यदि \(x=5+\sqrt{2}\) है तो (x-5) किस प्रकार की संख्या है?
If \(x=5+\sqrt{2}\), what type of number is (x-5)?
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A परिमेय / Rational
B पूर्णांक / Integer
C अपरिमेय / Irrational
D शून्य / Zero
Explanation opens after your attempt
Correct Answer
C. अपरिमेय / Irrational
Step 1
Concept
\(x-5=\sqrt{2}\), which is irrational. First remove rational terms and simplify.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय / Irrational. \(x-5=\sqrt{2}\), which is irrational. First remove rational terms and simplify.
Step 3
Exam Tip
\(x-5=\sqrt{2}\) है जो अपरिमेय है। पहले परिमेय पदों को हटाकर सरल करें।
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यदि \(x=\sqrt{12}+2\sqrt{3}\) है तो (x) का सरल रूप क्या होगा?
If \(x=\sqrt{12}+2\sqrt{3}\), what will be the simplified form of (x)?
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A \(2\sqrt{3}\)
B \(3\sqrt{3}\)
C \(\sqrt{15}\)
D \(4\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
D. \(4\sqrt{3}\)
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) so the total is \(4\sqrt{3}\). Simplify before adding like radicals.
Step 2
Why this answer is correct
The correct answer is D. \(4\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\) so the total is \(4\sqrt{3}\). Simplify before adding like radicals.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) इसलिए कुल \(4\sqrt{3}\) मिलता है। समान मूलों को जोड़ने से पहले सरल करें।
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\(\sqrt{48}+\sqrt{27}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{48}+\sqrt{27}\)?
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A \(7\sqrt{3}\)
B \(5\sqrt{3}\)
C \(\sqrt{75}\)
D \(9\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(7\sqrt{3}\)
Step 1
Concept
\(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the sum is \(7\sqrt{3}\). Simplify before adding like radicals.
Step 2
Why this answer is correct
The correct answer is A. \(7\sqrt{3}\). \(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the sum is \(7\sqrt{3}\). Simplify before adding like radicals.
Step 3
Exam Tip
\(\sqrt{48}=4\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए योग \(7\sqrt{3}\) है। समान मूलों को जोड़ने से पहले सरल करें।
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(\(\sqrt{15}+\sqrt{6}\)\(\sqrt{15}-\sqrt{6}\)) का मान क्या है?
What is the value of (\(\sqrt{15}+\sqrt{6}\)\(\sqrt{15}-\sqrt{6}\))?
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A (9)
B (21)
C \(\sqrt{90}\)
D \(2\sqrt{15}\)
Explanation opens after your attempt
Step 1
Concept
This is the \(a^2-b^2\) form so the value is (15-6=9). Conjugate multiplication removes radicals.
Step 2
Why this answer is correct
The correct answer is A. (9). This is the \(a^2-b^2\) form so the value is (15-6=9). Conjugate multiplication removes radicals.
Step 3
Exam Tip
यह \(a^2-b^2\) रूप है इसलिए मान (15-6=9) है। संयुग्मी गुणन में मूल हट जाता है।
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(\(\sqrt{19}\)2 +8) का मान क्या है?
What is the value of (\(\sqrt{19}\)2 +8)?
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A (11)
B (27)
C \(\sqrt{27}\)
D \(19\sqrt{8}\)
Explanation opens after your attempt
Step 1
Concept
(\(\sqrt{19}\)2 =19), so the value is (27). Squaring removes the square root.
Step 2
Why this answer is correct
The correct answer is B. (27). (\(\sqrt{19}\)2 =19), so the value is (27). Squaring removes the square root.
Step 3
Exam Tip
(\(\sqrt{19}\)2 =19), इसलिए मान (27) है। वर्ग करने पर वर्गमूल हट जाता है।
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\(\frac{4}{\sqrt{3}+1}\) का परिमेयकृत रूप कौन-सा है?
Which is the rationalised form of \(\frac{4}{\sqrt{3}+1}\)?
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A (2\(\sqrt{3}+1\))
B \(4\sqrt{3}-4\)
C \(2\sqrt{3}-2\)
D \(\sqrt{3}-1\)
Explanation opens after your attempt
Correct Answer
C. \(2\sqrt{3}-2\)
Step 1
Concept
Multiplying by the conjugate gives (\frac{4\(\sqrt{3}-1\)}{2}=2\sqrt{3}-2). Simplify the whole fraction after rationalising.
Step 2
Why this answer is correct
The correct answer is C. \(2\sqrt{3}-2\). Multiplying by the conjugate gives (\frac{4\(\sqrt{3}-1\)}{2}=2\sqrt{3}-2). Simplify the whole fraction after rationalising.
Step 3
Exam Tip
संयुग्मी से गुणा करने पर (\frac{4\(\sqrt{3}-1\)}{2}=2\sqrt{3}-2) मिलता है। हर को परिमेय बनाते समय पूरा भिन्न सरल करें।
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\(\frac{1}{\sqrt{6}+1}\) को परिमेयकृत करने पर क्या मिलेगा?
