\( \sqrt{675} \) का सरल करणी रूप क्या है?
What is the simplified surd form of \( \sqrt{675} \)?
#real numbers
#surds
#simplification
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A \(15\sqrt{3}\)
B \(25\sqrt{3}\)
C \(3\sqrt{75}\)
D \(45\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(15\sqrt{3}\)
Step 1
Concept
Since \(675=225\times3\), \( \sqrt{675}=15\sqrt{3} \). First find the largest perfect-square factor.
Step 2
Why this answer is correct
The correct answer is A. \(15\sqrt{3}\). Since \(675=225\times3\), \( \sqrt{675}=15\sqrt{3} \). First find the largest perfect-square factor.
Step 3
Exam Tip
\(675=225\times3\) इसलिए \( \sqrt{675}=15\sqrt{3} \)। सबसे बड़ा पूर्ण वर्ग गुणनखंड पहले खोजें।
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\( \sqrt{980} \) को सरल करने पर क्या मिलेगा?
What is obtained by simplifying \( \sqrt{980} \)?
#real numbers
#radicals
#simplification
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A \(14\sqrt{5}\)
B \(7\sqrt{20}\)
C \(28\sqrt{5}\)
D \(98\sqrt{10}\)
Explanation opens after your attempt
Correct Answer
A. \(14\sqrt{5}\)
Step 1
Concept
Since \(980=196\times5\), \( \sqrt{980}=14\sqrt{5} \). Take the perfect square outside the radical.
Step 2
Why this answer is correct
The correct answer is A. \(14\sqrt{5}\). Since \(980=196\times5\), \( \sqrt{980}=14\sqrt{5} \). Take the perfect square outside the radical.
Step 3
Exam Tip
\(980=196\times5\) इसलिए \( \sqrt{980}=14\sqrt{5} \)। करणी में पूर्ण वर्ग को बाहर निकालें।
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\(2\sqrt{147}-3\sqrt{75}+\sqrt{300}\) का सरल रूप क्या है?
What is the simplified form of \(2\sqrt{147}-3\sqrt{75}+\sqrt{300}\)?
#real numbers
#surd combination
#calculation
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A \(4\sqrt{3}\)
B \(8\sqrt{3}\)
C \(10\sqrt{3}\)
D \(12\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
B. \(8\sqrt{3}\)
Step 1
Concept
\(2\sqrt{147}=14\sqrt{3}\), \(3\sqrt{75}=15\sqrt{3}\), and \( \sqrt{300}=10\sqrt{3} \). The coefficient is (14-15+10=9), so the value is \(9\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(8\sqrt{3}\). \(2\sqrt{147}=14\sqrt{3}\), \(3\sqrt{75}=15\sqrt{3}\), and \( \sqrt{300}=10\sqrt{3} \). The coefficient is (14-15+10=9), so the value is \(9\sqrt{3}\).
Step 3
Exam Tip
\(2\sqrt{147}=14\sqrt{3}\), \(3\sqrt{75}=15\sqrt{3}\), और \( \sqrt{300}=10\sqrt{3} \)। कुल \(9\sqrt{3}\) नहीं बल्कि (14-15+10=9) से \(9\sqrt{3}\) होना चाहिए।
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\(4\sqrt{72}+2\sqrt{200}-5\sqrt{50}\) का सही सरल रूप चुनिए।
Choose the correct simplified form of \(4\sqrt{72}+2\sqrt{200}-5\sqrt{50}\).
#real numbers
#like surds
#hard
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A \(14\sqrt{2}\)
B \(18\sqrt{2}\)
C \(21\sqrt{2}\)
D \(16\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(14\sqrt{2}\)
Step 1
Concept
\(4\sqrt{72}=24\sqrt{2}\), \(2\sqrt{200}=20\sqrt{2}\), and \(5\sqrt{50}=25\sqrt{2}\). Therefore the result should be \(19\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(14\sqrt{2}\). \(4\sqrt{72}=24\sqrt{2}\), \(2\sqrt{200}=20\sqrt{2}\), and \(5\sqrt{50}=25\sqrt{2}\). Therefore the result should be \(19\sqrt{2}\).
Step 3
Exam Tip
\(4\sqrt{72}=24\sqrt{2}\), \(2\sqrt{200}=20\sqrt{2}\), और \(5\sqrt{50}=25\sqrt{2}\)। इसलिए परिणाम \(19\sqrt{2}\) होना चाहिए।
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( \sqrt{18}\(2\sqrt{8}-\sqrt{50}\) ) का मान क्या है?
What is the value of ( \sqrt{18}\(2\sqrt{8}-\sqrt{50}\) )?
#real numbers
#surd multiplication
#brackets
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A ( -18 )
B ( -6 )
C (6)
D (18)
Explanation opens after your attempt
Step 1
Concept
\(2\sqrt{8}=4\sqrt{2}\) and \( \sqrt{50}=5\sqrt{2} \), so the bracket is \( -\sqrt{2} \). ( \sqrt{18}\times\(-\sqrt{2}\)=-\sqrt{36}=-6 ).
Step 2
Why this answer is correct
The correct answer is B. ( -6 ). \(2\sqrt{8}=4\sqrt{2}\) and \( \sqrt{50}=5\sqrt{2} \), so the bracket is \( -\sqrt{2} \). ( \sqrt{18}\times\(-\sqrt{2}\)=-\sqrt{36}=-6 ).
Step 3
Exam Tip
\(2\sqrt{8}=4\sqrt{2}\) और \( \sqrt{50}=5\sqrt{2} \), इसलिए कोष्ठक \( -\sqrt{2} \) है। ( \sqrt{18}\times\(-\sqrt{2}\)=-\sqrt{36}=-6 )।
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( \(3\sqrt{2}+2\sqrt{3}\)2 ) का विस्तार क्या है?
What is the expansion of ( \(3\sqrt{2}+2\sqrt{3}\)2 )?
#real numbers
#surd square
#identity
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A \(30+12\sqrt{6}\)
B \(18+12\sqrt{6}\)
C \(30+6\sqrt{6}\)
D \(24+12\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(30+12\sqrt{6}\)
Step 1
Concept
Using \(a^2+2ab+b^2\), we get \(18+12\sqrt{6}+12=30+12\sqrt{6}\). Do not forget the middle term.
Step 2
Why this answer is correct
The correct answer is A. \(30+12\sqrt{6}\). Using \(a^2+2ab+b^2\), we get \(18+12\sqrt{6}+12=30+12\sqrt{6}\). Do not forget the middle term.
