यदि \(x=2+\sqrt{3}\) है तो (x-2) किस प्रकार की संख्या है?
If \(x=2+\sqrt{3}\), what type of number is (x-2)?
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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
\(x-2=\sqrt{3}\), which is irrational. First cancel matching rational terms in such questions.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. \(x-2=\sqrt{3}\), which is irrational. First cancel matching rational terms in such questions.
Step 3
Exam Tip
\(x-2=\sqrt{3}\) है जो अपरिमेय है। प्रश्न में पहले समान परिमेय पद हटाएं।
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\(\sqrt{18}+\sqrt{8}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{18}+\sqrt{8}\)?
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A \(3\sqrt{2}\)
B \(4\sqrt{2}\)
C \(5\sqrt{2}\)
D \(\sqrt{26}\)
Explanation opens after your attempt
Correct Answer
C. \(5\sqrt{2}\)
Step 1
Concept
\(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the sum is \(5\sqrt{2}\). Simplify radicals first.
Step 2
Why this answer is correct
The correct answer is C. \(5\sqrt{2}\). \(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the sum is \(5\sqrt{2}\). Simplify radicals first.
Step 3
Exam Tip
\(\sqrt{18}=3\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\), इसलिए योग \(5\sqrt{2}\) है। पहले मूलों को सरल करें।
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(\(\sqrt{7}\)2 +3) का मान किस प्रकार की संख्या है?
What type of number is the value of (\(\sqrt{7}\)2 +3)?
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A परिमेय / Rational
B अपरिमेय / Irrational
C न वास्तविक / Not real
D अनिर्धारित / Undefined
Explanation opens after your attempt
Correct Answer
A. परिमेय / Rational
Step 1
Concept
(\(\sqrt{7}\)2 +3=10), which is rational. The square of an irrational can be rational.
Step 2
Why this answer is correct
The correct answer is A. परिमेय / Rational. (\(\sqrt{7}\)2 +3=10), which is rational. The square of an irrational can be rational.
Step 3
Exam Tip
(\(\sqrt{7}\)2 +3=10) है जो परिमेय है। अपरिमेय का वर्ग परिमेय हो सकता है।
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\(\frac{1}{2+\sqrt{3}}\) को सरल करने पर क्या मिलता है?
What is obtained after simplifying \(\frac{1}{2+\sqrt{3}}\)?
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A \(2+\sqrt{3}\)
B \(\sqrt{3}-2\)
C \(1+\sqrt{3}\)
D \(2-\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
D. \(2-\sqrt{3}\)
Step 1
Concept
Rationalising the denominator gives \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\). Remember to multiply by the conjugate.
Step 2
Why this answer is correct
The correct answer is D. \(2-\sqrt{3}\). Rationalising the denominator gives \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\). Remember to multiply by the conjugate.
Step 3
Exam Tip
हर का परिमेयकरण करने पर \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\) मिलता है। संयुग्मी से गुणा करना याद रखें।
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यदि \(a=\sqrt{5}+1\) और \(b=\sqrt{5}-1\) हैं तो (a+b) क्या है?
If \(a=\sqrt{5}+1\) and \(b=\sqrt{5}-1\), what is (a+b)?
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A (2)
B \(2\sqrt{5}\)
C \(\sqrt{10}\)
D (5)
Explanation opens after your attempt
Correct Answer
B. \(2\sqrt{5}\)
Step 1
Concept
\(a+b=2\sqrt{5}\), which is irrational. Add like radical terms.
Step 2
Why this answer is correct
The correct answer is B. \(2\sqrt{5}\). \(a+b=2\sqrt{5}\), which is irrational. Add like radical terms.
Step 3
Exam Tip
\(a+b=2\sqrt{5}\) है जो अपरिमेय है। समान मूल वाले पदों को जोड़ें।
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\(\sqrt{50}-\sqrt{2}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{50}-\sqrt{2}\)?
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A (5)
B \(3\sqrt{2}\)
C \(4\sqrt{2}\)
D \(\sqrt{48}\)
Explanation opens after your attempt
Correct Answer
C. \(4\sqrt{2}\)
Step 1
Concept
\(\sqrt{50}=5\sqrt{2}\), so the difference is \(4\sqrt{2}\). Subtracting like radicals is the easy method.
Step 2
Why this answer is correct
The correct answer is C. \(4\sqrt{2}\). \(\sqrt{50}=5\sqrt{2}\), so the difference is \(4\sqrt{2}\). Subtracting like radicals is the easy method.
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\), इसलिए अंतर \(4\sqrt{2}\) है। समान मूलों को घटाना आसान तरीका है।
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\(\sqrt{3}+\sqrt{12}+\sqrt{27}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{3}+\sqrt{12}+\sqrt{27}\)?
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A \(4\sqrt{3}\)
B \(5\sqrt{3}\)
C \(7\sqrt{3}\)
D \(6\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
D. \(6\sqrt{3}\)
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the total is \(6\sqrt{3}\). First convert all terms into like radicals.