What is obtained by rationalising \(\frac{1}{\sqrt{6}+1}\)?
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A \(\frac{\sqrt{6}+1}{5}\)
B \(\sqrt{6}+1\)
C \(\frac{\sqrt{6}-1}{5}\)
D \(\frac{1}{\sqrt{6}-1}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{\sqrt{6}-1}{5}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (6-1=5). So the rationalised form is \(\frac{\sqrt{6}-1}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{\sqrt{6}-1}{5}\). Multiplying by the conjugate makes the denominator (6-1=5). So the rationalised form is \(\frac{\sqrt{6}-1}{5}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (6-1=5) बनता है। इसलिए परिमेयकृत रूप \(\frac{\sqrt{6}-1}{5}\) है।
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यदि \(a=\sqrt{8}-\sqrt{2}\) है तो (a) किसके बराबर है?
If \(a=\sqrt{8}-\sqrt{2}\), what is (a) equal to?
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A \(3\sqrt{2}\)
B \(\sqrt{2}\)
C \(\sqrt{6}\)
D (2)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{2}\)
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\) so \(2\sqrt{2}-\sqrt{2}=\sqrt{2}\). Subtract coefficients of like radicals.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). \(\sqrt{8}=2\sqrt{2}\) so \(2\sqrt{2}-\sqrt{2}=\sqrt{2}\). Subtract coefficients of like radicals.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\) इसलिए \(2\sqrt{2}-\sqrt{2}=\sqrt{2}\) है। समान मूलों के गुणांक घटाएं।
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यदि \(a=3\sqrt{2}+4\) और \(b=3\sqrt{2}-4\) हैं तो (a+b) क्या है?
If \(a=3\sqrt{2}+4\) and \(b=3\sqrt{2}-4\), what is (a+b)?
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A (8)
B \(6\sqrt{2}\)
C (18)
D \(3\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
B. \(6\sqrt{2}\)
Step 1
Concept
The constant terms cancel and \(3\sqrt{2}+3\sqrt{2}=6\sqrt{2}\). Add like radicals.
Step 2
Why this answer is correct
The correct answer is B. \(6\sqrt{2}\). The constant terms cancel and \(3\sqrt{2}+3\sqrt{2}=6\sqrt{2}\). Add like radicals.
Step 3
Exam Tip
स्थिर पद कट जाते हैं और \(3\sqrt{2}+3\sqrt{2}=6\sqrt{2}\) मिलता है। समान मूलों को जोड़ें।
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(\(5+\sqrt{3}\)\(5-\sqrt{3}\)) का मान क्या है?
What is the value of (\(5+\sqrt{3}\)\(5-\sqrt{3}\))?
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A (28)
B (22)
C \(10\sqrt{3}\)
D \(25+\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
This is \(a^2-b^2\), so the value is (25-3=22). Conjugate multiplication removes the radical.
Step 2
Why this answer is correct
The correct answer is B. (22). This is \(a^2-b^2\), so the value is (25-3=22). Conjugate multiplication removes the radical.
Step 3
Exam Tip
यह \(a^2-b^2\) है इसलिए मान (25-3=22) है। संयुग्मी गुणन में मूल हट जाता है।
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दशमलव \(0.414141\ldots\) और \(0.4141141114\ldots\) में कौन-सा अपरिमेय है?
Which is irrational between \(0.414141\ldots\) and \(0.4141141114\ldots\)?
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A \(0.4141141114\ldots\)
B \(0.414141\ldots\)
C दोनों परिमेय हैं / Both are rational
D दोनों सांत हैं / Both are terminating
Explanation opens after your attempt
Correct Answer
A. \(0.4141141114\ldots\)
Step 1
Concept
\(0.4141141114\ldots\) has no fixed repetition so it is irrational. A repeating decimal is rational.
Step 2
Why this answer is correct
The correct answer is A. \(0.4141141114\ldots\). \(0.4141141114\ldots\) has no fixed repetition so it is irrational. A repeating decimal is rational.
Step 3
Exam Tip
\(0.4141141114\ldots\) में निश्चित दोहराव नहीं है इसलिए यह अपरिमेय है। आवर्ती दशमलव परिमेय होता है।
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\(\sqrt{98}-\sqrt{50}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{98}-\sqrt{50}\)?
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A \(12\sqrt{2}\)
B \(\sqrt{48}\)
C \(2\sqrt{2}\)
D \(4\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
C. \(2\sqrt{2}\)
Step 1
Concept
\(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the difference is \(2\sqrt{2}\). First take out perfect-square factors.
Step 2
Why this answer is correct
The correct answer is C. \(2\sqrt{2}\). \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the difference is \(2\sqrt{2}\). First take out perfect-square factors.
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए अंतर \(2\sqrt{2}\) है। पहले पूर्ण वर्ग गुणनखंड निकालें।
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\(\sqrt{7}+\sqrt{28}+\sqrt{63}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{7}+\sqrt{28}+\sqrt{63}\)?