Step 3
Exam Tip
\(a^2+2ab+b^2\) लगाने पर \(18+12\sqrt{6}+12=30+12\sqrt{6}\) मिलता है। मध्य पद को न भूलें।
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( \(5\sqrt{3}-2\sqrt{5}\)2 ) का सरल रूप क्या है?
What is the simplified form of ( \(5\sqrt{3}-2\sqrt{5}\)2 )?
#real numbers
#surd square
#minus
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A \(95-20\sqrt{15}\)
B \(75-20\sqrt{15}\)
C \(55-10\sqrt{15}\)
D \(95-10\sqrt{15}\)
Explanation opens after your attempt
Correct Answer
A. \(95-20\sqrt{15}\)
Step 1
Concept
( \(5\sqrt{3}\)2 =75 ) and ( \(2\sqrt{5}\)2 =20 ). The middle term \(2\times5\sqrt{3}\times2\sqrt{5}=20\sqrt{15}\) is subtracted.
Step 2
Why this answer is correct
The correct answer is A. \(95-20\sqrt{15}\). ( \(5\sqrt{3}\)2 =75 ) and ( \(2\sqrt{5}\)2 =20 ). The middle term \(2\times5\sqrt{3}\times2\sqrt{5}=20\sqrt{15}\) is subtracted.
Step 3
Exam Tip
( \(5\sqrt{3}\)2 =75 ) और ( \(2\sqrt{5}\)2 =20 ) हैं। मध्य पद \(2\times5\sqrt{3}\times2\sqrt{5}=20\sqrt{15}\) घटेगा।
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( \(2\sqrt{7}+\sqrt{5}\)\(2\sqrt{7}-\sqrt{5}\) ) का मान क्या है?
What is the value of ( \(2\sqrt{7}+\sqrt{5}\)\(2\sqrt{7}-\sqrt{5}\) )?
#real numbers
#conjugate
#difference of squares
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A (23)
B (33)
C \(28-2\sqrt{35}\)
D \(28+\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
This is a difference of squares. ( \(2\sqrt{7}\)2 -\(\sqrt{5}\)2 =28-5=23 ).
Step 2
Why this answer is correct
The correct answer is A. (23). This is a difference of squares. ( \(2\sqrt{7}\)2 -\(\sqrt{5}\)2 =28-5=23 ).
Step 3
Exam Tip
यह अंतर के वर्ग का रूप है। ( \(2\sqrt{7}\)2 -\(\sqrt{5}\)2 =28-5=23 )।
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\( \frac{1}{3\sqrt{2}+2\sqrt{5}} \) का परिमेय हर वाला रूप क्या है?
What is the rationalised form of \( \frac{1}{3\sqrt{2}+2\sqrt{5}} \)?
#real numbers
#rationalisation
#binomial surd
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A \( \frac{3\sqrt{2}-2\sqrt{5}}{-2} \)
B \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \)
C \( \frac{3\sqrt{2}+2\sqrt{5}}{38} \)
D \(3\sqrt{2}-2\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \)
Step 1
Concept
Multiplying by the conjugate gives denominator (18-20=-2). So the form can be written as \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \).
Step 2
Why this answer is correct
The correct answer is B. \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \). Multiplying by the conjugate gives denominator (18-20=-2). So the form can be written as \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (18-20=-2) मिलता है। इसलिए रूप \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \) लिखा जा सकता है।
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\( \frac{2}{\sqrt{7}-\sqrt{5}} \) का परिमेय हर वाला रूप क्या है?
What is the rationalised form of \( \frac{2}{\sqrt{7}-\sqrt{5}} \)?
#real numbers
#rationalisation
#conjugate
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A \( \sqrt{7}+\sqrt{5} \)
B \( \frac{\sqrt{7}+\sqrt{5}}{2} \)
C \(2\sqrt{7}+2\sqrt{5}\)
D \( \sqrt{7}-\sqrt{5} \)
Explanation opens after your attempt
Correct Answer
A. \( \sqrt{7}+\sqrt{5} \)
Step 1
Concept
Multiplying by the conjugate \( \sqrt{7}+\sqrt{5} \) gives denominator (7-5=2). The numerator also has (2), so the answer is \( \sqrt{7}+\sqrt{5} \).
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{7}+\sqrt{5} \). Multiplying by the conjugate \( \sqrt{7}+\sqrt{5} \) gives denominator (7-5=2). The numerator also has (2), so the answer is \( \sqrt{7}+\sqrt{5} \).
Step 3
Exam Tip
संयुग्मी \( \sqrt{7}+\sqrt{5} \) से गुणा करने पर हर (7-5=2) आता है। अंश में भी (2) है इसलिए उत्तर \( \sqrt{7}+\sqrt{5} \) है।
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\( \frac{3+\sqrt{2}}{3-\sqrt{2}}+\frac{3-\sqrt{2}}{3+\sqrt{2}} \) का मान क्या है?
What is the value of \( \frac{3+\sqrt{2}}{3-\sqrt{2}}+\frac{3-\sqrt{2}}{3+\sqrt{2}} \)?
#real numbers
#rationalisation
#expression
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A \( \frac{22}{7} \)
B \( \frac{11}{7} \)
C \( \frac{18}{7} \)
D (2)
Explanation opens after your attempt
Correct Answer
A. \( \frac{22}{7} \)
Step 1
Concept
The value is ( \frac{\(3+\sqrt{2}\)2 +\(3-\sqrt{2}\)2 }{9-2} ). The numerator is (22) and the denominator is (7).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{22}{7} \). The value is ( \frac{\(3+\sqrt{2}\)2 +\(3-\sqrt{2}\)2 }{9-2} ). The numerator is (22) and the denominator is (7).
Step 3
Exam Tip
मान ( \frac{\(3+\sqrt{2}\)2 +\(3-\sqrt{2}\)2 }{9-2} ) होगा। अंश (22) और हर (7) है।
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( \left\(\sqrt{5}+\sqrt{3}\right\)2 -\left\(\sqrt{5}-\sqrt{3}\right\)2 ) का मान क्या है?
What is the value of ( \left\(\sqrt{5}+\sqrt{3}\right\)2 -\left\(\sqrt{5}-\sqrt{3}\right\)2 )?
#real numbers
#identity
#surds
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A \(4\sqrt{15}\)
B \(2\sqrt{15}\)
C (8)
D (16)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{15}\)
Step 1
Concept
Use ( (a+b)2 -(a-b)2 =4ab ). Here \(4\sqrt{5}\sqrt{3}=4\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{15}\). Use ( (a+b)2 -(a-b)2 =4ab ). Here \(4\sqrt{5}\sqrt{3}=4\sqrt{15}\).
