यदि \(a=3+\sqrt{5}\) और \(b=3-\sqrt{5}\) हैं, तो \(a^2-b^2\) का मान क्या है?
If \(a=3+\sqrt{5}\) and \(b=3-\sqrt{5}\), what is the value of \(a^2-b^2\)?
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A \(12\sqrt{5}\)
B \(6\sqrt{5}\)
C (18)
D \(4\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(12\sqrt{5}\)
Step 1
Concept
(a-2 -b-2 =(a-b)(a+b)), where \(a-b=2\sqrt{5}\) and (a+b=6). So the value is \(12\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{5}\). (a-2 -b-2 =(a-b)(a+b)), where \(a-b=2\sqrt{5}\) and (a+b=6). So the value is \(12\sqrt{5}\).
Step 3
Exam Tip
(a-2 -b-2 =(a-b)(a+b)), जहाँ \(a-b=2\sqrt{5}\) और (a+b=6) है। इसलिए मान \(12\sqrt{5}\) है।
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यदि \(x=\frac{1}{\sqrt{2}+1}\) है, तो \(x+\sqrt{2}\) का मान क्या है?
If \(x=\frac{1}{\sqrt{2}+1}\), what is the value of \(x+\sqrt{2}\)?
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A (1)
B \(2\sqrt{2}-1\)
C (2)
D \(\sqrt{2}+1\)
Explanation opens after your attempt
Step 1
Concept
\(\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\). So (x+\sqrt{2}=\(\sqrt{2}-1\)+\sqrt{2}=2\sqrt{2}-1).
Step 2
Why this answer is correct
The correct answer is A. (1). \(\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\). So (x+\sqrt{2}=\(\sqrt{2}-1\)+\sqrt{2}=2\sqrt{2}-1).
Step 3
Exam Tip
\(\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\) होता है। इसलिए \(x+\sqrt{2}=2\sqrt{2}-1\) नहीं, बल्कि (\(\sqrt{2}-1\)+\sqrt{2}=2\sqrt{2}-1) है।
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\(\frac{2+\sqrt{3}}{2-\sqrt{3}}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\)?
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A \(7+4\sqrt{3}\)
B \(7-4\sqrt{3}\)
C \(1+\sqrt{3}\)
D \(4+\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(7+4\sqrt{3}\)
Step 1
Concept
Multiplying by the conjugate \(2+\sqrt{3}\) makes the denominator (1) and numerator (\(2+\sqrt{3}\)2 ). So the answer is \(7+4\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(7+4\sqrt{3}\). Multiplying by the conjugate \(2+\sqrt{3}\) makes the denominator (1) and numerator (\(2+\sqrt{3}\)2 ). So the answer is \(7+4\sqrt{3}\).
Step 3
Exam Tip
हर के संयुग्मी \(2+\sqrt{3}\) से गुणा करने पर हर (1) और अंश (\(2+\sqrt{3}\)2 ) बनता है। इसलिए उत्तर \(7+4\sqrt{3}\) है।
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\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{2}-1}\) का मान क्या है?
What is the value of \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{2}-1}\)?
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A (2)
B \(2\sqrt{2}\)
C \(\sqrt{2}\)
D (4)
Explanation opens after your attempt
Correct Answer
B. \(2\sqrt{2}\)
Step 1
Concept
The first term becomes \(\sqrt{2}-1\) and the second becomes \(\sqrt{2}+1\). Their sum is \(2\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(2\sqrt{2}\). The first term becomes \(\sqrt{2}-1\) and the second becomes \(\sqrt{2}+1\). Their sum is \(2\sqrt{2}\).
Step 3
Exam Tip
पहला पद \(\sqrt{2}-1\) और दूसरा \(\sqrt{2}+1\) बनता है। योग \(2\sqrt{2}\) है।
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यदि \(x=\sqrt{6}+\sqrt{2}\) है, तो \(x^2\) का सही मान क्या है?
If \(x=\sqrt{6}+\sqrt{2}\), what is the correct value of \(x^2\)?
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A \(8+4\sqrt{3}\)
B \(8+\sqrt{12}\)
C \(4+2\sqrt{3}\)
D \(12+4\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(8+4\sqrt{3}\)
Step 1
Concept
\(x^2=6+2+2\sqrt{12}=8+4\sqrt{3}\). Simplify the middle term correctly while squaring.
Step 2
Why this answer is correct
The correct answer is A. \(8+4\sqrt{3}\). \(x^2=6+2+2\sqrt{12}=8+4\sqrt{3}\). Simplify the middle term correctly while squaring.
Step 3
Exam Tip
\(x^2=6+2+2\sqrt{12}=8+4\sqrt{3}\) है। वर्ग करते समय मध्य पद को सही सरल करें।
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यदि \(a=5+\sqrt{3}\) और \(b=5-\sqrt{3}\) हैं, तो \(a^2-b^2\) का मान क्या है?
If \(a=5+\sqrt{3}\) and \(b=5-\sqrt{3}\), what is the value of \(a^2-b^2\)?
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A \(20\sqrt{3}\)
B \(10\sqrt{3}\)
C (22)
D (100)
Explanation opens after your attempt
Correct Answer
A. \(20\sqrt{3}\)
Step 1
Concept
(a-2 -b-2 =(a-b)(a+b)), where \(a-b=2\sqrt{3}\) and (a+b=10). So the value is \(20\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(20\sqrt{3}\). (a-2 -b-2 =(a-b)(a+b)), where \(a-b=2\sqrt{3}\) and (a+b=10). So the value is \(20\sqrt{3}\).