Step 2
Why this answer is correct
The correct answer is D. \(6\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the total is \(6\sqrt{3}\). First convert all terms into like radicals.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए कुल \(6\sqrt{3}\) है। सभी पदों को पहले समान मूल में बदलें।
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दशमलव \(1.02002000200002\ldots\) किस प्रकार की संख्या दर्शाता है?
What type of number does decimal \(1.02002000200002\ldots\) represent?
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A सांत परिमेय / Terminating rational
B अपरिमेय / Irrational
C आवर्ती परिमेय / Repeating rational
D पूर्णांक / Integer
Explanation opens after your attempt
Correct Answer
B. अपरिमेय / Irrational
Step 1
Concept
It has no fixed repeating block, so it is irrational. Avoid treating a non-repeating decimal as repeating.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. It has no fixed repeating block, so it is irrational. Avoid treating a non-repeating decimal as repeating.
Step 3
Exam Tip
इसमें निश्चित दोहराने वाला खंड नहीं है इसलिए यह अपरिमेय है। अनावर्ती दशमलव को आवर्ती समझने से बचें।
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\(\frac{\sqrt{27}}{\sqrt{3}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{27}}{\sqrt{3}}\)?
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A (3)
B \(\sqrt{24}\)
C (9)
D \(3\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{9}=3\). In division take the ratio of the numbers inside roots.
Step 2
Why this answer is correct
The correct answer is A. (3). \(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{9}=3\). In division take the ratio of the numbers inside roots.
Step 3
Exam Tip
\(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{9}=3\) है। भाग में अंदर की संख्याओं का अनुपात लें।
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\(\sqrt{45}+\sqrt{20}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{45}+\sqrt{20}\)?
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A \(3\sqrt{5}\)
B \(4\sqrt{5}\)
C \(7\sqrt{5}\)
D \(5\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
D. \(5\sqrt{5}\)
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the sum is \(5\sqrt{5}\). First take out perfect-square factors.
Step 2
Why this answer is correct
The correct answer is D. \(5\sqrt{5}\). \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the sum is \(5\sqrt{5}\). First take out perfect-square factors.
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\), इसलिए योग \(5\sqrt{5}\) है। पहले पूर्ण वर्ग गुणनखंड निकालें।
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\(\sqrt{2}\) और \(\sqrt{8}\) के बीच कौन-सी संख्या है?
Which number lies between \(\sqrt{2}\) and \(\sqrt{8}\)?
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A (1)
B (3)
C \(\sqrt{5}\)
D (4)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{5}\)
Step 1
Concept
Since (2<5<8), \(\sqrt{5}\) lies between them. Compare square roots using the numbers inside.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{5}\). Since (2<5<8), \(\sqrt{5}\) lies between them. Compare square roots using the numbers inside.
Step 3
Exam Tip
क्योंकि (2<5<8), इसलिए \(\sqrt{5}\) इनके बीच है। वर्गमूल की तुलना अंदर की संख्याओं से करें।
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यदि \(x=\sqrt{11}\) है तो \(x^2-1\) का मान क्या है?
If \(x=\sqrt{11}\), what is the value of \(x^2-1\)?
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A (9)
B (10)
C \(\sqrt{10}\)
D \(11\sqrt{11}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=11\), so \(x^2-1=10\). First square the given radical.
Step 2
Why this answer is correct
The correct answer is B. (10). \(x^2=11\), so \(x^2-1=10\). First square the given radical.
Step 3
Exam Tip
\(x^2=11\) इसलिए \(x^2-1=10\) है। पहले दिए गए मूल का वर्ग निकालें।
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कौन-सा विकल्प अपरिमेय संख्या है?
Which option is an irrational number?
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A \(0.101001000100001\ldots\)
B \(0.454545\ldots\)
C (0.75)
D (11 / 13)
Explanation opens after your attempt
Correct Answer
A. \(0.101001000100001\ldots\)
Step 1
Concept
The first decimal is non-terminating and non-repeating, so it is irrational. Repeating decimals are rational.
Step 2
Why this answer is correct
The correct answer is A. \(0.101001000100001\ldots\). The first decimal is non-terminating and non-repeating, so it is irrational. Repeating decimals are rational.
Step 3
Exam Tip
पहला दशमलव असांत और अनावर्ती है इसलिए अपरिमेय है। आवर्ती दशमलव परिमेय होते हैं।
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\(\frac{2}{\sqrt{5}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{2}{\sqrt{5}}\)?
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A \(\frac{2\sqrt{5}}{5}\)
B \(2\sqrt{5}\)
C \(\frac{\sqrt{5}}{2}\)
D \(\frac{5}{2\sqrt{5}}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{2\sqrt{5}}{5}\)
Step 1
Concept
Multiply by \(\sqrt{5}\) to make the denominator rational. The result is \(\frac{2\sqrt{5}}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2\sqrt{5}}{5}\). Multiply by \(\sqrt{5}\) to make the denominator rational. The result is \(\frac{2\sqrt{5}}{5}\).
Step 3
Exam Tip
हर को परिमेय बनाने के लिए \(\sqrt{5}\) से गुणा करें। परिणाम \(\frac{2\sqrt{5}}{5}\) है।
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(\(\sqrt{10}+\sqrt{2}\)2 ) का प्रसार कौन-सा है?