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A \(4\sqrt{7}\)
B \(5\sqrt{7}\)
C \(6\sqrt{7}\)
D \(7\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
C. \(6\sqrt{7}\)
Step 1
Concept
\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the total is \(6\sqrt{7}\). Convert all terms to like radicals.
Step 2
Why this answer is correct
The correct answer is C. \(6\sqrt{7}\). \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the total is \(6\sqrt{7}\). Convert all terms to like radicals.
Step 3
Exam Tip
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए कुल \(6\sqrt{7}\) है। सभी पदों को समान मूल में बदलें।
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दशमलव \(0.404004000400004\ldots\) किस प्रकार की संख्या है?
What type of number is the decimal \(0.404004000400004\ldots\)?
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A सांत परिमेय / Terminating rational
B आवर्ती परिमेय / Repeating rational
C पूर्णांक / Integer
D अपरिमेय / Irrational
Explanation opens after your attempt
Correct Answer
D. अपरिमेय / Irrational
Step 1
Concept
This decimal has no fixed repeating block, so it is irrational. Identify non-terminating non-repeating decimals.
Step 2
Why this answer is correct
The correct answer is D. अपरिमेय / Irrational. This decimal has no fixed repeating block, so it is irrational. Identify non-terminating non-repeating decimals.
Step 3
Exam Tip
इस दशमलव में निश्चित दोहराने वाला खंड नहीं है इसलिए यह अपरिमेय है। असांत अनावर्ती दशमलव को पहचानें।
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\(\frac{\sqrt{98}-\sqrt{18}}{\sqrt{2}}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sqrt{98}-\sqrt{18}}{\sqrt{2}}\)?
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A (2)
B (4)
C (6)
D (8)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\) so division gives (4). First convert the numerator into like radicals.
Step 2
Why this answer is correct
The correct answer is B. (4). \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\) so division gives (4). First convert the numerator into like radicals.
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\) इसलिए भाग देने पर (4) मिलता है। पहले अंश को समान मूल में बदलें।
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\(\frac{\sqrt{108}}{\sqrt{3}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{108}}{\sqrt{3}}\)?
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A (6)
B \(\sqrt{105}\)
C (36)
D \(6\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{36}=6\). Combine radicals in division and simplify.
Step 2
Why this answer is correct
The correct answer is A. (6). \(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{36}=6\). Combine radicals in division and simplify.
Step 3
Exam Tip
\(\frac{\sqrt{108}}{\sqrt{3}}=\sqrt{36}=6\) है। भाग में मूलों को एक साथ लिखकर सरल करें।
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(\(2+\sqrt{5}\)2 ) का प्रसार कौन-सा है?
Which is the expansion of (\(2+\sqrt{5}\)2 )?
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A \(9+4\sqrt{5}\)
B \(4+\sqrt{5}\)
C \(9+2\sqrt{5}\)
D \(7+4\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(9+4\sqrt{5}\)
Step 1
Concept
(\(2+\sqrt{5}\)2 =4+4\sqrt{5}+5=9+4\sqrt{5}). Do not forget the middle term while squaring.
Step 2
Why this answer is correct
The correct answer is A. \(9+4\sqrt{5}\). (\(2+\sqrt{5}\)2 =4+4\sqrt{5}+5=9+4\sqrt{5}). Do not forget the middle term while squaring.
Step 3
Exam Tip
(\(2+\sqrt{5}\)2 =4+4\sqrt{5}+5=9+4\sqrt{5}) है। वर्ग करते समय मध्य पद न भूलें।
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\(\sqrt{162}+\sqrt{72}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{162}+\sqrt{72}\)?
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A \(9\sqrt{2}\)
B \(15\sqrt{2}\)
C \(6\sqrt{2}\)
D \(21\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
B. \(15\sqrt{2}\)
Step 1
Concept
\(\sqrt{162}=9\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so the sum is \(15\sqrt{2}\). Add coefficients of like radicals.
Step 2
Why this answer is correct
The correct answer is B. \(15\sqrt{2}\). \(\sqrt{162}=9\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so the sum is \(15\sqrt{2}\). Add coefficients of like radicals.
Step 3
Exam Tip
\(\sqrt{162}=9\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\), इसलिए योग \(15\sqrt{2}\) है। समान मूलों के गुणांक जोड़ें।
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यदि \(\sqrt{k}\) संख्या (8) और (9) के बीच है तो (k) के लिए कौन-सा मान संभव है?
If \(\sqrt{k}\) lies between (8) and (9), which value of (k) is possible?
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A (63)
B (64)
C (73)
D (82)
Explanation opens after your attempt
Step 1
Concept
Since (64<73<81), \(8<\sqrt{73}<9\). Decide square-root bounds using squares.
Step 2
Why this answer is correct
The correct answer is C. (73). Since (64<73<81), \(8<\sqrt{73}<9\). Decide square-root bounds using squares.
Step 3
Exam Tip
क्योंकि (64<73<81) इसलिए \(8<\sqrt{73}<9\)। वर्गमूल की सीमा वर्गों से तय करें।
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\(\sqrt{21}\) और \(\sqrt{30}\) के बीच कौन-सी संख्या है?