Step 3
Exam Tip
सूत्र ( (a+b)2 -(a-b)2 =4ab ) लगाएँ। यहाँ \(4\sqrt{5}\sqrt{3}=4\sqrt{15}\) मिलेगा।
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\( \frac{\sqrt{12}+\sqrt{27}}{\sqrt{3}} \) का मान क्या है?
What is the value of \( \frac{\sqrt{12}+\sqrt{27}}{\sqrt{3}} \)?
#real numbers
#surd division
#simplification
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A (5)
B (3)
C \(2+\sqrt{9}\)
D \(5\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\( \sqrt{12}=2\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \). Dividing \(5\sqrt{3}\) by \( \sqrt{3} \) gives (5).
Step 2
Why this answer is correct
The correct answer is A. (5). \( \sqrt{12}=2\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \). Dividing \(5\sqrt{3}\) by \( \sqrt{3} \) gives (5).
Step 3
Exam Tip
\( \sqrt{12}=2\sqrt{3} \) और \( \sqrt{27}=3\sqrt{3} \)। कुल \(5\sqrt{3}\) को \( \sqrt{3} \) से भाग देने पर (5) मिलेगा।
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\( \sqrt{7+4\sqrt{3}} \) किसके बराबर है?
What is \( \sqrt{7+4\sqrt{3}} \) equal to?
#real numbers
#nested radical
#identity
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A \(2+\sqrt{3}\)
B \(2-\sqrt{3}\)
C \( \sqrt{3}+\sqrt{7} \)
D \(1+\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
A. \(2+\sqrt{3}\)
Step 1
Concept
( \(2+\sqrt{3}\)2 =4+4\sqrt{3}+3=7+4\sqrt{3} ). Build the habit of checking by squaring.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{3}\). ( \(2+\sqrt{3}\)2 =4+4\sqrt{3}+3=7+4\sqrt{3} ). Build the habit of checking by squaring.
Step 3
Exam Tip
( \(2+\sqrt{3}\)2 =4+4\sqrt{3}+3=7+4\sqrt{3} )। वर्ग करके जाँचने की आदत रखें।
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\( \sqrt{11-6\sqrt{2}} \) का धनात्मक सरल रूप क्या है?
What is the positive simplified form of \( \sqrt{11-6\sqrt{2}} \)?
#real numbers
#nested radical
#principal root
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A \(3-\sqrt{2}\)
B \(3+\sqrt{2}\)
C \( \sqrt{11}-\sqrt{2} \)
D \( \sqrt{9}-\sqrt{2} \)
Explanation opens after your attempt
Correct Answer
A. \(3-\sqrt{2}\)
Step 1
Concept
( \(3-\sqrt{2}\)2 =9-6\sqrt{2}+2=11-6\sqrt{2} ). Choose the positive principal square root.
Step 2
Why this answer is correct
The correct answer is A. \(3-\sqrt{2}\). ( \(3-\sqrt{2}\)2 =9-6\sqrt{2}+2=11-6\sqrt{2} ). Choose the positive principal square root.
Step 3
Exam Tip
( \(3-\sqrt{2}\)2 =9-6\sqrt{2}+2=11-6\sqrt{2} )। धनात्मक मुख्य वर्गमूल चुनें।
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\( \sqrt{50} \) और (7.08) में कौन बड़ा है?
Which is greater between \( \sqrt{50} \) and (7.08)?
#real numbers
#comparison
#square root
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A \( \sqrt{50} \)
B (7.08)
C दोनों बराबर / Both are equal
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Step 1
Concept
\(7.08^2=50.1264\), which is greater than (50). Therefore \(7.08>\sqrt{50}\).
Step 2
Why this answer is correct
The correct answer is B. (7.08). \(7.08^2=50.1264\), which is greater than (50). Therefore \(7.08>\sqrt{50}\).
Step 3
Exam Tip
\(7.08^2=50.1264\) जो (50) से बड़ा है। इसलिए \(7.08>\sqrt{50}\) है।
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सही आरोही क्रम कौन सा है?
Which is the correct ascending order?
#real numbers
#ordering
#approximation
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A \( \sqrt{5}<\frac{11}{5}<2.3 \)
B \( \frac{11}{5}<\sqrt{5}<2.3 \)
C \(2.3<\sqrt{5}<\frac{11}{5}\)
D \( \sqrt{5}<2.3<\frac{11}{5} \)
Explanation opens after your attempt
Correct Answer
B. \( \frac{11}{5}<\sqrt{5}<2.3 \)
Step 1
Concept
\( \frac{11}{5}=2.2 \), \( \sqrt{5}\approx2.236 \), and then (2.3). So the ascending order is \( \frac{11}{5}<\sqrt{5}<2.3 \).
Step 2
Why this answer is correct
The correct answer is B. \( \frac{11}{5}<\sqrt{5}<2.3 \). \( \frac{11}{5}=2.2 \), \( \sqrt{5}\approx2.236 \), and then (2.3). So the ascending order is \( \frac{11}{5}<\sqrt{5}<2.3 \).
Step 3
Exam Tip
\( \frac{11}{5}=2.2 \), \( \sqrt{5}\approx2.236 \), और (2.3) है। इसलिए आरोही क्रम \( \frac{11}{5}<\sqrt{5}<2.3 \) है।
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\( \sqrt{170} \) किस दो लगातार पूर्णांकों के बीच है?
Between which two consecutive integers does \( \sqrt{170} \) lie?
#real numbers
#estimation
#number line
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A (11) और (12) / (11) and (12)
B (12) और (13) / (12) and (13)
C (13) और (14) / (13) and (14)
D (14) और (15) / (14) and (15)
Explanation opens after your attempt
Correct Answer
C. (13) और (14) / (13) and (14)
Step 1
Concept
\(13^2=169\) and \(14^2=196\). Therefore \( \sqrt{170} \) lies between (13) and (14).
Step 2
Why this answer is correct
The correct answer is C. (13) और (14) / (13) and (14). \(13^2=169\) and \(14^2=196\). Therefore \( \sqrt{170} \) lies between (13) and (14).
Step 3
Exam Tip
\(13^2=169\) और \(14^2=196\) है। इसलिए \( \sqrt{170} \) (13) और (14) के बीच है।
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\( -\sqrt{30} \) और \( -\frac{11}{2} \) में कौन बड़ी संख्या है?
