\(\frac{\sqrt{5}+2}{\sqrt{5}-2}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{\sqrt{5}+2}{\sqrt{5}-2}\)?
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A \(9+4\sqrt{5}\)
B \(9-4\sqrt{5}\)
C \(5+2\sqrt{5}\)
D \(1+\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(9+4\sqrt{5}\)
Step 1
Concept
The denominator conjugate is \(\sqrt{5}+2\) and the denominator becomes (1). So the value is (\(\sqrt{5}+2\)2 =9+4\sqrt{5}).
Step 2
Why this answer is correct
The correct answer is A. \(9+4\sqrt{5}\). The denominator conjugate is \(\sqrt{5}+2\) and the denominator becomes (1). So the value is (\(\sqrt{5}+2\)2 =9+4\sqrt{5}).
Step 3
Exam Tip
हर का संयुग्मी \(\sqrt{5}+2\) है और हर (1) बनता है। इसलिए मान (\(\sqrt{5}+2\)2 =9+4\sqrt{5}) है।
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यदि \(a=5+\sqrt{6}\) और \(b=5-\sqrt{6}\) हैं तो \(a^2-b^2\) का मान क्या है?
If \(a=5+\sqrt{6}\) and \(b=5-\sqrt{6}\), what is the value of \(a^2-b^2\)?
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A \(10\sqrt{6}\)
B (19)
C \(20\sqrt{6}\)
D (50)
Explanation opens after your attempt
Correct Answer
C. \(20\sqrt{6}\)
Step 1
Concept
(a-2 -b-2 =(a-b)(a+b)) where \(a-b=2\sqrt{6}\) and (a+b=10). So the value is \(20\sqrt{6}\).
Step 2
Why this answer is correct
The correct answer is C. \(20\sqrt{6}\). (a-2 -b-2 =(a-b)(a+b)) where \(a-b=2\sqrt{6}\) and (a+b=10). So the value is \(20\sqrt{6}\).
Step 3
Exam Tip
(a-2 -b-2 =(a-b)(a+b)) है जहाँ \(a-b=2\sqrt{6}\) और (a+b=10) है। इसलिए मान \(20\sqrt{6}\) है।
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\(\sqrt{200}-\sqrt{72}+\sqrt{18}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{200}-\sqrt{72}+\sqrt{18}\)?
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A \(5\sqrt{2}\)
B \(7\sqrt{2}\)
C \(9\sqrt{2}\)
D \(7\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
D. \(7\sqrt{2}\)
Step 1
Concept
\(\sqrt{200}=10\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). The result is \(7\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is D. \(7\sqrt{2}\). \(\sqrt{200}=10\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). The result is \(7\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{200}=10\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\) है। परिणाम \(7\sqrt{2}\) है।
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यदि \(p=\frac{1}{\sqrt{11}+3}\) है तो (p) का सरल रूप क्या है?
If \(p=\frac{1}{\sqrt{11}+3}\), what is the simplified form of (p)?
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A \(\sqrt{11}-3\)
B \(\frac{\sqrt{11}-3}{2}\)
C \(\frac{\sqrt{11}+3}{2}\)
D \(3-\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{\sqrt{11}-3}{2}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (11-9=2). So \(p=\frac{\sqrt{11}-3}{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\sqrt{11}-3}{2}\). Multiplying by the conjugate makes the denominator (11-9=2). So \(p=\frac{\sqrt{11}-3}{2}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (11-9=2) बनता है। इसलिए \(p=\frac{\sqrt{11}-3}{2}\) है।
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(\(\sqrt{13}+\sqrt{5}\)\(\sqrt{13}-\sqrt{5}\)) का मान किस प्रकार की संख्या है?
What type of number is the value of (\(\sqrt{13}+\sqrt{5}\)\(\sqrt{13}-\sqrt{5}\))?
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A अपरिमेय (8) / Irrational (8)
B अपरिमेय \(\sqrt{65}\) / Irrational \(\sqrt{65}\)
C परिमेय (8) / Rational (8)
D परिमेय (18) / Rational (18)
Explanation opens after your attempt
Correct Answer
C. परिमेय (8) / Rational (8)
Step 1
Concept
Conjugate multiplication gives (13-5=8). (8) is a rational number.