Which number lies between \(\sqrt{21}\) and \(\sqrt{30}\)?
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A (4)
B \(\sqrt{19}\)
C \(\sqrt{25}\)
D (6)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{25}\)
Step 1
Concept
Since (21<25<30), \(\sqrt{25}\) lies between them. Compare square roots using the numbers inside.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{25}\). Since (21<25<30), \(\sqrt{25}\) lies between them. Compare square roots using the numbers inside.
Step 3
Exam Tip
क्योंकि (21<25<30), इसलिए \(\sqrt{25}\) इनके बीच है। वर्गमूलों की तुलना अंदर की संख्याओं से करें।
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यदि \(z=\sqrt{17}\) है तो \(z^2-9\) का मान क्या है?
If \(z=\sqrt{17}\), what is the value of \(z^2-9\)?
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A (26)
B (8)
C \(\sqrt{8}\)
D \(17\sqrt{9}\)
Explanation opens after your attempt
Step 1
Concept
\(z^2=17\), so \(z^2-9=8\). First square the given radical.
Step 2
Why this answer is correct
The correct answer is B. (8). \(z^2=17\), so \(z^2-9=8\). First square the given radical.
Step 3
Exam Tip
\(z^2=17\), इसलिए \(z^2-9=8\) है। पहले दिए गए मूल का वर्ग करें।
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\(\sqrt{243}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{243}\)?
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A \(3\sqrt{27}\)
B \(27\sqrt{3}\)
C \(9\sqrt{3}\)
D \(81\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
C. \(9\sqrt{3}\)
Step 1
Concept
\(\sqrt{243}=\sqrt{81\times3}=9\sqrt{3}\). Choose the largest perfect-square factor.
Step 2
Why this answer is correct
The correct answer is C. \(9\sqrt{3}\). \(\sqrt{243}=\sqrt{81\times3}=9\sqrt{3}\). Choose the largest perfect-square factor.
Step 3
Exam Tip
\(\sqrt{243}=\sqrt{81\times3}=9\sqrt{3}\) है। सबसे बड़ा पूर्ण वर्ग गुणनखंड चुनें।
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\(\frac{7}{\sqrt{3}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{7}{\sqrt{3}}\)?
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A \(\frac{7}{3}\)
B \(7\sqrt{3}\)
C \(\frac{7\sqrt{3}}{3}\)
D \(\frac{\sqrt{3}}{7}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{7\sqrt{3}}{3}\)
Step 1
Concept
Multiplying numerator and denominator by \(\sqrt{3}\) gives \(\frac{7\sqrt{3}}{3}\). Rationalisation removes the radical from the denominator.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{7\sqrt{3}}{3}\). Multiplying numerator and denominator by \(\sqrt{3}\) gives \(\frac{7\sqrt{3}}{3}\). Rationalisation removes the radical from the denominator.
Step 3
Exam Tip
अंश और हर को \(\sqrt{3}\) से गुणा करने पर \(\frac{7\sqrt{3}}{3}\) मिलता है। परिमेयकरण में हर से मूल हटता है।
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(\(\sqrt{12}+\sqrt{3}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{12}+\sqrt{3}\)2 )?
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A (15)
B (27)
C \(12+3\sqrt{3}\)
D (9)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\), so the bracket is \(3\sqrt{3}\) and its square is (27). Simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is B. (27). \(\sqrt{12}=2\sqrt{3}\), so the bracket is \(3\sqrt{3}\) and its square is (27). Simplify the bracket first.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\), इसलिए कोष्ठक \(3\sqrt{3}\) है और वर्ग (27) है। पहले कोष्ठक सरल करें।
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\(9+\sqrt{14}\) में अपरिमेय भाग कौन-सा है?
What is the irrational part in \(9+\sqrt{14}\)?
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A (9)
B \(\sqrt{14}\)
C (14)
D \(9\sqrt{14}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{14}\)
Step 1
Concept
(9) is rational and \(\sqrt{14}\) is the irrational part. Identify the radical term in a mixed form.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{14}\). (9) is rational and \(\sqrt{14}\) is the irrational part. Identify the radical term in a mixed form.
Step 3
Exam Tip
(9) परिमेय है और \(\sqrt{14}\) अपरिमेय भाग है। मिश्रित रूप में मूल वाला पद पहचानें।
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\(\sqrt{147}+\sqrt{75}-\sqrt{27}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{147}+\sqrt{75}-\sqrt{27}\)?
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A \(9\sqrt{3}\)
B \(7\sqrt{3}\)
C \(11\sqrt{3}\)
D \(15\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(9\sqrt{3}\)
Step 1
Concept
\(\sqrt{147}=7\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the answer is \(9\sqrt{3}\). Add and subtract coefficients of like radicals.
Step 2
Why this answer is correct
The correct answer is A. \(9\sqrt{3}\). \(\sqrt{147}=7\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the answer is \(9\sqrt{3}\). Add and subtract coefficients of like radicals.