Which is the expansion of (\(\sqrt{10}+\sqrt{2}\)2 )?
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A \(12+2\sqrt{20}\)
B \(12+4\sqrt{5}\)
C \(10+2\sqrt{2}\)
D \(8+4\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(12+4\sqrt{5}\)
Step 1
Concept
The expansion is \(10+2+2\sqrt{20}=12+4\sqrt{5}\). Do not forget to simplify the middle term.
Step 2
Why this answer is correct
The correct answer is B. \(12+4\sqrt{5}\). The expansion is \(10+2+2\sqrt{20}=12+4\sqrt{5}\). Do not forget to simplify the middle term.
Step 3
Exam Tip
विस्तार \(10+2+2\sqrt{20}=12+4\sqrt{5}\) है। मध्य पद को सरल करना न भूलें।
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किस विकल्प का मान परिमेय है?
Which option has a rational value?
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A \(\sqrt{2}+\sqrt{3}\)
B \(\sqrt{7}+2\)
C (\(\sqrt{13}\)2 )
D \(\sqrt{5}-1\)
Explanation opens after your attempt
Correct Answer
C. (\(\sqrt{13}\)2 )
Step 1
Concept
(\(\sqrt{13}\)2 =13), which is rational. Squaring can remove the radical.
Step 2
Why this answer is correct
The correct answer is C. (\(\sqrt{13}\)2 ). (\(\sqrt{13}\)2 =13), which is rational. Squaring can remove the radical.
Step 3
Exam Tip
(\(\sqrt{13}\)2 =13) है जो परिमेय है। वर्ग करने से मूल हट सकता है।
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\(4+\sqrt{6}\) का अपरिमेय भाग कौन-सा है?
What is the irrational part of \(4+\sqrt{6}\)?
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A (4)
B (6)
C \(4\sqrt{6}\)
D \(\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
D. \(\sqrt{6}\)
Step 1
Concept
(4) is rational and \(\sqrt{6}\) is irrational. In a mixed form the radical term is the irrational part.
Step 2
Why this answer is correct
The correct answer is D. \(\sqrt{6}\). (4) is rational and \(\sqrt{6}\) is irrational. In a mixed form the radical term is the irrational part.
Step 3
Exam Tip
(4) परिमेय है और \(\sqrt{6}\) अपरिमेय है। मिश्रित रूप में मूल वाला पद अपरिमेय भाग होता है।
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\(\sqrt{28}+\sqrt{63}-\sqrt{7}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{28}+\sqrt{63}-\sqrt{7}\)?
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A \(3\sqrt{7}\)
B \(5\sqrt{7}\)
C \(4\sqrt{7}\)
D \(7\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
C. \(4\sqrt{7}\)
Step 1
Concept
\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so \(2\sqrt{7}+3\sqrt{7}-\sqrt{7}=4\sqrt{7}\). Add and subtract like radicals.
Step 2
Why this answer is correct
The correct answer is C. \(4\sqrt{7}\). \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so \(2\sqrt{7}+3\sqrt{7}-\sqrt{7}=4\sqrt{7}\). Add and subtract like radicals.
Step 3
Exam Tip
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए \(2\sqrt{7}+3\sqrt{7}-\sqrt{7}=4\sqrt{7}\) है। समान मूलों को जोड़ें और घटाएं।
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यदि \(\sqrt{m}=6\) है तो (m) का मान क्या है?
If \(\sqrt{m}=6\), what is the value of (m)?
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A (36)
B (12)
C (6)
D \(\sqrt{6}\)
Explanation opens after your attempt
Step 1
Concept
Squaring both sides gives (m=36). Identifying perfect squares is essential.
Step 2
Why this answer is correct
The correct answer is A. (36). Squaring both sides gives (m=36). Identifying perfect squares is essential.
Step 3
Exam Tip
दोनों पक्षों का वर्ग करने पर (m=36) मिलता है। पूर्ण वर्ग पहचानना आवश्यक है।
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यदि (n) पूर्ण वर्ग नहीं है और (n>0) है तो \(\sqrt{n}\) सामान्यतः कैसी संख्या होगी?
If (n) is not a perfect square and (n>0), what type of number will \(\sqrt{n}\) generally be?
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A पूर्णांक / Integer
B सांत दशमलव / Terminating decimal
C परिमेय / Rational
D अपरिमेय / Irrational
Explanation opens after your attempt
Correct Answer
D. अपरिमेय / Irrational
Step 1
Concept
The square root of a positive integer is irrational when it is not a perfect square. This is the main rule for identifying square roots.
Step 2
Why this answer is correct
The correct answer is D. अपरिमेय / Irrational. The square root of a positive integer is irrational when it is not a perfect square. This is the main rule for identifying square roots.
Step 3
Exam Tip
धनात्मक पूर्ण संख्या का वर्गमूल अपरिमेय होता है जब वह पूर्ण वर्ग नहीं होती। यही वर्गमूल पहचान का मुख्य नियम है।
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\(\sqrt{12}\times\sqrt{27}\) का मान क्या है?
What is the value of \(\sqrt{12}\times\sqrt{27}\)?