Step 2
Why this answer is correct
The correct answer is C. परिमेय (8) / Rational (8). Conjugate multiplication gives (13-5=8). (8) is a rational number.
Step 3
Exam Tip
संयुग्मी गुणन से (13-5=8) मिलता है। (8) परिमेय संख्या है।
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यदि \(x=\sqrt{10}-\sqrt{2}\) है तो \(x^2\) का मान कौन-सा है?
If \(x=\sqrt{10}-\sqrt{2}\), which is the value of \(x^2\)?
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A \(12-4\sqrt{5}\)
B (8)
C \(12+4\sqrt{5}\)
D \(\sqrt{8}\)
Explanation opens after your attempt
Correct Answer
A. \(12-4\sqrt{5}\)
Step 1
Concept
(\(\sqrt{10}-\sqrt{2}\)2 =10+2-2\sqrt{20}=12-4\sqrt{5}). Watch the sign of the middle term.
Step 2
Why this answer is correct
The correct answer is A. \(12-4\sqrt{5}\). (\(\sqrt{10}-\sqrt{2}\)2 =10+2-2\sqrt{20}=12-4\sqrt{5}). Watch the sign of the middle term.
Step 3
Exam Tip
(\(\sqrt{10}-\sqrt{2}\)2 =10+2-2\sqrt{20}=12-4\sqrt{5}) है। मध्य पद का चिन्ह ध्यान रखें।
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\(\frac{3}{\sqrt{7}+\sqrt{4}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{3}{\sqrt{7}+\sqrt{4}}\)?
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A (3\(\sqrt{7}-2\))
B \(\sqrt{7}-2\)
C \(\frac{\sqrt{7}-2}{3}\)
D \(3\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{7}-2\)
Step 1
Concept
The denominator is \(\sqrt{7}+2\) and the conjugate makes it (7-4=3). So the answer is \(\sqrt{7}-2\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{7}-2\). The denominator is \(\sqrt{7}+2\) and the conjugate makes it (7-4=3). So the answer is \(\sqrt{7}-2\).
Step 3
Exam Tip
हर \(\sqrt{7}+2\) है और संयुग्मी से हर (7-4=3) बनता है। इसलिए उत्तर \(\sqrt{7}-2\) है।
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\(\sqrt{6+\sqrt{7}}\times\sqrt{6+\sqrt{7}}\) का मान क्या है?
What is the value of \(\sqrt{6+\sqrt{7}}\times\sqrt{6+\sqrt{7}}\)?
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A (36+7)
B \(\sqrt{13}\)
C \(6-\sqrt{7}\)
D \(6+\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
D. \(6+\sqrt{7}\)
Step 1
Concept
Multiplying the same square root by itself gives the number inside. Therefore the value is \(6+\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is D. \(6+\sqrt{7}\). Multiplying the same square root by itself gives the number inside. Therefore the value is \(6+\sqrt{7}\).
Step 3
Exam Tip
एक ही वर्गमूल को अपने आप से गुणा करने पर अंदर की संख्या मिलती है। इसलिए मान \(6+\sqrt{7}\) है।
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यदि \(A=\sqrt{98}+\sqrt{162}\) और \(B=16\sqrt{2}\) हैं तो कौन-सा कथन सही है?
If \(A=\sqrt{98}+\sqrt{162}\) and \(B=16\sqrt{2}\), which statement is correct?
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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
\(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{162}=9\sqrt{2}\), so \(A=16\sqrt{2}\). Hence (A=B).
Step 2
Why this answer is correct
The correct answer is C. (A=B). \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{162}=9\sqrt{2}\), so \(A=16\sqrt{2}\). Hence (A=B).
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{162}=9\sqrt{2}\), इसलिए \(A=16\sqrt{2}\) है। अतः (A=B) है।
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यदि \(q=\sqrt{3}+2\) है तो \(q^2-4q\) का मान क्या है?
If \(q=\sqrt{3}+2\), what is the value of \(q^2-4q\)?
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A (-1)
B (1)
C \(\sqrt{3}\)
D (7)
Explanation opens after your attempt
Step 1
Concept
\(q^2=7+4\sqrt{3}\) and \(4q=8+4\sqrt{3}\). Subtracting gives (-1).
