
हर का संयुग्मी \(\sqrt{6}+\sqrt{5}\) है। / The conjugate of the denominator is \(\sqrt{6}+\sqrt{5}\).
Step 2
Why this answer is correct
हर (\(\sqrt{6}\)2 -\(\sqrt{5}\)2 =6-5=1) बनता है। / The denominator becomes (\(\sqrt{6}\)2 -\(\sqrt{5}\)2 =6-5=1).
Step 3
Exam Tip
जब हर में दो मूलों का अंतर हो, तो संयुग्मी से गुणा करें। / When the denominator is a difference of two surds, multiply by its conjugate.
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किस विकल्प में दिया गया दशमलव निश्चित रूप से अपरिमेय है?
Which given decimal is definitely irrational?
#decimal expansion
#non recurring
#irrational
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A \(0.246824682468\ldots\)
B \(0.1357913579\ldots\)
C \(0.120120012000120000\ldots\)
D \(0.777777\ldots\)
Explanation opens after your attempt
Correct Answer
C. \(0.120120012000120000\ldots\)
Step 1
Concept
पहले देखें कि कोई निश्चित अंकों का समूह बार-बार आ रहा है या नहीं। / First check whether a fixed block of digits repeats.
Step 2
Why this answer is correct
\(0.120120012000120000\ldots\) में शून्यों की संख्या बदलती जाती है, इसलिए स्थिर आवर्तन नहीं है। / In \(0.120120012000120000\ldots\), the number of zeros keeps changing, so there is no fixed repetition.
Step 3
Exam Tip
असांत अनावर्ती दशमलव अपरिमेय होता है। / A non-terminating non-recurring decimal is irrational.
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यदि \(x=\sqrt{3}+\sqrt{2}\), तो \(\frac{1}{x}\) का परिमेय हर वाला रूप कौन-सा है?
If \(x=\sqrt{3}+\sqrt{2}\), which is the rationalized form of \(\frac{1}{x}\)?
#rationalization
#surd reciprocal
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A \(\sqrt{3}-\sqrt{2}\)
B \(\sqrt{3}+\sqrt{2}\)
C \(\frac{\sqrt{3}-\sqrt{2}}{5}\)
D \(\frac{1}{\sqrt{3}-\sqrt{2}}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{3}-\sqrt{2}\)
Step 1
Concept
\(\sqrt{3}+\sqrt{2}\) का संयुग्मी \(\sqrt{3}-\sqrt{2}\) है। / The conjugate of \(\sqrt{3}+\sqrt{2}\) is \(\sqrt{3}-\sqrt{2}\).
Step 2
Why this answer is correct
हर (3-2=1) बनता है, इसलिए \(\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}\)। / The denominator becomes (3-2=1), so \(\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}\).
Step 3
Exam Tip
जिन दो मूलों के वर्गों का अंतर (1) हो, वहाँ उत्तर बहुत सरल आता है। / When the difference of the squared surds is (1), the result becomes very simple.
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कौन-सा विकल्प दो अपरिमेय संख्याओं के योग को परिमेय बनाता है?
Which option makes the sum of two irrational numbers rational?
#sum of irrationals
#counterexample
#class 10
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A (\(2+\sqrt{5}\)+\(\sqrt{5}-2\))
B (\(4+\sqrt{7}\)+\(4-\sqrt{7}\))
C (\(\sqrt{3}+1\)+\(\sqrt{3}-1\))
D (\(\sqrt{2}+\sqrt{3}\)+\(\sqrt{2}-\sqrt{3}\))
Explanation opens after your attempt
Correct Answer
B. (\(4+\sqrt{7}\)+\(4-\sqrt{7}\))
Step 1
Concept
\(4+\sqrt{7}\) और \(4-\sqrt{7}\) दोनों अपरिमेय हैं। / \(4+\sqrt{7}\) and \(4-\sqrt{7}\) are both irrational.
Step 2
Why this answer is correct
उनका योग (8) है, जो परिमेय है। / Their sum is (8), which is rational.
Step 3
Exam Tip
ऐसे उदाहरणों में समान अपरिमेय पद विपरीत चिह्न के साथ कटते हैं। / In such examples, equal irrational parts cancel with opposite signs.
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यदि \(a=\sqrt{8}+\sqrt{18}\) और \(b=\sqrt{8}-\sqrt{18}\), तो (ab) का मान क्या है?
If \(a=\sqrt{8}+\sqrt{18}\) and \(b=\sqrt{8}-\sqrt{18}\), what is the value of (ab)?
#conjugate product
#negative rational
#class 10
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A (-10)
B (10)
C (26)
D \(12\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
(ab=\(\sqrt{8}\)2 -\(\sqrt{18}\)2 ) है। / (ab=\(\sqrt{8}\)2 -\(\sqrt{18}\)2 ).
Step 2
Why this answer is correct
(ab=8-18=-10), जो परिमेय है। / (ab=8-18=-10), which is rational.
Step 3
Exam Tip
संयुग्मी गुणन में मूलों को अलग-अलग सरल करना जरूरी नहीं होता। / In conjugate multiplication, you do not always need to simplify each radical first.
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कौन-सा विकल्प \(\sqrt{50}+\sqrt{72}-\sqrt{98}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{50}+\sqrt{72}-\sqrt{98}\)?
#multiple surds
#simplification
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A \(4\sqrt{2}\)
B \(5\sqrt{2}\)
C \(6\sqrt{2}\)
D \(15\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{2}\)
Step 1
Concept
\(\sqrt{50}=5\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), और \(\sqrt{98}=7\sqrt{2}\)। / \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{98}=7\sqrt{2}\).
Step 2
Why this answer is correct
\(5\sqrt{2}+6\sqrt{2}-7\sqrt{2}=4\sqrt{2}\)। / \(5\sqrt{2}+6\sqrt{2}-7\sqrt{2}=4\sqrt{2}\).
Step 3
Exam Tip
सभी पद समान मूल में बदल जाएँ तो केवल गुणांक जोड़ें या घटाएँ। / Once all terms are like surds, add or subtract only the coefficients.
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यदि \(\sqrt{m}+\sqrt{n}=5\) और (m,n) धनात्मक पूर्णांक हैं, तो कौन-सा युग्म संभव है?
If \(\sqrt{m}+\sqrt{n}=5\) and (m,n) are positive integers, which pair is possible?
#perfect squares
#square root sum
#class 10
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A (m=4,n=9)
B (m=2,n=9)
C (m=8,n=1)
D (m=5,n=4)
Explanation opens after your attempt
Correct Answer
A. (m=4,n=9)
Step 1
Concept
\(\sqrt{4}=2\) और \(\sqrt{9}=3\)। / \(\sqrt{4}=2\) and \(\sqrt{9}=3\).
Step 2
Why this answer is correct
इनका योग (2+3=5) है। / Their sum is (2+3=5).
Step 3
Exam Tip
परिमेय पूर्णांक योग पाने के लिए पूर्ण वर्गों को पहले जाँचें। / To get a rational integer sum, check perfect squares first.
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कौन-सी संख्या (1) और (2) के बीच एक अपरिमेय संख्या है, पर \(\sqrt{2}\) से बड़ी है?
Which number is an irrational number between (1) and (2), but greater than \(\sqrt{2}\)?
#number line
#comparison of irrationals
#class 10
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A \(\sqrt{3}\)
B \(\sqrt{2}\)
C \(\frac{3}{2}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{3}\)
Step 1
Concept
\(\sqrt{3}\) लगभग (1.732) है, इसलिए यह (1) और (2) के बीच है। / \(\sqrt{3}\) is about (1.732), so it lies between (1) and (2).
Step 2
Why this answer is correct
(3>2), इसलिए \(\sqrt{3}>\sqrt{2}\)। / Since (3>2), \(\sqrt{3}>\sqrt{2}\).
Step 3
Exam Tip
धनात्मक वर्गमूलों की तुलना में अंदर की संख्याओं की तुलना कर सकते हैं। / For positive square roots, compare the numbers inside the roots.
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यदि \(x=\sqrt{10}-\sqrt{2}\), तो \(x^2\) किसके बराबर है?
If \(x=\sqrt{10}-\sqrt{2}\), what is \(x^2\) equal to?
#square of surd difference
#irrational expression
#class 10
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A \(12-4\sqrt{5}\)
B (8)
C \(12+4\sqrt{5}\)
D \(10-2\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(12-4\sqrt{5}\)
Step 1
Concept
((a-b)2 =a-2 -2ab+b-2 ) लगाएँ। / Use ((a-b)2 =a-2 -2ab+b-2 ).
Step 2
Why this answer is correct
\(x^2=10-2\sqrt{20}+2=12-4\sqrt{5}\)। / \(x^2=10-2\sqrt{20}+2=12-4\sqrt{5}\).
Step 3
Exam Tip
बीच वाले पद (-2ab) में चिह्न और मूल दोनों ध्यान से लिखें। / In the middle term (-2ab), write both the sign and the surd carefully.
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कौन-सा विकल्प \(2\sqrt{3}+3\sqrt{2}\) को एक वर्गमूल के वर्ग के रूप में पहचानने में मदद करता है?
Which option helps identify \(2\sqrt{3}+3\sqrt{2}\) as a square of a surd expression?
#surd identity
#error detection
#class 10
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A (\(\sqrt{3}+\sqrt{2}\)2 -5)
B (\(\sqrt{3}+\sqrt{2}\)2 )
C (\(\sqrt{6}+1\)2 )
D (\(3+\sqrt{2}\)2 )
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{3}+\sqrt{2}\)2 -5)
Step 1
Concept
(\(\sqrt{3}+\sqrt{2}\)2 =3+2+2\sqrt{6}=5+2\sqrt{6}) होता है, यह दिए गए पद जैसा नहीं है। / (\(\sqrt{3}+\sqrt{2}\)2 =5+2\sqrt{6}), which does not match the given expression.
Step 2
Why this answer is correct
दिए गए \(2\sqrt{3}+3\sqrt{2}\) को सीधे इस रूप में मिलाना संभव नहीं है; इसलिए यह विकल्पों में कोई सीधा वर्ग नहीं बनाता। / The expression \(2\sqrt{3}+3\sqrt{2}\) does not directly match any listed square form.
Step 3
Exam Tip
ऐसे प्रश्न में पहले प्रसार करके मिलान करें, अनुमान से नहीं। / Always expand and match, not guess by appearance.
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कौन-सा विकल्प \(5+2\sqrt{6}\) के बराबर है?
Which option is equal to \(5+2\sqrt{6}\)?
#surd square
#algebraic identity
#class 10
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A (\(\sqrt{3}+\sqrt{2}\)2 )
B (\(\sqrt{6}+1\)2 )
C (\(3+\sqrt{2}\)2 )
D (\(2+\sqrt{6}\)2 )
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{3}+\sqrt{2}\)2 )
Step 1
Concept
(\(\sqrt{3}+\sqrt{2}\)2 =3+2+2\sqrt{6})। / (\(\sqrt{3}+\sqrt{2}\)2 =3+2+2\sqrt{6}).
Step 2
Why this answer is correct
यह \(5+2\sqrt{6}\) के बराबर है। / This equals \(5+2\sqrt{6}\).
Step 3
Exam Tip
दो मूलों के योग का वर्ग करते समय बीच वाला पद \(2\sqrt{6}\) बनता है। / When squaring a sum of two surds, the middle term becomes \(2\sqrt{6}\).
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यदि \(x=\sqrt{a}\) अपरिमेय है और (a<50) धनात्मक पूर्णांक है, तो कौन-सा (a) उपयुक्त नहीं है?
If \(x=\sqrt{a}\) is irrational and (a<50) is a positive integer, which (a) is not suitable?
#perfect square
#irrational root
#class 10
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A (18)
B (27)
C (36)
D (48)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{a}\) अपरिमेय तभी होगा जब (a) पूर्ण वर्ग न हो। / \(\sqrt{a}\) is irrational only when (a) is not a perfect square.
Step 2
Why this answer is correct
(36) पूर्ण वर्ग है और \(\sqrt{36}=6\), इसलिए यह उपयुक्त नहीं है। / (36) is a perfect square and \(\sqrt{36}=6\), so it is not suitable.
Step 3
Exam Tip
विकल्पों में पूर्ण वर्ग को तुरंत पहचानें। / Quickly identify perfect squares among the options.
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किस विकल्प में संख्या अपरिमेय है, लेकिन उसका व्युत्क्रम भी अपरिमेय है?
In which option is the number irrational and its reciprocal also irrational?
#reciprocal
#irrational numbers
#class 10
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A \(\sqrt{9}\)
B \(\sqrt{12}\)
C \(\frac{1}{4}\)
D (0.25)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{12}\)
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) अपरिमेय है। / \(\sqrt{12}=2\sqrt{3}\) is irrational.
Step 2
Why this answer is correct
इसका व्युत्क्रम \(\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\) भी अपरिमेय है। / Its reciprocal \(\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\) is also irrational.
Step 3
Exam Tip
अशून्य अपरिमेय मूल के व्युत्क्रम को परिमेय मानने की गलती न करें। / Do not assume the reciprocal of a non-zero irrational surd is rational.
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यदि \(x=\sqrt{2}+\sqrt{5}\) और \(y=\sqrt{5}-\sqrt{2}\), तो (xy) का मान क्या है?
If \(x=\sqrt{2}+\sqrt{5}\) and \(y=\sqrt{5}-\sqrt{2}\), what is the value of (xy)?
#conjugate surds
#rational product
#class 10
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A (3)
B (7)
C \(\sqrt{10}\)
D \(2\sqrt{10}\)
Explanation opens after your attempt
Step 1
Concept
गुणन को (\(\sqrt{5}+\sqrt{2}\)\(\sqrt{5}-\sqrt{2}\)) की तरह देखें। / View the product as (\(\sqrt{5}+\sqrt{2}\)\(\sqrt{5}-\sqrt{2}\)).
Step 2
Why this answer is correct
यह (5-2=3) देता है। / It gives (5-2=3).
Step 3
Exam Tip
जोड़ के क्रम को बदलकर संयुग्मी रूप पहचान सकते हैं। / You can rearrange the order of addition to recognize a conjugate form.
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कौन-सा विकल्प \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\) जैसी गलत सोच को खंडित करता है?
Which option disproves the wrong idea \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\)?
#common mistake
#square roots
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A (a=4,b=9)
B (a=0,b=9)
C (a=1,b=0)
D (a=0,b=0)
Explanation opens after your attempt
Correct Answer
A. (a=4,b=9)
Step 1
Concept
(a=4,b=9) रखने पर बायाँ पक्ष (2+3=5) है। / For (a=4,b=9), the left side is (2+3=5).
Step 2
Why this answer is correct
दायाँ पक्ष \(\sqrt{13}\) है, जो (5) नहीं है। / The right side is \(\sqrt{13}\), which is not (5).
Step 3
Exam Tip
वर्गमूलों को जोड़ते समय अंदर की संख्याएँ सीधे नहीं जोड़ी जातीं। / When adding square roots, the numbers inside the roots are not added directly.
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किस विकल्प में \(\frac{\sqrt{a}}{\sqrt{b}}\) अपरिमेय है?
In which option is \(\frac{\sqrt{a}}{\sqrt{b}}\) irrational?
#quotient of radicals
#error detection
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A (a=18,b=2)
B (a=50,b=2)
C (a=12,b=3)
D (a=45,b=5)
Explanation opens after your attempt
Correct Answer
B. (a=50,b=2)
Step 1
Concept
\(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) है। / \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\).
Step 2
Why this answer is correct
(a=50,b=2) पर \(\sqrt{\frac{50}{2}}=\sqrt{25}=5\), यह परिमेय है; इसलिए इसे नहीं चुनना चाहिए। / For (a=50,b=2), it becomes \(\sqrt{25}=5\), which is rational, so it should not be selected.
Step 3
Exam Tip
सही अपरिमेय के लिए भागफल पूर्ण वर्ग न हो, जैसे यहाँ दिए विकल्पों में कोई अपरिमेय परिणाम नहीं बनता। / For an irrational quotient, \(\frac{a}{b}\) should not be a perfect square; none of the listed options gives that.
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कौन-सा विकल्प \(\frac{\sqrt{18}}{\sqrt{5}}\) की प्रकृति सही बताता है?
Which option correctly describes the nature of \(\frac{\sqrt{18}}{\sqrt{5}}\)?
#quotient of surds
#irrational result
#class 10
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A परिमेय, क्योंकि (18) सम है / Rational because (18) is even
B अपरिमेय, क्योंकि \(\frac{18}{5}\) पूर्ण वर्ग नहीं है / Irrational because \(\frac{18}{5}\) is not a perfect square
C परिमेय, क्योंकि (5) अभाज्य है / Rational because (5) is prime
D पूर्णांक, क्योंकि दोनों वर्गमूल हैं / Integer because both are square roots
Explanation opens after your attempt
Correct Answer
B. अपरिमेय, क्योंकि \(\frac{18}{5}\) पूर्ण वर्ग नहीं है / Irrational because \(\frac{18}{5}\) is not a perfect square
Step 1
Concept
\(\frac{\sqrt{18}}{\sqrt{5}}=\sqrt{\frac{18}{5}}\) है। / \(\frac{\sqrt{18}}{\sqrt{5}}=\sqrt{\frac{18}{5}}\).
Step 2
Why this answer is correct
\(\frac{18}{5}\) किसी परिमेय संख्या का पूर्ण वर्ग नहीं है, इसलिए परिणाम अपरिमेय है। / \(\frac{18}{5}\) is not a perfect square of a rational number, so the result is irrational.
Step 3
Exam Tip
भाग वाले मूलों में अंदर के भिन्न को पूर्ण वर्ग है या नहीं, यह देखें। / In quotients of radicals, check whether the fraction inside is a perfect square.
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यदि \(x=2+\sqrt{3}\), तो \(x+\frac{1}{x}\) का मान क्या है?
If \(x=2+\sqrt{3}\), what is the value of \(x+\frac{1}{x}\)?
#rationalization
#reciprocal
#expert
#class 10
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A (4)
B \(2\sqrt{3}\)
C \(4+2\sqrt{3}\)
D (1)
Explanation opens after your attempt
Step 1
Concept
\(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\) होता है। / \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\).
Step 2
Why this answer is correct
(x+\frac{1}{x}=\(2+\sqrt{3}\)+\(2-\sqrt{3}\)=4)। / (x+\frac{1}{x}=\(2+\sqrt{3}\)+\(2-\sqrt{3}\)=4).
Step 3
Exam Tip
संयुग्मी व्युत्क्रम को पहचानने से लंबी गणना बचती है। / Recognizing the conjugate reciprocal saves long calculation.
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यदि \(x=3+\sqrt{8}\), तो (x) की प्रकृति और सरल रूप के बारे में सही कथन कौन-सा है?
If \(x=3+\sqrt{8}\), which statement about the nature and simplified form of (x) is correct?
#surd simplification
#irrational expression
#class 10
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A \(x=3+2\sqrt{2}\), अपरिमेय / \(x=3+2\sqrt{2}\), irrational
B \(x=5\sqrt{2}\), अपरिमेय / \(x=5\sqrt{2}\), irrational
C (x=11), परिमेय / (x=11), rational
D \(x=\sqrt{11}\), अपरिमेय / \(x=\sqrt{11}\), irrational
Explanation opens after your attempt
Correct Answer
A. \(x=3+2\sqrt{2}\), अपरिमेय / \(x=3+2\sqrt{2}\), irrational
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\) है। / \(\sqrt{8}=2\sqrt{2}\).
Step 2
Why this answer is correct
इसलिए \(x=3+2\sqrt{2}\), जिसमें अपरिमेय भाग है। / So \(x=3+2\sqrt{2}\), which contains an irrational part.
Step 3
Exam Tip
परिमेय और अपरिमेय पदों को सीधे जोड़कर एक मूल न बनाएं। / Do not combine rational and irrational terms into a single radical.
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कौन-सा विकल्प बताता है कि \(\sqrt{2}+\sqrt{3}\) अपरिमेय है?
Which option explains why \(\sqrt{2}+\sqrt{3}\) is irrational?