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A \(9\sqrt{3}\)
B (18)
C \(6\sqrt{3}\)
D \(\sqrt{39}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}\times\sqrt{27}=\sqrt{324}=18\). The product of two irrationals can be rational.
Step 2
Why this answer is correct
The correct answer is B. (18). \(\sqrt{12}\times\sqrt{27}=\sqrt{324}=18\). The product of two irrationals can be rational.
Step 3
Exam Tip
\(\sqrt{12}\times\sqrt{27}=\sqrt{324}=18\) है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है।
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\(\sqrt{2}+\sqrt{3}\) के बारे में सही कथन कौन-सा है?
Which statement is correct about \(\sqrt{2}+\sqrt{3}\)?
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A यह परिमेय है / It is rational
B यह पूर्णांक है / It is an integer
C यह अपरिमेय है / It is irrational
D यह शून्य है / It is zero
Explanation opens after your attempt
Correct Answer
C. यह अपरिमेय है / It is irrational
Step 1
Concept
\(\sqrt{2}\) and \(\sqrt{3}\) are different irrational radicals and their sum is irrational. It is not written as \(\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is C. यह अपरिमेय है / It is irrational. \(\sqrt{2}\) and \(\sqrt{3}\) are different irrational radicals and their sum is irrational. It is not written as \(\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{2}\) और \(\sqrt{3}\) अलग अपरिमेय मूल हैं और उनका योग अपरिमेय है। इसे \(\sqrt{5}\) नहीं लिखते।
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\(\sqrt{200}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{200}\)?
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A \(10\sqrt{2}\)
B \(20\sqrt{2}\)
C \(5\sqrt{8}\)
D \(100\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(10\sqrt{2}\)
Step 1
Concept
\(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}\). Using a large perfect-square factor makes the solution shorter.
Step 2
Why this answer is correct
The correct answer is A. \(10\sqrt{2}\). \(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}\). Using a large perfect-square factor makes the solution shorter.
Step 3
Exam Tip
\(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}\) है। बड़े पूर्ण वर्ग गुणनखंड से हल छोटा होता है।
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\(\frac{\sqrt{80}}{\sqrt{5}}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sqrt{80}}{\sqrt{5}}\)?
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A (8)
B (4)
C \(\sqrt{75}\)
D (16)
Explanation opens after your attempt
Step 1
Concept
\(\frac{\sqrt{80}}{\sqrt{5}}=\sqrt{16}=4\). Combine roots in division and simplify.
Step 2
Why this answer is correct
The correct answer is B. (4). \(\frac{\sqrt{80}}{\sqrt{5}}=\sqrt{16}=4\). Combine roots in division and simplify.
Step 3
Exam Tip
\(\frac{\sqrt{80}}{\sqrt{5}}=\sqrt{16}=4\) है। भाग में मूलों को मिलाकर सरल करें।
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कौन-सा विकल्प असांत और आवर्ती दशमलव है?
Which option is a non-terminating and repeating decimal?
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A \(0.123123123\ldots\)
B \(0.1010010001\ldots\)
C \(\sqrt{2}\) का दशमलव / Decimal of \(\sqrt{2}\)
D (0.25)
Explanation opens after your attempt
Correct Answer
A. \(0.123123123\ldots\)
Step 1
Concept
(123) repeats again and again, so it is repeating and rational. A repeating decimal is not irrational.
Step 2
Why this answer is correct
The correct answer is A. \(0.123123123\ldots\). (123) repeats again and again, so it is repeating and rational. A repeating decimal is not irrational.
Step 3
Exam Tip
(123) बार-बार दोहरता है इसलिए यह आवर्ती और परिमेय है। आवर्ती दशमलव अपरिमेय नहीं होता।
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\(5\sqrt{3}-2\sqrt{3}\) का परिणाम क्या है?
What is the result of \(5\sqrt{3}-2\sqrt{3}\)?
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A \(7\sqrt{3}\)
B \(\sqrt{3}\)
C \(3\sqrt{3}\)
D (3)
Explanation opens after your attempt
Correct Answer
C. \(3\sqrt{3}\)
Step 1
Concept
Subtracting coefficients of like radicals gives \(3\sqrt{3}\). It is irrational.
Step 2
Why this answer is correct
The correct answer is C. \(3\sqrt{3}\). Subtracting coefficients of like radicals gives \(3\sqrt{3}\). It is irrational.
Step 3
Exam Tip
समान मूलों के गुणांक घटाने पर \(3\sqrt{3}\) मिलता है। यह अपरिमेय है।
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(\sqrt{3}\(\sqrt{3}+\sqrt{12}\)) का मान क्या है?
What is the value of (\sqrt{3}\(\sqrt{3}+\sqrt{12}\))?
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A (9)
B (3+6)
C \(3+3\sqrt{4}\)
D (9)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\), so the bracket is \(3\sqrt{3}\) and the product is (9). Simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is D. (9). \(\sqrt{12}=2\sqrt{3}\), so the bracket is \(3\sqrt{3}\) and the product is (9). Simplify the bracket first.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\), इसलिए अंदर \(3\sqrt{3}\) है और गुणनफल (9) है। पहले कोष्ठक को सरल करें।
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\(\sqrt{17}\) और \(\sqrt{18}\) की तुलना में कौन-सा कथन सही है?