Step 3
Exam Tip
(a-2 -b-2 =(a-b)(a+b)), जहाँ \(a-b=2\sqrt{3}\) और (a+b=10) है। इसलिए मान \(20\sqrt{3}\) है।
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\(\sqrt{45}+\sqrt{20}-\sqrt{80}\) किसके बराबर है?
What is \(\sqrt{45}+\sqrt{20}-\sqrt{80}\) equal to?
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A \(\sqrt{5}\)
B \(3\sqrt{5}\)
C \(5\sqrt{5}\)
D \(-\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{5}\)
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\), \(\sqrt{20}=2\sqrt{5}\), and \(\sqrt{80}=4\sqrt{5}\). Therefore the result is \(\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}\). \(\sqrt{45}=3\sqrt{5}\), \(\sqrt{20}=2\sqrt{5}\), and \(\sqrt{80}=4\sqrt{5}\). Therefore the result is \(\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\), \(\sqrt{20}=2\sqrt{5}\) और \(\sqrt{80}=4\sqrt{5}\) है। इसलिए परिणाम \(\sqrt{5}\) है।
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(\(\sqrt{10}+\sqrt{6}\)\(\sqrt{10}-\sqrt{6}\)) का मान किस प्रकार की संख्या है?
What type of number is the value of (\(\sqrt{10}+\sqrt{6}\)\(\sqrt{10}-\sqrt{6}\))?
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A अपरिमेय (4) / Irrational (4)
B परिमेय (4) / Rational (4)
C अपरिमेय (16) / Irrational (16)
D परिमेय \(\sqrt{60}\) / Rational \(\sqrt{60}\)
Explanation opens after your attempt
Correct Answer
B. परिमेय (4) / Rational (4)
Step 1
Concept
Conjugate multiplication gives (10-6=4). It is a rational number.
Step 2
Why this answer is correct
The correct answer is B. परिमेय (4) / Rational (4). Conjugate multiplication gives (10-6=4). It is a rational number.
Step 3
Exam Tip
संयुग्मी गुणन से (10-6=4) मिलता है। यह परिमेय संख्या है।
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\(\sqrt{48}+\sqrt{75}-\sqrt{27}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{48}+\sqrt{75}-\sqrt{27}\)?
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A \(6\sqrt{3}\)
B \(4\sqrt{3}\)
C \(5\sqrt{3}\)
D \(7\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{3}\)
Step 1
Concept
\(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{27}=3\sqrt{3}\). Therefore the result is \(6\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{3}\). \(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{27}=3\sqrt{3}\). Therefore the result is \(6\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\) है। इसलिए परिणाम \(6\sqrt{3}\) है।
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यदि \(p=\frac{1}{\sqrt{5}+2}\) है, तो (p) का सरल रूप कौन-सा है?
If \(p=\frac{1}{\sqrt{5}+2}\), which is the simplified form of (p)?
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A \(\sqrt{5}-2\)
B \(\frac{\sqrt{5}-2}{9}\)
C \(\sqrt{5}+2\)
D \(2-\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{5}-2\)
Step 1
Concept
Multiplying by the conjugate \(\sqrt{5}-2\) makes the denominator (5-4=1). So \(p=\sqrt{5}-2\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}-2\). Multiplying by the conjugate \(\sqrt{5}-2\) makes the denominator (5-4=1). So \(p=\sqrt{5}-2\).
Step 3
Exam Tip
संयुग्मी \(\sqrt{5}-2\) से गुणा करने पर हर (5-4=1) होता है। इसलिए \(p=\sqrt{5}-2\) है।
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\(\frac{\sqrt{72}+\sqrt{50}}{\sqrt{2}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{72}+\sqrt{50}}{\sqrt{2}}\)?
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A (11)
B \(6\sqrt{2}\)
C \(5\sqrt{2}\)
D (1)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{72}=6\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the numerator is \(11\sqrt{2}\). Dividing by \(\sqrt{2}\) gives (11).
Step 2
Why this answer is correct
The correct answer is A. (11). \(\sqrt{72}=6\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the numerator is \(11\sqrt{2}\). Dividing by \(\sqrt{2}\) gives (11).
Step 3
Exam Tip
\(\sqrt{72}=6\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए अंश \(11\sqrt{2}\) है। \(\sqrt{2}\) से भाग देने पर (11) मिलता है।
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\(\sqrt{3+\sqrt{5}}\) किस कथन के अनुसार पहचाना जा सकता है?
How can \(\sqrt{3+\sqrt{5}}\) be identified according to the given statements?
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A यह निश्चित रूप से परिमेय है / It is definitely rational
B यह वास्तविक संख्या है और इसका वर्ग \(3+\sqrt{5}\) है / It is real and its square is \(3+\sqrt{5}\)
C यह \(3+\sqrt{5}\) के बराबर है / It is equal to \(3+\sqrt{5}\)
D यह शून्य है / It is zero
Explanation opens after your attempt
Correct Answer
B. यह वास्तविक संख्या है और इसका वर्ग \(3+\sqrt{5}\) है / It is real and its square is \(3+\sqrt{5}\)
Step 1
Concept
The square root of a positive number is real and its square gives the same number. Here \(3+\sqrt{5}\) is positive.