Step 3
Exam Tip
\(\sqrt{147}=7\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए उत्तर \(9\sqrt{3}\) है। समान मूलों के गुणांक जोड़ें और घटाएं।
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यदि \(\sqrt{m}=11\) है तो (m) का मान क्या है?
If \(\sqrt{m}=11\), what is the value of (m)?
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A (22)
B (121)
C (11)
D \(\sqrt{11}\)
Explanation opens after your attempt
Step 1
Concept
Squaring both sides gives (m=121). Remembering perfect squares is useful.
Step 2
Why this answer is correct
The correct answer is B. (121). Squaring both sides gives (m=121). Remembering perfect squares is useful.
Step 3
Exam Tip
दोनों पक्षों का वर्ग करने पर (m=121) मिलता है। पूर्ण वर्गों को याद रखना उपयोगी है।
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यदि (n) धनात्मक पूर्ण संख्या है और (n) पूर्ण वर्ग नहीं है तो \(\sqrt{n}\) कैसी संख्या होगी?
If (n) is a positive integer and (n) is not a perfect square, what type of number will \(\sqrt{n}\) be?
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A परिमेय / Rational
B पूर्णांक / Integer
C अपरिमेय / Irrational
D सांत दशमलव / Terminating decimal
Explanation opens after your attempt
Correct Answer
C. अपरिमेय / Irrational
Step 1
Concept
The square root of a positive integer is irrational when it is not a perfect square. Check for a perfect square first.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय / Irrational. The square root of a positive integer is irrational when it is not a perfect square. Check for a perfect square first.
Step 3
Exam Tip
पूर्ण वर्ग न होने पर धनात्मक पूर्ण संख्या का वर्गमूल अपरिमेय होता है। पहले पूर्ण वर्ग की जाँच करें।
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\(\sqrt{24}\times\sqrt{54}\) का मान क्या है?
What is the value of \(\sqrt{24}\times\sqrt{54}\)?
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A (36)
B \(\sqrt{78}\)
C \(12\sqrt{3}\)
D (1296)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{24}\times\sqrt{54}=\sqrt{1296}=36\). The product of two irrationals can be rational.
Step 2
Why this answer is correct
The correct answer is A. (36). \(\sqrt{24}\times\sqrt{54}=\sqrt{1296}=36\). The product of two irrationals can be rational.
Step 3
Exam Tip
\(\sqrt{24}\times\sqrt{54}=\sqrt{1296}=36\) है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है।
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\(\sqrt{13}+\sqrt{17}\) के बारे में सही कथन कौन-सा है?
Which statement is correct about \(\sqrt{13}+\sqrt{17}\)?
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A यह \(\sqrt{30}\) के बराबर है / It is equal to \(\sqrt{30}\)
B यह परिमेय है / It is rational
C यह अपरिमेय है / It is irrational
D यह (30) है / It is (30)
Explanation opens after your attempt
Correct Answer
C. यह अपरिमेय है / It is irrational
Step 1
Concept
\(\sqrt{13}\) and \(\sqrt{17}\) are different irrational radicals and their sum is irrational. Different radicals are not added directly.
Step 2
Why this answer is correct
The correct answer is C. यह अपरिमेय है / It is irrational. \(\sqrt{13}\) and \(\sqrt{17}\) are different irrational radicals and their sum is irrational. Different radicals are not added directly.
Step 3
Exam Tip
\(\sqrt{13}\) और \(\sqrt{17}\) अलग अपरिमेय मूल हैं और उनका योग अपरिमेय है। अलग मूलों को सीधे नहीं जोड़ा जाता।
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\(\sqrt{392}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{392}\)?
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A \(14\sqrt{2}\)
B \(28\sqrt{2}\)
C \(7\sqrt{8}\)
D \(196\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(14\sqrt{2}\)
Step 1
Concept
\(\sqrt{392}=\sqrt{196\times2}=14\sqrt{2}\). A large perfect-square factor makes the solution easier.
Step 2
Why this answer is correct
The correct answer is A. \(14\sqrt{2}\). \(\sqrt{392}=\sqrt{196\times2}=14\sqrt{2}\). A large perfect-square factor makes the solution easier.
Step 3
Exam Tip
\(\sqrt{392}=\sqrt{196\times2}=14\sqrt{2}\) है। बड़े पूर्ण वर्ग गुणनखंड से हल सरल होता है।
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\(\frac{\sqrt{180}}{\sqrt{5}}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sqrt{180}}{\sqrt{5}}\)?
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A (18)
B (6)
C \(\sqrt{175}\)
D (36)
Explanation opens after your attempt
Step 1
Concept
\(\frac{\sqrt{180}}{\sqrt{5}}=\sqrt{36}=6\). In division of roots take the quotient inside.
Step 2
Why this answer is correct
The correct answer is B. (6). \(\frac{\sqrt{180}}{\sqrt{5}}=\sqrt{36}=6\). In division of roots take the quotient inside.
Step 3
Exam Tip
\(\frac{\sqrt{180}}{\sqrt{5}}=\sqrt{36}=6\) है। मूलों के भाग में अंदर का भागफल लें।
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\(11\sqrt{3}-4\sqrt{3}\) का परिणाम क्या है?