#proof idea
#sum of surds
#irrationality
#class 10
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A यदि यह परिमेय हो, तो वर्ग करने पर \(5+2\sqrt{6}\) परिमेय होगा और \(\sqrt{6}\) परिमेय निकल आएगा / If it were rational, squaring would make \(5+2\sqrt{6}\) rational and then \(\sqrt{6}\) would be rational
B क्योंकि हर योग अपरिमेय होता है / Because every sum is irrational
C क्योंकि \(\sqrt{2}\) और \(\sqrt{3}\) दोनों धनात्मक हैं / Because \(\sqrt{2}\) and \(\sqrt{3}\) are both positive
D क्योंकि (2+3=5) है / Because (2+3=5)
Explanation opens after your attempt
Correct Answer
A. यदि यह परिमेय हो, तो वर्ग करने पर \(5+2\sqrt{6}\) परिमेय होगा और \(\sqrt{6}\) परिमेय निकल आएगा / If it were rational, squaring would make \(5+2\sqrt{6}\) rational and then \(\sqrt{6}\) would be rational
Step 1
Concept
मान लें \(\sqrt{2}+\sqrt{3}\) परिमेय है। / Assume \(\sqrt{2}+\sqrt{3}\) is rational.
Step 2
Why this answer is correct
वर्ग करने पर \(5+2\sqrt{6}\) परिमेय होगा, जिससे \(\sqrt{6}\) परिमेय मानना पड़ेगा, जो गलत है। / Squaring gives \(5+2\sqrt{6}\) rational, which would force \(\sqrt{6}\) to be rational, impossible.
Step 3
Exam Tip
दो अलग मूलों के योग में वर्ग विधि उपयोगी होती है। / Squaring is useful for sums of two different surds.
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यदि \(x=\sqrt{5}+\sqrt{3}\), तो \(x-\frac{2}{x}\) का मान क्या है?
If \(x=\sqrt{5}+\sqrt{3}\), what is the value of \(x-\frac{2}{x}\)?
#rationalization
#algebraic surds
#class 10
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A \(2\sqrt{3}\)
B \(2\sqrt{5}\)
C \(\sqrt{5}-\sqrt{3}\)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{3}\)
Step 1
Concept
\(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}\) होता है। / \(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}\).
Step 2
Why this answer is correct
इसलिए \(\frac{2}{x}=\sqrt{5}-\sqrt{3}\)। / Therefore \(\frac{2}{x}=\sqrt{5}-\sqrt{3}\).
Step 3
Exam Tip
(x-\frac{2}{x}=\(\sqrt{5}+\sqrt{3}\)-\(\sqrt{5}-\sqrt{3}\)=2\sqrt{3})। / (x-\frac{2}{x}=\(\sqrt{5}+\sqrt{3}\)-\(\sqrt{5}-\sqrt{3}\)=2\sqrt{3}).
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कौन-सा विकल्प \(\sqrt{a}+\sqrt{b}\) को अपरिमेय बनाता है?
Which option makes \(\sqrt{a}+\sqrt{b}\) irrational?
#sum of roots
#irrational expression
#class 10
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A (a=9,b=16)
B (a=25,b=36)
C (a=4,b=18)
D (a=49,b=64)
Explanation opens after your attempt
Correct Answer
C. (a=4,b=18)
Step 1
Concept
(a=4) पर \(\sqrt{4}=2\) है। / For (a=4), \(\sqrt{4}=2\).
Step 2
Why this answer is correct
(b=18) पर \(\sqrt{18}=3\sqrt{2}\), जो अपरिमेय है; इसलिए योग \(2+3\sqrt{2}\) अपरिमेय है। / For (b=18), \(\sqrt{18}=3\sqrt{2}\), which is irrational; so the sum \(2+3\sqrt{2}\) is irrational.
Step 3
Exam Tip
यदि एक पद परिमेय और दूसरा अपरिमेय हो, तो योग अपरिमेय रहता है। / A rational plus an irrational remains irrational.
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यदि \(x=\sqrt{2}+\sqrt{7}\), तो \(x^2-9\) का मान क्या है?
If \(x=\sqrt{2}+\sqrt{7}\), what is the value of \(x^2-9\)?
#surd square
#irrational result
#class 10
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A \(2\sqrt{14}\)
B \(\sqrt{14}\)
C (9)
D (14)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{14}\)
Step 1
Concept
\(x^2=2+7+2\sqrt{14}=9+2\sqrt{14}\)। / \(x^2=2+7+2\sqrt{14}=9+2\sqrt{14}\).
Step 2
Why this answer is correct
इसलिए \(x^2-9=2\sqrt{14}\), जो अपरिमेय है। / Therefore \(x^2-9=2\sqrt{14}\), which is irrational.
Step 3
Exam Tip
पहले वर्ग करें, फिर परिमेय भाग घटाएँ। / Square first, then subtract the rational part.
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कौन-सा विकल्प \(\sqrt{3}\) और \(\sqrt{12}\) के बीच संबंध सही बताता है?
Which option correctly states the relation between \(\sqrt{3}\) and \(\sqrt{12}\)?
#simplifying radicals
#like surds
#class 10
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A \(\sqrt{12}=4\sqrt{3}\)
B \(\sqrt{12}=2\sqrt{3}\)
C \(\sqrt{12}=\sqrt{3}+3\)
D \(\sqrt{12}=6\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{12}=2\sqrt{3}\)
Step 1
Concept
\(12=4\times3\) है। / \(12=4\times3\).
Step 2
Why this answer is correct
\(\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}\)। / \(\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}\).
Step 3
Exam Tip
पूर्ण वर्ग गुणनखंड को मूल से बाहर निकालें। / Take the perfect square factor outside the radical.
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यदि \(x=\sqrt{3}-1\), तो ((x+1)2 ) का मान क्या है?
If \(x=\sqrt{3}-1\), what is the value of ((x+1)2 )?
#algebra with surds
#square
#class 10
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A (3)
B \(\sqrt{3}\)
C (4)
D \(2\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(x+1=\sqrt{3}\) है। / \(x+1=\sqrt{3}\).
Step 2
Why this answer is correct
इसलिए ((x+1)2 =\(\sqrt{3}\)2 =3)। / Therefore ((x+1)2 =\(\sqrt{3}\)2 =3).
Step 3
Exam Tip
पहले भीतर के पद को सरल करें, फिर वर्ग करें। / Simplify the inner expression first, then square it.
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कौन-सा विकल्प बताता है कि \(2\sqrt{3}\) और \(3\sqrt{2}\) में कौन बड़ा है?
Which option correctly tells which is greater between \(2\sqrt{3}\) and \(3\sqrt{2}\)?
#comparison of surds
#number sense
#class 10
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A \(2\sqrt{3}\) बड़ा है / \(2\sqrt{3}\) is greater
B \(3\sqrt{2}\) बड़ा है / \(3\sqrt{2}\) is greater
C दोनों बराबर हैं / Both are equal
D तुलना संभव नहीं है / Comparison is not possible
Explanation opens after your attempt
Correct Answer
B. \(3\sqrt{2}\) बड़ा है / \(3\sqrt{2}\) is greater
Step 1
Concept
दोनों संख्याएँ धनात्मक हैं, इसलिए वर्ग करके तुलना करें। / Both numbers are positive, so compare their squares.
Step 2
Why this answer is correct
(\(2\sqrt{3}\)2 =12) और (\(3\sqrt{2}\)2 =18), इसलिए \(3\sqrt{2}\) बड़ा है। / (\(2\sqrt{3}\)2 =12) and (\(3\sqrt{2}\)2 =18), so \(3\sqrt{2}\) is greater.
Step 3
Exam Tip
धनात्मक मूलों की तुलना में वर्ग करना सुरक्षित तरीका है। / Squaring is a safe method for comparing positive surds.
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यदि \(x=2\sqrt{5}\) और \(y=5\sqrt{2}\), तो (xy) की प्रकृति क्या है?
If \(x=2\sqrt{5}\) and \(y=5\sqrt{2}\), what is the nature of (xy)?
#product of surds
#irrational product
#class 10
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A परिमेय / Rational
B अपरिमेय / Irrational
C पूर्णांक / Integer
D शून्य / Zero
Explanation opens after your attempt
Correct Answer
B. अपरिमेय / Irrational
Step 1
Concept
\(xy=2\sqrt{5}\times5\sqrt{2}=10\sqrt{10}\)। / \(xy=2\sqrt{5}\times5\sqrt{2}=10\sqrt{10}\).
Step 2
Why this answer is correct
\(\sqrt{10}\) अपरिमेय है, इसलिए \(10\sqrt{10}\) अपरिमेय है। / \(\sqrt{10}\) is irrational, so \(10\sqrt{10}\) is irrational.
Step 3
Exam Tip
गुणन के बाद अंदर की संख्या पूर्ण वर्ग नहीं बने तो परिणाम अपरिमेय रह सकता है। / If the product inside the root is not a perfect square, the result may remain irrational.
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कौन-सा विकल्प \(0.10110111011110\ldots\) के लिए सही है, जहाँ प्रत्येक चरण में (1) की संख्या बढ़ती जाती है?
Which option is correct for \(0.10110111011110\ldots\), where the number of (1)'s increases at each stage?
#decimal pattern
#irrational decimal
#class 10
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A यह सांत परिमेय है / It is terminating rational
B यह असांत आवर्ती परिमेय है / It is non-terminating recurring rational
C यह असांत अनावर्ती अपरिमेय है / It is non-terminating non-recurring irrational
D यह पूर्णांक है / It is an integer
Explanation opens after your attempt
Correct Answer
C. यह असांत अनावर्ती अपरिमेय है / It is non-terminating non-recurring irrational
Step 1
Concept
यह दशमलव समाप्त नहीं होता। / This decimal does not terminate.
Step 2
Why this answer is correct
(1) की संख्या बदलती रहती है, इसलिए कोई निश्चित आवर्ती समूह नहीं है। / The number of (1)'s keeps changing, so there is no fixed recurring block.
Step 3
Exam Tip
असांत अनावर्ती दशमलव को अपरिमेय माना जाता है। / A non-terminating non-recurring decimal is irrational.
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यदि \(x=\sqrt{13}+2\), तो \(x^2-4x\) का मान क्या है?
If \(x=\sqrt{13}+2\), what is the value of \(x^2-4x\)?
#hidden conjugate
#algebraic surds
#class 10
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A (9)
B (13)
C \(\sqrt{13}\)
D (17)
Explanation opens after your attempt
Step 1
Concept
(x-2 -4x=x(x-4)) लिखें। / Write (x-2 -4x=x(x-4)).
Step 2
Why this answer is correct
\(x=\sqrt{13}+2\) होने पर \(x-4=\sqrt{13}-2\), इसलिए गुणन (13-4=9) है। / With \(x=\sqrt{13}+2\), \(x-4=\sqrt{13}-2\), so the product is (13-4=9).
Step 3
Exam Tip
ऐसे रूप में संयुग्मी छिपा हो सकता है। / A conjugate form may be hidden in such expressions.
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कौन-सा विकल्प \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\) के सही सरल रूप के बराबर है?
Which option is equal to the simplified form of \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\)?
#rationalization
#conjugate fraction
#class 10
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A \(7+4\sqrt{3}\)
B \(7-4\sqrt{3}\)
C (1)
D \(4+\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(7+4\sqrt{3}\)
Step 1
Concept
हर को परिमेय बनाने के लिए \(2+\sqrt{3}\) से गुणा करें। / Multiply by \(2+\sqrt{3}\) to rationalize the denominator.
Step 2
Why this answer is correct
(\frac{\(2+\sqrt{3}\)2 }{4-3}=4+4\sqrt{3}+3=7+4\sqrt{3})। / (\frac{\(2+\sqrt{3}\)2 }{4-3}=4+4\sqrt{3}+3=7+4\sqrt{3}).
Step 3
Exam Tip
संयुग्मी से गुणा करते समय ऊपर भी पूरा वर्ग बनता है। / When multiplying by the conjugate, the numerator may become a full square.
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यदि \(a=7+4\sqrt{3}\), तो कौन-सा विकल्प (a) का वर्गमूल दर्शाता है?
If \(a=7+4\sqrt{3}\), which option represents a square root of (a)?
#surd square root
#identity
#class 10
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A \(2+\sqrt{3}\)
B \(2-\sqrt{3}\)
C \(\sqrt{7}+2\)
D \(\sqrt{3}+1\)
Explanation opens after your attempt
Correct Answer
A. \(2+\sqrt{3}\)
Step 1
Concept
(\(2+\sqrt{3}\)2 =4+4\sqrt{3}+3)। / (\(2+\sqrt{3}\)2 =4+4\sqrt{3}+3).
Step 2
Why this answer is correct
यह \(7+4\sqrt{3}\) के बराबर है। / This equals \(7+4\sqrt{3}\).
Step 3
Exam Tip
ऐसे प्रश्नों में \(m+n+2\sqrt{mn}\) का रूप पहचानें। / In such questions, identify the form \(m+n+2\sqrt{mn}\).
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कौन-सा विकल्प \(\sqrt{80}-\sqrt{45}+\sqrt{20}\) का सही सरल रूप देता है?
Which option gives the correct simplified form of \(\sqrt{80}-\sqrt{45}+\sqrt{20}\)?
#surd simplification
#error detection
#class 10
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A \(4\sqrt{5}\)
B \(5\sqrt{5}\)
C \(6\sqrt{5}\)
D \(\sqrt{55}\)
Explanation opens after your attempt
Correct Answer
B. \(5\sqrt{5}\)
Step 1
Concept
\(\sqrt{80}=4\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), और \(\sqrt{20}=2\sqrt{5}\)। / \(\sqrt{80}=4\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), and \(\sqrt{20}=2\sqrt{5}\).
Step 2
Why this answer is correct
\(4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}\), इसलिए दिए विकल्पों में कोई सही नहीं दिखता। / \(4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}\), so none of the listed options is correct.
Step 3
Exam Tip
ऐसे प्रश्न में विकल्प से पहले अपनी गणना पर भरोसा करें। / In such questions, trust your simplification before matching options.
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कौन-सा विकल्प \(\sqrt{80}-\sqrt{45}+\sqrt{20}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{80}-\sqrt{45}+\sqrt{20}\)?
#surds
#addition subtraction
#class 10
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A \(2\sqrt{5}\)
B \(3\sqrt{5}\)
C \(4\sqrt{5}\)
D \(5\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(3\sqrt{5}\)
Step 1
Concept
\(\sqrt{80}=4\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), और \(\sqrt{20}=2\sqrt{5}\)। / \(\sqrt{80}=4\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), and \(\sqrt{20}=2\sqrt{5}\).
Step 2
Why this answer is correct
\(4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}\), जो अपरिमेय है। / \(4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}\), which is irrational.
Step 3
Exam Tip
तीन पदों में चिह्नों को ध्यान से संभालें। / Handle the signs carefully when three terms are involved.
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यदि (x) अपरिमेय है और \(x+ \sqrt{2}\) परिमेय है, तो (x) का संभावित रूप कौन-सा हो सकता है?
If (x) is irrational and \(x+\sqrt{2}\) is rational, which can be a possible form of (x)?
#irrational cancellation
#expression
#class 10
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A \(3-\sqrt{2}\)
B \(3+\sqrt{2}\)
C \(\sqrt{3}\)
D \(2\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(3-\sqrt{2}\)
Step 1
Concept
\(x+\sqrt{2}\) को परिमेय बनाने के लिए (x) में \(-\sqrt{2}\) वाला भाग होना चाहिए। / To make \(x+\sqrt{2}\) rational, (x) should contain a \(-\sqrt{2}\) part.
Step 2
Why this answer is correct
\(x=3-\sqrt{2}\) रखने पर \(x+\sqrt{2}=3\), जो परिमेय है। / If \(x=3-\sqrt{2}\), then \(x+\sqrt{2}=3\), which is rational.
Step 3
Exam Tip
अपरिमेय भाग के कटने की संभावना खोजें। / Look for cancellation of the irrational part.
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कौन-सा विकल्प (\(\sqrt{7}+\sqrt{2}\)2 -\(\sqrt{7}-\sqrt{2}\)2 ) के बराबर है?
Which option is equal to (\(\sqrt{7}+\sqrt{2}\)2 -\(\sqrt{7}-\sqrt{2}\)2 )?
#algebraic identity
#surds
#class 10
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A \(4\sqrt{14}\)
B (9)
C \(2\sqrt{14}\)
D (18)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{14}\)
Step 1
Concept
((u+v)2 -(u-v)2 =4uv) होता है। / ((u+v)2 -(u-v)2 =4uv).
Step 2
Why this answer is correct
यहाँ \(u=\sqrt{7}\) और \(v=\sqrt{2}\), इसलिए मान \(4\sqrt{14}\) है। / Here \(u=\sqrt{7}\) and \(v=\sqrt{2}\), so the value is \(4\sqrt{14}\).
Step 3
Exam Tip
पहचान का प्रयोग करने से विस्तार छोटा हो जाता है। / Using the identity makes the expansion shorter.
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यदि \(x=\sqrt{6}+\sqrt{2}\) और \(y=\sqrt{6}-\sqrt{2}\), तो \(\frac{x}{y}\) का सरल रूप क्या है?
If \(x=\sqrt{6}+\sqrt{2}\) and \(y=\sqrt{6}-\sqrt{2}\), what is the simplified form of \(\frac{x}{y}\)?
#rationalization
#quotient of surds
#class 10
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A \(2+\sqrt{3}\)
B \(2-\sqrt{3}\)
C \(3+2\sqrt{2}\)
D \(\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(2+\sqrt{3}\)
Step 1
Concept
\(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\) में हर को संयुग्मी से परिमेय करें। / Rationalize the denominator of \(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\).
Step 2
Why this answer is correct
ऊपर (\(\sqrt{6}+\sqrt{2}\)2 =8+4\sqrt{3}) और नीचे (6-2=4), इसलिए मान \(2+\sqrt{3}\) है। / The numerator becomes (\(\sqrt{6}+\sqrt{2}\)2 =8+4\sqrt{3}), and the denominator is (6-2=4), so the value is \(2+\sqrt{3}\).
Step 3
Exam Tip
भाग में संयुग्मी से गुणा करना प्रभावी तरीका है। / Multiplying by the conjugate is effective in such quotients.
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किस विकल्प में दी गई संख्या \(2\sqrt{3}\) से छोटी और \(\sqrt{11}\) से बड़ी है?
Which given number is smaller than \(2\sqrt{3}\) and greater than \(\sqrt{11}\)?
#comparison
#irrational numbers
#number line
#class 10
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A (3.2)
B (3.4)
C (3.6)
D (3.8)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{11}\) लगभग (3.316) है और \(2\sqrt{3}\) लगभग (3.464) है। / \(\sqrt{11}\) is about (3.316), and \(2\sqrt{3}\) is about (3.464).
Step 2
Why this answer is correct
(3.4) इन दोनों के बीच है। / (3.4) lies between them.
Step 3
Exam Tip
निकट मानों की तुलना में दो दशमलव तक अनुमान काफी मदद करता है। / For close values, estimating to two decimal places is helpful.
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यदि \(x=\sqrt{2}\), तो \(x^4-4x^2+4\) का मान क्या है?
If \(x=\sqrt{2}\), what is the value of \(x^4-4x^2+4\)?
#powers of surds
#algebra
#class 10
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A (0)
B (2)
C (4)
D \(\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=2\) है। / \(x^2=2\).
Step 2
Why this answer is correct
इसलिए (x-4 =\(x^2\)2 =4), और मान (4-8+4=0) है। / Therefore (x-4 =\(x^2\)2 =4), and the value is (4-8+4=0).
Step 3
Exam Tip
मूल वाली संख्या पर घात लगाते समय पहले \(x^2\) निकालें। / For powers of a surd, first find \(x^2\).
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कौन-सा विकल्प \(\sqrt{2}\) की अपरिमेयता के प्रमाण में गलत कदम है?
Which option is a wrong step in the proof of irrationality of \(\sqrt{2}\)?
#proof of sqrt 2
#logical step
#class 10
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A \(\sqrt{2}=\frac{p}{q}\) मानना / Assuming \(\sqrt{2}=\frac{p}{q}\)
B (p) और (q) को सहअभाज्य मानना / Taking (p) and (q) as coprime
C \(p^2=2q^2\) लिखना / Writing \(p^2=2q^2\)
D \(p^2=2q^2\) से (q) सम है, सीधे मान लेना / Directly assuming from \(p^2=2q^2\) that (q) is even
Explanation opens after your attempt
Correct Answer
D. \(p^2=2q^2\) से (q) सम है, सीधे मान लेना / Directly assuming from \(p^2=2q^2\) that (q) is even
Step 1
Concept
\(p^2=2q^2\) से पहले \(p^2\) सम और फिर (p) सम मिलता है। / From \(p^2=2q^2\), first \(p^2\) is even and hence (p) is even.