Step 2
Why this answer is correct
The correct answer is B. यह वास्तविक संख्या है और इसका वर्ग \(3+\sqrt{5}\) है / It is real and its square is \(3+\sqrt{5}\). The square root of a positive number is real and its square gives the same number. Here \(3+\sqrt{5}\) is positive.
Step 3
Exam Tip
किसी धनात्मक संख्या का वर्गमूल वास्तविक होता है और उसका वर्ग वही संख्या देता है। यहाँ \(3+\sqrt{5}\) धनात्मक है।
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यदि \(p=\sqrt{12}-\sqrt{3}\) है, तो \(p^2\) का मान क्या है?
If \(p=\sqrt{12}-\sqrt{3}\), what is the value of \(p^2\)?
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A (3)
B (9)
C (6)
D \(\sqrt{9}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\), so \(p=\sqrt{3}\). Therefore \(p^2=3\).
Step 2
Why this answer is correct
The correct answer is A. (3). \(\sqrt{12}=2\sqrt{3}\), so \(p=\sqrt{3}\). Therefore \(p^2=3\).
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\), इसलिए \(p=\sqrt{3}\) है। अतः \(p^2=3\) होगा।
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यदि \(m=\sqrt{8}+\sqrt{18}\) और \(n=\sqrt{50}\) हैं, तो (m-n) का मान क्या है?
If \(m=\sqrt{8}+\sqrt{18}\) and \(n=\sqrt{50}\), what is the value of (m-n)?
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A (0)
B \(\sqrt{2}\)
C \(2\sqrt{2}\)
D \(5\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), and \(\sqrt{50}=5\sqrt{2}\). Therefore (m-n=0).
Step 2
Why this answer is correct
The correct answer is A. (0). \(\sqrt{8}=2\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), and \(\sqrt{50}=5\sqrt{2}\). Therefore (m-n=0).
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\) है। इसलिए (m-n=0) मिलता है।
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(\(\sqrt{5}+\sqrt{2}\)2 -\(\sqrt{5}-\sqrt{2}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{5}+\sqrt{2}\)2 -\(\sqrt{5}-\sqrt{2}\)2 )?
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A \(4\sqrt{10}\)
B (7)
C \(2\sqrt{10}\)
D (14)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{10}\)
Step 1
Concept
The first square adds \(2\sqrt{10}\) and the second subtracts \(2\sqrt{10}\). The difference is \(4\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{10}\). The first square adds \(2\sqrt{10}\) and the second subtracts \(2\sqrt{10}\). The difference is \(4\sqrt{10}\).
Step 3
Exam Tip
पहले वर्ग में \(2\sqrt{10}\) जुड़ता है और दूसरे में \(2\sqrt{10}\) घटता है। अंतर \(4\sqrt{10}\) है।
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\(\frac{3+\sqrt{5}}{3-\sqrt{5}}\) का सरल रूप क्या है?
What is the simplified form of \(\frac{3+\sqrt{5}}{3-\sqrt{5}}\)?
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A \(\frac{7+3\sqrt{5}}{2}\)
B \(7+3\sqrt{5}\)
C \(\frac{3+\sqrt{5}}{4}\)
D \(\frac{7-3\sqrt{5}}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{7+3\sqrt{5}}{2}\)
Step 1
Concept
Multiplying by the conjugate gives denominator (9-5=4) and numerator (\(3+\sqrt{5}\)2 =14+6\sqrt{5}). So the answer is \(\frac{7+3\sqrt{5}}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{7+3\sqrt{5}}{2}\). Multiplying by the conjugate gives denominator (9-5=4) and numerator (\(3+\sqrt{5}\)2 =14+6\sqrt{5}). So the answer is \(\frac{7+3\sqrt{5}}{2}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (9-5=4) और अंश (\(3+\sqrt{5}\)2 =14+6\sqrt{5}) मिलता है। इसलिए उत्तर \(\frac{7+3\sqrt{5}}{2}\) है।
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यदि \(x=\sqrt{3}+\sqrt{2}\) और \(y=\sqrt{3}-\sqrt{2}\) हैं, तो \(\frac{x}{y}\) का मान क्या है?
If \(x=\sqrt{3}+\sqrt{2}\) and \(y=\sqrt{3}-\sqrt{2}\), what is the value of \(\frac{x}{y}\)?
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A \(5+2\sqrt{6}\)
B \(1+\sqrt{6}\)
C \(5-2\sqrt{6}\)
D (6)
Explanation opens after your attempt
Correct Answer
A. \(5+2\sqrt{6}\)
Step 1
Concept
The conjugate of the denominator is \(\sqrt{3}+\sqrt{2}\) and the denominator becomes (1). So the value is (\(\sqrt{3}+\sqrt{2}\)2 =5+2\sqrt{6}).
Step 2
Why this answer is correct
The correct answer is A. \(5+2\sqrt{6}\). The conjugate of the denominator is \(\sqrt{3}+\sqrt{2}\) and the denominator becomes (1). So the value is (\(\sqrt{3}+\sqrt{2}\)2 =5+2\sqrt{6}).
Step 3
Exam Tip
हर का संयुग्मी \(\sqrt{3}+\sqrt{2}\) है और हर (1) बनता है। इसलिए मान (\(\sqrt{3}+\sqrt{2}\)2 =5+2\sqrt{6}) है।
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किस विकल्प का मान अपरिमेय है?
Which option has an irrational value?