Which statement is correct when comparing \(\sqrt{17}\) and \(\sqrt{18}\)?
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A \(\sqrt{17}<\sqrt{18}\)
B \(\sqrt{17}=\sqrt{18}\)
C \(\sqrt{17}>\sqrt{18}\)
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{17}<\sqrt{18}\)
Step 1
Concept
Since (17<18), \(\sqrt{17}<\sqrt{18}\). Compare positive roots using the numbers inside.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{17}<\sqrt{18}\). Since (17<18), \(\sqrt{17}<\sqrt{18}\). Compare positive roots using the numbers inside.
Step 3
Exam Tip
क्योंकि (17<18), इसलिए \(\sqrt{17}<\sqrt{18}\)। धनात्मक मूलों की तुलना अंदर की संख्या से करें।
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(\sqrt{2}\times\(3\sqrt{2}+1\)) का सरल रूप क्या है?
What is the simplified form of (\sqrt{2}\times\(3\sqrt{2}+1\))?
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A \(3+\sqrt{2}\)
B \(6+\sqrt{2}\)
C \(6+2\sqrt{2}\)
D \(3\sqrt{2}+2\)
Explanation opens after your attempt
Correct Answer
B. \(6+\sqrt{2}\)
Step 1
Concept
Distributing gives \(3\times2+\sqrt{2}=6+\sqrt{2}\). Multiply radical terms carefully.
Step 2
Why this answer is correct
The correct answer is B. \(6+\sqrt{2}\). Distributing gives \(3\times2+\sqrt{2}=6+\sqrt{2}\). Multiply radical terms carefully.
Step 3
Exam Tip
वितरण करने पर \(3\times2+\sqrt{2}=6+\sqrt{2}\) मिलता है। मूल का गुणन ध्यान से करें।
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\(\frac{3}{\sqrt{2}+1}\) का परिमेयकृत रूप कौन-सा है?
Which is the rationalised form of \(\frac{3}{\sqrt{2}+1}\)?
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A (3\(\sqrt{2}+1\))
B \(3\sqrt{2}+3\)
C \(3\sqrt{2}-3\)
D \(\frac{3}{\sqrt{2}-1}\)
Explanation opens after your attempt
Correct Answer
C. \(3\sqrt{2}-3\)
Step 1
Concept
Multiplying by the conjugate \(\sqrt{2}-1\) gives (3\(\sqrt{2}-1\)). So the form is \(3\sqrt{2}-3\).
Step 2
Why this answer is correct
The correct answer is C. \(3\sqrt{2}-3\). Multiplying by the conjugate \(\sqrt{2}-1\) gives (3\(\sqrt{2}-1\)). So the form is \(3\sqrt{2}-3\).
Step 3
Exam Tip
हर के संयुग्मी \(\sqrt{2}-1\) से गुणा करने पर (3\(\sqrt{2}-1\)) मिलता है। इसलिए रूप \(3\sqrt{2}-3\) है।
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\(7+\sqrt{13}\) और \(7-\sqrt{13}\) का योग क्या है?
What is the sum of \(7+\sqrt{13}\) and \(7-\sqrt{13}\)?
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A (14)
B \(2\sqrt{13}\)
C (7)
D (13)
Explanation opens after your attempt
Step 1
Concept
The irrational terms cancel and the sum is (14). The sum of conjugate numbers can be rational.
Step 2
Why this answer is correct
The correct answer is A. (14). The irrational terms cancel and the sum is (14). The sum of conjugate numbers can be rational.
Step 3
Exam Tip
अपरिमेय पद कट जाते हैं और योग (14) है। संयुग्मी संख्याओं का योग परिमेय हो सकता है।
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\(7+\sqrt{13}\) और \(7-\sqrt{13}\) का गुणनफल क्या है?
What is the product of \(7+\sqrt{13}\) and \(7-\sqrt{13}\)?
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A \(49+\sqrt{13}\)
B (36)
C (62)
D \(14\sqrt{13}\)
Explanation opens after your attempt
Step 1
Concept
The product is (72 -\(\sqrt{13}\)2 =49-13=36). Use \(a^2-b^2\) for conjugate multiplication.
Step 2
Why this answer is correct
The correct answer is B. (36). The product is (72 -\(\sqrt{13}\)2 =49-13=36). Use \(a^2-b^2\) for conjugate multiplication.
Step 3
Exam Tip
गुणनफल (72 -\(\sqrt{13}\)2 =49-13=36) है। संयुग्मी गुणन में \(a^2-b^2\) लगाएं।
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यदि \(p=4+\sqrt{5}\) है तो (p) किस प्रकार की संख्या है?
If \(p=4+\sqrt{5}\), what type of number is (p)?
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A पूर्णांक / Integer
B परिमेय / Rational
C असांत आवर्ती / Non-terminating repeating
D अपरिमेय / Irrational
Explanation opens after your attempt
Correct Answer
D. अपरिमेय / Irrational
Step 1
Concept
Adding irrational \(\sqrt{5}\) to rational (4) gives an irrational number. The irrational part remains.