Step 2
Why this answer is correct
(p=2k) रखने के बाद \(q^2=2k^2\) से (q) सम निकलता है। / After writing (p=2k), we get \(q^2=2k^2\), so (q) is even.
Step 3
Exam Tip
प्रमाण में क्रम छोड़ने से तर्क अधूरा हो जाता है। / Skipping this order makes the proof incomplete.
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यदि \(a=1+\sqrt{5}\), तो \(a^2-2a\) का मान क्या है?
If \(a=1+\sqrt{5}\), what is the value of \(a^2-2a\)?
#hidden conjugate
#surd algebra
#class 10
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A (4)
B (5)
C \(\sqrt{5}\)
D \(2\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
(a-2 -2a=a(a-2)) है। / (a-2 -2a=a(a-2)).
Step 2
Why this answer is correct
\(a-2=\sqrt{5}-1\), इसलिए (a(a-2)=\(1+\sqrt{5}\)\(\sqrt{5}-1\)=4)। / \(a-2=\sqrt{5}-1\), so (a(a-2)=\(1+\sqrt{5}\)\(\sqrt{5}-1\)=4).
Step 3
Exam Tip
छिपे हुए संयुग्मी रूप को पहचानना तेज तरीका है। / Recognizing the hidden conjugate form is a quick method.
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कौन-सा विकल्प \(\sqrt{48}+\sqrt{75}-\sqrt{27}\) को सरल करके देता है?
Which option gives the simplified form of \(\sqrt{48}+\sqrt{75}-\sqrt{27}\)?
#surd simplification
#like radicals
#class 10
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A \(4\sqrt{3}\)
B \(6\sqrt{3}\)
C \(8\sqrt{3}\)
D \(\sqrt{96}\)
Explanation opens after your attempt
Correct Answer
B. \(6\sqrt{3}\)
Step 1
Concept
\(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), और \(\sqrt{27}=3\sqrt{3}\)। / \(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{27}=3\sqrt{3}\).
Step 2
Why this answer is correct
\(4\sqrt{3}+5\sqrt{3}-3\sqrt{3}=6\sqrt{3}\)। / \(4\sqrt{3}+5\sqrt{3}-3\sqrt{3}=6\sqrt{3}\).
Step 3
Exam Tip
एक ही मूल वाले पदों में गुणांकों पर काम करें। / For like surds, work with the coefficients.
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यदि \(x=\sqrt{3}+\sqrt{2}\), तो (\(x-\sqrt{3}\)\(x-\sqrt{2}\)) का मान क्या है?
If \(x=\sqrt{3}+\sqrt{2}\), what is the value of (\(x-\sqrt{3}\)\(x-\sqrt{2}\))?
#substitution
#surd product
#class 10
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A \(\sqrt{6}\)
B (1)
C (5)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{6}\)
Step 1
Concept
\(x-\sqrt{3}=\sqrt{2}\) और \(x-\sqrt{2}=\sqrt{3}\)। / \(x-\sqrt{3}=\sqrt{2}\) and \(x-\sqrt{2}=\sqrt{3}\).
Step 2
Why this answer is correct
उनका गुणन \(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) है। / Their product is \(\sqrt{2}\times\sqrt{3}=\sqrt{6}\).
Step 3
Exam Tip
पहले छोटे-छोटे कोष्ठकों को सरल करें। / Simplify the small brackets first.
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कौन-सा विकल्प \(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}\)?
#series of surds
#simplification
#class 10
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A \(10\sqrt{2}\)
B \(8\sqrt{2}\)
C \(12\sqrt{2}\)
D (60)
Explanation opens after your attempt
Correct Answer
A. \(10\sqrt{2}\)
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), और \(\sqrt{32}=4\sqrt{2}\)। / \(\sqrt{8}=2\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), and \(\sqrt{32}=4\sqrt{2}\).
Step 2
Why this answer is correct
कुल योग \(1\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}=10\sqrt{2}\) है। / The total is \(1\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}=10\sqrt{2}\).
Step 3
Exam Tip
क्रमबद्ध मूलों में गुणांक का पैटर्न पहचानें। / In ordered surds, identify the coefficient pattern.
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यदि \(x=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\), तो (x) का मान और प्रकृति क्या है?
If \(x=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\), what is the value and nature of (x)?
#conjugate surds
#rationalization
#irrational numbers
#class 10
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A (12), परिमेय / (12), rational
B \(6\sqrt{35}\), अपरिमेय / \(6\sqrt{35}\), irrational
C (24), परिमेय / (24), rational
D \(\sqrt{35}\), अपरिमेय / \(\sqrt{35}\), irrational
Explanation opens after your attempt
Correct Answer
A. (12), परिमेय / (12), rational
Step 1
Concept
पहले दोनों भिन्नों का साझा रूप देखें और \(a=\sqrt{7}+\sqrt{5}\) तथा \(b=\sqrt{7}-\sqrt{5}\) मानें। / First observe the common structure and take \(a=\sqrt{7}+\sqrt{5}\) and \(b=\sqrt{7}-\sqrt{5}\).
Step 2
Why this answer is correct
\(\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\) होगा। यहाँ \(a^2+b^2=24\) और (ab=2) इसलिए (x=12) है। / \(\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\). Here \(a^2+b^2=24\) and (ab=2), so (x=12).
Step 3
Exam Tip
संयुग्मी मूलों वाले भिन्नों में सीधे लंबा प्रसार करने के बजाय (a) और (b) रखकर हल करें। / For fractions with conjugate surds, use substitution instead of expanding everything directly.
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यदि \(x=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{10}-\sqrt{6}}\), तो (x) का सरल रूप क्या है?
If \(x=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{10}-\sqrt{6}}\), what is the simplified form of (x)?
#rationalization
#conjugate surds
#irrational numbers
#class 10
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A \(4+\sqrt{15}\)
B \(4-\sqrt{15}\)
C \(16+\sqrt{60}\)
D \(\sqrt{15}\)
Explanation opens after your attempt
Correct Answer
A. \(4+\sqrt{15}\)
Step 1
Concept
हर को परिमेय बनाने के लिए \(\sqrt{10}+\sqrt{6}\) से गुणा करें। / Multiply by \(\sqrt{10}+\sqrt{6}\) to rationalize the denominator.
Step 2
Why this answer is correct
ऊपर (\(\sqrt{10}+\sqrt{6}\)2 =16+2\sqrt{60}) और नीचे (10-6=4) मिलता है, इसलिए मान \(4+\sqrt{15}\) है। / The numerator becomes (\(\sqrt{10}+\sqrt{6}\)2 =16+2\sqrt{60}) and the denominator is (10-6=4), so the value is \(4+\sqrt{15}\).
Step 3
Exam Tip
संयुग्मी वाले भिन्नों में हर को पहले साफ करें। / In conjugate fractions, clear the denominator first.
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यदि \(x=\sqrt{5}+\sqrt{2}\) और \(y=\sqrt{5}-\sqrt{2}\), तो \(x^2+y^2\) का मान क्या है?
If \(x=\sqrt{5}+\sqrt{2}\) and \(y=\sqrt{5}-\sqrt{2}\), what is the value of \(x^2+y^2\)?
#conjugate surds
#square identity
#class 10
#expert
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A (14)
B (10)
C \(4\sqrt{10}\)
D \(7+2\sqrt{10}\)
Explanation opens after your attempt
Step 1
Concept
(x) और (y) संयुग्मी रूप में हैं। / (x) and (y) are conjugates.
Step 2
Why this answer is correct
(\(\sqrt{5}+\sqrt{2}\)2 +\(\sqrt{5}-\sqrt{2}\)2 ) में बीच के अपरिमेय पद कट जाते हैं और मान (2(5+2)=14) मिलता है। / In (\(\sqrt{5}+\sqrt{2}\)2 +\(\sqrt{5}-\sqrt{2}\)2 ), the middle irrational terms cancel and the value is (2(5+2)=14).
Step 3
Exam Tip
दो संयुग्मी वर्गों का योग लेते समय बीच वाले पद नहीं बचते। / When adding squares of conjugates, the middle terms vanish.
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कौन-सा विकल्प \(\frac{\sqrt{12}+\sqrt{27}}{\sqrt{3}}\) का सही मान देता है?
Which option gives the correct value of \(\frac{\sqrt{12}+\sqrt{27}}{\sqrt{3}}\)?
#surd simplification
#rational result
#class 10
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A (5)
B \(\sqrt{39}\)
C \(3\sqrt{3}\)
D (15)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\) लिखें। / Write \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\).
Step 2
Why this answer is correct
ऊपर का योग \(5\sqrt{3}\) है, इसलिए \(\frac{5\sqrt{3}}{\sqrt{3}}=5\)। / The numerator is \(5\sqrt{3}\), so \(\frac{5\sqrt{3}}{\sqrt{3}}=5\).
Step 3
Exam Tip
भाग से पहले समान मूल वाले पदों को जोड़ें। / Combine like surds before division.
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यदि \(\sqrt{m}+\sqrt{n}=7\) और (m,n) धनात्मक पूर्णांक हैं, तो कौन-सा युग्म निश्चित रूप से संभव है?
If \(\sqrt{m}+\sqrt{n}=7\) and (m,n) are positive integers, which pair is definitely possible?
#perfect squares
#square root sum
#real numbers
#class 10
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A (m=9,n=16)
B (m=8,n=25)
C (m=12,n=9)
D (m=7,n=36)
Explanation opens after your attempt
Correct Answer
A. (m=9,n=16)
Step 1
Concept
\(\sqrt{9}=3\) और \(\sqrt{16}=4\) हैं। / \(\sqrt{9}=3\) and \(\sqrt{16}=4\).
Step 2
Why this answer is correct
इनका योग (3+4=7) है, इसलिए यह युग्म शर्त पूरी करता है। / Their sum is (3+4=7), so this pair satisfies the condition.
Step 3
Exam Tip
पूर्णांक योग पाने के लिए पहले पूर्ण वर्ग वाले विकल्प जाँचें। / To get an integer sum, first check the perfect-square options.
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यदि \(x=2+\sqrt{3}\), तो \(x^2+\frac{1}{x^2}\) का मान क्या है?
If \(x=2+\sqrt{3}\), what is the value of \(x^2+\frac{1}{x^2}\)?
#reciprocal surds
#rationalization
#class 10
#expert
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A (14)
B (10)
C \(8\sqrt{3}\)
D (4)
Explanation opens after your attempt
Step 1
Concept
\(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\) होता है। / \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\).
Step 2
Why this answer is correct
इसलिए \(x+\frac{1}{x}=4\), अतः (x-2 +\frac{1}{x-2 }=(4)2 -2=14)। / Hence \(x+\frac{1}{x}=4\), so \(x^2+\frac{1}{x^2}=4^2-2=14\).
Step 3
Exam Tip
पहले \(x+\frac{1}{x}\) निकालना लंबी गणना बचाता है। / Finding \(x+\frac{1}{x}\) first saves long calculation.
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कौन-सा विकल्प \(0.303003000300003\ldots\) की प्रकृति सही बताता है?
Which option correctly describes the nature of \(0.303003000300003\ldots\)?
#irrational decimal
#non recurring
#decimal expansion
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A सांत परिमेय / Terminating rational
B असांत आवर्ती परिमेय / Non-terminating recurring rational
C असांत अनावर्ती अपरिमेय / Non-terminating non-recurring irrational
D ऋणात्मक पूर्णांक / Negative integer
Explanation opens after your attempt
Correct Answer
C. असांत अनावर्ती अपरिमेय / Non-terminating non-recurring irrational
Step 1
Concept
यह दशमलव समाप्त नहीं होता। / This decimal does not terminate.
Step 2
Why this answer is correct
इसमें शून्यों की संख्या बदलती रहती है, इसलिए कोई निश्चित आवर्ती समूह नहीं बनता। / The number of zeros keeps changing, so no fixed repeating block is formed.
Step 3
Exam Tip
असांत और अनावर्ती दशमलव अपरिमेय होता है। / A non-terminating non-recurring decimal is irrational.
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यदि \(a=\sqrt{6}+\sqrt{2}\) और \(b=\sqrt{6}-\sqrt{2}\), तो \(a^2-b^2\) का मान क्या है?
If \(a=\sqrt{6}+\sqrt{2}\) and \(b=\sqrt{6}-\sqrt{2}\), what is the value of \(a^2-b^2\)?
#algebraic identity
#surds
#irrational numbers
#class 10
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A \(8\sqrt{3}\)
B \(4\sqrt{3}\)
C (8)
D (16)
Explanation opens after your attempt
Correct Answer
A. \(8\sqrt{3}\)
Step 1
Concept
(a-2 -b-2 =(a-b)(a+b)) लगाएँ। / Use (a-2 -b-2 =(a-b)(a+b)).
Step 2
Why this answer is correct
\(a-b=2\sqrt{2}\) और \(a+b=2\sqrt{6}\), इसलिए गुणन \(4\sqrt{12}=8\sqrt{3}\) है। / \(a-b=2\sqrt{2}\) and \(a+b=2\sqrt{6}\), so the product is \(4\sqrt{12}=8\sqrt{3}\).
Step 3
Exam Tip
पहचान सूत्र से हल तेज और साफ होता है। / Identities make the solution quicker and cleaner.
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किस विकल्प में (x) अपरिमेय है, पर \(x+\frac{1}{x}\) परिमेय है?
In which option is (x) irrational but \(x+\frac{1}{x}\) rational?
#reciprocal
#conjugate surds
#rational sum
#class 10
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A \(x=3+\sqrt{8}\)
B \(x=\sqrt{5}\)
C \(x=1+\sqrt{3}\)
D \(x=2\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
A. \(x=3+\sqrt{8}\)
Step 1
Concept
\(3+\sqrt{8}=3+2\sqrt{2}\) अपरिमेय है। / \(3+\sqrt{8}=3+2\sqrt{2}\) is irrational.
Step 2
Why this answer is correct
इसका व्युत्क्रम \(3-\sqrt{8}\) है, क्योंकि (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=1)। इसलिए योग (6) परिमेय है। / Its reciprocal is \(3-\sqrt{8}\), because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=1). Hence the sum is (6), which is rational.
Step 3
Exam Tip
जिन संयुग्मियों का गुणन (1) हो, वहाँ व्युत्क्रम तुरंत मिल सकता है। / When conjugates multiply to (1), the reciprocal is easy to identify.
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कौन-सा विकल्प \(\sqrt{18}+\sqrt{50}-\sqrt{8}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{18}+\sqrt{50}-\sqrt{8}\)?
#surds
#simplification
#like radicals
#class 10
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A \(6\sqrt{2}\)
B \(5\sqrt{2}\)
C \(10\sqrt{2}\)
D \(\sqrt{60}\)
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{2}\)
Step 1
Concept
\(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), और \(\sqrt{8}=2\sqrt{2}\)। / \(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\).
Step 2
Why this answer is correct
\(3\sqrt{2}+5\sqrt{2}-2\sqrt{2}=6\sqrt{2}\)। / \(3\sqrt{2}+5\sqrt{2}-2\sqrt{2}=6\sqrt{2}\).
Step 3
Exam Tip
चिह्नों को ध्यान से रखकर गुणांक जोड़ें या घटाएँ। / Keep the signs carefully while adding or subtracting coefficients.
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यदि \(x=\sqrt{7}-\sqrt{3}\), तो \(x^2\) किसके बराबर है?
If \(x=\sqrt{7}-\sqrt{3}\), what is \(x^2\) equal to?
#square of surds
#algebraic identity
#class 10
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A \(10-2\sqrt{21}\)
B (4)
C \(10+2\sqrt{21}\)
D \(7-3\sqrt{21}\)
Explanation opens after your attempt
Correct Answer
A. \(10-2\sqrt{21}\)
Step 1
Concept
((a-b)2 =a-2 -2ab+b-2 ) का प्रयोग करें। / Use ((a-b)2 =a-2 -2ab+b-2 ).
Step 2
Why this answer is correct
\(x^2=7-2\sqrt{21}+3=10-2\sqrt{21}\)। / \(x^2=7-2\sqrt{21}+3=10-2\sqrt{21}\).
Step 3
Exam Tip
अंतर के वर्ग में बीच वाले पद का ऋण चिह्न न भूलें। / Do not forget the negative sign of the middle term in the square of a difference.
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कौन-सा विकल्प \(\sqrt{2}+\sqrt{8}+\sqrt{32}+\sqrt{128}\) का सरल रूप है?
Which option is the simplified form of \(\sqrt{2}+\sqrt{8}+\sqrt{32}+\sqrt{128}\)?
#series of surds
#pattern
#class 10
#expert
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A \(15\sqrt{2}\)
B \(14\sqrt{2}\)
C \(10\sqrt{2}\)
D (170)
Explanation opens after your attempt
Correct Answer
A. \(15\sqrt{2}\)
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), \(\sqrt{32}=4\sqrt{2}\), और \(\sqrt{128}=8\sqrt{2}\)। / \(\sqrt{8}=2\sqrt{2}\), \(\sqrt{32}=4\sqrt{2}\), and \(\sqrt{128}=8\sqrt{2}\).
Step 2
Why this answer is correct
योग ((1+2+4+8)\sqrt{2}=15\sqrt{2}) है। / The sum is ((1+2+4+8)\sqrt{2}=15\sqrt{2}).
Step 3
Exam Tip
गुणनखंडों में पूर्ण वर्गों का क्रम पहचानें। / Recognize the pattern of perfect-square factors.
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यदि (p) और (q) सहअभाज्य पूर्णांक हैं तथा \(\sqrt{5}=\frac{p}{q}\) मान लिया जाए, तो विरोध किस रूप में मिलता है?
If (p) and (q) are coprime integers and \(\sqrt{5}=\frac{p}{q}\) is assumed, in what form does the contradiction appear?
#irrationality proof
#coprime
#prime factor
#class 10
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A (p) और (q) दोनों (5) से विभाज्य निकलते हैं / Both (p) and (q) turn out divisible by (5)
B (p) और (q) दोनों (2) से विभाज्य निकलते हैं / Both (p) and (q) turn out divisible by (2)
C (p) और (q) दोनों शून्य हो जाते हैं / Both (p) and (q) become zero
D (p) और (q) दोनों परिमेय नहीं रहते / Both (p) and (q) stop being rational
Explanation opens after your attempt
Correct Answer
A. (p) और (q) दोनों (5) से विभाज्य निकलते हैं / Both (p) and (q) turn out divisible by (5)
Step 1
Concept
\(\sqrt{5}=\frac{p}{q}\) मानकर वर्ग करने से \(p^2=5q^2\) मिलता है। / Assuming \(\sqrt{5}=\frac{p}{q}\) and squaring gives \(p^2=5q^2\).
Step 2
Why this answer is correct
इससे (p) और फिर (q) दोनों (5) से विभाज्य निकलते हैं, जो सहअभाज्य होने के विरुद्ध है। / This makes both (p) and (q) divisible by (5), contradicting that they are coprime.
Step 3
Exam Tip
प्रमाण में साझा गुणनखंड मिलना ही मुख्य विरोध है। / Finding a common factor is the key contradiction in the proof.
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कौन-सा विकल्प \(4\sqrt{3}\) और \(3\sqrt{5}\) की सही तुलना करता है?
Which option correctly compares \(4\sqrt{3}\) and \(3\sqrt{5}\)?
#comparison of surds
#number sense
#class 10
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A \(4\sqrt{3}>3\sqrt{5}\)
B \(4\sqrt{3}<3\sqrt{5}\)
C दोनों बराबर हैं / Both are equal
D तुलना संभव नहीं है / Comparison is not possible
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{3}>3\sqrt{5}\)
Step 1
Concept
दोनों संख्याएँ धनात्मक हैं, इसलिए वर्ग करके तुलना करें। / Both numbers are positive, so compare their squares.