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A (\(\sqrt{11}\)2 )
B (\(\sqrt{8}\)\(\sqrt{2}\))
C \(\sqrt{7}+\sqrt{28}\)
D (\(2+\sqrt{3}\)\(2-\sqrt{3}\))
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{7}+\sqrt{28}\)
Step 1
Concept
\(\sqrt{7}+\sqrt{28}=3\sqrt{7}\), which is irrational. In the other options the radicals simplify to rational values.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{7}+\sqrt{28}\). \(\sqrt{7}+\sqrt{28}=3\sqrt{7}\), which is irrational. In the other options the radicals simplify to rational values.
Step 3
Exam Tip
\(\sqrt{7}+\sqrt{28}=3\sqrt{7}\) है जो अपरिमेय है। बाकी विकल्पों में मूल हटकर परिमेय मान मिलते हैं।
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यदि \(r=\sqrt{20}+\sqrt{45}\) है, तो (r) किसके बराबर है?
If \(r=\sqrt{20}+\sqrt{45}\), what is (r) equal to?
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A \(5\sqrt{5}\)
B \(7\sqrt{5}\)
C \(9\sqrt{5}\)
D \(13\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(5\sqrt{5}\)
Step 1
Concept
\(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\). Therefore \(r=5\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(5\sqrt{5}\). \(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\). Therefore \(r=5\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{20}=2\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\) है। इसलिए \(r=5\sqrt{5}\) है।
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\(\sqrt{6+\sqrt{11}}\) के बारे में कौन-सा कथन निश्चित रूप से सही है?
Which statement is definitely true about \(\sqrt{6+\sqrt{11}}\)?
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A यह वास्तविक नहीं है / It is not real
B इसका वर्ग \(6+\sqrt{11}\) है / Its square is \(6+\sqrt{11}\)
C यह \(6+\sqrt{11}\) के बराबर है / It equals \(6+\sqrt{11}\)
D यह परिमेय पूर्णांक है / It is a rational integer
Explanation opens after your attempt
Correct Answer
B. इसका वर्ग \(6+\sqrt{11}\) है / Its square is \(6+\sqrt{11}\)
Step 1
Concept
Since \(6+\sqrt{11}\) is positive, its square root is real. Squaring it gives the inside number \(6+\sqrt{11}\).
Step 2
Why this answer is correct
The correct answer is B. इसका वर्ग \(6+\sqrt{11}\) है / Its square is \(6+\sqrt{11}\). Since \(6+\sqrt{11}\) is positive, its square root is real. Squaring it gives the inside number \(6+\sqrt{11}\).
Step 3
Exam Tip
क्योंकि \(6+\sqrt{11}\) धनात्मक है, इसका वर्गमूल वास्तविक है। वर्ग करने पर अंदर की संख्या \(6+\sqrt{11}\) मिलती है।
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यदि \(u=\sqrt{80}-\sqrt{45}\) और \(v=\sqrt{5}\) हैं, तो (u+v) का मान क्या है?
If \(u=\sqrt{80}-\sqrt{45}\) and \(v=\sqrt{5}\), what is the value of (u+v)?
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A \(2\sqrt{5}\)
B \(3\sqrt{5}\)
C \(\sqrt{5}\)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{5}\)
Step 1
Concept
\(\sqrt{80}=4\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so \(u=\sqrt{5}\). Hence \(u+v=2\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{5}\). \(\sqrt{80}=4\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so \(u=\sqrt{5}\). Hence \(u+v=2\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{80}=4\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\), इसलिए \(u=\sqrt{5}\) है। अतः \(u+v=2\sqrt{5}\) है।
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\(\frac{5}{\sqrt{6}-1}\) का परिमेयकृत रूप कौन-सा है?
Which is the rationalised form of \(\frac{5}{\sqrt{6}-1}\)?
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A \(\sqrt{6}+1\)
B (5\(\sqrt{6}+1\))
C (5\(\sqrt{6}-1\))
D (\frac{5\(\sqrt{6}+1\)}{5})
Explanation opens after your attempt
Correct Answer
D. (\frac{5\(\sqrt{6}+1\)}{5})
Step 1
Concept
Multiplying by the conjugate makes the denominator (6-1=5). So the form is (\frac{5\(\sqrt{6}+1\)}{5}), that is \(\sqrt{6}+1\).
Step 2
Why this answer is correct
The correct answer is D. (\frac{5\(\sqrt{6}+1\)}{5}). Multiplying by the conjugate makes the denominator (6-1=5). So the form is (\frac{5\(\sqrt{6}+1\)}{5}), that is \(\sqrt{6}+1\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (6-1=5) बनता है। इसलिए रूप (\frac{5\(\sqrt{6}+1\)}{5}), यानी \(\sqrt{6}+1\), है।
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(\(\sqrt{12}+\sqrt{27}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{12}+\sqrt{27}\)2 )?
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A (75)
B (45)
C \(27+12\sqrt{3}\)
D (39)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the sum is \(5\sqrt{3}\). Its square is (75).
Step 2
Why this answer is correct
The correct answer is A. (75). \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the sum is \(5\sqrt{3}\). Its square is (75).
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए योग \(5\sqrt{3}\) है। इसका वर्ग (75) है।
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कौन-सा दशमलव अपरिमेय संख्या को दर्शा सकता है?
Which decimal can represent an irrational number?