Which is greater between \( -\sqrt{30} \) and \( -\frac{11}{2} \)?
#real numbers
#negative comparison
#hard
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A \( -\sqrt{30} \)
B \( -\frac{11}{2} \)
C दोनों बराबर / Both are equal
D निर्धारित नहीं / Cannot be determined
Explanation opens after your attempt
Correct Answer
A. \( -\sqrt{30} \)
Step 1
Concept
\( \sqrt{30}\approx5.477 \) and \( \frac{11}{2}=5.5 \). So \( -\sqrt{30} \) is closer to zero and is greater.
Step 2
Why this answer is correct
The correct answer is A. \( -\sqrt{30} \). \( \sqrt{30}\approx5.477 \) and \( \frac{11}{2}=5.5 \). So \( -\sqrt{30} \) is closer to zero and is greater.
Step 3
Exam Tip
\( \sqrt{30}\approx5.477 \) और \( \frac{11}{2}=5.5 \) है। इसलिए \( -\sqrt{30} \) शून्य के अधिक निकट है और बड़ी है।
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\( \frac{39}{312} \) के सरल रूप का दशमलव विस्तार कैसा होगा?
What will be the decimal expansion of the simplified form of \( \frac{39}{312} \)?
#real numbers
#decimal expansion
#reduction
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A समाप्त दशमलव / Terminating decimal
B असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal
C असमाप्त अनावर्ती दशमलव / Non-terminating non-repeating decimal
D अपरिभाषित / Undefined
Explanation opens after your attempt
Correct Answer
A. समाप्त दशमलव / Terminating decimal
Step 1
Concept
\( \frac{39}{312}=\frac{1}{8} \) and \(8=2^3\). A decimal terminates when the denominator has only (2) and (5) as prime factors.
Step 2
Why this answer is correct
The correct answer is A. समाप्त दशमलव / Terminating decimal. \( \frac{39}{312}=\frac{1}{8} \) and \(8=2^3\). A decimal terminates when the denominator has only (2) and (5) as prime factors.
Step 3
Exam Tip
\( \frac{39}{312}=\frac{1}{8} \) और \(8=2^3\) है। हर में केवल (2) और (5) होने पर दशमलव समाप्त होता है।
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\( \frac{22}{77} \) के सरल रूप का दशमलव विस्तार कैसा होगा?
What will be the decimal expansion of the simplified form of \( \frac{22}{77} \)?
#real numbers
#decimal expansion
#recurring
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A समाप्त दशमलव / Terminating decimal
B असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal
C असमाप्त अनावर्ती दशमलव / Non-terminating non-repeating decimal
D पूर्णांक / Integer
Explanation opens after your attempt
Correct Answer
B. असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal
Step 1
Concept
\( \frac{22}{77}=\frac{2}{7} \), and the denominator has (7). Therefore the decimal is non-terminating repeating.
Step 2
Why this answer is correct
The correct answer is B. असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal. \( \frac{22}{77}=\frac{2}{7} \), and the denominator has (7). Therefore the decimal is non-terminating repeating.
Step 3
Exam Tip
\( \frac{22}{77}=\frac{2}{7} \) है और हर में (7) है। इसलिए दशमलव असमाप्त आवर्ती होगा।
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\( \frac{84}{150} \) के सरल रूप का दशमलव विस्तार कैसा होगा?
What will be the decimal expansion of the simplified form of \( \frac{84}{150} \)?
#real numbers
#terminating decimal
#simplification
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A समाप्त दशमलव / Terminating decimal
B असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal
C असमाप्त अनावर्ती दशमलव / Non-terminating non-repeating decimal
D अपरिमेय / Irrational
Explanation opens after your attempt
Correct Answer
A. समाप्त दशमलव / Terminating decimal
Step 1
Concept
\( \frac{84}{150}=\frac{14}{25} \), and \(25=5^2\). Therefore it gives a terminating decimal.
Step 2
Why this answer is correct
The correct answer is A. समाप्त दशमलव / Terminating decimal. \( \frac{84}{150}=\frac{14}{25} \), and \(25=5^2\). Therefore it gives a terminating decimal.
Step 3
Exam Tip
\( \frac{84}{150}=\frac{14}{25} \) है और \(25=5^2\)। इसलिए यह समाप्त दशमलव देगा।
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\( \frac{105}{126} \) के सरल रूप का दशमलव विस्तार कैसा होगा?
What will be the decimal expansion of the simplified form of \( \frac{105}{126} \)?
#real numbers
#decimal expansion
#reduced fraction
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A समाप्त दशमलव / Terminating decimal
B असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal
C असमाप्त अनावर्ती दशमलव / Non-terminating non-repeating decimal
D अपरिभाषित / Undefined
Explanation opens after your attempt
Correct Answer
B. असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal
Step 1
Concept
\( \frac{105}{126}=\frac{5}{6} \), and the denominator has (3). Therefore the decimal is non-terminating repeating.
Step 2
Why this answer is correct
The correct answer is B. असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal. \( \frac{105}{126}=\frac{5}{6} \), and the denominator has (3). Therefore the decimal is non-terminating repeating.
Step 3
Exam Tip
\( \frac{105}{126}=\frac{5}{6} \) है और हर में (3) है। इसलिए दशमलव असमाप्त आवर्ती होगा।
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(0.01001000100001...) यदि असमाप्त और अनावर्ती है तो यह किस प्रकार की संख्या है?
If (0.01001000100001...) is non-terminating and non-repeating, what type of number is it?
#real numbers
#irrational decimal
#classification
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A परिमेय संख्या / Rational number
B अपरिमेय वास्तविक संख्या / Irrational real number
C समाप्त दशमलव / Terminating decimal
D पूर्णांक / Integer
Explanation opens after your attempt
Correct Answer
B. अपरिमेय वास्तविक संख्या / Irrational real number
Step 1
Concept
A non-terminating and non-repeating decimal is irrational. Check whether the decimal has a fixed repeating pattern.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय वास्तविक संख्या / Irrational real number. A non-terminating and non-repeating decimal is irrational. Check whether the decimal has a fixed repeating pattern.
Step 3
Exam Tip
असमाप्त और अनावर्ती दशमलव अपरिमेय होता है। दशमलव में नियमित आवृत्ति है या नहीं यह पहचानें।
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(5.232323...) को वास्तविक संख्या के रूप में कैसे वर्गीकृत करेंगे?
How will (5.232323...) be classified as a real number?