Step 2
Why this answer is correct
(\(4\sqrt{3}\)2 =48) और (\(3\sqrt{5}\)2 =45), इसलिए \(4\sqrt{3}\) बड़ा है। / (\(4\sqrt{3}\)2 =48) and (\(3\sqrt{5}\)2 =45), so \(4\sqrt{3}\) is greater.
Step 3
Exam Tip
धनात्मक मूलों की तुलना में वर्ग करना सुरक्षित रहता है। / Squaring is safe for comparing positive surds.
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यदि \(x=\sqrt{11}+\sqrt{7}\), तो \(x^2-18\) का मान क्या है?
If \(x=\sqrt{11}+\sqrt{7}\), what is the value of \(x^2-18\)?
#surd square
#irrational expression
#class 10
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A \(2\sqrt{77}\)
B \(\sqrt{77}\)
C (18)
D (77)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{77}\)
Step 1
Concept
\(x^2=11+7+2\sqrt{77}=18+2\sqrt{77}\)। / \(x^2=11+7+2\sqrt{77}=18+2\sqrt{77}\).
Step 2
Why this answer is correct
इसलिए \(x^2-18=2\sqrt{77}\), जो अपरिमेय है। / Therefore \(x^2-18=2\sqrt{77}\), which is irrational.
Step 3
Exam Tip
दो अलग मूलों के योग का वर्ग करते समय बीच वाला पद मुख्य होता है। / In the square of a sum of different surds, the middle term is the key.
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किस विकल्प में दो अपरिमेय संख्याओं का गुणनफल परिमेय लेकिन योग अपरिमेय है?
In which option do two irrational numbers have a rational product but an irrational sum?
#product of irrationals
#sum of irrationals
#class 10
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A \(\sqrt{12}\) और \(\sqrt{3}\) / \(\sqrt{12}\) and \(\sqrt{3}\)
B \(\sqrt{5}\) और \(-\sqrt{5}\) / \(\sqrt{5}\) and \(-\sqrt{5}\)
C \(\sqrt{2}\) और \(\sqrt{8}\) / \(\sqrt{2}\) and \(\sqrt{8}\)
D \(\sqrt{7}\) और \(\sqrt{28}\) / \(\sqrt{7}\) and \(\sqrt{28}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{12}\) और \(\sqrt{3}\) / \(\sqrt{12}\) and \(\sqrt{3}\)
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{3}\) दोनों अपरिमेय हैं। / \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{3}\) are both irrational.
Step 2
Why this answer is correct
उनका गुणन \(\sqrt{36}=6\) परिमेय है, और योग \(3\sqrt{3}\) अपरिमेय है। / Their product is \(\sqrt{36}=6\), which is rational, and their sum is \(3\sqrt{3}\), which is irrational.
Step 3
Exam Tip
योग और गुणन की प्रकृति अलग-अलग जाँचें। / Check the nature of the sum and product separately.
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यदि \(x=\frac{\sqrt{3}}{\sqrt{2}}\), तो \(x^2\) और (x) की प्रकृति के बारे में सही कथन कौन-सा है?
If \(x=\frac{\sqrt{3}}{\sqrt{2}}\), which statement about \(x^2\) and (x) is correct?
#square of irrational
#quotient of surds
#class 10
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A (x) अपरिमेय है और \(x^2\) परिमेय है / (x) is irrational and \(x^2\) is rational
B (x) परिमेय है और \(x^2\) परिमेय है / (x) is rational and \(x^2\) is rational
C (x) अपरिमेय है और \(x^2\) अपरिमेय है / (x) is irrational and \(x^2\) is irrational
D (x=0) है / (x=0)
Explanation opens after your attempt
Correct Answer
A. (x) अपरिमेय है और \(x^2\) परिमेय है / (x) is irrational and \(x^2\) is rational
Step 1
Concept
\(x=\sqrt{\frac{3}{2}}\) है, जो अपरिमेय है क्योंकि \(\frac{3}{2}\) परिमेय पूर्ण वर्ग नहीं है। / \(x=\sqrt{\frac{3}{2}}\), which is irrational because \(\frac{3}{2}\) is not a perfect square of a rational number.
Step 2
Why this answer is correct
\(x^2=\frac{3}{2}\), जो परिमेय है। / \(x^2=\frac{3}{2}\), which is rational.
Step 3
Exam Tip
किसी अपरिमेय संख्या का वर्ग कभी-कभी परिमेय हो सकता है। / The square of an irrational number can sometimes be rational.
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कौन-सा विकल्प \(\frac{1}{\sqrt{5}+\sqrt{2}}\) का परिमेय हर वाला रूप है?
Which option is the rationalized form of \(\frac{1}{\sqrt{5}+\sqrt{2}}\)?
#rationalization
#denominator
#conjugate surds
#class 10
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A \(\frac{\sqrt{5}-\sqrt{2}}{3}\)
B \(\frac{\sqrt{5}+\sqrt{2}}{3}\)
C \(\sqrt{5}-\sqrt{2}\)
D \(\frac{1}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{\sqrt{5}-\sqrt{2}}{3}\)
Step 1
Concept
हर का संयुग्मी \(\sqrt{5}-\sqrt{2}\) है। / The conjugate of the denominator is \(\sqrt{5}-\sqrt{2}\).
Step 2
Why this answer is correct
हर (5-2=3) बनता है, इसलिए रूप \(\frac{\sqrt{5}-\sqrt{2}}{3}\) है। / The denominator becomes (5-2=3), so the form is \(\frac{\sqrt{5}-\sqrt{2}}{3}\).
Step 3
Exam Tip
दो मूलों के योग में संयुग्मी का चिह्न बदलता है। / For a sum of two surds, the conjugate changes the sign between them.
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यदि \(a=\sqrt{2}+\sqrt{3}+\sqrt{5}\), तो \(a^2\) में कौन-सा अपरिमेय पद अवश्य आएगा?
If \(a=\sqrt{2}+\sqrt{3}+\sqrt{5}\), which irrational term must appear in \(a^2\)?
#square of multiple surds
#irrational terms
#class 10
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A \(2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\)
B (2+3+5)
C \(\sqrt{30}\)
D (10)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\)
Step 1
Concept
तीन पदों के वर्ग में अलग-अलग वर्गों के साथ दो-दो पदों के गुणन भी आते हैं। / In the square of three terms, pairwise products appear along with individual squares.
Step 2
Why this answer is correct
इसलिए \(a^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\) होगा। / Thus \(a^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\).
Step 3
Exam Tip
कई मूलों के योग का वर्ग करते समय सभी जोड़ीदार गुणन लिखें। / While squaring a sum of many surds, write all pairwise products.
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किस विकल्प में \(\sqrt{a}+\sqrt{b}\) परिमेय है?
In which option is \(\sqrt{a}+\sqrt{b}\) rational?
#perfect squares
#rational sum
#class 10
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A (a=20,b=45)
B (a=25,b=49)
C (a=18,b=50)
D (a=12,b=27)
Explanation opens after your attempt
Correct Answer
B. (a=25,b=49)
Step 1
Concept
(25) और (49) दोनों पूर्ण वर्ग हैं। / (25) and (49) are both perfect squares.
Step 2
Why this answer is correct
\(\sqrt{25}+\sqrt{49}=5+7=12\), जो परिमेय है। / \(\sqrt{25}+\sqrt{49}=5+7=12\), which is rational.
Step 3
Exam Tip
परिमेय योग के लिए दोनों वर्गमूलों को अलग-अलग जाँचें। / For a rational sum, check both square roots separately.
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यदि \(x=\sqrt{3}+\sqrt{2}\), तो \(x^4-10x^2+1\) का मान क्या है?
If \(x=\sqrt{3}+\sqrt{2}\), what is the value of \(x^4-10x^2+1\)?
#advanced surd identity
#algebra
#class 10
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A (0)
B (1)
C (10)
D \(4\sqrt{6}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=5+2\sqrt{6}\) है। / \(x^2=5+2\sqrt{6}\).
Step 2
Why this answer is correct
\(x^2+\frac{1}{x^2}=10\) की पहचान से \(x^4-10x^2+1=0\) मिलता है। / Using the identity \(x^2+\frac{1}{x^2}=10\), we get \(x^4-10x^2+1=0\).
Step 3
Exam Tip
ऐसे प्रश्नों में (x) और उसके संयुग्मी व्युत्क्रम का संबंध पहचानें। / In such questions, recognize the relation between (x) and its conjugate reciprocal.
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कौन-सा विकल्प \(\sqrt{a}\times\sqrt{b}\) को अपरिमेय बनाता है?
Which option makes \(\sqrt{a}\times\sqrt{b}\) irrational?
#product of radicals
#irrational result
#class 10
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A (a=2,b=18)
B (a=3,b=12)
C (a=5,b=20)
D (a=6,b=15)
Explanation opens after your attempt
Correct Answer
D. (a=6,b=15)
Step 1
Concept
\(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) होता है। / \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\).
Step 2
Why this answer is correct
(a=6,b=15) पर (ab=90), जो पूर्ण वर्ग नहीं है, इसलिए \(\sqrt{90}\) अपरिमेय है। / For (a=6,b=15), (ab=90), which is not a perfect square, so \(\sqrt{90}\) is irrational.
Step 3
Exam Tip
गुणन में अंदर का गुणनफल पूर्ण वर्ग है या नहीं, यह मुख्य जाँच है। / In multiplication, the key check is whether the product inside the root is a perfect square.
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यदि \(x=5-\sqrt{24}\), तो \(\frac{1}{x}\) का सही रूप कौन-सा है?
If \(x=5-\sqrt{24}\), which is the correct form of \(\frac{1}{x}\)?
#reciprocal
#conjugate surds
#rationalization
#class 10
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A \(5+\sqrt{24}\)
B \(5-\sqrt{24}\)
C \(\frac{5+\sqrt{24}}{49}\)
D \(\frac{5-\sqrt{24}}{25}\)
Explanation opens after your attempt
Correct Answer
A. \(5+\sqrt{24}\)
Step 1
Concept
(\(5-\sqrt{24}\)\(5+\sqrt{24}\)=25-24=1)। / (\(5-\sqrt{24}\)\(5+\sqrt{24}\)=25-24=1).
Step 2
Why this answer is correct
इसलिए \(5+\sqrt{24}\), \(5-\sqrt{24}\) का व्युत्क्रम है। / Therefore \(5+\sqrt{24}\) is the reciprocal of \(5-\sqrt{24}\).
Step 3
Exam Tip
यदि संयुग्मी गुणन (1) दे, तो व्युत्क्रम सीधे संयुग्मी होता है। / If conjugates multiply to (1), the reciprocal is directly the conjugate.
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किस विकल्प में दिया गया कथन सही है?
Which given statement is correct?
#properties of irrational numbers
#closure
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A हर अपरिमेय संख्या का वर्ग अपरिमेय होता है / The square of every irrational number is irrational
B दो अपरिमेय संख्याओं का योग कभी परिमेय नहीं हो सकता / The sum of two irrational numbers can never be rational
C अशून्य परिमेय संख्या से अपरिमेय संख्या को गुणा करने पर परिणाम अपरिमेय होता है / Multiplying an irrational number by a non-zero rational number gives an irrational number
D दो अपरिमेय संख्याओं का भागफल हमेशा अपरिमेय होता है / The quotient of two irrational numbers is always irrational
Explanation opens after your attempt
Correct Answer
C. अशून्य परिमेय संख्या से अपरिमेय संख्या को गुणा करने पर परिणाम अपरिमेय होता है / Multiplying an irrational number by a non-zero rational number gives an irrational number
Step 1
Concept
अशून्य परिमेय गुणक अपरिमेयता को समाप्त नहीं करता। / A non-zero rational multiplier does not remove irrationality.
Step 2
Why this answer is correct
यदि परिणाम परिमेय मानें, तो अपरिमेय संख्या परिमेय बन जाएगी, जो विरोध है। / If the product were rational, the irrational number would become rational, a contradiction.
Step 3
Exam Tip
दो अपरिमेय संख्याओं वाले हमेशा वाले कथनों से सावधान रहें। / Be careful with universal statements about two irrational numbers.
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यदि \(x=\sqrt{8}+\sqrt{18}\), तो \(\frac{x}{\sqrt{2}}\) का मान क्या है?
If \(x=\sqrt{8}+\sqrt{18}\), what is the value of \(\frac{x}{\sqrt{2}}\)?
#surd simplification
#division
#class 10
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A (5)
B (6)
C (10)
D \(\sqrt{26}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\)। / \(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\).
Step 2
Why this answer is correct
\(x=5\sqrt{2}\), इसलिए \(\frac{x}{\sqrt{2}}=5\)। / \(x=5\sqrt{2}\), so \(\frac{x}{\sqrt{2}}=5\).
Step 3
Exam Tip
समान मूल वाले पदों को जोड़ने के बाद भाग देना आसान होता है। / Division is easier after combining like surds.
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कौन-सा विकल्प \(\sqrt{3}+\sqrt{6}\) और \(\sqrt{12}\) के बीच सही तुलना देता है?
Which option gives the correct comparison between \(\sqrt{3}+\sqrt{6}\) and \(\sqrt{12}\)?
#comparison of irrationals
#surds
#class 10
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A \(\sqrt{3}+\sqrt{6}>\sqrt{12}\)
B \(\sqrt{3}+\sqrt{6}<\sqrt{12}\)
C दोनों बराबर हैं / Both are equal
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{3}+\sqrt{6}>\sqrt{12}\)
Step 1
Concept
सभी पद धनात्मक हैं और \(\sqrt{6}>0\)। / All terms are positive and \(\sqrt{6}>0\).
Step 2
Why this answer is correct
\(\sqrt{3}+\sqrt{6}\), \(\sqrt{3}\) से बड़ा है और \(\sqrt{12}=2\sqrt{3}\) है; संख्यात्मक रूप से \(\sqrt{6}>\sqrt{3}\), इसलिए योग \(2\sqrt{3}\) से बड़ा है। / Since \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{6}>\sqrt{3}\), the sum \(\sqrt{3}+\sqrt{6}\) is greater than \(2\sqrt{3}\).
Step 3
Exam Tip
तुलना में समान मूल में बदलना और धनात्मकता देखना मदद करता है। / For comparison, convert what you can and use positivity.
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यदि \(a=3+\sqrt{5}\), तो \(a^2-6a\) का मान क्या है?
If \(a=3+\sqrt{5}\), what is the value of \(a^2-6a\)?
#hidden conjugate
#algebraic surds
#class 10
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A (-4)
B (4)
C (5)
D \(6\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
(a-2 -6a=a(a-6)) है। / (a-2 -6a=a(a-6)).
Step 2
Why this answer is correct
\(a-6=\sqrt{5}-3\), इसलिए (a(a-6)=\(3+\sqrt{5}\)\(\sqrt{5}-3\)=5-9=-4)। / \(a-6=\sqrt{5}-3\), so (a(a-6)=\(3+\sqrt{5}\)\(\sqrt{5}-3\)=5-9=-4).
Step 3
Exam Tip
छिपे हुए संयुग्मी रूप को पहचानें। / Recognize the hidden conjugate form.
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कौन-सा विकल्प \(1.202002000200002\ldots\) के लिए सही है?
Which option is correct for \(1.202002000200002\ldots\)?
#decimal expansion
#non recurring
#class 10
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A यह परिमेय है क्योंकि (2) बार-बार आता है / It is rational because (2) appears repeatedly
B यह अपरिमेय है क्योंकि दशमलव असांत और अनावर्ती है / It is irrational because the decimal is non-terminating and non-recurring
C यह सांत दशमलव है / It is a terminating decimal
D यह पूर्ण वर्ग है / It is a perfect square
Explanation opens after your attempt
Correct Answer
B. यह अपरिमेय है क्योंकि दशमलव असांत और अनावर्ती है / It is irrational because the decimal is non-terminating and non-recurring
Step 1
Concept
केवल एक अंक का बार-बार आना आवर्तन नहीं कहलाता, जब तक निश्चित समूह न दोहराए। / Reappearance of a digit alone is not recurrence unless a fixed block repeats.
Step 2
Why this answer is correct
यहाँ शून्यों की संख्या बदल रही है, इसलिए दशमलव अनावर्ती है। / Here the number of zeros changes, so the decimal is non-recurring.
Step 3
Exam Tip
असांत अनावर्ती दशमलव को अपरिमेय पहचानें। / Identify a non-terminating non-recurring decimal as irrational.
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यदि \(x=\sqrt{13}+\sqrt{12}\), तो (x\cdot\(\sqrt{13}-\sqrt{12}\)) का मान क्या है?
If \(x=\sqrt{13}+\sqrt{12}\), what is the value of (x\cdot\(\sqrt{13}-\sqrt{12}\))?
#conjugate product
#difference of squares
#class 10
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A (1)
B (25)
C \(\sqrt{156}\)
D \(2\sqrt{13}\)
Explanation opens after your attempt
Step 1
Concept
यह संयुग्मी गुणन है। / This is a conjugate product.
Step 2
Why this answer is correct
(\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=13-12=1)। / (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=13-12=1).
Step 3
Exam Tip
ऐसे रूपों में विस्तार करने से पहले अंतर के वर्ग को पहचानें। / In such forms, identify the difference of squares before expanding.
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किस विकल्प में \(\sqrt{a}-\sqrt{b}\) परिमेय है, जबकि दोनों वर्गमूल अलग-अलग हैं?
In which option is \(\sqrt{a}-\sqrt{b}\) rational while the two square roots are different?
#rational difference
#perfect squares
#class 10
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A (a=25,b=9)
B (a=18,b=8)
C (a=20,b=5)
D (a=27,b=12)
Explanation opens after your attempt
Correct Answer
A. (a=25,b=9)
Step 1
Concept
\(\sqrt{25}=5\) और \(\sqrt{9}=3\) हैं। / \(\sqrt{25}=5\) and \(\sqrt{9}=3\).
Step 2
Why this answer is correct
अंतर (5-3=2) परिमेय है। / The difference is (5-3=2), which is rational.
Step 3
Exam Tip
परिमेय अंतर पाने का सरल तरीका पूर्ण वर्गों के वर्गमूल लेना है। / A simple way to get a rational difference is to use square roots of perfect squares.
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यदि \(x=\sqrt{2}+\sqrt{3}\), तो \(x^2-2\sqrt{6}\) का मान क्या है?
If \(x=\sqrt{2}+\sqrt{3}\), what is the value of \(x^2-2\sqrt{6}\)?
#surd square
#cancellation
#class 10
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A (5)
B (1)
C \(\sqrt{6}\)
D (2)
Explanation opens after your attempt
Step 1
Concept
(x-2 =\(\sqrt{2}+\sqrt{3}\)2 =5+2\sqrt{6})। / (x-2 =\(\sqrt{2}+\sqrt{3}\)2 =5+2\sqrt{6}).
Step 2
Why this answer is correct
इसमें से \(2\sqrt{6}\) घटाने पर (5) बचता है। / Subtracting \(2\sqrt{6}\) leaves (5).
Step 3
Exam Tip
वर्ग करने के बाद समान अपरिमेय पदों को काटें। / After squaring, cancel like irrational terms.
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कौन-सा विकल्प \(\frac{\sqrt{75}-\sqrt{12}}{\sqrt{3}}\) का सही मान है?
Which option is the correct value of \(\frac{\sqrt{75}-\sqrt{12}}{\sqrt{3}}\)?
#surd division
#simplification
#class 10
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A (3)
B (5)
C \(\sqrt{63}\)
D (7)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\) हैं। / \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\).
Step 2
Why this answer is correct
ऊपर का अंतर \(3\sqrt{3}\) है, इसलिए भाग देने पर (3) मिलता है। / The numerator becomes \(3\sqrt{3}\), so division gives (3).
Step 3
Exam Tip
घटाव के बाद ही हर से भाग दें। / Subtract first, then divide by the denominator.
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यदि \(x=\sqrt{5}-2\), तो \(x+\frac{1}{x}\) का मान क्या है?
If \(x=\sqrt{5}-2\), what is the value of \(x+\frac{1}{x}\)?
#reciprocal
#surd conjugate
#class 10
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A \(2\sqrt{5}\)
B \(\sqrt{5}\)
C (4)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{5}\)
Step 1
Concept
\(\frac{1}{\sqrt{5}-2}=\sqrt{5}+2\), क्योंकि (\(\sqrt{5}-2\)\(\sqrt{5}+2\)=1)। / \(\frac{1}{\sqrt{5}-2}=\sqrt{5}+2\), because (\(\sqrt{5}-2\)\(\sqrt{5}+2\)=1).