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A \(4.252525\ldots\)
B \(4.25000\ldots\)
C \(4.251251251\ldots\)
D \(4.25025002500025\ldots\)
Explanation opens after your attempt
Correct Answer
D. \(4.25025002500025\ldots\)
Step 1
Concept
An irrational decimal is non-terminating and non-repeating. The fourth option has no fixed repetition.
Step 2
Why this answer is correct
The correct answer is D. \(4.25025002500025\ldots\). An irrational decimal is non-terminating and non-repeating. The fourth option has no fixed repetition.
Step 3
Exam Tip
अपरिमेय दशमलव असांत और अनावर्ती होता है। चौथे विकल्प में कोई निश्चित दोहराव नहीं है।
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यदि \(a=\sqrt{7}+\sqrt{3}\) और \(b=\sqrt{7}-\sqrt{3}\) हैं, तो (ab) और (a+b) का सही युग्म कौन-सा है?
If \(a=\sqrt{7}+\sqrt{3}\) and \(b=\sqrt{7}-\sqrt{3}\), which is the correct pair of (ab) and (a+b)?
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A (4), \(2\sqrt{7}\)
B (10), \(2\sqrt{3}\)
C \(4\sqrt{21}\), \(2\sqrt{7}\)
D (7), (3)
Explanation opens after your attempt
Correct Answer
A. (4), \(2\sqrt{7}\)
Step 1
Concept
(ab=7-3=4) and \(a+b=2\sqrt{7}\). In conjugates check product and sum separately.
Step 2
Why this answer is correct
The correct answer is A. (4), \(2\sqrt{7}\). (ab=7-3=4) and \(a+b=2\sqrt{7}\). In conjugates check product and sum separately.
Step 3
Exam Tip
(ab=7-3=4) और \(a+b=2\sqrt{7}\) है। संयुग्मी में गुणन और योग अलग-अलग देखें।
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(\sqrt{2}\left\(\sqrt{50}-\sqrt{18}\right\)) का मान क्या है?
What is the value of (\sqrt{2}\left\(\sqrt{50}-\sqrt{18}\right\))?
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A (4)
B \(2\sqrt{2}\)
C (8)
D \(\sqrt{32}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the bracket is \(2\sqrt{2}\). Multiplying by \(\sqrt{2}\) gives (4).
Step 2
Why this answer is correct
The correct answer is A. (4). \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so the bracket is \(2\sqrt{2}\). Multiplying by \(\sqrt{2}\) gives (4).
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए कोष्ठक \(2\sqrt{2}\) है। \(\sqrt{2}\) से गुणा करने पर (4) मिलता है।
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\(\frac{2}{\sqrt{7}+\sqrt{5}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{2}{\sqrt{7}+\sqrt{5}}\)?
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A \(\sqrt{7}-\sqrt{5}\)
B \(\sqrt{7}+\sqrt{5}\)
C \(\frac{\sqrt{7}-\sqrt{5}}{2}\)
D \(2\sqrt{35}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{7}-\sqrt{5}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (7-5=2). So (\frac{2\(\sqrt{7}-\sqrt{5}\)}{2}=\sqrt{7}-\sqrt{5}).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{7}-\sqrt{5}\). Multiplying by the conjugate makes the denominator (7-5=2). So (\frac{2\(\sqrt{7}-\sqrt{5}\)}{2}=\sqrt{7}-\sqrt{5}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (7-5=2) बनता है। इसलिए (\frac{2\(\sqrt{7}-\sqrt{5}\)}{2}=\sqrt{7}-\sqrt{5}) है।
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यदि \(x=2\sqrt{3}-\sqrt{12}\) है, तो (x) का मान क्या है?
If \(x=2\sqrt{3}-\sqrt{12}\), what is the value of (x)?
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A (0)
B \(\sqrt{3}\)
C \(4\sqrt{3}\)
D (2)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\), so \(2\sqrt{3}-2\sqrt{3}=0\). Simplifying the radical first is necessary.
Step 2
Why this answer is correct
The correct answer is A. (0). \(\sqrt{12}=2\sqrt{3}\), so \(2\sqrt{3}-2\sqrt{3}=0\). Simplifying the radical first is necessary.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\), इसलिए \(2\sqrt{3}-2\sqrt{3}=0\) है। पहले मूल को सरल करना जरूरी है।
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\(\sqrt{3+\sqrt{2}}\times\sqrt{3+\sqrt{2}}\) का मान क्या है?
What is the value of \(\sqrt{3+\sqrt{2}}\times\sqrt{3+\sqrt{2}}\)?
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A \(3+\sqrt{2}\)
B (9+2)
C \(\sqrt{5}\)
D \(3-\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(3+\sqrt{2}\)
Step 1
Concept
Multiplying the same square root by itself gives the number inside. Therefore the value is \(3+\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(3+\sqrt{2}\). Multiplying the same square root by itself gives the number inside. Therefore the value is \(3+\sqrt{2}\).
Step 3
Exam Tip
एक ही वर्गमूल का अपने आप से गुणन अंदर की संख्या देता है। इसलिए मान \(3+\sqrt{2}\) है।
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\(\sqrt{98}+\sqrt{72}-\sqrt{8}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{98}+\sqrt{72}-\sqrt{8}\)?
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A \(11\sqrt{2}\)
B \(9\sqrt{2}\)
C \(13\sqrt{2}\)
D \(15\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(11\sqrt{2}\)
Step 1
Concept
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\). Therefore the result is \(11\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(11\sqrt{2}\). \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\). Therefore the result is \(11\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\) है। इसलिए परिणाम \(11\sqrt{2}\) है।
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यदि \(t=\sqrt{5}+2\) है, तो \(t+\frac{1}{t}\) का मान क्या है?