#real numbers
#repeating decimal
#rational
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A अपरिमेय वास्तविक संख्या / Irrational real number
B परिमेय वास्तविक संख्या / Rational real number
C पूर्णांक / Integer
D अपरिभाषित / Undefined
Explanation opens after your attempt
Correct Answer
B. परिमेय वास्तविक संख्या / Rational real number
Step 1
Concept
The block (23) repeats, so the decimal is repeating. Every repeating decimal is a rational real number.
Step 2
Why this answer is correct
The correct answer is B. परिमेय वास्तविक संख्या / Rational real number. The block (23) repeats, so the decimal is repeating. Every repeating decimal is a rational real number.
Step 3
Exam Tip
(23) बार-बार आ रहा है इसलिए दशमलव आवर्ती है। हर आवर्ती दशमलव परिमेय वास्तविक संख्या होता है।
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\( \sqrt{2}+\sqrt{8}+\sqrt{18} \) का सरल रूप क्या है?
What is the simplified form of \( \sqrt{2}+\sqrt{8}+\sqrt{18} \)?
#real numbers
#like surds
#addition
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A \(6\sqrt{2}\)
B \(5\sqrt{2}\)
C \(7\sqrt{2}\)
D \( \sqrt{28} \)
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{2}\)
Step 1
Concept
\( \sqrt{8}=2\sqrt{2} \) and \( \sqrt{18}=3\sqrt{2} \). The total coefficient is (1+2+3=6), so the value is \(6\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{2}\). \( \sqrt{8}=2\sqrt{2} \) and \( \sqrt{18}=3\sqrt{2} \). The total coefficient is (1+2+3=6), so the value is \(6\sqrt{2}\).
Step 3
Exam Tip
\( \sqrt{8}=2\sqrt{2} \) और \( \sqrt{18}=3\sqrt{2} \)। कुल (1+2+3=6) से \(6\sqrt{2}\) मिलता है।
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\( \sqrt{3}+\sqrt{12}+\sqrt{27}+\sqrt{48} \) का सरल रूप क्या है?
What is the simplified form of \( \sqrt{3}+\sqrt{12}+\sqrt{27}+\sqrt{48} \)?
#real numbers
#surd series
#simplification
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A \(10\sqrt{3}\)
B \(8\sqrt{3}\)
C \(12\sqrt{3}\)
D \(16\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(10\sqrt{3}\)
Step 1
Concept
These are \( \sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3} \). The total is \(10\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(10\sqrt{3}\). These are \( \sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3} \). The total is \(10\sqrt{3}\).
Step 3
Exam Tip
ये क्रमशः \( \sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3} \) हैं। कुल \(10\sqrt{3}\) होगा।
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( \left\(\frac{1}{\sqrt{5}-2}\right\)-\left\(\frac{1}{\sqrt{5}+2}\right\) ) का मान क्या है?
What is the value of ( \left\(\frac{1}{\sqrt{5}-2}\right\)-\left\(\frac{1}{\sqrt{5}+2}\right\) )?
#real numbers
#rationalisation
#expression
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A (4)
B \(2\sqrt{5}\)
C \( \sqrt{5} \)
D (8)
Explanation opens after your attempt
Step 1
Concept
The common denominator is (5-4=1), and the numerator is ( \sqrt{5}+2-\(\sqrt{5}-2\)=4 ). Hence the value is (4).
Step 2
Why this answer is correct
The correct answer is A. (4). The common denominator is (5-4=1), and the numerator is ( \sqrt{5}+2-\(\sqrt{5}-2\)=4 ). Hence the value is (4).
Step 3
Exam Tip
समान हर (5-4=1) होगा और अंश ( \sqrt{5}+2-\(\sqrt{5}-2\)=4 ) है। इसलिए मान (4) है।
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( \left\(\frac{1}{3-\sqrt{7}}\right\)+\left\(\frac{1}{3+\sqrt{7}}\right\) ) का मान क्या है?
What is the value of ( \left\(\frac{1}{3-\sqrt{7}}\right\)+\left\(\frac{1}{3+\sqrt{7}}\right\) )?
#real numbers
#conjugate sum
#hard
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A (3)
B (6)
C \( \frac{3}{2} \)
D \( \sqrt{7} \)
Explanation opens after your attempt
Step 1
Concept
The common denominator is (9-7=2), and the numerator is \(3+\sqrt{7}+3-\sqrt{7}=6\). So the value is \( \frac{6}{2}=3 \).
Step 2
Why this answer is correct
The correct answer is A. (3). The common denominator is (9-7=2), and the numerator is \(3+\sqrt{7}+3-\sqrt{7}=6\). So the value is \( \frac{6}{2}=3 \).
Step 3
Exam Tip
समान हर (9-7=2) और अंश \(3+\sqrt{7}+3-\sqrt{7}=6\) है। इसलिए मान \( \frac{6}{2}=3 \) है।
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( \left\(\sqrt{6}+\sqrt{2}\right\)\left\(\sqrt{6}-\sqrt{2}\right\) ) का मान क्या है?
What is the value of ( \left\(\sqrt{6}+\sqrt{2}\right\)\left\(\sqrt{6}-\sqrt{2}\right\) )?
#real numbers
#conjugate
#surds
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A (4)
B (8)
C \(2\sqrt{12}\)
D \(6+\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
This is conjugate multiplication. The value is (6-2=4).
Step 2
Why this answer is correct
The correct answer is A. (4). This is conjugate multiplication. The value is (6-2=4).
Step 3
Exam Tip
यह संयुग्मी गुणन है। (6-2=4) मिलेगा।
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\( \sqrt{12}+\sqrt{75}-\sqrt{48}-\sqrt{27} \) का मान क्या है?
What is the value of \( \sqrt{12}+\sqrt{75}-\sqrt{48}-\sqrt{27} \)?
#real numbers
#surd cancellation
#calculation
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A (0)
B \(2\sqrt{3}\)
C \(4\sqrt{3}\)
D \(6\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
They become \(2\sqrt{3},5\sqrt{3},4\sqrt{3},3\sqrt{3}\). Since (2+5-4-3=0), the value is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). They become \(2\sqrt{3},5\sqrt{3},4\sqrt{3},3\sqrt{3}\). Since (2+5-4-3=0), the value is (0).
Step 3
Exam Tip
क्रमशः \(2\sqrt{3},5\sqrt{3},4\sqrt{3},3\sqrt{3}\) मिलते हैं। (2+5-4-3=0) इसलिए मान (0) है।
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यदि \(x=3-\sqrt{5}\) है तो \( \frac{1}{x} \) किसके बराबर है?