Step 2
Why this answer is correct
इसलिए (x+\frac{1}{x}=\(\sqrt{5}-2\)+\(\sqrt{5}+2\)=2\sqrt{5})। / Hence (x+\frac{1}{x}=\(\sqrt{5}-2\)+\(\sqrt{5}+2\)=2\sqrt{5}).
Step 3
Exam Tip
जहाँ संयुग्मी गुणन (1) दे, वहाँ व्युत्क्रम तुरंत मिल जाता है। / When conjugates multiply to (1), the reciprocal is immediate.
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कौन-सा विकल्प (2) और (3) के बीच दो अपरिमेय संख्याओं का सही युग्म है?
Which option is a correct pair of two irrational numbers between (2) and (3)?
#irrational between integers
#number line
#class 10
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A \(\sqrt{5},\sqrt{8}\)
B \(\sqrt{4},\sqrt{9}\)
C \(\frac{5}{2},\sqrt{6}\)
D \(\sqrt{10},\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{5},\sqrt{8}\)
Step 1
Concept
\(2=\sqrt{4}\) और \(3=\sqrt{9}\) हैं। / \(2=\sqrt{4}\) and \(3=\sqrt{9}\).
Step 2
Why this answer is correct
(5) और (8) दोनों (4) और (9) के बीच हैं तथा पूर्ण वर्ग नहीं हैं। इसलिए \(\sqrt{5}\) और \(\sqrt{8}\) दोनों अपरिमेय हैं और (2) से (3) के बीच हैं। / (5) and (8) lie between (4) and (9) and are not perfect squares. Therefore \(\sqrt{5}\) and \(\sqrt{8}\) are irrational numbers between (2) and (3).
Step 3
Exam Tip
बीच के अपूर्ण वर्गों से ऐसे युग्म बनते हैं। / Non-perfect squares between two square numbers give such pairs.
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यदि \(a=\sqrt{3}+\sqrt{2}\) और \(b=\sqrt{3}-\sqrt{2}\), तो \(\frac{a-b}{a+b}\) का मान क्या है?
If \(a=\sqrt{3}+\sqrt{2}\) and \(b=\sqrt{3}-\sqrt{2}\), what is the value of \(\frac{a-b}{a+b}\)?
#surd ratio
#rationalization
#class 10
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A \(\frac{\sqrt{6}}{3}\)
B \(\frac{\sqrt{2}}{\sqrt{3}}\)
C \(\sqrt{6}\)
D \(\frac{1}{\sqrt{6}}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{\sqrt{6}}{3}\)
Step 1
Concept
\(a-b=2\sqrt{2}\) और \(a+b=2\sqrt{3}\)। / \(a-b=2\sqrt{2}\) and \(a+b=2\sqrt{3}\).
Step 2
Why this answer is correct
\(\frac{a-b}{a+b}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\)। / \(\frac{a-b}{a+b}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\).
Step 3
Exam Tip
अंत में हर को परिमेय बनाना न भूलें। / Do not forget to rationalize the denominator at the end.
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किस विकल्प में \(\sqrt{a}+\sqrt{b}\) अपरिमेय है, पर (\(\sqrt{a}+\sqrt{b}\)\(\sqrt{a}-\sqrt{b}\)) परिमेय है?
In which option is \(\sqrt{a}+\sqrt{b}\) irrational but (\(\sqrt{a}+\sqrt{b}\)\(\sqrt{a}-\sqrt{b}\)) rational?
#conjugate product
#irrational sum
#class 10
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A (a=7,b=2)
B (a=9,b=4)
C (a=16,b=25)
D (a=36,b=49)
Explanation opens after your attempt
Correct Answer
A. (a=7,b=2)
Step 1
Concept
(a=7,b=2) पर \(\sqrt{7}+\sqrt{2}\) अपरिमेय है। / For (a=7,b=2), \(\sqrt{7}+\sqrt{2}\) is irrational.
Step 2
Why this answer is correct
गुणन (\(\sqrt{7}\)2 -\(\sqrt{2}\)2 =7-2=5) परिमेय है। / The product is (\(\sqrt{7}\)2 -\(\sqrt{2}\)2 =7-2=5), which is rational.
Step 3
Exam Tip
संयुग्मी गुणन अपरिमेय योग को भी परिमेय गुणनफल दे सकता है। / A conjugate product can give a rational result even when the sum is irrational.
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यदि \(x=\sqrt{2}+\sqrt{5}\), तो (\(x-\sqrt{2}\)2 ) का मान क्या है?
If \(x=\sqrt{2}+\sqrt{5}\), what is the value of (\(x-\sqrt{2}\)2 )?
#substitution
#surd square
#class 10
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A (5)
B (2)
C \(\sqrt{5}\)
D \(7+2\sqrt{10}\)
Explanation opens after your attempt
Step 1
Concept
\(x-\sqrt{2}=\sqrt{5}\) है। / \(x-\sqrt{2}=\sqrt{5}\).
Step 2
Why this answer is correct
इसलिए (\(x-\sqrt{2}\)2 =\(\sqrt{5}\)2 =5)। / Therefore (\(x-\sqrt{2}\)2 =\(\sqrt{5}\)2 =5).
Step 3
Exam Tip
पूरे वर्ग को फैलाने से पहले कोष्ठक के अंदर सरल करें। / Simplify inside the bracket before expanding the square.
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कौन-सा विकल्प \(\sqrt{2}+\sqrt{3}\) के व्युत्क्रम को सही बताता है?
Which option correctly gives the reciprocal of \(\sqrt{2}+\sqrt{3}\)?
#reciprocal of surd
#conjugate
#class 10
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A \(\sqrt{3}-\sqrt{2}\)
B \(\sqrt{2}-\sqrt{3}\)
C \(\frac{\sqrt{2}+\sqrt{3}}{5}\)
D \(\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{3}-\sqrt{2}\)
Step 1
Concept
(\(\sqrt{2}+\sqrt{3}\)\(\sqrt{3}-\sqrt{2}\)=3-2=1)। / (\(\sqrt{2}+\sqrt{3}\)\(\sqrt{3}-\sqrt{2}\)=3-2=1).
Step 2
Why this answer is correct
इसलिए \(\sqrt{3}-\sqrt{2}\) इसका व्युत्क्रम है। / Therefore \(\sqrt{3}-\sqrt{2}\) is its reciprocal.
Step 3
Exam Tip
व्युत्क्रम में संयुग्मी का क्रम और चिह्न सावधानी से रखें। / In reciprocals, keep the order and sign of the conjugate carefully.
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यदि \(x=1+\sqrt{2}+\sqrt{3}\), तो (x-1) की प्रकृति क्या है?
If \(x=1+\sqrt{2}+\sqrt{3}\), what is the nature of (x-1)?
#sum of surds
#irrationality
#class 10
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A अपरिमेय / Irrational
B परिमेय / Rational
C पूर्णांक / Integer
D सांत दशमलव / Terminating decimal
Explanation opens after your attempt
Correct Answer
A. अपरिमेय / Irrational
Step 1
Concept
\(x-1=\sqrt{2}+\sqrt{3}\) है। / \(x-1=\sqrt{2}+\sqrt{3}\).
Step 2
Why this answer is correct
\(\sqrt{2}+\sqrt{3}\) अपरिमेय है, क्योंकि इसे परिमेय मानने पर वर्ग करने से \(\sqrt{6}\) परिमेय निकलने का विरोध आता है। / \(\sqrt{2}+\sqrt{3}\) is irrational, because assuming it rational and squaring would force \(\sqrt{6}\) to be rational.
Step 3
Exam Tip
अलग-अलग मूलों का योग सावधानी से जाँचें। / Check sums of different surds carefully.
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कौन-सा विकल्प \(\sqrt{96}-\sqrt{54}+\sqrt{24}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{96}-\sqrt{54}+\sqrt{24}\)?
#surd simplification
#like radicals
#class 10
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A \(5\sqrt{6}\)
B \(4\sqrt{6}\)
C \(3\sqrt{6}\)
D \(\sqrt{66}\)
Explanation opens after your attempt
Correct Answer
A. \(5\sqrt{6}\)
Step 1
Concept
\(\sqrt{96}=4\sqrt{6}\), \(\sqrt{54}=3\sqrt{6}\), और \(\sqrt{24}=2\sqrt{6}\)। / \(\sqrt{96}=4\sqrt{6}\), \(\sqrt{54}=3\sqrt{6}\), and \(\sqrt{24}=2\sqrt{6}\).
Step 2
Why this answer is correct
\(4\sqrt{6}-3\sqrt{6}+2\sqrt{6}=3\sqrt{6}\), इसलिए सही मान \(3\sqrt{6}\) है। / \(4\sqrt{6}-3\sqrt{6}+2\sqrt{6}=3\sqrt{6}\), so the correct value is \(3\sqrt{6}\).
Step 3
Exam Tip
विकल्प मिलाते समय अपनी सरल गणना से मिलान करें। / Match the options with your simplified result carefully.
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यदि \(x=\sqrt{3}+\sqrt{5}\) और \(y=\sqrt{5}+\sqrt{7}\), तो (y-x) की प्रकृति क्या है?
If \(x=\sqrt{3}+\sqrt{5}\) and \(y=\sqrt{5}+\sqrt{7}\), what is the nature of (y-x)?
#surd subtraction
#irrational result
#class 10
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A अपरिमेय / Irrational
B परिमेय / Rational
C पूर्णांक / Integer
D शून्य / Zero
Explanation opens after your attempt
Correct Answer
A. अपरिमेय / Irrational
Step 1
Concept
(y-x=\(\sqrt{5}+\sqrt{7}\)-\(\sqrt{3}+\sqrt{5}\))। / (y-x=\(\sqrt{5}+\sqrt{7}\)-\(\sqrt{3}+\sqrt{5}\)).
Step 2
Why this answer is correct
\(\sqrt{5}\) कट जाता है और \(\sqrt{7}-\sqrt{3}\) बचता है, जो अपरिमेय है। / \(\sqrt{5}\) cancels and \(\sqrt{7}-\sqrt{3}\) remains, which is irrational.
Step 3
Exam Tip
समान पद कटने के बाद बचे हुए मूलों की प्रकृति देखें। / After like terms cancel, check the nature of the remaining surds.
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कौन-सा विकल्प \(3+\sqrt{2}\) और \(3-\sqrt{2}\) के बारे में सही है?
Which option is correct about \(3+\sqrt{2}\) and \(3-\sqrt{2}\)?
#conjugate irrationals
#rational product
#class 10
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A दोनों अपरिमेय हैं और उनका गुणनफल परिमेय है / Both are irrational and their product is rational
B दोनों परिमेय हैं और उनका योग अपरिमेय है / Both are rational and their sum is irrational
C एक परिमेय है और एक अपरिमेय है / One is rational and one is irrational
D दोनों का गुणनफल अपरिमेय है / Their product is irrational
Explanation opens after your attempt
Correct Answer
A. दोनों अपरिमेय हैं और उनका गुणनफल परिमेय है / Both are irrational and their product is rational
Step 1
Concept
\(3+\sqrt{2}\) और \(3-\sqrt{2}\) दोनों में अपरिमेय भाग है, इसलिए दोनों अपरिमेय हैं। / \(3+\sqrt{2}\) and \(3-\sqrt{2}\) both contain an irrational part, so both are irrational.
Step 2
Why this answer is correct
उनका गुणनफल (9-2=7) परिमेय है। / Their product is (9-2=7), which is rational.
Step 3
Exam Tip
संयुग्मी अपरिमेय संख्याओं का गुणनफल परिमेय हो सकता है। / Conjugate irrational numbers can have a rational product.
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यदि \(x=\sqrt{2}\), तो \(\frac{x^6-8}{x^2-2}\) के बारे में सही कथन कौन-सा है?
If \(x=\sqrt{2}\), which statement about \(\frac{x^6-8}{x^2-2}\) is correct?
#surd substitution
#undefined fraction
#class 10
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A यह अपरिभाषित है / It is undefined
B यह (0) है / It is (0)
C यह (8) है / It is (8)
D यह \(\sqrt{2}\) है / It is \(\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. यह अपरिभाषित है / It is undefined
Step 1
Concept
\(x=\sqrt{2}\) होने पर \(x^2=2\)। / For \(x=\sqrt{2}\), \(x^2=2\).
Step 2
Why this answer is correct
हर \(x^2-2=0\) हो जाता है, इसलिए भिन्न अपरिभाषित है। / The denominator \(x^2-2=0\), so the fraction is undefined.
Step 3
Exam Tip
भिन्न का मान निकालने से पहले हर शून्य तो नहीं, यह जरूर जाँचें। / Before evaluating a fraction, always check whether the denominator becomes zero.
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कौन-सा विकल्प \(\sqrt{45}\) और \(2\sqrt{5}\) के अंतर को सही बताता है?
Which option correctly gives the difference between \(\sqrt{45}\) and \(2\sqrt{5}\)?
#surd subtraction
#irrational number
#class 10
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A \(\sqrt{5}\)
B \(3\sqrt{5}\)
C \(5\sqrt{5}\)
D (7)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{5}\)
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\) है। / \(\sqrt{45}=3\sqrt{5}\).
Step 2
Why this answer is correct
अंतर \(3\sqrt{5}-2\sqrt{5}=\sqrt{5}\), जो अपरिमेय है। / The difference is \(3\sqrt{5}-2\sqrt{5}=\sqrt{5}\), which is irrational.
Step 3
Exam Tip
समान मूल वाले पदों में केवल गुणांक घटाएँ। / For like surds, subtract only the coefficients.
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यदि \(x=\sqrt{5}+\sqrt{2}\), तो (\(x^2-7\)2 ) का मान क्या है?
If \(x=\sqrt{5}+\sqrt{2}\), what is the value of (\(x^2-7\)2 )?
#surd square
#algebraic expression
#class 10
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A (40)
B (20)
C (10)
D \(4\sqrt{10}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=5+2+2\sqrt{10}=7+2\sqrt{10}\)। / \(x^2=5+2+2\sqrt{10}=7+2\sqrt{10}\).
Step 2
Why this answer is correct
इसलिए \(x^2-7=2\sqrt{10}\), और इसका वर्ग (40) है। / Thus \(x^2-7=2\sqrt{10}\), and its square is (40).
Step 3
Exam Tip
पहले परिमेय भाग अलग करें, फिर वर्ग करें। / First isolate the irrational part, then square it.
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किस विकल्प में \(\sqrt{p}\) अपरिमेय है और (p) अभाज्य भी है?
In which option is \(\sqrt{p}\) irrational and (p) also prime?
#prime number
#irrational square root
#class 10
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A (p=11)
B (p=16)
C (p=21)
D (p=25)
Explanation opens after your attempt
Step 1
Concept
(11) अभाज्य संख्या है। / (11) is a prime number.
Step 2
Why this answer is correct
कोई अभाज्य संख्या पूर्ण वर्ग नहीं होती, इसलिए \(\sqrt{11}\) अपरिमेय है। / A prime number is not a perfect square, so \(\sqrt{11}\) is irrational.
Step 3
Exam Tip
अभाज्य संख्या के वर्गमूल पर सीधे अपूर्ण वर्ग का विचार लगाएँ। / For the square root of a prime, use the non-perfect-square idea directly.
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यदि \(x=4+\sqrt{15}\) और \(y=4-\sqrt{15}\), तो \(x^2+y^2\) का मान क्या है?
If \(x=4+\sqrt{15}\) and \(y=4-\sqrt{15}\), what is the value of \(x^2+y^2\)?
#conjugate squares
#rational result
#class 10
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A (62)
B (2)
C \(32\sqrt{15}\)
D (31)
Explanation opens after your attempt
Step 1
Concept
(x) और (y) संयुग्मी हैं। / (x) and (y) are conjugates.
Step 2
Why this answer is correct
(x-2 +y-2 =2\(4^2+15\)=2(31)=62)। / (x-2 +y-2 =2\(4^2+15\)=2(31)=62).
Step 3
Exam Tip
संयुग्मी वर्गों के योग में अपरिमेय पद कट जाते हैं। / In the sum of squares of conjugates, irrational terms cancel.
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कौन-सा विकल्प \(\sqrt{2}+\sqrt{3}+\sqrt{6}\) के परिमेय होने के दावे को गलत दिखाने में मदद करता है?
Which option helps show that the claim \(\sqrt{2}+\sqrt{3}+\sqrt{6}\) is rational is false?
#irrational sum
#conceptual proof
#class 10
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A विभिन्न अपूर्ण वर्गों के मूल स्वतंत्र अपरिमेय भाग देते हैं / Roots of different non-perfect squares give independent irrational parts
B सभी वर्गमूल हमेशा परिमेय होते हैं / All square roots are always rational
C तीन अपरिमेय संख्याओं का योग हमेशा परिमेय होता है / The sum of three irrational numbers is always rational
D \(\sqrt{6}=\sqrt{2}+\sqrt{3}\) होता है / \(\sqrt{6}=\sqrt{2}+\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. विभिन्न अपूर्ण वर्गों के मूल स्वतंत्र अपरिमेय भाग देते हैं / Roots of different non-perfect squares give independent irrational parts
Step 1
Concept
\(\sqrt{2}\), \(\sqrt{3}\), और \(\sqrt{6}\) अलग-अलग अपूर्ण वर्गों से जुड़े हैं। / \(\sqrt{2}\), \(\sqrt{3}\), and \(\sqrt{6}\) are linked to different non-perfect squares.
Step 2
Why this answer is correct
इनके अपरिमेय भाग सामान्य जोड़ से पूरी तरह नहीं कटते, इसलिए योग परिमेय नहीं बनता। / Their irrational parts do not cancel through ordinary addition, so the sum is not rational.
Step 3
Exam Tip
ऐसे दावों में गलत पहचान जैसे \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) से बचें। / Avoid false identities such as \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\).
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यदि \(x=\sqrt{7}+2\), तो ((x-2)(x+2)) का मान क्या है?
If \(x=\sqrt{7}+2\), what is the value of ((x-2)(x+2))?
#substitution
#surd algebra
#class 10
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A \(7+4\sqrt{7}\)
B (7)
C (11)
D \(3+4\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
A. \(7+4\sqrt{7}\)
Step 1
Concept
((x-2)=\sqrt{7}) और ((x+2)=\sqrt{7}+4)। / ((x-2)=\sqrt{7}) and ((x+2)=\sqrt{7}+4).
Step 2
Why this answer is correct
गुणन (\sqrt{7}\(\sqrt{7}+4\)=7+4\sqrt{7}) है। / The product is (\sqrt{7}\(\sqrt{7}+4\)=7+4\sqrt{7}).
Step 3
Exam Tip
सीधे सूत्र लगाने से पहले (x) का दिया हुआ मान ध्यान से रखें। / Before applying an identity directly, substitute the given value of (x) carefully.
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कौन-सा विकल्प \(\sqrt{2}+\sqrt{18}-\sqrt{50}+\sqrt{98}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{2}+\sqrt{18}-\sqrt{50}+\sqrt{98}\)?
#long surd expression
#simplification
#class 10
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A \(6\sqrt{2}\)
B \(4\sqrt{2}\)
C \(8\sqrt{2}\)
D (148)
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{2}\)
Step 1
Concept
\(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), और \(\sqrt{98}=7\sqrt{2}\)। / \(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{98}=7\sqrt{2}\).
Step 2
Why this answer is correct
\(1\sqrt{2}+3\sqrt{2}-5\sqrt{2}+7\sqrt{2}=6\sqrt{2}\)। / \(1\sqrt{2}+3\sqrt{2}-5\sqrt{2}+7\sqrt{2}=6\sqrt{2}\).
Step 3
Exam Tip
लंबे मूल वाले प्रश्न में गुणांक अलग लिखकर जोड़ना आसान रहता है। / In long surd expressions, write the coefficients separately and add them.
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यदि \(x=\sqrt{6}+\sqrt{5}\), तो \(x^3+\frac{1}{x^3}\) का मान और प्रकृति क्या है?
If \(x=\sqrt{6}+\sqrt{5}\), what is the value and nature of \(x^3+\frac{1}{x^3}\)?