If \(t=\sqrt{5}+2\), what is the value of \(t+\frac{1}{t}\)?
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A \(2\sqrt{5}\)
B (4)
C \(2\sqrt{5}+4\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{5}\)
Step 1
Concept
\(\frac{1}{\sqrt{5}+2}=\sqrt{5}-2\). Therefore the sum is \(2\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{5}\). \(\frac{1}{\sqrt{5}+2}=\sqrt{5}-2\). Therefore the sum is \(2\sqrt{5}\).
Step 3
Exam Tip
\(\frac{1}{\sqrt{5}+2}=\sqrt{5}-2\) है। इसलिए योग \(2\sqrt{5}\) है।
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यदि \(s=\sqrt{27}+\sqrt{12}\) है, तो \(\frac{s}{\sqrt{3}}\) का मान क्या है?
If \(s=\sqrt{27}+\sqrt{12}\), what is the value of \(\frac{s}{\sqrt{3}}\)?
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A (5)
B (3)
C (7)
D (15)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so \(s=5\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (5).
Step 2
Why this answer is correct
The correct answer is A. (5). \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so \(s=5\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (5).
Step 3
Exam Tip
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\), इसलिए \(s=5\sqrt{3}\) है। \(\sqrt{3}\) से भाग देने पर (5) मिलता है।
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\(\frac{\sqrt{3}-1}{\sqrt{3}+1}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)?
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A \(2-\sqrt{3}\)
B \(2+\sqrt{3}\)
C \(\sqrt{3}-2\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(2-\sqrt{3}\)
Step 1
Concept
Multiplying by the conjugate gives (\frac{\(\sqrt{3}-1\)2 }{2}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}). Rationalise the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{3}\). Multiplying by the conjugate gives (\frac{\(\sqrt{3}-1\)2 }{2}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}). Rationalise the denominator.
Step 3
Exam Tip
संयुग्मी से गुणा करने पर (\frac{\(\sqrt{3}-1\)2 }{2}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}) मिलता है। हर का परिमेयकरण करें।
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यदि \(x=\sqrt{2}+\sqrt{8}\) और \(y=\sqrt{18}\) हैं, तो (x-y) क्या है?
If \(x=\sqrt{2}+\sqrt{8}\) and \(y=\sqrt{18}\), what is (x-y)?
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A (0)
B \(\sqrt{2}\)
C \(2\sqrt{2}\)
D \(4\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
\(x=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\) and \(y=3\sqrt{2}\). Therefore (x-y=0).
Step 2
Why this answer is correct
The correct answer is A. (0). \(x=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\) and \(y=3\sqrt{2}\). Therefore (x-y=0).
Step 3
Exam Tip
\(x=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\) और \(y=3\sqrt{2}\) है। इसलिए (x-y=0) है।
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किस विकल्प में परिणाम परिमेय नहीं है?
In which option is the result not rational?
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A \(\sqrt{32}\div\sqrt{2}\)
B \(\sqrt{18}\div\sqrt{2}\)
C \(\sqrt{45}\div\sqrt{5}\)
D \(\sqrt{20}\div\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
D. \(\sqrt{20}\div\sqrt{2}\)
Step 1
Concept
The first three options give (4), (3), and (3) respectively. The fourth is \(\sqrt{10}\), which is irrational.
Step 2
Why this answer is correct
The correct answer is D. \(\sqrt{20}\div\sqrt{2}\). The first three options give (4), (3), and (3) respectively. The fourth is \(\sqrt{10}\), which is irrational.
Step 3
Exam Tip
पहले तीन विकल्प क्रमशः (4), (3) और (3) देते हैं। चौथा \(\sqrt{10}\) है जो अपरिमेय है।
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(\(\sqrt{13}+2\)2 -\(\sqrt{13}-2\)2 ) का मान क्या है?
What is the value of (\(\sqrt{13}+2\)2 -\(\sqrt{13}-2\)2 )?
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A \(8\sqrt{13}\)
B (17)
C \(4\sqrt{13}\)
D (52)
Explanation opens after your attempt
Correct Answer
A. \(8\sqrt{13}\)
Step 1
Concept
Use ((a+b)2 -(a-b)2 =4ab). Here \(a=\sqrt{13}\) and (b=2), so the value is \(8\sqrt{13}\).
Step 2
Why this answer is correct
The correct answer is A. \(8\sqrt{13}\). Use ((a+b)2 -(a-b)2 =4ab). Here \(a=\sqrt{13}\) and (b=2), so the value is \(8\sqrt{13}\).
Step 3
Exam Tip
पहचान ((a+b)2 -(a-b)2 =4ab) लगाएं। यहाँ \(a=\sqrt{13}\) और (b=2), इसलिए मान \(8\sqrt{13}\) है।
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यदि एक आयत की लंबाई \(3+\sqrt{2}\) और चौड़ाई \(3-\sqrt{2}\) है, तो क्षेत्रफल क्या होगा?
If a rectangle has length \(3+\sqrt{2}\) and breadth \(3-\sqrt{2}\), what will be its area?
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A (7)
B (9)
C (11)
D \(6\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
Area is (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7). Multiplying conjugate dimensions can give a rational area.