Step 2
Why this answer is correct
The correct answer is A. (-1). \(q^2=7+4\sqrt{3}\) and \(4q=8+4\sqrt{3}\). Subtracting gives (-1).
Step 3
Exam Tip
\(q^2=7+4\sqrt{3}\) और \(4q=8+4\sqrt{3}\) है। घटाने पर (-1) मिलता है।
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किस दशमलव को अपरिमेय संख्या का उदाहरण माना जा सकता है?
Which decimal can be considered an example of an irrational number?
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A \(2.474747\ldots\)
B \(2.47000\ldots\)
C \(2.471471471\ldots\)
D \(2.47047004700047\ldots\)
Explanation opens after your attempt
Correct Answer
D. \(2.47047004700047\ldots\)
Step 1
Concept
An irrational decimal is non-terminating and non-repeating. The fourth option has no fixed repeating block.
Step 2
Why this answer is correct
The correct answer is D. \(2.47047004700047\ldots\). An irrational decimal is non-terminating and non-repeating. The fourth option has no fixed repeating block.
Step 3
Exam Tip
अपरिमेय दशमलव असांत और अनावर्ती होता है। चौथे विकल्प में निश्चित दोहराने वाला खंड नहीं है।
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(\(\sqrt{5}+\sqrt{3}\)2 -\(\sqrt{5}-\sqrt{3}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{5}+\sqrt{3}\)2 -\(\sqrt{5}-\sqrt{3}\)2 )?
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A \(4\sqrt{15}\)
B (8)
C \(2\sqrt{15}\)
D (15)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{15}\)
Step 1
Concept
Use the identity ((a+b)2 -(a-b)2 =4ab). Here the value is \(4\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{15}\). Use the identity ((a+b)2 -(a-b)2 =4ab). Here the value is \(4\sqrt{15}\).
Step 3
Exam Tip
पहचान ((a+b)2 -(a-b)2 =4ab) लगाएं। यहाँ मान \(4\sqrt{15}\) है।
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यदि \(t=\sqrt{17}+4\) है तो \(t+\frac{1}{t}\) का मान क्या है?
If \(t=\sqrt{17}+4\), what is the value of \(t+\frac{1}{t}\)?
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A \(2\sqrt{17}\)
B (8)
C \(\sqrt{17}\)
D \(2\sqrt{17}+8\)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{17}\)
Step 1
Concept
\(\frac{1}{\sqrt{17}+4}=\sqrt{17}-4\) because the denominator becomes (17-16=1). So the sum is \(2\sqrt{17}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{17}\). \(\frac{1}{\sqrt{17}+4}=\sqrt{17}-4\) because the denominator becomes (17-16=1). So the sum is \(2\sqrt{17}\).
Step 3
Exam Tip
\(\frac{1}{\sqrt{17}+4}=\sqrt{17}-4\) है क्योंकि हर (17-16=1) बनता है। इसलिए योग \(2\sqrt{17}\) है।
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यदि \(c=\sqrt{15}+\sqrt{6}\) और \(d=\sqrt{15}-\sqrt{6}\) हैं तो (cd) और (c-d) का सही युग्म कौन-सा है?
If \(c=\sqrt{15}+\sqrt{6}\) and \(d=\sqrt{15}-\sqrt{6}\), which is the correct pair of (cd) and (c-d)?
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A (9), \(2\sqrt{6}\)
B (21), \(2\sqrt{15}\)
C \(9\sqrt{90}\), \(2\sqrt{6}\)
D (15), (6)
Explanation opens after your attempt
Correct Answer
A. (9), \(2\sqrt{6}\)
Step 1
Concept
(cd=15-6=9) and \(c-d=2\sqrt{6}\). Find both values separately in a conjugate pair.
Step 2
Why this answer is correct
The correct answer is A. (9), \(2\sqrt{6}\). (cd=15-6=9) and \(c-d=2\sqrt{6}\). Find both values separately in a conjugate pair.
Step 3
Exam Tip
(cd=15-6=9) और \(c-d=2\sqrt{6}\) है। संयुग्मी युग्म में दोनों मान अलग-अलग निकालें।
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\(\sqrt{10+\sqrt{21}}\) का वर्ग किसके बराबर है?
What is the square of \(\sqrt{10+\sqrt{21}}\) equal to?