#reciprocal surds
#cubic identity
#irrational numbers
#class 10
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A \(42\sqrt{6}\), अपरिमेय / \(42\sqrt{6}\), irrational
B (42), परिमेय / (42), rational
C \(36\sqrt{5}\), अपरिमेय / \(36\sqrt{5}\), irrational
D \(11\sqrt{6}\), अपरिमेय / \(11\sqrt{6}\), irrational
Explanation opens after your attempt
Correct Answer
A. \(42\sqrt{6}\), अपरिमेय / \(42\sqrt{6}\), irrational
Step 1
Concept
(\(\sqrt{6}+\sqrt{5}\)\(\sqrt{6}-\sqrt{5}\)=1), इसलिए \(\frac{1}{x}=\sqrt{6}-\sqrt{5}\)। / (\(\sqrt{6}+\sqrt{5}\)\(\sqrt{6}-\sqrt{5}\)=1), so \(\frac{1}{x}=\sqrt{6}-\sqrt{5}\).
Step 2
Why this answer is correct
\(x+\frac{1}{x}=2\sqrt{6}\), अतः (x-3 +\frac{1}{x-3 }=\(2\sqrt{6}\)3 -3\(2\sqrt{6}\)=42\sqrt{6})। / \(x+\frac{1}{x}=2\sqrt{6}\), hence (x-3 +\frac{1}{x-3 }=\(2\sqrt{6}\)3 -3\(2\sqrt{6}\)=42\sqrt{6}).
Step 3
Exam Tip
घन वाले प्रश्नों में पहले \(x+\frac{1}{x}\) निकालना आसान तरीका है। / In cube-type questions, finding \(x+\frac{1}{x}\) first is the easier method.
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यदि (a) एक परिमेय संख्या है और \(a\neq 0\) है तो \(a\sqrt{3}\) के बारे में कौन सा निष्कर्ष सही है?
If (a) is a rational number and \(a\neq 0\) then which conclusion about \(a\sqrt{3}\) is correct?
#irrational numbers
#real numbers
#class 10
#expert
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A यह सदैव परिमेय है / It is always rational
B यह सदैव अपरिमेय है / It is always irrational
C यह कभी परिमेय कभी अपरिमेय हो सकता है / It may be rational or irrational
D यह केवल पूर्णांक होने पर परिमेय है / It is rational only when it is an integer
Explanation opens after your attempt
Correct Answer
B. यह सदैव अपरिमेय है / It is always irrational
Step 1
Concept
(a) परिमेय और शून्य नहीं है। / (a) is rational and non-zero.
Step 2
Why this answer is correct
यदि \(a\sqrt{3}\) परिमेय मान लें तो \(\sqrt{3}\) भी परिमेय मिलेगा जो गलत है। / If \(a\sqrt{3}\) were rational then \(\sqrt{3}\) would also become rational which is false.
Step 3
Exam Tip
परीक्षा में शून्य से गुणा वाले विशेष मामले को अलग याद रखें। / In exams always check the special case of multiplication by zero.
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कौन सी संख्या निश्चित रूप से अपरिमेय है?
Which number is definitely irrational?
#irrational numbers
#simplification
#surds
#class 10
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A \(0\times \sqrt{5}\)
B \(\sqrt{9}\)
C \(\sqrt{2}+\sqrt{8}\)
D \(\sqrt{16}-\sqrt{4}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{2}+\sqrt{8}\)
Step 1
Concept
सरल करें \(\sqrt{8}=2\sqrt{2}\)। / Simplify \(\sqrt{8}=2\sqrt{2}\).
Step 2
Why this answer is correct
\(\sqrt{2}+\sqrt{8}=3\sqrt{2}\) है और \(\sqrt{2}\) अपरिमेय है। / \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\) and \(\sqrt{2}\) is irrational.
Step 3
Exam Tip
वर्गमूलों को सरल किए बिना उत्तर जल्दी न चुनें। / Do not choose the answer before simplifying square roots.
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यदि \(\sqrt{n}\) परिमेय है और (n) धनात्मक पूर्णांक है तो (n) के बारे में क्या सही है?
If \(\sqrt{n}\) is rational and (n) is a positive integer then what is true about (n)?
#perfect square
#irrational numbers
#conceptual
#class 10
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A (n) अभाज्य है / (n) is prime
B (n) विषम है / (n) is odd
C (n) पूर्ण वर्ग है / (n) is a perfect square
D (n) अपरिमेय है / (n) is irrational
Explanation opens after your attempt
Correct Answer
C. (n) पूर्ण वर्ग है / (n) is a perfect square
Step 1
Concept
धनात्मक पूर्णांक का वर्गमूल परिमेय तभी होता है जब वह पूर्ण वर्ग हो। / The square root of a positive integer is rational only when the integer is a perfect square.
Step 2
Why this answer is correct
जैसे \(\sqrt{25}=5\) परिमेय है। / For example \(\sqrt{25}=5\).
Step 3
Exam Tip
अभाज्य संख्या का वर्गमूल सामान्यतः अपरिमेय होता है। / The square root of a prime number is usually irrational.
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यदि \(x=\sqrt{2}+\sqrt{3}\) है तो (x) के बारे में सही निष्कर्ष क्या है?
If \(x=\sqrt{2}+\sqrt{3}\) then what is the correct conclusion about (x)?
#irrational numbers
#proof based
#surds
#class 10
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A (x) परिमेय है / (x) is rational
B (x) अपरिमेय है / (x) is irrational
C (x) पूर्णांक है / (x) is an integer
D (x=5) है / (x=5)
Explanation opens after your attempt
Correct Answer
B. (x) अपरिमेय है / (x) is irrational
Step 1
Concept
मान लें \(\sqrt{2}+\sqrt{3}\) परिमेय है। / Suppose \(\sqrt{2}+\sqrt{3}\) is rational.
Step 2
Why this answer is correct
वर्ग करने पर \(5+2\sqrt{6}\) परिमेय होना चाहिए इसलिए \(\sqrt{6}\) परिमेय मिलेगा जो गलत है। / Squaring gives \(5+2\sqrt{6}\) so \(\sqrt{6}\) would be rational which is false.
Step 3
Exam Tip
दो अलग अपरिमेय संख्याओं के योग को सीधे परिमेय या अपरिमेय न मानें। / Do not decide the sum of two different irrational numbers without reasoning.
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कौन सा मान परिमेय है?
Which value is rational?
#rational vs irrational
#square root
#class 10
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A \(\sqrt{50}\)
B \(\sqrt{72}\)
C \(\sqrt{49}\)
D \(\sqrt{45}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{49}\)
Step 1
Concept
पूर्ण वर्ग का वर्गमूल परिमेय होता है। / The square root of a perfect square is rational.
Step 2
Why this answer is correct
\(\sqrt{49}=7\) है इसलिए यह परिमेय है। / \(\sqrt{49}=7\) so it is rational.
Step 3
Exam Tip
संख्या के अंदर पूर्ण वर्ग गुणनखंड देखकर सरल करें। / First look for perfect-square factors inside the radical.
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यदि (p) और (q) सहअभाज्य धनात्मक पूर्णांक हैं और \(\sqrt{2}=\frac{p}{q}\) मान लिया जाए तो विरोधाभास कहां बनता है?
If (p) and (q) are coprime positive integers and \(\sqrt{2}=\frac{p}{q}\) is assumed then where does the contradiction arise?
#proof
#irrationality
#sqrt2
#class 10
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A (p) और (q) दोनों विषम मिलते हैं / Both (p) and (q) become odd
B (p) और (q) दोनों सम मिलते हैं / Both (p) and (q) become even
C (p) अभाज्य नहीं मिलता / (p) is not prime
D (q=1) मिल जाता है / (q=1) is obtained
Explanation opens after your attempt
Correct Answer
B. (p) और (q) दोनों सम मिलते हैं / Both (p) and (q) become even
Step 1
Concept
\(\sqrt{2}=\frac{p}{q}\) से \(p^2=2q^2\) मिलता है। / From \(\sqrt{2}=\frac{p}{q}\) we get \(p^2=2q^2\).
Step 2
Why this answer is correct
इससे (p) सम और फिर (q) भी सम मिलता है। / This makes (p) even and then (q) even.
Step 3
Exam Tip
सहअभाज्य संख्याएं दोनों सम नहीं हो सकतीं इसलिए मान्यता गलत है। / Coprime numbers cannot both be even so the assumption is false.
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कौन सा कथन \(\sqrt{5}\) के लिए सही है?
Which statement is correct for \(\sqrt{5}\)?
#decimal expansion
#irrational numbers
#class 10
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A यह समाप्त दशमलव है / It is a terminating decimal
B यह आवर्ती दशमलव है / It is a recurring decimal
C यह अनावर्ती और अनंत दशमलव है / It is non-recurring and non-terminating decimal
D यह प्राकृतिक संख्या है / It is a natural number
Explanation opens after your attempt
Correct Answer
C. यह अनावर्ती और अनंत दशमलव है / It is non-recurring and non-terminating decimal
Step 1
Concept
\(\sqrt{5}\) पूर्ण वर्ग का वर्गमूल नहीं है। / \(\sqrt{5}\) is not the square root of a perfect square.
Step 2
Why this answer is correct
इसलिए यह अपरिमेय है और इसका दशमलव विस्तार अनंत अनावर्ती होता है। / So it is irrational and its decimal expansion is non-terminating and non-recurring.
Step 3
Exam Tip
दशमलव रूप से पहचानते समय आवृत्ति पर ध्यान दें। / While using decimal form check whether repetition exists.
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कौन सा गुणनफल अपरिमेय है?
Which product is irrational?
#product
#irrational numbers
#class 10
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A \(0\cdot \sqrt{11}\)
B \(\sqrt{8}\cdot \sqrt{2}\)
C \(\frac{3}{5}\cdot \sqrt{13}\)
D \(\sqrt{3}\cdot \sqrt{12}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{3}{5}\cdot \sqrt{13}\)
Step 1
Concept
\(\frac{3}{5}\) शून्य रहित परिमेय है। / \(\frac{3}{5}\) is a non-zero rational number.
Step 2
Why this answer is correct
शून्य रहित परिमेय संख्या से \(\sqrt{13}\) को गुणा करने पर अपरिमेय संख्या मिलती है। / Multiplying it by \(\sqrt{13}\) gives an irrational number.
Step 3
Exam Tip
गुणनफल में पहले देखें कि वर्गमूल मिलकर पूर्ण वर्ग तो नहीं बना रहे। / In products check first whether square roots combine to a perfect square.
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\(\sqrt{2}\) और \(\sqrt{8}\) के बीच संबंध क्या है?
What is the relation between \(\sqrt{2}\) and \(\sqrt{8}\)?
#surds
#irrational numbers
#comparison
#class 10
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A दोनों परिमेय हैं / Both are rational
B दोनों अपरिमेय हैं और \(\sqrt{8}=2\sqrt{2}\) / Both are irrational and \(\sqrt{8}=2\sqrt{2}\)
C पहली परिमेय और दूसरी अपरिमेय है / First is rational and second is irrational
D दोनों बराबर हैं / Both are equal
Explanation opens after your attempt
Correct Answer
B. दोनों अपरिमेय हैं और \(\sqrt{8}=2\sqrt{2}\) / Both are irrational and \(\sqrt{8}=2\sqrt{2}\)
Step 1
Concept
\(8=4\cdot 2\) है इसलिए \(\sqrt{8}=2\sqrt{2}\)। / Since \(8=4\cdot 2\) we have \(\sqrt{8}=2\sqrt{2}\).
Step 2
Why this answer is correct
\(\sqrt{2}\) अपरिमेय है और उसका दुगुना भी अपरिमेय है। / \(\sqrt{2}\) is irrational and its double is also irrational.
Step 3
Exam Tip
समान मूल वाली संख्याओं को सरल रूप में तुलना करें। / Compare like radicals after simplifying them.
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यदि \(\sqrt{m}\) अपरिमेय है और (r) शून्य रहित परिमेय है तो \(\frac{\sqrt{m}}{r}\) कैसा होगा?
If \(\sqrt{m}\) is irrational and (r) is a non-zero rational number then what type is \(\frac{\sqrt{m}}{r}\)?
#division
#irrational numbers
#properties
#class 10
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A सदैव परिमेय / Always rational
B सदैव अपरिमेय / Always irrational
C सदैव पूर्णांक / Always integer
D कभी परिभाषित नहीं / Never defined
Explanation opens after your attempt
Correct Answer
B. सदैव अपरिमेय / Always irrational
Step 1
Concept
शून्य रहित परिमेय से भाग देना उसी के व्युत्क्रम से गुणा करना है। / Dividing by a non-zero rational number is the same as multiplying by its reciprocal.
Step 2
Why this answer is correct
अपरिमेय संख्या को शून्य रहित परिमेय से गुणा करने पर अपरिमेय संख्या मिलती है। / An irrational number multiplied by a non-zero rational number remains irrational.
Step 3
Exam Tip
भाग के प्रश्नों को गुणा में बदलकर सोचें। / Convert division questions into multiplication for easier reasoning.
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कौन सा विकल्प \(\sqrt{18}-\sqrt{8}\) का सही प्रकार बताता है?
Which option correctly describes \(\sqrt{18}-\sqrt{8}\)?
#surds
#subtraction
#irrational numbers
#class 10
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A परिमेय क्योंकि उत्तर (1) है / Rational because the answer is (1)
B अपरिमेय क्योंकि उत्तर \(\sqrt{2}\) है / Irrational because the answer is \(\sqrt{2}\)
C पूर्णांक क्योंकि वर्गमूल घटते हैं / Integer because square roots subtract
D शून्य क्योंकि (18-8=10) है / Zero because (18-8=10)
Explanation opens after your attempt
Correct Answer
B. अपरिमेय क्योंकि उत्तर \(\sqrt{2}\) है / Irrational because the answer is \(\sqrt{2}\)
Step 1
Concept
\(\sqrt{18}=3\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\)। / \(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\).
Step 2
Why this answer is correct
अंतर \(\sqrt{2}\) है जो अपरिमेय है। / The difference is \(\sqrt{2}\) which is irrational.
Step 3
Exam Tip
वर्गमूल घटाते समय भीतर की संख्याओं को सीधे न घटाएं। / Do not subtract the numbers inside square roots directly.
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किस दशमलव विस्तार से अपरिमेय संख्या पहचानी जाती है?
Which decimal expansion identifies an irrational number?
#decimal expansion
#irrational numbers
#class 10
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A समाप्त दशमलव / Terminating decimal
B अनंत आवर्ती दशमलव / Non-terminating recurring decimal
C अनंत अनावर्ती दशमलव / Non-terminating non-recurring decimal
D केवल एक अंकीय दशमलव / One-digit decimal only
Explanation opens after your attempt
Correct Answer
C. अनंत अनावर्ती दशमलव / Non-terminating non-recurring decimal
Step 1
Concept
परिमेय संख्याओं का दशमलव समाप्त या आवर्ती होता है। / Rational numbers have terminating or recurring decimals.
Step 2
Why this answer is correct
अपरिमेय संख्याओं का दशमलव अनंत और अनावर्ती होता है। / Irrational numbers have non-terminating and non-recurring decimals.
Step 3
Exam Tip
दशमलव में बार-बार आने वाला समूह दिखाई दे तो वह परिमेय हो सकता है। / If a repeating block is visible the number may be rational.
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कौन सा उदाहरण दिखाता है कि दो अपरिमेय संख्याओं का गुणनफल परिमेय हो सकता है?
Which example shows that the product of two irrational numbers can be rational?
#product of irrationals
#examples
#class 10
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A \(\sqrt{2}\cdot \sqrt{3}\)
B \(\sqrt{5}\cdot \sqrt{5}\)
C \(\sqrt{7}\cdot 2\)
D \(\sqrt{11}\cdot \sqrt{2}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{5}\cdot \sqrt{5}\)
Step 1
Concept
\(\sqrt{5}\) अपरिमेय है। / \(\sqrt{5}\) is irrational.
Step 2
Why this answer is correct
\(\sqrt{5}\cdot \sqrt{5}=5\) परिमेय है। / \(\sqrt{5}\cdot \sqrt{5}=5\) which is rational.
Step 3
Exam Tip
दो समान अपरिमेय वर्गमूलों का गुणनफल भीतर की संख्या बन जाता है। / The product of two identical irrational square roots becomes the number inside.
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यदि \(x=3+\sqrt{2}\) है तो (x) के बारे में क्या सही है?
If \(x=3+\sqrt{2}\) then what is true about (x)?
#rational plus irrational
#class 10
#real numbers
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A (x) परिमेय है / (x) is rational
B (x) अपरिमेय है / (x) is irrational
C (x=5) है / (x=5)
D (x) प्राकृतिक संख्या है / (x) is a natural number
Explanation opens after your attempt
Correct Answer
B. (x) अपरिमेय है / (x) is irrational
Step 1
Concept
(3) परिमेय है और \(\sqrt{2}\) अपरिमेय है। / (3) is rational and \(\sqrt{2}\) is irrational.
Step 2
Why this answer is correct
परिमेय और अपरिमेय का योग अपरिमेय होता है। / The sum of a rational and an irrational number is irrational.
Step 3
Exam Tip
किसी पूर्णांक में अपरिमेय संख्या जोड़ने से वह पूर्णांक नहीं रहता। / Adding an irrational number to an integer does not give an integer.
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कौन सा विकल्प \(\frac{1}{\sqrt{3}}\) के बारे में सही है?
Which option is correct about \(\frac{1}{\sqrt{3}}\)?
#reciprocal
#irrational numbers
#class 10
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A यह परिमेय है / It is rational
B यह अपरिमेय है / It is irrational
C यह पूर्णांक है / It is an integer
D यह (0) है / It is (0)
Explanation opens after your attempt
Correct Answer
B. यह अपरिमेय है / It is irrational
Step 1
Concept
यदि \(\frac{1}{\sqrt{3}}\) परिमेय हो तो उसका व्युत्क्रम \(\sqrt{3}\) भी परिमेय होगा। / If \(\frac{1}{\sqrt{3}}\) were rational then its reciprocal \(\sqrt{3}\) would be rational.
Step 2
Why this answer is correct
\(\sqrt{3}\) अपरिमेय है इसलिए दी गई संख्या भी अपरिमेय है। / \(\sqrt{3}\) is irrational so the given number is irrational.
Step 3
Exam Tip
अपरिमेय हर देखकर उसे अपने आप परिमेय न मानें। / A denominator with an irrational radical does not make the value rational automatically.
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\(\sqrt{48}\) को सरल करने पर संख्या किस प्रकार की है?
After simplifying \(\sqrt{48}\) what type of number is it?
#simplifying radicals
#irrational numbers
#class 10
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A परिमेय क्योंकि (48) सम है / Rational because (48) is even
B अपरिमेय क्योंकि \(\sqrt{48}=4\sqrt{3}\) / Irrational because \(\sqrt{48}=4\sqrt{3}\)
C पूर्ण वर्ग क्योंकि \(48=6\times 8\) / Perfect square because \(48=6\times 8\)
D शून्य क्योंकि वर्गमूल नहीं निकलता / Zero because square root cannot be taken
Explanation opens after your attempt
Correct Answer
B. अपरिमेय क्योंकि \(\sqrt{48}=4\sqrt{3}\) / Irrational because \(\sqrt{48}=4\sqrt{3}\)
Step 1
Concept
\(48=16\cdot 3\) है। / \(48=16\cdot 3\).
Step 2
Why this answer is correct
\(\sqrt{48}=4\sqrt{3}\) और \(\sqrt{3}\) अपरिमेय है। / \(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{3}\) is irrational.
Step 3
Exam Tip
सम संख्या का वर्गमूल परिमेय होगा यह जरूरी नहीं। / The square root of an even number need not be rational.
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कौन सा युग्म दोनों अपरिमेय संख्याएं देता है लेकिन उनका भाग परिमेय है?
Which pair gives two irrational numbers but their quotient is rational?