Step 2
Why this answer is correct
The correct answer is A. (7). Area is (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7). Multiplying conjugate dimensions can give a rational area.
Step 3
Exam Tip
क्षेत्रफल (\(3+\sqrt{2}\)\(3-\sqrt{2}\)=9-2=7) है। संयुग्मी आयामों का गुणन परिमेय क्षेत्रफल दे सकता है।
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\(\sqrt{200}-\sqrt{128}+\sqrt{18}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{200}-\sqrt{128}+\sqrt{18}\)?
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A \(5\sqrt{2}\)
B \(3\sqrt{2}\)
C \(7\sqrt{2}\)
D \(\sqrt{90}\)
Explanation opens after your attempt
Correct Answer
A. \(5\sqrt{2}\)
Step 1
Concept
\(\sqrt{200}=10\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). Therefore the result is \(5\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(5\sqrt{2}\). \(\sqrt{200}=10\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). Therefore the result is \(5\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{200}=10\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\) है। इसलिए परिणाम \(5\sqrt{2}\) है।
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\(\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{3}-\sqrt{2}}\) का मान क्या है?
What is the value of \(\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{3}-\sqrt{2}}\)?
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A \(2\sqrt{3}\)
B \(2\sqrt{2}\)
C (2)
D (5)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{3}\)
Step 1
Concept
The first term becomes \(\sqrt{3}-\sqrt{2}\) and the second becomes \(\sqrt{3}+\sqrt{2}\). Their sum is \(2\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{3}\). The first term becomes \(\sqrt{3}-\sqrt{2}\) and the second becomes \(\sqrt{3}+\sqrt{2}\). Their sum is \(2\sqrt{3}\).
Step 3
Exam Tip
पहला पद \(\sqrt{3}-\sqrt{2}\) और दूसरा \(\sqrt{3}+\sqrt{2}\) बनता है। योग \(2\sqrt{3}\) है।
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यदि \(z=\sqrt{11}-\sqrt{7}\) है, तो \(z^2\) का मान कौन-सा है?
If \(z=\sqrt{11}-\sqrt{7}\), which is the value of \(z^2\)?
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A \(18-2\sqrt{77}\)
B (4)
C \(18+2\sqrt{77}\)
D \(\sqrt{4}\)
Explanation opens after your attempt
Correct Answer
A. \(18-2\sqrt{77}\)
Step 1
Concept
(\(\sqrt{11}-\sqrt{7}\)2 =11+7-2\sqrt{77}). The middle term stays negative.
Step 2
Why this answer is correct
The correct answer is A. \(18-2\sqrt{77}\). (\(\sqrt{11}-\sqrt{7}\)2 =11+7-2\sqrt{77}). The middle term stays negative.
Step 3
Exam Tip
(\(\sqrt{11}-\sqrt{7}\)2 =11+7-2\sqrt{77}) है। मध्य पद का चिन्ह ऋणात्मक रहेगा।
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कौन-सा विकल्प \(\sqrt{50}+\sqrt{8}\) के बराबर है?
Which option is equal to \(\sqrt{50}+\sqrt{8}\)?
#number-systems
#irrational-numbers
#hard
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A \(7\sqrt{2}\)
B \(3\sqrt{2}\)
C \(\sqrt{58}\)
D \(10\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(7\sqrt{2}\)
Step 1
Concept
\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\). So the sum is \(7\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(7\sqrt{2}\). \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\). So the sum is \(7\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\) है। इसलिए योग \(7\sqrt{2}\) है।
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यदि \(q=\sqrt{3}+1\) है, तो \(q^2-2q\) का मान क्या है?
If \(q=\sqrt{3}+1\), what is the value of \(q^2-2q\)?
#number-systems
#irrational-numbers
#hard
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A (2)
B (1)
C \(\sqrt{3}\)
D (3)
Explanation opens after your attempt
Step 1
Concept
\(q^2=4+2\sqrt{3}\) and \(2q=2\sqrt{3}+2\). Subtracting gives (2).
Step 2
Why this answer is correct
The correct answer is A. (2). \(q^2=4+2\sqrt{3}\) and \(2q=2\sqrt{3}+2\). Subtracting gives (2).
Step 3
Exam Tip
\(q^2=4+2\sqrt{3}\) और \(2q=2\sqrt{3}+2\) है। घटाने पर (2) मिलता है।
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\(\sqrt{6}\) और \(\sqrt{7}\) के बीच कौन-सी संख्या निश्चित रूप से आती है?
Which number definitely lies between \(\sqrt{6}\) and \(\sqrt{7}\)?
#number-systems
#irrational-numbers
#hard
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A \(\sqrt{6.5}\)
B (3)
C \(\sqrt{5}\)
D \(\sqrt{8}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{6.5}\)
Step 1
Concept
Since (6<6.5<7), \(\sqrt{6.5}\) lies between them. For positive square roots compare the numbers inside.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{6.5}\). Since (6<6.5<7), \(\sqrt{6.5}\) lies between them. For positive square roots compare the numbers inside.
Step 3
Exam Tip
क्योंकि (6<6.5<7), इसलिए \(\sqrt{6.5}\) दोनों के बीच होगा। धनात्मक वर्गमूलों में अंदर की संख्या से तुलना करें।
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यदि \(A=\sqrt{28}+\sqrt{63}\) और \(B=5\sqrt{7}\) हैं, तो कौन-सा कथन सही है?