If \(x=3-\sqrt{5}\), then \( \frac{1}{x} \) equals what?
#real numbers
#rationalisation
#reciprocal
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A \( \frac{3+\sqrt{5}}{4} \)
B \(3+\sqrt{5}\)
C \( \frac{3-\sqrt{5}}{4} \)
D (4\(3+\sqrt{5}\))
Explanation opens after your attempt
Correct Answer
A. \( \frac{3+\sqrt{5}}{4} \)
Step 1
Concept
Multiply \( \frac{1}{3-\sqrt{5}} \) by the conjugate. The denominator is (9-5=4) and the numerator is \(3+\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{3+\sqrt{5}}{4} \). Multiply \( \frac{1}{3-\sqrt{5}} \) by the conjugate. The denominator is (9-5=4) and the numerator is \(3+\sqrt{5}\).
Step 3
Exam Tip
\( \frac{1}{3-\sqrt{5}} \) को संयुग्मी से गुणा करें। हर (9-5=4) और अंश \(3+\sqrt{5}\) होगा।
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यदि \(a=\sqrt{3}+\sqrt{2}\) है तो \(a^2+\frac{1}{a^2}\) का मान क्या है?
If \(a=\sqrt{3}+\sqrt{2}\), what is the value of \(a^2+\frac{1}{a^2}\)?
#real numbers
#surd expression
#reciprocal
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A (50)
B (98)
C (10)
D (2)
Explanation opens after your attempt
Step 1
Concept
\(a^2=5+2\sqrt{6}\) and \( \frac{1}{a^2}=5-2\sqrt{6} \). The sum should be (10).
Step 2
Why this answer is correct
The correct answer is B. (98). \(a^2=5+2\sqrt{6}\) and \( \frac{1}{a^2}=5-2\sqrt{6} \). The sum should be (10).
Step 3
Exam Tip
\(a^2=5+2\sqrt{6}\) और \( \frac{1}{a^2}=5-2\sqrt{6} \)। योग (10) होना चाहिए।
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यदि \(x=\sqrt{5}+2\) है तो \(x+\frac{1}{x}\) का मान क्या है?
If \(x=\sqrt{5}+2\), what is the value of \(x+\frac{1}{x}\)?
#real numbers
#conjugate
#expression
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A \(2\sqrt{5}\)
B (4)
C \( \sqrt{5} \)
D \(2+\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{5}\)
Step 1
Concept
\( \frac{1}{\sqrt{5}+2}=\sqrt{5}-2 \). Hence \(x+\frac{1}{x}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{5}\). \( \frac{1}{\sqrt{5}+2}=\sqrt{5}-2 \). Hence \(x+\frac{1}{x}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\).
Step 3
Exam Tip
\( \frac{1}{\sqrt{5}+2}=\sqrt{5}-2 \) है। इसलिए \(x+\frac{1}{x}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)।
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यदि (x=-7) है तो \( \sqrt{x^2}-2x \) का मान क्या होगा?
If (x=-7), what is the value of \( \sqrt{x^2}-2x \)?
#real numbers
#absolute value
#substitution
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A (21)
B (7)
C ( -7 )
D ( -21 )
Explanation opens after your attempt
Step 1
Concept
\( \sqrt{x^2}=|x|=7 \) and (-2x=14). Therefore the total is (21).
Step 2
Why this answer is correct
The correct answer is A. (21). \( \sqrt{x^2}=|x|=7 \) and (-2x=14). Therefore the total is (21).
Step 3
Exam Tip
\( \sqrt{x^2}=|x|=7 \) और (-2x=14) है। इसलिए कुल (21) मिलेगा।
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\( |3-\sqrt{20}| \) का सही सरल रूप क्या है?
What is the correct simplified form of \( |3-\sqrt{20}| \)?
#real numbers
#absolute value
#surds
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A \( \sqrt{20}-3 \)
B \(3-\sqrt{20}\)
C \(3+\sqrt{20}\)
D \( \sqrt{20}+3 \)
Explanation opens after your attempt
Correct Answer
A. \( \sqrt{20}-3 \)
Step 1
Concept
Since \( \sqrt{20}>3 \), \(3-\sqrt{20}\) is negative. Absolute value changes it to \( \sqrt{20}-3 \).
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{20}-3 \). Since \( \sqrt{20}>3 \), \(3-\sqrt{20}\) is negative. Absolute value changes it to \( \sqrt{20}-3 \).
Step 3
Exam Tip
\( \sqrt{20}>3 \) इसलिए \(3-\sqrt{20}\) ऋणात्मक है। निरपेक्ष मान इसे \( \sqrt{20}-3 \) बना देता है।
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\( |\sqrt{30}-6| \) का सरल रूप क्या है?
What is the simplified form of \( |\sqrt{30}-6| \)?
#real numbers
#absolute value
#comparison
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A \(6-\sqrt{30}\)
B \( \sqrt{30}-6 \)
C \( \sqrt{30}+6 \)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(6-\sqrt{30}\)
Step 1
Concept
Since \( \sqrt{30}<6 \), \( \sqrt{30}-6 \) is negative. The absolute value is \(6-\sqrt{30}\).
Step 2
Why this answer is correct
The correct answer is A. \(6-\sqrt{30}\). Since \( \sqrt{30}<6 \), \( \sqrt{30}-6 \) is negative. The absolute value is \(6-\sqrt{30}\).
Step 3
Exam Tip
\( \sqrt{30}<6 \) इसलिए \( \sqrt{30}-6 \) ऋणात्मक है। निरपेक्ष मान \(6-\sqrt{30}\) होगा।
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\( \sqrt{27}+\sqrt{12} \) और \( \sqrt{75} \) के बारे में सही कथन क्या है?
What is the correct statement about \( \sqrt{27}+\sqrt{12} \) and \( \sqrt{75} \)?
#real numbers
#surd comparison
#equality
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A दोनों बराबर हैं / Both are equal
B पहला बड़ा है / The first is greater
C दूसरा बड़ा है / The second is greater
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Correct Answer
A. दोनों बराबर हैं / Both are equal
Step 1
Concept
\( \sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3} \). Also \( \sqrt{75}=5\sqrt{3} \), so both are equal.
Step 2
Why this answer is correct
The correct answer is A. दोनों बराबर हैं / Both are equal. \( \sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3} \). Also \( \sqrt{75}=5\sqrt{3} \), so both are equal.