What is the result of \(11\sqrt{3}-4\sqrt{3}\)?
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A \(15\sqrt{3}\)
B \(7\sqrt{3}\)
C (7)
D \(\sqrt{8}\)
Explanation opens after your attempt
Correct Answer
B. \(7\sqrt{3}\)
Step 1
Concept
Subtracting coefficients of like radicals gives \(7\sqrt{3}\). It is irrational because \(\sqrt{3}\) remains.
Step 2
Why this answer is correct
The correct answer is B. \(7\sqrt{3}\). Subtracting coefficients of like radicals gives \(7\sqrt{3}\). It is irrational because \(\sqrt{3}\) remains.
Step 3
Exam Tip
समान मूलों के गुणांक घटाने पर \(7\sqrt{3}\) मिलता है। यह अपरिमेय है क्योंकि \(\sqrt{3}\) बचता है।
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(\sqrt{5}\(\sqrt{45}+\sqrt{80}\)) का मान क्या है?
What is the value of (\sqrt{5}\(\sqrt{45}+\sqrt{80}\))?
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A (35)
B \(7\sqrt{5}\)
C (25)
D \(5\sqrt{125}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\), so the bracket is \(7\sqrt{5}\) and the product is (35). Simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is A. (35). \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\), so the bracket is \(7\sqrt{5}\) and the product is (35). Simplify the bracket first.
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{80}=4\sqrt{5}\), इसलिए कोष्ठक \(7\sqrt{5}\) और गुणनफल (35) है। पहले कोष्ठक सरल करें।
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\(\sqrt{43}\) और \(\sqrt{47}\) की तुलना में कौन-सा कथन सही है?
Which statement is correct when comparing \(\sqrt{43}\) and \(\sqrt{47}\)?
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A \(\sqrt{43}>\sqrt{47}\)
B \(\sqrt{43}=\sqrt{47}\)
C \(\sqrt{43}<\sqrt{47}\)
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{43}<\sqrt{47}\)
Step 1
Concept
Since (43<47), \(\sqrt{43}<\sqrt{47}\). For positive roots compare the numbers inside.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{43}<\sqrt{47}\). Since (43<47), \(\sqrt{43}<\sqrt{47}\). For positive roots compare the numbers inside.
Step 3
Exam Tip
क्योंकि (43<47), इसलिए \(\sqrt{43}<\sqrt{47}\)। धनात्मक मूलों में अंदर की संख्याओं की तुलना करें।
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(\sqrt{7}\times\(2\sqrt{7}-3\)) का सरल रूप क्या है?
What is the simplified form of (\sqrt{7}\times\(2\sqrt{7}-3\))?
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A \(14-3\sqrt{7}\)
B \(2-3\sqrt{7}\)
C (14-3)
D \(6\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
A. \(14-3\sqrt{7}\)
Step 1
Concept
Distributing gives \(2\times7-3\sqrt{7}=14-3\sqrt{7}\). Keep radical and rational terms separate.
Step 2
Why this answer is correct
The correct answer is A. \(14-3\sqrt{7}\). Distributing gives \(2\times7-3\sqrt{7}=14-3\sqrt{7}\). Keep radical and rational terms separate.
Step 3
Exam Tip
वितरण करने पर \(2\times7-3\sqrt{7}=14-3\sqrt{7}\) मिलता है। मूल और परिमेय पद अलग रखें।
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\(\frac{4}{\sqrt{5}+1}\) का परिमेयकृत रूप कौन-सा है?
Which is the rationalised form of \(\frac{4}{\sqrt{5}+1}\)?
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A \(4\sqrt{5}+4\)
B \(\sqrt{5}-1\)
C \(4\sqrt{5}-4\)
D \(\frac{4}{\sqrt{5}-1}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{5}-1\)
Step 1
Concept
Multiplying by the conjugate gives (\frac{4\(\sqrt{5}-1\)}{5-1}=\sqrt{5}-1). Make the denominator rational.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{5}-1\). Multiplying by the conjugate gives (\frac{4\(\sqrt{5}-1\)}{5-1}=\sqrt{5}-1). Make the denominator rational.
Step 3
Exam Tip
संयुग्मी से गुणा करने पर (\frac{4\(\sqrt{5}-1\)}{5-1}=\sqrt{5}-1) मिलता है। हर को परिमेय बनाएं।
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\(10+\sqrt{29}\) और \(10-\sqrt{29}\) का योग क्या है?
What is the sum of \(10+\sqrt{29}\) and \(10-\sqrt{29}\)?
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A (20)
B \(2\sqrt{29}\)
C (10)
D (29)
Explanation opens after your attempt
Step 1
Concept
The irrational terms cancel and the sum is (20). The sum of conjugate numbers is rational.
Step 2
Why this answer is correct
The correct answer is A. (20). The irrational terms cancel and the sum is (20). The sum of conjugate numbers is rational.