If \(A=\sqrt{28}+\sqrt{63}\) and \(B=5\sqrt{7}\), which statement is correct?
#number-systems
#irrational-numbers
#hard
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A (A=B)
B (A>B)
C (A<B)
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so \(A=5\sqrt{7}\). Hence (A=B).
Step 2
Why this answer is correct
The correct answer is A. (A=B). \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so \(A=5\sqrt{7}\). Hence (A=B).
Step 3
Exam Tip
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए \(A=5\sqrt{7}\) है। अतः (A=B) है।
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\(\frac{\sqrt{45}-\sqrt{20}}{\sqrt{5}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{45}-\sqrt{20}}{\sqrt{5}}\)?
#number-systems
#irrational-numbers
#hard
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A (1)
B (5)
C \(\sqrt{5}\)
D (3)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the numerator is \(\sqrt{5}\). Dividing gives (1).
Step 2
Why this answer is correct
The correct answer is A. (1). \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the numerator is \(\sqrt{5}\). Dividing gives (1).
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\), इसलिए अंश \(\sqrt{5}\) है। भाग देने पर (1) मिलता है।
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\(\sqrt{2+\sqrt{3}}\) का वर्ग किसके बराबर है?
What is the square of \(\sqrt{2+\sqrt{3}}\) equal to?
#number-systems
#irrational-numbers
#hard
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A \(2+\sqrt{3}\)
B (4+3)
C \(\sqrt{5}\)
D \(2-\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(2+\sqrt{3}\)
Step 1
Concept
The square of a square root gives the number inside. So (\left\(\sqrt{2+\sqrt{3}}\right\)2 =2+\sqrt{3}).
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{3}\). The square of a square root gives the number inside. So (\left\(\sqrt{2+\sqrt{3}}\right\)2 =2+\sqrt{3}).
Step 3
Exam Tip
वर्गमूल का वर्ग अंदर की संख्या देता है। इसलिए (\left\(\sqrt{2+\sqrt{3}}\right\)2 =2+\sqrt{3}) है।
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यदि \(x=\sqrt{5}+\sqrt{3}\) है, तो \(x^2-8\) का मान क्या है?
If \(x=\sqrt{5}+\sqrt{3}\), what is the value of \(x^2-8\)?
#number-systems
#irrational-numbers
#hard
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A \(2\sqrt{15}\)
B (8)
C \(\sqrt{15}\)
D (15)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{15}\)
Step 1
Concept
\(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\). Therefore \(x^2-8=2\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{15}\). \(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\). Therefore \(x^2-8=2\sqrt{15}\).
Step 3
Exam Tip
\(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\) है। इसलिए \(x^2-8=2\sqrt{15}\) है।
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किस विकल्प में दो अपरिमेय संख्याओं का योग परिमेय है?
In which option is the sum of two irrational numbers rational?
#number-systems
#irrational-numbers
#hard
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A (\sqrt{11}+\(-\sqrt{11}\))
B \(\sqrt{2}+\sqrt{3}\)
C \(\sqrt{5}+\sqrt{20}\)
D \(\sqrt{7}+2\)
Explanation opens after your attempt
Correct Answer
A. (\sqrt{11}+\(-\sqrt{11}\))
Step 1
Concept
\(\sqrt{11}\) and \(-\sqrt{11}\) are both irrational and their sum is (0). (0) is rational.
Step 2
Why this answer is correct
The correct answer is A. (\sqrt{11}+\(-\sqrt{11}\)). \(\sqrt{11}\) and \(-\sqrt{11}\) are both irrational and their sum is (0). (0) is rational.
Step 3
Exam Tip
\(\sqrt{11}\) और \(-\sqrt{11}\) दोनों अपरिमेय हैं और योग (0) है। (0) परिमेय संख्या है।
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(\(\sqrt{8}+\sqrt{18}\)\(\sqrt{8}-\sqrt{18}\)) का मान क्या है?
What is the value of (\(\sqrt{8}+\sqrt{18}\)\(\sqrt{8}-\sqrt{18}\))?
#number-systems
#irrational-numbers
#hard
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A (-10)
B (10)
C (26)
D \(2\sqrt{26}\)
Explanation opens after your attempt
Step 1
Concept
This is the \(a^2-b^2\) form, so the value is (8-18=-10). In conjugate multiplication take the difference of squares directly.
Step 2
Why this answer is correct
The correct answer is A. (-10). This is the \(a^2-b^2\) form, so the value is (8-18=-10). In conjugate multiplication take the difference of squares directly.
Step 3
Exam Tip
यह \(a^2-b^2\) रूप है, इसलिए मान (8-18=-10) है। संयुग्मी गुणन में सीधे वर्गों का अंतर लें।
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यदि \(a=\sqrt{12}+\sqrt{75}\) और \(b=7\sqrt{3}\) हैं, तो (a-b) का मान क्या है?
If \(a=\sqrt{12}+\sqrt{75}\) and \(b=7\sqrt{3}\), what is the value of (a-b)?
#number-systems
#irrational-numbers
#hard
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A (0)
B \(\sqrt{3}\)
C \(2\sqrt{3}\)
D \(5\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so \(a=7\sqrt{3}\). Hence (a-b=0).
Step 2
Why this answer is correct
The correct answer is A. (0). \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so \(a=7\sqrt{3}\). Hence (a-b=0).
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\), इसलिए \(a=7\sqrt{3}\) है। अतः (a-b=0) है।
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