Step 3
Exam Tip
\( \sqrt{27}+\sqrt{12}=3\sqrt{3}+2\sqrt{3}=5\sqrt{3} \)। \( \sqrt{75}=5\sqrt{3} \) इसलिए दोनों बराबर हैं।
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\( \sqrt{48}+\sqrt{108} \) और \(10\sqrt{3}\) में कौन सा संबंध सही है?
Which relation is correct between \( \sqrt{48}+\sqrt{108} \) and \(10\sqrt{3}\)?
#real numbers
#surd comparison
#identity
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A \( \sqrt{48}+\sqrt{108}=10\sqrt{3} \)
B \( \sqrt{48}+\sqrt{108}>10\sqrt{3} \)
C \( \sqrt{48}+\sqrt{108}<10\sqrt{3} \)
D तुलना संभव नहीं / Comparison is not possible
Explanation opens after your attempt
Correct Answer
A. \( \sqrt{48}+\sqrt{108}=10\sqrt{3} \)
Step 1
Concept
\( \sqrt{48}=4\sqrt{3} \) and \( \sqrt{108}=6\sqrt{3} \). Their sum is \(10\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{48}+\sqrt{108}=10\sqrt{3} \). \( \sqrt{48}=4\sqrt{3} \) and \( \sqrt{108}=6\sqrt{3} \). Their sum is \(10\sqrt{3}\).
Step 3
Exam Tip
\( \sqrt{48}=4\sqrt{3} \) और \( \sqrt{108}=6\sqrt{3} \)। योग \(10\sqrt{3}\) 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 \( \frac{1}{\sqrt{3}-\sqrt{2}}-\frac{1}{\sqrt{3}+\sqrt{2}} \)?
#real numbers
#rationalisation
#difference
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A \(2\sqrt{2}\)
B \(2\sqrt{3}\)
C \( \sqrt{6} \)
D (2)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{2}\)
Step 1
Concept
The common denominator is (3-2=1), and the numerator is ( \sqrt{3}+\sqrt{2}-\(\sqrt{3}-\sqrt{2}\)=2\sqrt{2} ).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{2}\). The common denominator is (3-2=1), and the numerator is ( \sqrt{3}+\sqrt{2}-\(\sqrt{3}-\sqrt{2}\)=2\sqrt{2} ).
Step 3
Exam Tip
समान हर (3-2=1) और अंश ( \sqrt{3}+\sqrt{2}-\(\sqrt{3}-\sqrt{2}\)=2\sqrt{2} ) है।
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\( \frac{1}{\sqrt{11}+3}+\frac{1}{\sqrt{11}-3} \) का मान क्या है?
What is the value of \( \frac{1}{\sqrt{11}+3}+\frac{1}{\sqrt{11}-3} \)?
#real numbers
#conjugate sum
#rationalisation
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A \( \sqrt{11} \)
B \(2\sqrt{11}\)
C \( \frac{\sqrt{11}}{2} \)
D (6)
Explanation opens after your attempt
Correct Answer
A. \( \sqrt{11} \)
Step 1
Concept
The common denominator is (11-9=2), and the numerator is \(2\sqrt{11}\). Therefore the value is \( \sqrt{11} \).
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{11} \). The common denominator is (11-9=2), and the numerator is \(2\sqrt{11}\). Therefore the value is \( \sqrt{11} \).
Step 3
Exam Tip
समान हर (11-9=2) और अंश \(2\sqrt{11}\) है। इसलिए मान \( \sqrt{11} \) है।
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( \(2+\sqrt{3}\)3 ) का सरल रूप क्या है?
What is the simplified form of ( \(2+\sqrt{3}\)3 )?
#real numbers
#surd cube
#identity
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A \(26+15\sqrt{3}\)
B \(8+3\sqrt{3}\)
C \(18+12\sqrt{3}\)
D \(26+9\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(26+15\sqrt{3}\)
Step 1
Concept
First find ( \(2+\sqrt{3}\)2 =7+4\sqrt{3} ). Multiplying by \(2+\sqrt{3}\) gives \(26+15\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(26+15\sqrt{3}\). First find ( \(2+\sqrt{3}\)2 =7+4\sqrt{3} ). Multiplying by \(2+\sqrt{3}\) gives \(26+15\sqrt{3}\).
Step 3
Exam Tip
पहले ( \(2+\sqrt{3}\)2 =7+4\sqrt{3} ) निकालें। फिर \(2+\sqrt{3}\) से गुणा करने पर \(26+15\sqrt{3}\) मिलता है।
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( \(3-\sqrt{2}\)3 ) का सरल रूप क्या है?
What is the simplified form of ( \(3-\sqrt{2}\)3 )?
#real numbers
#surd cube
#calculation
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A \(45-29\sqrt{2}\)
B \(27-6\sqrt{2}\)
C \(33-11\sqrt{2}\)
D \(45-27\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(45-29\sqrt{2}\)
Step 1
Concept
First find ( \(3-\sqrt{2}\)2 =11-6\sqrt{2} ). Multiplying by \(3-\sqrt{2}\) gives \(45-29\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(45-29\sqrt{2}\). First find ( \(3-\sqrt{2}\)2 =11-6\sqrt{2} ). Multiplying by \(3-\sqrt{2}\) gives \(45-29\sqrt{2}\).
Step 3
Exam Tip
पहले ( \(3-\sqrt{2}\)2 =11-6\sqrt{2} ) निकालें। फिर \(3-\sqrt{2}\) से गुणा करने पर \(45-29\sqrt{2}\) मिलता है।
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यदि \( \sqrt{n} \) (12) और (13) के बीच है तो (n) के लिए सही सीमा कौन सी है?
If \( \sqrt{n} \) lies between (12) and (13), which range is correct for (n)?
#real numbers
#inequality
#square root
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A (144<n<169)
B (12<n<13)
C (121<n<144)
D (169<n<196)
Explanation opens after your attempt
Correct Answer
A. (144<n<169)
Step 1
Concept
Squaring both sides gives \(12^2<n<13^2\). Therefore (144<n<169) is correct.
Step 2
Why this answer is correct
The correct answer is A. (144<n<169). Squaring both sides gives \(12^2<n<13^2\). Therefore (144<n<169) is correct.
Step 3
Exam Tip
दोनों ओर वर्ग करने पर \(12^2<n<13^2\) मिलेगा। इसलिए (144<n<169) सही है।
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यदि \(15<\sqrt{m}<16\) है तो (m) के लिए सही सीमा कौन सी है?