Step 3
Exam Tip
अपरिमेय पद कट जाते हैं और योग (20) है। संयुग्मी संख्याओं का योग परिमेय होता है।
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\(10+\sqrt{29}\) और \(10-\sqrt{29}\) का गुणनफल क्या है?
What is the product of \(10+\sqrt{29}\) and \(10-\sqrt{29}\)?
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A (129)
B (71)
C \(100+\sqrt{29}\)
D \(20\sqrt{29}\)
Explanation opens after your attempt
Step 1
Concept
The product is (102 -\(\sqrt{29}\)2 =100-29=71). Use \(a^2-b^2\) in conjugate multiplication.
Step 2
Why this answer is correct
The correct answer is B. (71). The product is (102 -\(\sqrt{29}\)2 =100-29=71). Use \(a^2-b^2\) in conjugate multiplication.
Step 3
Exam Tip
गुणनफल (102 -\(\sqrt{29}\)2 =100-29=71) है। संयुग्मी गुणन में \(a^2-b^2\) लगाएं।
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यदि \(q=7-\sqrt{11}\) है तो (q) किस प्रकार की संख्या है?
If \(q=7-\sqrt{11}\), what type of number is (q)?
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A परिमेय / Rational
B पूर्णांक / Integer
C अपरिमेय / Irrational
D सांत दशमलव / Terminating decimal
Explanation opens after your attempt
Correct Answer
C. अपरिमेय / Irrational
Step 1
Concept
Subtracting irrational \(\sqrt{11}\) from rational (7) gives an irrational number. The irrational part remains.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय / Irrational. Subtracting irrational \(\sqrt{11}\) from rational (7) gives an irrational number. The irrational part remains.
Step 3
Exam Tip
परिमेय (7) में से अपरिमेय \(\sqrt{11}\) घटाने पर अपरिमेय संख्या मिलती है। अपरिमेय भाग बचा रहता है।
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संख्या रेखा पर \(\sqrt{10}\) बनाने के लिए किस समकोण त्रिभुज का कर्ण उपयोग हो सकता है?
To construct \(\sqrt{10}\) on the number line, which right triangle hypotenuse can be used?
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A भुजाएँ (1) और (3) / Legs (1) and (3)
B भुजाएँ (2) और (2) / Legs (2) and (2)
C भुजाएँ (1) और (2) / Legs (1) and (2)
D भुजाएँ (3) और (3) / Legs (3) and (3)
Explanation opens after your attempt
Correct Answer
A. भुजाएँ (1) और (3) / Legs (1) and (3)
Step 1
Concept
In a right triangle, the hypotenuse is \(\sqrt{1^2+3^2}=\sqrt{10}\). Pythagoras theorem is used in construction.
Step 2
Why this answer is correct
The correct answer is A. भुजाएँ (1) और (3) / Legs (1) and (3). In a right triangle, the hypotenuse is \(\sqrt{1^2+3^2}=\sqrt{10}\). Pythagoras theorem is used in construction.
Step 3
Exam Tip
समकोण त्रिभुज में कर्ण \(\sqrt{1^2+3^2}=\sqrt{10}\) होगा। निर्माण में पाइथागोरस प्रमेय लगती है।
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\(\sqrt{6}\times\sqrt{24}\) किस प्रकार की संख्या है?
What type of number is \(\sqrt{6}\times\sqrt{24}\)?
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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
\(\sqrt{6}\times\sqrt{24}=\sqrt{144}=12\), 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. \(\sqrt{6}\times\sqrt{24}=\sqrt{144}=12\), which is rational. The product of two irrationals can be rational.
Step 3
Exam Tip
\(\sqrt{6}\times\sqrt{24}=\sqrt{144}=12\) है जो परिमेय है। दो अपरिमेयों का गुणनफल परिमेय हो सकता है।
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\(\sqrt{96}\div\sqrt{6}\) का मान क्या है?
What is the value of \(\sqrt{96}\div\sqrt{6}\)?
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A \(\sqrt{90}\)
B (4)
C \(\sqrt{16}\)
D (16)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{96}\div\sqrt{6}=\sqrt{16}=4\). In division of roots take the quotient inside.
Step 2
Why this answer is correct
The correct answer is B. (4). \(\sqrt{96}\div\sqrt{6}=\sqrt{16}=4\). In division of roots take the quotient inside.
Step 3
Exam Tip
\(\sqrt{96}\div\sqrt{6}=\sqrt{16}=4\) है। मूलों के भाग में अंदर का भागफल लें।
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यदि एक वर्ग का क्षेत्रफल (72) वर्ग इकाई है तो उसकी भुजा का सरल रूप क्या होगा?
If the area of a square is (72) square units, what will be the simplified form of its side?
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A \(36\sqrt{2}\)
B (12)
C \(6\sqrt{2}\)
D \(2\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
C. \(6\sqrt{2}\)
Step 1
Concept
The side will be \(\sqrt{72}=6\sqrt{2}\). In a square, side equals the square root of area.
Step 2
Why this answer is correct
The correct answer is C. \(6\sqrt{2}\). The side will be \(\sqrt{72}=6\sqrt{2}\). In a square, side equals the square root of area.