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A \(10-\sqrt{21}\)
B \(\sqrt{31}\)
C (100+21)
D \(10+\sqrt{21}\)
Explanation opens after your attempt
Correct Answer
D. \(10+\sqrt{21}\)
Step 1
Concept
The square of a square root gives the number inside. So (\left\(\sqrt{10+\sqrt{21}}\right\)2 =10+\sqrt{21}).
Step 2
Why this answer is correct
The correct answer is D. \(10+\sqrt{21}\). The square of a square root gives the number inside. So (\left\(\sqrt{10+\sqrt{21}}\right\)2 =10+\sqrt{21}).
Step 3
Exam Tip
वर्गमूल का वर्ग अंदर की संख्या देता है। इसलिए (\left\(\sqrt{10+\sqrt{21}}\right\)2 =10+\sqrt{21}) है।
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यदि एक आयत की लंबाई \(\sqrt{14}+\sqrt{5}\) और चौड़ाई \(\sqrt{14}-\sqrt{5}\) है तो क्षेत्रफल क्या होगा?
If a rectangle has length \(\sqrt{14}+\sqrt{5}\) and breadth \(\sqrt{14}-\sqrt{5}\), what will be its area?
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A (19)
B (9)
C \(\sqrt{70}\)
D \(2\sqrt{14}\)
Explanation opens after your attempt
Step 1
Concept
Area is (\(\sqrt{14}+\sqrt{5}\)\(\sqrt{14}-\sqrt{5}\)=14-5=9). Conjugate dimensions can give a rational area.
Step 2
Why this answer is correct
The correct answer is B. (9). Area is (\(\sqrt{14}+\sqrt{5}\)\(\sqrt{14}-\sqrt{5}\)=14-5=9). Conjugate dimensions can give a rational area.
Step 3
Exam Tip
क्षेत्रफल (\(\sqrt{14}+\sqrt{5}\)\(\sqrt{14}-\sqrt{5}\)=14-5=9) है। संयुग्मी आयामों से परिमेय क्षेत्रफल मिल सकता है।
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यदि \(y=\sqrt{8}+\sqrt{18}\) है तो \(\frac{y}{\sqrt{2}}\) का मान क्या है?
If \(y=\sqrt{8}+\sqrt{18}\), what is the value of \(\frac{y}{\sqrt{2}}\)?
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A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so \(y=5\sqrt{2}\). Dividing gives (5).
Step 2
Why this answer is correct
The correct answer is C. (5). \(\sqrt{8}=2\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\), so \(y=5\sqrt{2}\). Dividing gives (5).
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए \(y=5\sqrt{2}\) है। भाग देने पर (5) मिलता है।
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यदि \(m=\sqrt{125}-\sqrt{45}\) और \(n=2\sqrt{5}\) हैं तो (m-n) क्या है?
If \(m=\sqrt{125}-\sqrt{45}\) and \(n=2\sqrt{5}\), what is (m-n)?
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A (0)
B \(\sqrt{5}\)
C \(3\sqrt{5}\)
D \(5\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so \(m=2\sqrt{5}\). Hence (m-n=0).
Step 2
Why this answer is correct
The correct answer is A. (0). \(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so \(m=2\sqrt{5}\). Hence (m-n=0).
Step 3
Exam Tip
\(\sqrt{125}=5\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\), इसलिए \(m=2\sqrt{5}\) है। अतः (m-n=0) है।
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यदि \(z=\sqrt{19}-\sqrt{10}\) है तो \(z^2\) का मान कौन-सा है?
If \(z=\sqrt{19}-\sqrt{10}\), which is the value of \(z^2\)?
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A \(29+2\sqrt{190}\)
B (9)
C \(29-2\sqrt{190}\)
D \(\sqrt{9}\)
Explanation opens after your attempt
Correct Answer
C. \(29-2\sqrt{190}\)
Step 1
Concept
(\(\sqrt{19}-\sqrt{10}\)2 =19+10-2\sqrt{190}). Keep the middle term negative.
Step 2
Why this answer is correct
The correct answer is C. \(29-2\sqrt{190}\). (\(\sqrt{19}-\sqrt{10}\)2 =19+10-2\sqrt{190}). Keep the middle term negative.