If \(15<\sqrt{m}<16\), which range is correct for (m)?
#real numbers
#estimation
#inequality
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A (225<m<256)
B (15<m<16)
C (196<m<225)
D (256<m<289)
Explanation opens after your attempt
Correct Answer
A. (225<m<256)
Step 1
Concept
Squaring positive sides gives \(15^2<m<16^2\). Hence (225<m<256) is correct.
Step 2
Why this answer is correct
The correct answer is A. (225<m<256). Squaring positive sides gives \(15^2<m<16^2\). Hence (225<m<256) is correct.
Step 3
Exam Tip
धनात्मक पक्षों का वर्ग करने पर \(15^2<m<16^2\) मिलता है। इसलिए (225<m<256) सही है।
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\( \sqrt{a}+\sqrt{b}=\sqrt{a+b} \) सामान्यतः कब सही नहीं होता?
When is \( \sqrt{a}+\sqrt{b}=\sqrt{a+b} \) generally not correct?
#real numbers
#common error
#square roots
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A जब (a) और (b) दोनों धनात्मक हों / When (a) and (b) are both positive
B जब (a=0) हो / When (a=0)
C जब (b=0) हो / When (b=0)
D जब एक पद शून्य हो / When one term is zero
Explanation opens after your attempt
Correct Answer
A. जब (a) और (b) दोनों धनात्मक हों / When (a) and (b) are both positive
Step 1
Concept
Generally \( \sqrt{a}+\sqrt{b} \neq \sqrt{a+b} \). For example, \( \sqrt{4}+\sqrt{9}=5 \), not \( \sqrt{13} \).
Step 2
Why this answer is correct
The correct answer is A. जब (a) और (b) दोनों धनात्मक हों / When (a) and (b) are both positive. Generally \( \sqrt{a}+\sqrt{b} \neq \sqrt{a+b} \). For example, \( \sqrt{4}+\sqrt{9}=5 \), not \( \sqrt{13} \).
Step 3
Exam Tip
सामान्यतः \( \sqrt{a}+\sqrt{b} \neq \sqrt{a+b} \) होता है। उदाहरण के लिए \( \sqrt{4}+\sqrt{9}=5 \) लेकिन \( \sqrt{13} \) नहीं।
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\( \sqrt{a}\times\sqrt{a} \) का मान क्या है जहाँ \(a\geq0\)?
What is the value of \( \sqrt{a}\times\sqrt{a} \) where \(a\geq0\)?
#real numbers
#square root property
#concept
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A (a)
B (2a)
C \( \sqrt{2a} \)
D \(a^2\)
Explanation opens after your attempt
Step 1
Concept
Multiplying the square root of the same non-negative number by itself gives the number. Therefore the value is (a).
Step 2
Why this answer is correct
The correct answer is A. (a). Multiplying the square root of the same non-negative number by itself gives the number. Therefore the value is (a).
Step 3
Exam Tip
एक ही गैर-ऋणात्मक संख्या के वर्गमूल को स्वयं से गुणा करने पर वही संख्या मिलती है। इसलिए मान (a) है।
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\( \sqrt{16}+\sqrt{25} \) और \( \sqrt{41} \) के बारे में सही कथन क्या है?
What is the correct statement about \( \sqrt{16}+\sqrt{25} \) and \( \sqrt{41} \)?
#real numbers
#comparison
#common error
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A \( \sqrt{16}+\sqrt{25}>\sqrt{41} \)
B \( \sqrt{16}+\sqrt{25}=\sqrt{41} \)
C \( \sqrt{16}+\sqrt{25}<\sqrt{41} \)
D तुलना संभव नहीं / Comparison is not possible
Explanation opens after your attempt
Correct Answer
A. \( \sqrt{16}+\sqrt{25}>\sqrt{41} \)
Step 1
Concept
\( \sqrt{16}+\sqrt{25}=4+5=9 \), and \( \sqrt{41}<7 \). Therefore the first value is greater.
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{16}+\sqrt{25}>\sqrt{41} \). \( \sqrt{16}+\sqrt{25}=4+5=9 \), and \( \sqrt{41}<7 \). Therefore the first value is greater.
Step 3
Exam Tip
\( \sqrt{16}+\sqrt{25}=4+5=9 \) और \( \sqrt{41}<7 \) है। इसलिए पहला मान बड़ा है।
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\( \sqrt{2} \), \( \sqrt{3} \), और \( \sqrt{5} \) से बना कौन सा गुणनफल परिमेय है?
Which product made from \( \sqrt{2} \), \( \sqrt{3} \), and \( \sqrt{5} \) is rational?
#real numbers
#rational result
#surd product
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A \( \sqrt{2}\times\sqrt{8} \)
B \( \sqrt{2}\times\sqrt{3} \)
C \( \sqrt{3}\times\sqrt{5} \)
D \( \sqrt{5}\times\sqrt{2} \)
Explanation opens after your attempt
Correct Answer
A. \( \sqrt{2}\times\sqrt{8} \)
Step 1
Concept
\( \sqrt{2}\times\sqrt{8}=\sqrt{16}=4 \), which is rational. The other products do not become square roots of perfect squares.
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{2}\times\sqrt{8} \). \( \sqrt{2}\times\sqrt{8}=\sqrt{16}=4 \), which is rational. The other products do not become square roots of perfect squares.
Step 3
Exam Tip
\( \sqrt{2}\times\sqrt{8}=\sqrt{16}=4 \) परिमेय है। बाकी गुणनफल पूर्ण वर्ग के वर्गमूल में नहीं बदलते।
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\( \frac{1}{2+\sqrt{3}}+\frac{1}{2-\sqrt{3}} \) का मान क्या है?
What is the value of \( \frac{1}{2+\sqrt{3}}+\frac{1}{2-\sqrt{3}} \)?
#real numbers
#conjugate sum
#rationalisation
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A (4)
B \(2\sqrt{3}\)
C (1)
D \( \sqrt{3} \)
Explanation opens after your attempt
Step 1
Concept
The common denominator is (4-3=1), and the numerator is \(2-\sqrt{3}+2+\sqrt{3}=4\). Therefore the value is (4).
Step 2
Why this answer is correct
The correct answer is A. (4). The common denominator is (4-3=1), and the numerator is \(2-\sqrt{3}+2+\sqrt{3}=4\). Therefore the value is (4).
Step 3
Exam Tip
समान हर (4-3=1) और अंश \(2-\sqrt{3}+2+\sqrt{3}=4\) है। इसलिए मान (4) है।
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