Step 2
Why this answer is correct
The correct answer is D. अपरिमेय / Irrational. Adding irrational \(\sqrt{5}\) to rational (4) gives an irrational number. The irrational part remains.
Step 3
Exam Tip
परिमेय (4) में अपरिमेय \(\sqrt{5}\) जोड़ने से अपरिमेय संख्या मिलती है। अपरिमेय भाग बचा रहता है।
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\(\sqrt{2}\) को संख्या रेखा पर बनाने के लिए किस लंबाई का कर्ण उपयोग होता है?
To construct \(\sqrt{2}\) on the number line, what hypotenuse length is used?
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A (1) और (2) भुजाओं वाला कर्ण / Hypotenuse with legs (1) and (2)
B (1) और (1) भुजाओं वाला कर्ण / Hypotenuse with legs (1) and (1)
C (2) और (2) भुजाओं वाला कर्ण / Hypotenuse with legs (2) and (2)
D (3) और (1) भुजाओं वाला कर्ण / Hypotenuse with legs (3) and (1)
Explanation opens after your attempt
Correct Answer
B. (1) और (1) भुजाओं वाला कर्ण / Hypotenuse with legs (1) and (1)
Step 1
Concept
In a right triangle, legs (1) and (1) give hypotenuse \(\sqrt{2}\). Use Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is B. (1) और (1) भुजाओं वाला कर्ण / Hypotenuse with legs (1) and (1). In a right triangle, legs (1) and (1) give hypotenuse \(\sqrt{2}\). Use Pythagoras theorem.
Step 3
Exam Tip
समकोण त्रिभुज में (1) और (1) भुजाओं का कर्ण \(\sqrt{2}\) होता है। पाइथागोरस प्रमेय का उपयोग करें।
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\(\sqrt{2}\times\sqrt{8}\) किस प्रकार की संख्या है?
What type of number is \(\sqrt{2}\times\sqrt{8}\)?
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A अपरिमेय / Irrational
B परिमेय / Rational
C अनावर्ती दशमलव / Non-repeating decimal
D ऋणात्मक / Negative
Explanation opens after your attempt
Correct Answer
B. परिमेय / Rational
Step 1
Concept
\(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. The product of two irrationals can be rational.
Step 2
Why this answer is correct
The correct answer is B. परिमेय / Rational. \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. The product of two irrationals can be rational.
Step 3
Exam Tip
\(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\) है जो परिमेय है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है।
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\(\sqrt{20}\div\sqrt{2}\) का मान क्या है?
What is the value of \(\sqrt{20}\div\sqrt{2}\)?
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A (5)
B \(\sqrt{10}\)
C (10)
D \(\sqrt{18}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{10}\)
Step 1
Concept
\(\sqrt{20}\div\sqrt{2}=\sqrt{10}\), which is irrational. Write the roots in division as one root.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{10}\). \(\sqrt{20}\div\sqrt{2}=\sqrt{10}\), which is irrational. Write the roots in division as one root.
Step 3
Exam Tip
\(\sqrt{20}\div\sqrt{2}=\sqrt{10}\) है जो अपरिमेय है। भाग में मूलों को एक मूल में लिखें।
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यदि एक वर्ग का क्षेत्रफल (18) वर्ग इकाई है तो उसकी भुजा का सरल रूप क्या होगा?
If the area of a square is (18) square units, what will be the simplified form of its side?
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A \(9\sqrt{2}\)
B (6)
C \(3\sqrt{2}\)
D \(2\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
C. \(3\sqrt{2}\)
Step 1
Concept
The side will be \(\sqrt{18}=3\sqrt{2}\). In a square, side equals the square root of area.
Step 2
Why this answer is correct
The correct answer is C. \(3\sqrt{2}\). The side will be \(\sqrt{18}=3\sqrt{2}\). In a square, side equals the square root of area.
Step 3
Exam Tip
भुजा \(\sqrt{18}=3\sqrt{2}\) होगी। वर्ग में भुजा क्षेत्रफल के वर्गमूल के बराबर होती है।
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\(\frac{1}{3-\sqrt{2}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{1}{3-\sqrt{2}}\)?
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A \(\frac{3+\sqrt{2}}{7}\)
B \(\frac{3-\sqrt{2}}{7}\)
C \(3+\sqrt{2}\)
D \(\frac{1}{3+\sqrt{2}}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{3+\sqrt{2}}{7}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (9-2=7). So the form is \(\frac{3+\sqrt{2}}{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3+\sqrt{2}}{7}\). Multiplying by the conjugate makes the denominator (9-2=7). So the form is \(\frac{3+\sqrt{2}}{7}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (9-2=7) बनता है। इसलिए रूप \(\frac{3+\sqrt{2}}{7}\) है।
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\(\sqrt{3}+\sqrt{27}\) और \(\sqrt{48}\) के बारे में सही कथन कौन-सा है?
Which statement is correct about \(\sqrt{3}+\sqrt{27}\) and \(\sqrt{48}\)?
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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
C. दोनों बराबर हैं / Both are equal
Step 1
Concept
\(\sqrt{3}+\sqrt{27}=4\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\). Simplified form makes comparison easy.