Step 3
Exam Tip
भुजा \(\sqrt{72}=6\sqrt{2}\) होगी। वर्ग में भुजा क्षेत्रफल के वर्गमूल के बराबर होती है।
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\(\frac{1}{6-\sqrt{5}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{1}{6-\sqrt{5}}\)?
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A \(\frac{6+\sqrt{5}}{31}\)
B \(\frac{6-\sqrt{5}}{31}\)
C \(6+\sqrt{5}\)
D \(\frac{1}{6+\sqrt{5}}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{6+\sqrt{5}}{31}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (36-5=31). So the answer is \(\frac{6+\sqrt{5}}{31}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{6+\sqrt{5}}{31}\). Multiplying by the conjugate makes the denominator (36-5=31). So the answer is \(\frac{6+\sqrt{5}}{31}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (36-5=31) बनता है। इसलिए उत्तर \(\frac{6+\sqrt{5}}{31}\) है।
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\(\sqrt{8}+\sqrt{200}\) और \(12\sqrt{2}\) के बारे में सही कथन कौन-सा है?
Which statement is correct about \(\sqrt{8}+\sqrt{200}\) and \(12\sqrt{2}\)?
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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
\(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{200}=10\sqrt{2}\), so the sum is \(12\sqrt{2}\). Simplify before comparing.
Step 2
Why this answer is correct
The correct answer is C. दोनों बराबर हैं / Both are equal. \(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{200}=10\sqrt{2}\), so the sum is \(12\sqrt{2}\). Simplify before comparing.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{200}=10\sqrt{2}\), इसलिए योग \(12\sqrt{2}\) है। तुलना से पहले सरल करें।
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\(4\sqrt{11}+3\sqrt{44}\) का सरल रूप क्या है?
What is the simplified form of \(4\sqrt{11}+3\sqrt{44}\)?
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A \(10\sqrt{11}\)
B \(7\sqrt{11}\)
C \(13\sqrt{11}\)
D \(22\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
A. \(10\sqrt{11}\)
Step 1
Concept
\(\sqrt{44}=2\sqrt{11}\), so \(4\sqrt{11}+6\sqrt{11}=10\sqrt{11}\). Watch both coefficients and radicals carefully.
Step 2
Why this answer is correct
The correct answer is A. \(10\sqrt{11}\). \(\sqrt{44}=2\sqrt{11}\), so \(4\sqrt{11}+6\sqrt{11}=10\sqrt{11}\). Watch both coefficients and radicals carefully.
Step 3
Exam Tip
\(\sqrt{44}=2\sqrt{11}\), इसलिए \(4\sqrt{11}+6\sqrt{11}=10\sqrt{11}\) है। गुणांक और मूल दोनों ध्यान से देखें।
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यदि \(a=\sqrt{5}+3\) और \(b=\sqrt{5}-3\) हैं तो (ab) क्या है?
If \(a=\sqrt{5}+3\) and \(b=\sqrt{5}-3\), what is (ab)?
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A (-4)
B (4)
C (14)
D \(6\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
(ab=\(\sqrt{5}\)2 -32 =5-9=-4). A conjugate product can be rational.
Step 2
Why this answer is correct
The correct answer is A. (-4). (ab=\(\sqrt{5}\)2 -32 =5-9=-4). A conjugate product can be rational.
Step 3
Exam Tip
(ab=\(\sqrt{5}\)2 -32 =5-9=-4) है। संयुग्मी गुणनफल परिमेय हो सकता है।
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\(\sqrt{13}+\sqrt{208}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{13}+\sqrt{208}\)?
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A \(3\sqrt{13}\)
B \(5\sqrt{13}\)
C \(17\sqrt{13}\)
D \(\sqrt{221}\)
Explanation opens after your attempt
Correct Answer
B. \(5\sqrt{13}\)
Step 1
Concept
\(\sqrt{208}=4\sqrt{13}\), so the sum is \(5\sqrt{13}\). Simplify before adding like radicals.
Step 2
Why this answer is correct
The correct answer is B. \(5\sqrt{13}\). \(\sqrt{208}=4\sqrt{13}\), so the sum is \(5\sqrt{13}\). Simplify before adding like radicals.
Step 3
Exam Tip
\(\sqrt{208}=4\sqrt{13}\), इसलिए योग \(5\sqrt{13}\) है। समान मूल जोड़ने से पहले सरल करें।
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(\(4\sqrt{3}\)2 ) का मान क्या है?
What is the value of (\(4\sqrt{3}\)2 )?
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A (12)
B (24)
C (48)
D \(16\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
(\(4\sqrt{3}\)2 =16\times3=48). Square both the coefficient and the radical.
Step 2
Why this answer is correct
The correct answer is C. (48). (\(4\sqrt{3}\)2 =16\times3=48). Square both the coefficient and the radical.
Step 3
Exam Tip
(\(4\sqrt{3}\)2 =16\times3=48) है। गुणांक और मूल दोनों का वर्ग करें।
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