Step 3
Exam Tip
(\(\sqrt{19}-\sqrt{10}\)2 =19+10-2\sqrt{190}) है। मध्य पद का चिन्ह ऋणात्मक रखें।
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यदि \(x=\sqrt{2}+\sqrt{8}\) और \(y=\sqrt{18}\) हैं तो (x-y) का मान क्या है?
If \(x=\sqrt{2}+\sqrt{8}\) and \(y=\sqrt{18}\), what is the value of (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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\(\sqrt{7}\) और \(\sqrt{8}\) के बीच कौन-सी संख्या निश्चित रूप से आती है?
Which number definitely lies between \(\sqrt{7}\) and \(\sqrt{8}\)?
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A \(\sqrt{6.5}\)
B (3)
C \(\sqrt{7.5}\)
D \(\sqrt{9}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{7.5}\)
Step 1
Concept
Since (7<7.5<8), \(\sqrt{7.5}\) lies between them. Compare square roots using the numbers inside.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{7.5}\). Since (7<7.5<8), \(\sqrt{7.5}\) lies between them. Compare square roots using the numbers inside.
Step 3
Exam Tip
क्योंकि (7<7.5<8), इसलिए \(\sqrt{7.5}\) दोनों के बीच होगा। वर्गमूलों में अंदर की संख्या से तुलना करें।
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यदि \(u=\sqrt{20}+\sqrt{45}\) और \(v=5\sqrt{5}\) हैं तो कौन-सा कथन सही है?
If \(u=\sqrt{20}+\sqrt{45}\) and \(v=5\sqrt{5}\), which statement is correct?
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A (u>v)
B (u<v)
C (u=v)
D दोनों परिमेय हैं / Both are rational
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so \(u=5\sqrt{5}\). Hence (u=v).
Step 2
Why this answer is correct
The correct answer is C. (u=v). \(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so \(u=5\sqrt{5}\). Hence (u=v).
Step 3
Exam Tip
\(\sqrt{20}=2\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\), इसलिए \(u=5\sqrt{5}\) है। अतः (u=v) है।
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यदि \(x=4+\sqrt{15}\) है तो \(x^2-8x\) का मान क्या है?
If \(x=4+\sqrt{15}\), what is the value of \(x^2-8x\)?
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A (1)
B (-1)
C (15)
D \(2\sqrt{15}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=31+8\sqrt{15}\) and \(8x=32+8\sqrt{15}\). Subtracting gives (-1).
Step 2
Why this answer is correct
The correct answer is B. (-1). \(x^2=31+8\sqrt{15}\) and \(8x=32+8\sqrt{15}\). Subtracting gives (-1).
Step 3
Exam Tip
\(x^2=31+8\sqrt{15}\) और \(8x=32+8\sqrt{15}\) है। घटाने पर (-1) मिलता है।
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\(\sqrt{300}-\sqrt{108}+\sqrt{75}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{300}-\sqrt{108}+\sqrt{75}\)?
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A \(9\sqrt{3}\)
B \(7\sqrt{3}\)
C \(5\sqrt{3}\)
D \(\sqrt{267}\)
Explanation opens after your attempt
Correct Answer
A. \(9\sqrt{3}\)
Step 1
Concept
\(\sqrt{300}=10\sqrt{3}\), \(\sqrt{108}=6\sqrt{3}\), and \(\sqrt{75}=5\sqrt{3}\). Therefore the result is \(9\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(9\sqrt{3}\). \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{108}=6\sqrt{3}\), and \(\sqrt{75}=5\sqrt{3}\). Therefore the result is \(9\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{300}=10\sqrt{3}\), \(\sqrt{108}=6\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\) है। इसलिए परिणाम \(9\sqrt{3}\) है।
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यदि \(x=\sqrt{9+\sqrt{7}}\) है तो \(x^2-9\) का मान क्या है?
If \(x=\sqrt{9+\sqrt{7}}\), what is the value of \(x^2-9\)?
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A \(\sqrt{7}\)
B (7)
C \(\sqrt{16}\)
D (9)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{7}\)
Step 1
Concept
\(x^2=9+\sqrt{7}\). Therefore \(x^2-9=\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{7}\). \(x^2=9+\sqrt{7}\). Therefore \(x^2-9=\sqrt{7}\).