Step 2
Why this answer is correct
The correct answer is C. दोनों बराबर हैं / Both are equal. \(\sqrt{3}+\sqrt{27}=4\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\). Simplified form makes comparison easy.
Step 3
Exam Tip
\(\sqrt{3}+\sqrt{27}=4\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\) हैं। सरल रूप तुलना को आसान बनाता है।
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\(2\sqrt{5}+3\sqrt{20}\) का सरल रूप क्या है?
What is the simplified form of \(2\sqrt{5}+3\sqrt{20}\)?
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A \(6\sqrt{5}\)
B \(8\sqrt{5}\)
C \(10\sqrt{5}\)
D \(12\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(8\sqrt{5}\)
Step 1
Concept
\(\sqrt{20}=2\sqrt{5}\), so \(2\sqrt{5}+6\sqrt{5}=8\sqrt{5}\). Watch both coefficients and radicals carefully.
Step 2
Why this answer is correct
The correct answer is B. \(8\sqrt{5}\). \(\sqrt{20}=2\sqrt{5}\), so \(2\sqrt{5}+6\sqrt{5}=8\sqrt{5}\). Watch both coefficients and radicals carefully.
Step 3
Exam Tip
\(\sqrt{20}=2\sqrt{5}\), इसलिए \(2\sqrt{5}+6\sqrt{5}=8\sqrt{5}\) है। गुणांक और मूल दोनों ध्यान से देखें।
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किस विकल्प का दशमलव प्रसार सांत होगा?
Which option will have a terminating decimal expansion?
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A \(\sqrt{2}\)
B \(5+\sqrt{3}\)
C \(\pi\)
D \(\frac{7}{16}\)
Explanation opens after your attempt
Correct Answer
D. \(\frac{7}{16}\)
Step 1
Concept
\(\frac{7}{16}\) is rational and the denominator is \(2^4\), so the decimal will terminate. Check prime factors of the denominator.
Step 2
Why this answer is correct
The correct answer is D. \(\frac{7}{16}\). \(\frac{7}{16}\) is rational and the denominator is \(2^4\), so the decimal will terminate. Check prime factors of the denominator.
Step 3
Exam Tip
\(\frac{7}{16}\) परिमेय है और हर \(2^4\) है, इसलिए दशमलव सांत होगा। हर के अभाज्य गुणनखंड देखें।
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यदि \(a=\sqrt{2}+1\) और \(b=\sqrt{2}-1\) हैं तो (ab) क्या है?
If \(a=\sqrt{2}+1\) and \(b=\sqrt{2}-1\), what is (ab)?
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A (1)
B \(2\sqrt{2}\)
C (3)
D \(\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
(ab=\(\sqrt{2}\)2 -12 =1). The product of conjugate forms gives a rational result.
Step 2
Why this answer is correct
The correct answer is A. (1). (ab=\(\sqrt{2}\)2 -12 =1). The product of conjugate forms gives a rational result.
Step 3
Exam Tip
(ab=\(\sqrt{2}\)2 -12 =1) है। संयुग्मी रूप का गुणनफल परिमेय देता है।
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\(\sqrt{2}+\sqrt{32}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{2}+\sqrt{32}\)?
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A \(3\sqrt{2}\)
B \(5\sqrt{2}\)
C \(4\sqrt{2}\)
D \(\sqrt{34}\)
Explanation opens after your attempt
Correct Answer
B. \(5\sqrt{2}\)
Step 1
Concept
\(\sqrt{32}=4\sqrt{2}\), so the sum is \(5\sqrt{2}\). Simplify before adding like radicals.
Step 2
Why this answer is correct
The correct answer is B. \(5\sqrt{2}\). \(\sqrt{32}=4\sqrt{2}\), so the sum is \(5\sqrt{2}\). Simplify before adding like radicals.
Step 3
Exam Tip
\(\sqrt{32}=4\sqrt{2}\), इसलिए योग \(5\sqrt{2}\) है। समान मूलों को जोड़ने से पहले सरल करें।
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(\(2\sqrt{3}\)2 ) का मान क्या है?
What is the value of (\(2\sqrt{3}\)2 )?
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A (6)
B (8)
C (12)
D \(2\sqrt{9}\)
Explanation opens after your attempt
Step 1
Concept
(\(2\sqrt{3}\)2 =4\times3=12). Square both the coefficient and the radical.
Step 2
Why this answer is correct
The correct answer is C. (12). (\(2\sqrt{3}\)2 =4\times3=12). Square both the coefficient and the radical.
Step 3
Exam Tip
(\(2\sqrt{3}\)2 =4\times3=12) है। गुणांक और मूल दोनों का वर्ग करें।
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कौन-सा विकल्प हर का सही परिमेयकरण दिखाता है?
Which option shows correct rationalisation of the denominator?
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A \(\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}\)
B \(\frac{1}{\sqrt{7}}=\sqrt{7}\)
C \(\frac{1}{\sqrt{7}}=\frac{7}{\sqrt{7}}\)
D \(\frac{1}{\sqrt{7}}=7\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}\)
Step 1
Concept
Multiplying numerator and denominator by \(\sqrt{7}\) gives \(\frac{\sqrt{7}}{7}\). Rationalisation should not change the value.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{\sqrt{7}}=\frac{\sqrt{7}}{7}\). Multiplying numerator and denominator by \(\sqrt{7}\) gives \(\frac{\sqrt{7}}{7}\). Rationalisation should not change the value.