#quotient
#irrational numbers
#class 10
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A \(\sqrt{6}\) और \(\sqrt{3}\) / \(\sqrt{6}\) and \(\sqrt{3}\)
B \(\sqrt{18}\) और \(\sqrt{2}\) / \(\sqrt{18}\) and \(\sqrt{2}\)
C \(\sqrt{5}\) और \(\sqrt{7}\) / \(\sqrt{5}\) and \(\sqrt{7}\)
D \(\sqrt{10}\) और (2) / \(\sqrt{10}\) and (2)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{18}\) और \(\sqrt{2}\) / \(\sqrt{18}\) and \(\sqrt{2}\)
Step 1
Concept
\(\sqrt{18}\) और \(\sqrt{2}\) दोनों अपरिमेय हैं। / \(\sqrt{18}\) and \(\sqrt{2}\) are both irrational.
Step 2
Why this answer is correct
\(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\) परिमेय है। / \(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{9}=3\) which is rational.
Step 3
Exam Tip
भाग में मूल के अंदर अनुपात पूर्ण वर्ग बन रहा है या नहीं यह देखें। / In quotients check whether the ratio inside the radical becomes a perfect square.
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यदि \(5-\sqrt{6}\) को परिमेय माना जाए तो कौन सा गलत निष्कर्ष निकलेगा?
If \(5-\sqrt{6}\) is assumed rational then which false conclusion follows?
#contradiction
#irrational numbers
#class 10
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A (5) अपरिमेय है / (5) is irrational
B \(\sqrt{6}\) परिमेय है / \(\sqrt{6}\) is rational
C (6) पूर्ण वर्ग है / (6) is a perfect square
D (5=0) है / (5=0)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{6}\) परिमेय है / \(\sqrt{6}\) is rational
Step 1
Concept
मान लें \(5-\sqrt{6}\) परिमेय है। / Suppose \(5-\sqrt{6}\) is rational.
Step 2
Why this answer is correct
तब \(\sqrt{6}=5-\) वह परिमेय संख्या होगा इसलिए \(\sqrt{6}\) परिमेय मिल जाएगा। / Then \(\sqrt{6}=5-\) that rational number so \(\sqrt{6}\) would be rational.
Step 3
Exam Tip
विरोधाभास विधि में उस निष्कर्ष को पकड़ें जो ज्ञात तथ्य से टकराता है। / In contradiction proofs identify the result that clashes with a known fact.
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कौन सी संख्या (2) और (3) के बीच अपरिमेय है?
Which number is irrational and lies between (2) and (3)?
#irrational between numbers
#comparison
#class 10
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A \(\sqrt{4}\)
B \(\sqrt{5}\)
C \(\sqrt{9}\)
D \(\frac{5}{2}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{5}\)
Step 1
Concept
(4<5<9) है।
Step 2
Why this answer is correct
इसलिए \(2<\sqrt{5}<3\) और (5) पूर्ण वर्ग नहीं है इसलिए \(\sqrt{5}\) अपरिमेय है। / Therefore \(2<\sqrt{5}<3\) and since (5) is not a perfect square \(\sqrt{5}\) is irrational.
Step 3
Exam Tip
बीच की संख्या खोजते समय वर्गों से सीमा बनाएं। / Use squares to locate irrational square roots between integers.
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कौन सा कथन \(\sqrt{a}+\sqrt{a}\) के लिए सही है जब (a) पूर्ण वर्ग नहीं है?
Which statement is correct for \(\sqrt{a}+\sqrt{a}\) when (a) is not a perfect square?
#like radicals
#irrational numbers
#class 10
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A यह (2a) के बराबर परिमेय है / It equals (2a) and is rational
B यह \(2\sqrt{a}\) के बराबर अपरिमेय है / It equals \(2\sqrt{a}\) and is irrational
C यह \(\sqrt{2a}\) के बराबर है / It equals \(\sqrt{2a}\)
D यह (a) के बराबर है / It equals (a)
Explanation opens after your attempt
Correct Answer
B. यह \(2\sqrt{a}\) के बराबर अपरिमेय है / It equals \(2\sqrt{a}\) and is irrational
Step 1
Concept
समान पद जोड़ने पर \(\sqrt{a}+\sqrt{a}=2\sqrt{a}\)। / Like terms give \(\sqrt{a}+\sqrt{a}=2\sqrt{a}\).
Step 2
Why this answer is correct
(a) पूर्ण वर्ग नहीं है इसलिए \(\sqrt{a}\) अपरिमेय है और उसका दुगुना भी अपरिमेय है। / Since (a) is not a perfect square \(\sqrt{a}\) is irrational and its double is irrational.
Step 3
Exam Tip
समान वर्गमूलों को बीजगणितीय पदों की तरह जोड़ें। / Add like radicals like algebraic terms.
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कौन सा विकल्प (\(\sqrt{3}+1\)\(\sqrt{3}-1\)) का सही मान और प्रकार देता है?
Which option gives the correct value and type of (\(\sqrt{3}+1\)\(\sqrt{3}-1\))?
#conjugates
#rationalization
#irrational numbers
#class 10
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A (2) और परिमेय / (2) and rational
B (4) और परिमेय / (4) and rational
C \(\sqrt{3}\) और अपरिमेय / \(\sqrt{3}\) and irrational
D \(3+\sqrt{3}\) और अपरिमेय / \(3+\sqrt{3}\) and irrational
Explanation opens after your attempt
Correct Answer
A. (2) और परिमेय / (2) and rational
Step 1
Concept
यह अंतर के वर्ग का रूप है। / This is a difference of squares form.
Step 2
Why this answer is correct
(\(\sqrt{3}+1\)\(\sqrt{3}-1\)=3-1=2) जो परिमेय है। / (\(\sqrt{3}+1\)\(\sqrt{3}-1\)=3-1=2) which is rational.
Step 3
Exam Tip
संयुग्मी पदों का गुणनफल अक्सर वर्गमूल हटा देता है। / Product of conjugates often removes the radical.
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कौन सा विकल्प परिमेय और अपरिमेय संख्या के अंतर का सही सामान्य निष्कर्ष देता है?
Which option gives the correct general conclusion for the difference of a rational and an irrational number?
#properties
#irrational numbers
#proof
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A सदैव परिमेय / Always rational
B सदैव अपरिमेय / Always irrational
C सदैव शून्य / Always zero
D सदैव पूर्णांक / Always integer
Explanation opens after your attempt
Correct Answer
B. सदैव अपरिमेय / Always irrational
Step 1
Concept
मान लें परिमेय (r) और अपरिमेय (s) हैं। / Let (r) be rational and (s) be irrational.
Step 2
Why this answer is correct
यदि (r-s) परिमेय हो तो (s=r-(r-s)) परिमेय हो जाएगा जो गलत है। / If (r-s) were rational then (s=r-(r-s)) would be rational which is false.
Step 3
Exam Tip
परिमेय और अपरिमेय को जोड़ने या घटाने पर परिणाम अपरिमेय रहता है। / Adding or subtracting a rational and an irrational number gives an irrational number.
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\(\sqrt{75}+\sqrt{12}\) किसके बराबर है और उसका प्रकार क्या है?
What is \(\sqrt{75}+\sqrt{12}\) equal to and what is its type?
#surds addition
#irrational numbers
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A \(7\sqrt{3}\) और अपरिमेय / \(7\sqrt{3}\) and irrational
B (87) और परिमेय / (87) and rational
C \(\sqrt{87}\) और अपरिमेय / \(\sqrt{87}\) and irrational
D \(5\sqrt{3}\) और अपरिमेय / \(5\sqrt{3}\) and irrational
Explanation opens after your attempt
Correct Answer
A. \(7\sqrt{3}\) और अपरिमेय / \(7\sqrt{3}\) and irrational
Step 1
Concept
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\)। / \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\).
Step 2
Why this answer is correct
योग \(7\sqrt{3}\) है जो अपरिमेय है। / The sum is \(7\sqrt{3}\) which is irrational.
Step 3
Exam Tip
अलग-अलग वर्गमूलों को जोड़ने से पहले सरल रूप में बदलें। / Simplify radicals before adding them.
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कौन सा विकल्प बताता है कि \(\sqrt{p}\) अपरिमेय है जब (p) अभाज्य संख्या है?
Which option explains why \(\sqrt{p}\) is irrational when (p) is a prime number?
#prime numbers
#irrationality
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A क्योंकि (p) का कोई वर्ग गुणनखंड (1) के अलावा नहीं होता / Because (p) has no square factor except (1)
B क्योंकि हर अभाज्य संख्या सम होती है / Because every prime number is even
C क्योंकि (p) दशमलव संख्या है / Because (p) is a decimal number
D क्योंकि \(\sqrt{p}=p\) होता है / Because \(\sqrt{p}=p\)
Explanation opens after your attempt
Correct Answer
A. क्योंकि (p) का कोई वर्ग गुणनखंड (1) के अलावा नहीं होता / Because (p) has no square factor except (1)
Step 1
Concept
अभाज्य संख्या (p) पूर्ण वर्ग नहीं होती। / A prime number (p) is not a perfect square.
Step 2
Why this answer is correct
पूर्ण वर्ग न होने पर \(\sqrt{p}\) परिमेय नहीं हो सकता। / If it is not a perfect square then \(\sqrt{p}\) cannot be rational.
Step 3
Exam Tip
अभाज्य संख्या के वर्गमूल को परिमेय मानने से गुणनखंडों में विरोधाभास आता है। / Assuming the square root of a prime to be rational leads to a factor contradiction.
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यदि \(x=\sqrt{10}\) है तो \(x^2\) किस प्रकार की संख्या है?
If \(x=\sqrt{10}\) then what type of number is \(x^2\)?
#square of irrational
#class 10
#real numbers
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A अपरिमेय / Irrational
B परिमेय / Rational
C अनिर्धारित / Undefined
D नकारात्मक / Negative
Explanation opens after your attempt
Correct Answer
B. परिमेय / Rational
Step 1
Concept
\(x=\sqrt{10}\) अपरिमेय है। / \(x=\sqrt{10}\) is irrational.
Step 2
Why this answer is correct
(x-2 =\(\sqrt{10}\)2 =10) है जो परिमेय है। / (x-2 =\(\sqrt{10}\)2 =10) which is rational.
Step 3
Exam Tip
अपरिमेय संख्या का वर्ग कई बार परिमेय हो सकता है। / The square of an irrational number can be rational.
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कौन सा विकल्प \(\frac{\sqrt{45}}{3}\) का सही प्रकार बताता है?
Which option correctly describes \(\frac{\sqrt{45}}{3}\)?
#simplification
#quotient
#irrational numbers
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A परिमेय क्योंकि हर (3) है / Rational because denominator is (3)
B अपरिमेय क्योंकि यह \(\sqrt{5}\) के बराबर है / Irrational because it equals \(\sqrt{5}\)
C परिमेय क्योंकि (45) (3) से विभाज्य है / Rational because (45) is divisible by (3)
D पूर्णांक क्योंकि वर्गमूल हट जाता है / Integer because radical disappears
Explanation opens after your attempt
Correct Answer
B. अपरिमेय क्योंकि यह \(\sqrt{5}\) के बराबर है / Irrational because it equals \(\sqrt{5}\)
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\)। / \(\sqrt{45}=3\sqrt{5}\).
Step 2
Why this answer is correct
\(\frac{\sqrt{45}}{3}=\sqrt{5}\) है जो अपरिमेय है। / \(\frac{\sqrt{45}}{3}=\sqrt{5}\) which is irrational.
Step 3
Exam Tip
हर से भाग देने पर भी बचा हुआ वर्गमूल जांचना जरूरी है। / Even after division check the remaining radical.
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कौन सा विकल्प \(\sqrt{2}+\frac{1}{\sqrt{2}}\) का सही रूप और प्रकार देता है?
Which option gives the correct form and type of \(\sqrt{2}+\frac{1}{\sqrt{2}}\)?
#rationalization
#irrational numbers
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A \(\frac{3\sqrt{2}}{2}\) और अपरिमेय / \(\frac{3\sqrt{2}}{2}\) and irrational
B (2) और परिमेय / (2) and rational
C \(\sqrt{2}\) और अपरिमेय / \(\sqrt{2}\) and irrational
D \(\frac{1}{2}\) और परिमेय / \(\frac{1}{2}\) and rational
Explanation opens after your attempt
Correct Answer
A. \(\frac{3\sqrt{2}}{2}\) और अपरिमेय / \(\frac{3\sqrt{2}}{2}\) and irrational
Step 1
Concept
\(\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)। / \(\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\).
Step 2
Why this answer is correct
\(\sqrt{2}+\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}\) जो अपरिमेय है। / \(\sqrt{2}+\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}\) which is irrational.
Step 3
Exam Tip
जोड़ने से पहले हर को परिमेय बनाने की आदत रखें। / Rationalize the denominator before combining terms.
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कौन सी संख्या (1) और (2) के बीच अपरिमेय है?
Which number is irrational and lies between (1) and (2)?
#number line
#irrational numbers
#class 10
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A \(\sqrt{2}\)
B \(\sqrt{4}\)
C \(\frac{3}{2}\)
D (1.25)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}\)
Step 1
Concept
(1<2<4) है।
Step 2
Why this answer is correct
इसलिए \(1<\sqrt{2}<2\) और \(\sqrt{2}\) अपरिमेय है। / Hence \(1<\sqrt{2}<2\) and \(\sqrt{2}\) is irrational.
Step 3
Exam Tip
अपरिमेय संख्या को स्थान देने के लिए वर्ग करके तुलना करें। / To locate an irrational number compare squares.
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यदि \(a+b\sqrt{2}=0\) है जहां (a) और (b) परिमेय हैं तथा \(b\neq 0\) है तो क्या निष्कर्ष निकलेगा?
If \(a+b\sqrt{2}=0\) where (a) and (b) are rational and \(b\neq 0\) then what conclusion follows?
#linear form
#irrational numbers
#proof
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A \(\sqrt{2}=-\frac{a}{b}\) परिमेय होगा जो असंभव है / \(\sqrt{2}=-\frac{a}{b}\) would be rational which is impossible
B (a) अपरिमेय होगा / (a) would be irrational
C (b=0) नहीं हो सकता इसलिए उत्तर (1) है / Since \(b\neq 0\) the answer is (1)
D यह हर स्थिति में सही है / It is true in every case
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}=-\frac{a}{b}\) परिमेय होगा जो असंभव है / \(\sqrt{2}=-\frac{a}{b}\) would be rational which is impossible
Step 1
Concept
समीकरण से \(\sqrt{2}=-\frac{a}{b}\) मिलेगा। / The equation gives \(\sqrt{2}=-\frac{a}{b}\).
Step 2
Why this answer is correct
\(-\frac{a}{b}\) परिमेय है लेकिन \(\sqrt{2}\) अपरिमेय है। / \(-\frac{a}{b}\) is rational but \(\sqrt{2}\) is irrational.
Step 3
Exam Tip
ऐसे प्रश्नों में अपरिमेयता से विरोधाभास बनता है। / Such questions use irrationality to create a contradiction.
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कौन सा विकल्प \(\sqrt{32}+\sqrt{50}-\sqrt{18}\) का सही प्रकार बताता है?
Which option correctly describes \(\sqrt{32}+\sqrt{50}-\sqrt{18}\)?
#surds
#addition subtraction
#irrational numbers
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A यह \(6\sqrt{2}\) है और अपरिमेय है / It is \(6\sqrt{2}\) and irrational
B यह (64) है और परिमेय है / It is (64) and rational
C यह \(\sqrt{64}\) है और परिमेय है / It is \(\sqrt{64}\) and rational
D यह (0) है और परिमेय है / It is (0) and rational
Explanation opens after your attempt
Correct Answer
A. यह \(6\sqrt{2}\) है और अपरिमेय है / It is \(6\sqrt{2}\) and irrational
Step 1
Concept
\(\sqrt{32}=4\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\)। / \(\sqrt{32}=4\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\).
Step 2
Why this answer is correct
परिणाम \(6\sqrt{2}\) है जो अपरिमेय है। / The result is \(6\sqrt{2}\) which is irrational.
Step 3
Exam Tip
समान मूल वाले पदों के गुणांक जोड़ें और घटाएं। / Add and subtract coefficients of like radicals.
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कौन सा विकल्प दिखाता है कि अपरिमेय संख्या में अपरिमेय संख्या जोड़ने पर परिमेय परिणाम मिल सकता है?
Which option shows that adding an irrational number to an irrational number can give a rational result?
#counterexample
#irrational sum
#class 10
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A \(\sqrt{3}+\sqrt{3}=2\sqrt{3}\)
B (\sqrt{5}+\(2-\sqrt{5}\)=2)
C \(\sqrt{2}+\sqrt{8}=3\sqrt{2}\)
D \(\sqrt{7}+\sqrt{11}\) अपरिमेय है / \(\sqrt{7}+\sqrt{11}\) is irrational
Explanation opens after your attempt
Correct Answer
B. (\sqrt{5}+\(2-\sqrt{5}\)=2)
Step 1
Concept
\(\sqrt{5}\) अपरिमेय है और \(2-\sqrt{5}\) भी अपरिमेय है। / \(\sqrt{5}\) is irrational and \(2-\sqrt{5}\) is also irrational.
Step 2
Why this answer is correct
उनका योग (2) है जो परिमेय है। / Their sum is (2) which is rational.
Step 3
Exam Tip
दो अपरिमेय संख्याओं के योग के लिए एक ही नियम हर बार लागू नहीं होता। / There is no single always rule for the sum of two irrational numbers.
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कौन सी संख्या अपरिमेय है लेकिन उसका वर्ग परिमेय है?
Which number is irrational but its square is rational?
#square
#irrational numbers
#class 10
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A \(\frac{7}{3}\)
B \(\sqrt{11}\)
C \(\sqrt{12}+\sqrt{3}\)
D (0.25)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{11}\)
Step 1
Concept
\(\sqrt{11}\) अपरिमेय है क्योंकि (11) पूर्ण वर्ग नहीं है। / \(\sqrt{11}\) is irrational because (11) is not a perfect square.
Step 2
Why this answer is correct
(\(\sqrt{11}\)2 =11) परिमेय है। / (\(\sqrt{11}\)2 =11) which is rational.
Step 3
Exam Tip
वर्ग करने पर वर्गमूल हट सकता है। / Squaring may remove the radical.
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कौन सा विकल्प \(\sqrt{2}\sqrt{8}+\sqrt{3}\sqrt{12}\) का सही मान देता है?
Which option gives the correct value of \(\sqrt{2}\sqrt{8}+\sqrt{3}\sqrt{12}\)?
#radical products
#rational result
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A (10)
B \(\sqrt{10}\)
C \(4\sqrt{2}\)
D \(2\sqrt{6}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{2}\sqrt{8}=\sqrt{16}=4\)। / \(\sqrt{2}\sqrt{8}=\sqrt{16}=4\).
Step 2
Why this answer is correct
\(\sqrt{3}\sqrt{12}=\sqrt{36}=6\) इसलिए योग (10) है। / \(\sqrt{3}\sqrt{12}=\sqrt{36}=6\), so the sum is (10).
Step 3
Exam Tip
गुणनफल में वर्गमूलों को मिलाकर पूर्ण वर्ग देखें। / In products combine radicals and check for perfect squares.
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यदि \(x=\sqrt{3}-\sqrt{2}\) है तो (x) के बारे में सही कथन कौन सा है?
If \(x=\sqrt{3}-\sqrt{2}\) then which statement about (x) is correct?
#difference of irrationals
#proof
#class 10
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A (x) परिमेय है / (x) is rational
B (x) अपरिमेय है / (x) is irrational
C (x=1) है / (x=1)
D (x=0) है / (x=0)
Explanation opens after your attempt
Correct Answer
B. (x) अपरिमेय है / (x) is irrational
Step 1
Concept
मान लें \(\sqrt{3}-\sqrt{2}\) परिमेय है। / Suppose \(\sqrt{3}-\sqrt{2}\) is rational.
Step 2
Why this answer is correct
वर्ग करने पर \(5-2\sqrt{6}\) से \(\sqrt{6}\) परिमेय होना पड़ेगा जो गलत है। / Squaring gives \(5-2\sqrt{6}\), forcing \(\sqrt{6}\) to be rational which is false.
Step 3
Exam Tip
अलग-अलग वर्गमूलों का अंतर सीधे पूर्णांक नहीं माना जाता। / The difference of unlike radicals is not directly an integer.
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कौन सा विकल्प \(\frac{2}{\sqrt{5}+1}\) को परिमेय हर वाले रूप में सही लिखता है?
Which option correctly writes \(\frac{2}{\sqrt{5}+1}\) with a rational denominator?
#rationalization
#conjugate
#irrational numbers
#class 10
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A \(\frac{\sqrt{5}-1}{2}\)
B \(\frac{\sqrt{5}+1}{2}\)
C \(\sqrt{5}-1\)
D \(\frac{2}{\sqrt{5}-1}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{\sqrt{5}-1}{2}\)
Step 1
Concept
हर को \(\sqrt{5}-1\) से गुणा करें। / Multiply by the conjugate \(\sqrt{5}-1\).
Step 2
Why this answer is correct
(\frac{2\(\sqrt{5}-1\)}{5-1}=\frac{\sqrt{5}-1}{2})। / (\frac{2\(\sqrt{5}-1\)}{5-1}=\frac{\sqrt{5}-1}{2}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर परिमेय बन जाता है। / Multiplying by the conjugate makes the denominator rational.
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कौन सा विकल्प \(\sqrt{0.04}\) का सही प्रकार बताता है?
Which option correctly describes \(\sqrt{0.04}\)?
#decimal square root
#rational vs irrational
#class 10
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A अपरिमेय क्योंकि दशमलव है / Irrational because it is decimal
B परिमेय क्योंकि \(\sqrt{0.04}=0.2\) / Rational because \(\sqrt{0.04}=0.2\)
C अपरिमेय क्योंकि वर्गमूल है / Irrational because it is a square root
D प्राकृतिक संख्या / Natural number
Explanation opens after your attempt
Correct Answer
B. परिमेय क्योंकि \(\sqrt{0.04}=0.2\) / Rational because \(\sqrt{0.04}=0.2\)
Step 1
Concept
\(0.04=\frac{4}{100}\)। / \(0.04=\frac{4}{100}\).
Step 2
Why this answer is correct
\(\sqrt{0.04}=\frac{2}{10}=0.2\) परिमेय है। / \(\sqrt{0.04}=\frac{2}{10}=0.2\) which is rational.
Step 3
Exam Tip
हर वर्गमूल अपरिमेय नहीं होता। / Every square root is not irrational.
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कौन सा विकल्प \(\sqrt{\frac{2}{9}}\) का सही प्रकार बताता है?
Which option correctly describes \(\sqrt{\frac{2}{9}}\)?
#fraction square root
#irrational numbers
#class 10
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A परिमेय क्योंकि हर पूर्ण वर्ग है / Rational because denominator is a perfect square
B अपरिमेय क्योंकि \(\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}\) / Irrational because \(\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}\)
C पूर्णांक क्योंकि (9) पूर्ण वर्ग है / Integer because (9) is a perfect square
D शून्य क्योंकि अंश छोटा है / Zero because numerator is small
Explanation opens after your attempt
Correct Answer
B. अपरिमेय क्योंकि \(\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}\) / Irrational because \(\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}\)
Step 1
Concept
\(\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}\)। / \(\sqrt{\frac{2}{9}}=\frac{\sqrt{2}}{3}\).
Step 2
Why this answer is correct
\(\sqrt{2}\) अपरिमेय है और (3) से भाग देने पर भी अपरिमेय रहता है। / \(\sqrt{2}\) is irrational and remains irrational after division by (3).
Step 3
Exam Tip
भिन्न के अंश और हर दोनों को जांचें। / Check both numerator and denominator of the fraction.
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कौन सा विकल्प अपरिमेय संख्या का सही दशमलव उदाहरण है?
Which option is a correct decimal example of an irrational number?
#decimal expansion
#non recurring
#class 10
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A \(0.3333\ldots\)
B \(2.12112111211112\ldots\) जिसमें कोई आवर्ती समूह नहीं है / \(2.12112111211112\ldots\) with no repeating block
C (5.75)
D \(1.272727\ldots\)
Explanation opens after your attempt
Correct Answer
B. \(2.12112111211112\ldots\) जिसमें कोई आवर्ती समूह नहीं है / \(2.12112111211112\ldots\) with no repeating block
Step 1
Concept
समाप्त या आवर्ती दशमलव परिमेय होते हैं। / Terminating or recurring decimals are rational.
Step 2
Why this answer is correct
दिया गया दशमलव अनंत है और उसमें कोई स्थिर आवर्ती समूह नहीं है इसलिए वह अपरिमेय है। / The given decimal is non-terminating and has no fixed repeating block, so it is irrational.
Step 3
Exam Tip
केवल लंबा दशमलव देखकर नहीं बल्कि आवृत्ति देखकर निर्णय लें। / Decide by checking repetition, not just by seeing many digits.
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यदि \(\sqrt{x}\) अपरिमेय है तो निम्न में से कौन सा निष्कर्ष सदैव सही है?
If \(\sqrt{x}\) is irrational then which conclusion is always correct?
#perfect square
#logical reasoning
#class 10
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A (x) पूर्ण वर्ग नहीं है / (x) is not a perfect square
B (x) ऋणात्मक है / (x) is negative
C (x) अभाज्य ही है / (x) is necessarily prime
D (x=0) है / (x=0)
Explanation opens after your attempt
Correct Answer
A. (x) पूर्ण वर्ग नहीं है / (x) is not a perfect square
Step 1
Concept
पूर्ण वर्ग का वर्गमूल पूर्णांक होता है और परिमेय होता है। / The square root of a perfect square is an integer and rational.
Step 2
Why this answer is correct
\(\sqrt{x}\) अपरिमेय है तो (x) पूर्ण वर्ग नहीं हो सकता। / If \(\sqrt{x}\) is irrational then (x) cannot be a perfect square.
Step 3
Exam Tip
अपरिमेय वर्गमूल के लिए पूर्ण वर्ग की जांच सबसे पहले करें। / For irrational square roots first check perfect-square status.
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कौन सा विकल्प \(\sqrt{2}\) को संख्या रेखा पर रखने की उचित विधि से जुड़ा है?
Which option is related to a proper method of locating \(\sqrt{2}\) on the number line?
#number line
#pythagoras
#irrational numbers
#class 10
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A समकोण त्रिभुज की भुजाएं (1) और (1) लेकर कर्ण बनाना / Make a right triangle with legs (1) and (1) and use the hypotenuse
B किसी भी स्थान पर बिंदु लगा देना / Mark any point anywhere
C \(\sqrt{2}\) को (2) मान लेना / Take \(\sqrt{2}\) as (2)
D केवल (0) और (1) के बीच रखना / Place it only between (0) and (1)
Explanation opens after your attempt
Correct Answer
A. समकोण त्रिभुज की भुजाएं (1) और (1) लेकर कर्ण बनाना / Make a right triangle with legs (1) and (1) and use the hypotenuse
Step 1
Concept
\(1^2+1^2=2\) होता है। / \(1^2+1^2=2\).
Step 2
Why this answer is correct
इसलिए ऐसे समकोण त्रिभुज का कर्ण \(\sqrt{2}\) होगा। / So the hypotenuse of that right triangle is \(\sqrt{2}\).
Step 3
Exam Tip
संख्या रेखा पर अपरिमेय संख्या रखने में पाइथागोरस प्रमेय उपयोगी है। / The Pythagoras theorem helps locate irrational numbers on a number line.
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कौन सा कथन \(\sqrt{2}\cdot \sqrt{3}\) के बारे में सही है?
Which statement is correct about \(\sqrt{2}\cdot \sqrt{3}\)?
#radical multiplication
#irrational numbers
#class 10
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A यह (5) है और परिमेय है / It is (5) and rational
B यह \(\sqrt{6}\) है और अपरिमेय है / It is \(\sqrt{6}\) and irrational
C यह \(\sqrt{5}\) है और अपरिमेय है / It is \(\sqrt{5}\) and irrational
D यह (6) है और परिमेय है / It is (6) and rational
Explanation opens after your attempt
Correct Answer
B. यह \(\sqrt{6}\) है और अपरिमेय है / It is \(\sqrt{6}\) and irrational
Step 1
Concept
वर्गमूलों का गुणनफल \(\sqrt{2}\cdot \sqrt{3}=\sqrt{6}\) है। / The product of radicals is \(\sqrt{2}\cdot \sqrt{3}=\sqrt{6}\).
Step 2
Why this answer is correct
(6) पूर्ण वर्ग नहीं है इसलिए \(\sqrt{6}\) अपरिमेय है। / Since (6) is not a perfect square \(\sqrt{6}\) is irrational.
Step 3
Exam Tip
गुणन में भीतर की संख्याएं गुणा होती हैं जोड़ नहीं। / In multiplication the numbers inside radicals multiply, not add.
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कौन सा विकल्प \(4+\sqrt{9}\) का सही प्रकार बताता है?
Which option correctly describes \(4+\sqrt{9}\)?
#perfect square
#rational number
#class 10
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A अपरिमेय क्योंकि वर्गमूल है / Irrational because it has a square root
B परिमेय क्योंकि \(4+\sqrt{9}=7\) / Rational because \(4+\sqrt{9}=7\)
C अपरिमेय क्योंकि (9) विषम है / Irrational because (9) is odd
D परिमेय नहीं क्योंकि (4) जोड़ा गया है / Not rational because (4) is added
Explanation opens after your attempt
Correct Answer
B. परिमेय क्योंकि \(4+\sqrt{9}=7\) / Rational because \(4+\sqrt{9}=7\)
Step 1
Concept
\(\sqrt{9}=3\)। / \(\sqrt{9}=3\).
Step 2
Why this answer is correct
(4+3=7) परिमेय संख्या है। / (4+3=7), a rational number.
Step 3
Exam Tip
वर्गमूल देखकर तुरंत अपरिमेय न मानें। / Do not call every expression with a square root irrational immediately.
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यदि (r) परिमेय है और (s) अपरिमेय है तो (r+s=s) कब संभव है?
If (r) is rational and (s) is irrational then when is (r+s=s) possible?
#rational irrational
#equation reasoning
#class 10
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A जब (r=0) / When (r=0)
B जब (s=0) / When (s=0)
C जब (r=1) / When (r=1)
D कभी नहीं / Never
Explanation opens after your attempt
Correct Answer
A. जब (r=0) / When (r=0)
Step 1
Concept
(r+s=s) से दोनों ओर (s) घटाने पर (r=0) मिलता है। / From (r+s=s), subtract (s) from both sides to get (r=0).
Step 2
Why this answer is correct
(0) परिमेय है इसलिए स्थिति संभव है। / (0) is rational, so the condition is possible.
Step 3
Exam Tip
सरल समीकरण बनाकर संख्या के प्रकार की जांच करें। / Form a simple equation before judging number types.
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कौन सा विकल्प \(\sqrt{98}-\sqrt{50}\) का सही रूप और प्रकार देता है?
Which option gives the correct form and type of \(\sqrt{98}-\sqrt{50}\)?
#surds subtraction
#irrational numbers
#class 10
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A \(2\sqrt{2}\) और अपरिमेय / \(2\sqrt{2}\) and irrational
B \(\sqrt{48}\) और अपरिमेय / \(\sqrt{48}\) and irrational
C (48) और परिमेय / (48) and rational
D (0) और परिमेय / (0) and rational
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{2}\) और अपरिमेय / \(2\sqrt{2}\) and irrational
Step 1
Concept
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\)। / \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\).
Step 2
Why this answer is correct
अंतर \(2\sqrt{2}\) है जो अपरिमेय है। / The difference is \(2\sqrt{2}\), which is irrational.
Step 3
Exam Tip
वर्गमूलों के अंदर की संख्याओं को सीधे घटाना गलत है। / Directly subtracting numbers inside radicals is wrong.
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कौन सा विकल्प \(\sqrt{2}+\sqrt{18}\) और \(\sqrt{8}+\sqrt{12}\) की तुलना के लिए सही है?
Which option is correct for comparing \(\sqrt{2}+\sqrt{18}\) and \(\sqrt{8}+\sqrt{12}\)?
#comparison
#surds
#irrational numbers
#class 10
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A पहला बड़ा है / The first is greater
B दूसरा बड़ा है / The second is greater
C दोनों बराबर हैं / Both are equal
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Correct Answer
A. पहला बड़ा है / The first is greater
Step 1
Concept
\(\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\)। / \(\sqrt{2}+\sqrt{18}=4\sqrt{2}\).
Step 2
Why this answer is correct
\(\sqrt{8}+\sqrt{12}=2\sqrt{2}+2\sqrt{3}\)। तुलना में \(4\sqrt{2}\) लगभग (5.66) और दूसरा लगभग (6.29) लगता है लेकिन शुद्ध तुलना में \(2\sqrt{2}\) और \(2\sqrt{3}\) के कारण दूसरा बड़ा है। / \(\sqrt{8}+\sqrt{12}=2\sqrt{2}+2\sqrt{3}\). Since \(\sqrt{3}>\sqrt{2}\), the second expression is greater.
Step 3
Exam Tip
अनुमान और सरल रूप दोनों जांचें। / Simplify first and compare carefully.
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निम्न में से कौन सा विकल्प परीक्षा में अपरिमेय संख्या पहचानने की सबसे अच्छी पहली जांच है?
Which option is the best first check for identifying an irrational number in an exam?
#exam tip
#irrational numbers
#class 10
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A क्या वह पूर्ण वर्ग के वर्गमूल में बदल रही है / Whether it becomes the square root of a perfect square
B क्या उसमें बड़ा अंक है / Whether it has a large digit
C क्या वह पृष्ठ के बीच में लिखी है / Whether it is written in the middle of the page
D क्या वह केवल धनात्मक है / Whether it is only positive
Explanation opens after your attempt
Correct Answer
A. क्या वह पूर्ण वर्ग के वर्गमूल में बदल रही है / Whether it becomes the square root of a perfect square
Step 1
Concept
वर्गमूल वाले प्रश्नों में पहले देखें कि अंदर की संख्या पूर्ण वर्ग है या नहीं। / In square-root questions first check whether the number inside is a perfect square.
Step 2
Why this answer is correct
पूर्ण वर्ग हो तो वर्गमूल परिमेय हो सकता है और पूर्ण वर्ग न हो तो अक्सर अपरिमेय होता है। / A perfect square may give a rational square root while a non-perfect square often gives an irrational value.
Step 3
Exam Tip
पहचान वाले प्रश्नों में सरल करना सबसे सुरक्षित शुरुआत है। / Simplifying is the safest first step in identification questions.
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यदि \(x=\sqrt{7}+\sqrt{28}\) है तो (x) का सही सरल रूप और प्रकार क्या है?
If \(x=\sqrt{7}+\sqrt{28}\), what is the correct simplified form and type of (x)?
#irrational numbers
#surds
#simplification
#class 10
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A \(3\sqrt{7}\) और अपरिमेय / \(3\sqrt{7}\) and irrational
B \(\sqrt{35}\) और अपरिमेय / \(\sqrt{35}\) and irrational
C (35) और परिमेय / (35) and rational
D \(5\sqrt{7}\) और अपरिमेय / \(5\sqrt{7}\) and irrational
Explanation opens after your attempt
Correct Answer
A. \(3\sqrt{7}\) और अपरिमेय / \(3\sqrt{7}\) and irrational
Step 1
Concept
\(28=4\cdot 7\) इसलिए \(\sqrt{28}=2\sqrt{7}\)। / Since \(28=4\cdot 7\), \(\sqrt{28}=2\sqrt{7}\).
Step 2
Why this answer is correct
अब \(\sqrt{7}+2\sqrt{7}=3\sqrt{7}\) और \(\sqrt{7}\) अपरिमेय है। / Now \(\sqrt{7}+2\sqrt{7}=3\sqrt{7}\), and \(\sqrt{7}\) is irrational.
Step 3
Exam Tip
परीक्षा में समान वर्गमूल वाले पदों को गुणांक जोड़कर सरल करें। / In exams, combine like radicals by adding their coefficients.
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कौन सा विकल्प सिद्ध करता है कि \(\frac{3}{2}+\sqrt{5}\) अपरिमेय है?
Which option proves that \(\frac{3}{2}+\sqrt{5}\) is irrational?
#rational plus irrational
#proof
#class 10
#real numbers
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A क्योंकि \(\frac{3}{2}\) अपरिमेय है / Because \(\frac{3}{2}\) is irrational
B क्योंकि परिमेय संख्या में अपरिमेय संख्या जोड़ने पर परिणाम अपरिमेय होता है / Because adding an irrational number to a rational number gives an irrational result
C क्योंकि \(\sqrt{5}\) पूर्णांक है / Because \(\sqrt{5}\) is an integer
D क्योंकि हर भिन्न अपरिमेय होती है / Because every fraction is irrational
Explanation opens after your attempt
Correct Answer
B. क्योंकि परिमेय संख्या में अपरिमेय संख्या जोड़ने पर परिणाम अपरिमेय होता है / Because adding an irrational number to a rational number gives an irrational result
Step 1
Concept
\(\frac{3}{2}\) परिमेय संख्या है और \(\sqrt{5}\) अपरिमेय संख्या है। / \(\frac{3}{2}\) is rational and \(\sqrt{5}\) is irrational.
Step 2
Why this answer is correct
यदि उनका योग परिमेय मानें तो \(\sqrt{5}\) को दो परिमेय संख्याओं के अंतर के रूप में लिखना पड़ेगा जो असंभव है। / If their sum were rational, then \(\sqrt{5}\) would become the difference of two rational numbers, which is impossible.
Step 3
Exam Tip
परिमेय और अपरिमेय के योग वाले प्रश्नों में विरोधाभास विधि बहुत उपयोगी है। / For rational-plus-irrational questions, contradiction is a very useful method.
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कौन सी संख्या परिमेय है, जबकि उसमें अपरिमेय वर्गमूल दिखाई दे रहे हैं?
Which number is rational even though irrational square roots appear in it?
#conjugates
#irrational numbers
#rational result
#class 10
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A (\(\sqrt{6}+\sqrt{2}\)\(\sqrt{6}-\sqrt{2}\))
B \(\sqrt{6}+\sqrt{2}\)
C \(\sqrt{6}-\sqrt{2}\)
D \(\sqrt{12}+\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{6}+\sqrt{2}\)\(\sqrt{6}-\sqrt{2}\))
Step 1
Concept
पहला विकल्प संयुग्मी पदों का गुणनफल है। / The first option is a product of conjugate terms.
Step 2
Why this answer is correct
(\(\sqrt{6}+\sqrt{2}\)\(\sqrt{6}-\sqrt{2}\)=6-2=4), जो परिमेय है। / (\(\sqrt{6}+\sqrt{2}\)\(\sqrt{6}-\sqrt{2}\)=6-2=4), which is rational.
Step 3
Exam Tip
संयुग्मी पद पहचानने से वर्गमूल जल्दी हट जाते हैं। / Identifying conjugates helps remove radicals quickly.
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यदि (0<a<1) और (a) परिमेय है, तो \(\sqrt{a}\) कब निश्चित रूप से परिमेय होगी?
If (0<a<1) and (a) is rational, when will \(\sqrt{a}\) definitely be rational?
#rational fractions
#square root
#irrational numbers
#class 10
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A जब (a) को दो पूर्ण वर्गों के अनुपात के रूप में लिखा जा सके / When (a) can be written as a ratio of two perfect squares
B जब (a) केवल दशमलव में लिखा हो / When (a) is only written in decimal form
C जब (a) (1) से छोटा हो / When (a) is less than (1)
D जब (a) धनात्मक हो / When (a) is positive
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Correct Answer
A. जब (a) को दो पूर्ण वर्गों के अनुपात के रूप में लिखा जा सके / When (a) can be written as a ratio of two perfect squares
Step 1
Concept
किसी परिमेय भिन्न का वर्गमूल परिमेय तब होता है जब अंश और हर दोनों पूर्ण वर्ग बन सकें। / The square root of a rational fraction is rational when both numerator and denominator can be perfect squares.
Step 2
Why this answer is correct
जैसे \(\sqrt{\frac{4}{9}}=\frac{2}{3}\), इसलिए दो पूर्ण वर्गों का अनुपात सुरक्षित स्थिति है। / For example, \(\sqrt{\frac{4}{9}}=\frac{2}{3}\), so a ratio of two perfect squares is a safe condition.
Step 3
Exam Tip
केवल धनात्मक या (1) से छोटा होना परिमेय वर्गमूल की गारंटी नहीं देता। / Being positive or less than (1) does not guarantee a rational square root.
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