Step 3
Exam Tip
हर और अंश को \(\sqrt{7}\) से गुणा करने पर \(\frac{\sqrt{7}}{7}\) मिलता है। परिमेयकरण में मान नहीं बदलना चाहिए।
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यदि \(r=\sqrt{15}-\sqrt{6}\) और \(s=\sqrt{15}+\sqrt{6}\) हैं तो (rs) का मान क्या है?
If \(r=\sqrt{15}-\sqrt{6}\) and \(s=\sqrt{15}+\sqrt{6}\), what is the value of (rs)?
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A (21)
B (9)
C \(\sqrt{90}\)
D \(2\sqrt{15}\)
Explanation opens after your attempt
Step 1
Concept
In conjugate multiplication (rs=\(\sqrt{15}\)2 -\(\sqrt{6}\)2 =9). Use \(a^2-b^2\) in such questions.
Step 2
Why this answer is correct
The correct answer is B. (9). In conjugate multiplication (rs=\(\sqrt{15}\)2 -\(\sqrt{6}\)2 =9). Use \(a^2-b^2\) in such questions.
Step 3
Exam Tip
संयुग्मी गुणन में (rs=\(\sqrt{15}\)2 -\(\sqrt{6}\)2 =9) होता है। ऐसे प्रश्नों में \(a^2-b^2\) लगाएं।
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\(\frac{\sqrt{98}+\sqrt{50}}{\sqrt{2}}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sqrt{98}+\sqrt{50}}{\sqrt{2}}\)?
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A (12)
B (7)
C (5)
D \(\sqrt{148}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so division gives (12). Simplify the numerator first.
Step 2
Why this answer is correct
The correct answer is A. (12). \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so division gives (12). Simplify the numerator first.
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए भाग देने पर (12) मिलता है। पहले अंश को सरल करें।
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यदि \(\sqrt{k}\) संख्या (6) और (7) के बीच है, तो (k) के लिए कौन-सा मान संभव है?
If \(\sqrt{k}\) lies between (6) and (7), which value of (k) is possible?
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A (35)
B (49)
C (43)
D (64)
Explanation opens after your attempt
Step 1
Concept
Since (36<43<49), \(6<\sqrt{43}<7\). Decide bounds of square roots using squares.
Step 2
Why this answer is correct
The correct answer is C. (43). Since (36<43<49), \(6<\sqrt{43}<7\). Decide bounds of square roots using squares.
Step 3
Exam Tip
क्योंकि (36<43<49), इसलिए \(6<\sqrt{43}<7\)। वर्गमूल की सीमा वर्गों से तय करें।
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(\sqrt{5}\(2\sqrt{5}-3\sqrt{20}\)) का सरल रूप क्या है?
What is the simplified form of (\sqrt{5}\(2\sqrt{5}-3\sqrt{20}\))?
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A (-20)
B (20)
C \(-10\sqrt{5}\)
D \(10-3\sqrt{20}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{20}=2\sqrt{5}\), so inside becomes \(2\sqrt{5}-6\sqrt{5}=-4\sqrt{5}\) and the product is (-20). Simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is A. (-20). \(\sqrt{20}=2\sqrt{5}\), so inside becomes \(2\sqrt{5}-6\sqrt{5}=-4\sqrt{5}\) and the product is (-20). Simplify the bracket first.
Step 3
Exam Tip
\(\sqrt{20}=2\sqrt{5}\), इसलिए अंदर \(2\sqrt{5}-6\sqrt{5}=-4\sqrt{5}\) और गुणनफल (-20) है। पहले कोष्ठक सरल करें।
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कौन-सा विकल्प अपरिमेय संख्या का परिमेयकृत हर वाला बराबर रूप है?
Which option is an equal form of an irrational number with a rationalised denominator?
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A \(\frac{4}{\sqrt{11}}=\frac{4}{11}\)
B \(\frac{4}{\sqrt{11}}=\frac{\sqrt{11}}{4}\)
C \(\frac{4}{\sqrt{11}}=\frac{4\sqrt{11}}{11}\)
D \(\frac{4}{\sqrt{11}}=4\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{4}{\sqrt{11}}=\frac{4\sqrt{11}}{11}\)
Step 1
Concept
Multiplying numerator and denominator by \(\sqrt{11}\) gives \(\frac{4\sqrt{11}}{11}\). Rationalisation must keep the value equal.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{4}{\sqrt{11}}=\frac{4\sqrt{11}}{11}\). Multiplying numerator and denominator by \(\sqrt{11}\) gives \(\frac{4\sqrt{11}}{11}\). Rationalisation must keep the value equal.
Step 3
Exam Tip
अंश और हर को \(\sqrt{11}\) से गुणा करने पर \(\frac{4\sqrt{11}}{11}\) मिलता है। परिमेयकरण में बराबर मान रखना जरूरी है।
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