Step 3
Exam Tip
\(x^2=9+\sqrt{7}\) है। इसलिए \(x^2-9=\sqrt{7}\) होगा।
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\(\frac{\sqrt{72}+\sqrt{128}}{\sqrt{2}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{72}+\sqrt{128}}{\sqrt{2}}\)?
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A (12)
B (14)
C (16)
D (18)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{72}=6\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so the numerator is \(14\sqrt{2}\). Dividing gives (14).
Step 2
Why this answer is correct
The correct answer is B. (14). \(\sqrt{72}=6\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so the numerator is \(14\sqrt{2}\). Dividing gives (14).
Step 3
Exam Tip
\(\sqrt{72}=6\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\), इसलिए अंश \(14\sqrt{2}\) है। भाग देने पर (14) मिलता है।
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यदि \(r=\sqrt{7}+\sqrt{2}\) और \(s=\sqrt{7}-\sqrt{2}\) हैं तो \(r^2-s^2\) का मान क्या है?
If \(r=\sqrt{7}+\sqrt{2}\) and \(s=\sqrt{7}-\sqrt{2}\), what is the value of \(r^2-s^2\)?
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A \(4\sqrt{14}\)
B (9)
C \(2\sqrt{14}\)
D (14)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{14}\)
Step 1
Concept
(r-2 -s-2 =(r-s)(r+s)) where \(r-s=2\sqrt{2}\) and \(r+s=2\sqrt{7}\). So the value is \(4\sqrt{14}\).
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{14}\). (r-2 -s-2 =(r-s)(r+s)) where \(r-s=2\sqrt{2}\) and \(r+s=2\sqrt{7}\). So the value is \(4\sqrt{14}\).
Step 3
Exam Tip
(r-2 -s-2 =(r-s)(r+s)) है जहाँ \(r-s=2\sqrt{2}\) और \(r+s=2\sqrt{7}\) है। इसलिए मान \(4\sqrt{14}\) है।
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यदि \(P=\sqrt{242}+\sqrt{50}-\sqrt{98}\) है तो (P) किसके बराबर है?
If \(P=\sqrt{242}+\sqrt{50}-\sqrt{98}\), what is (P) equal to?
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A \(7\sqrt{2}\)
B \(9\sqrt{2}\)
C \(11\sqrt{2}\)
D \(13\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
B. \(9\sqrt{2}\)
Step 1
Concept
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{98}=7\sqrt{2}\). Therefore \(P=9\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(9\sqrt{2}\). \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{98}=7\sqrt{2}\). Therefore \(P=9\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{98}=7\sqrt{2}\) है। इसलिए \(P=9\sqrt{2}\) है।
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(\(\sqrt{28}+\sqrt{63}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{28}+\sqrt{63}\)2 )?
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A (175)
B (91)
C \(25\sqrt{7}\)
D (700)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the sum is \(5\sqrt{7}\). Its square is (175).
Step 2
Why this answer is correct
The correct answer is A. (175). \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the sum is \(5\sqrt{7}\). Its square is (175).
Step 3
Exam Tip
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए योग \(5\sqrt{7}\) है। इसका वर्ग (175) है।
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यदि \(w=\sqrt{13}+\sqrt{6}\) है तो \(w^2-19\) का मान क्या है?
If \(w=\sqrt{13}+\sqrt{6}\), what is the value of \(w^2-19\)?
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A \(\sqrt{78}\)
B \(2\sqrt{78}\)
C (19)
D (78)
Explanation opens after your attempt
Correct Answer
B. \(2\sqrt{78}\)
Step 1
Concept
\(w^2=13+6+2\sqrt{78}=19+2\sqrt{78}\). So \(w^2-19=2\sqrt{78}\).
Step 2
Why this answer is correct
The correct answer is B. \(2\sqrt{78}\). \(w^2=13+6+2\sqrt{78}=19+2\sqrt{78}\). So \(w^2-19=2\sqrt{78}\).
Step 3
Exam Tip
\(w^2=13+6+2\sqrt{78}=19+2\sqrt{78}\) है। इसलिए \(w^2-19=2\sqrt{78}\) है।
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(\(\sqrt{31}+\sqrt{12}\)\(\sqrt{31}-\sqrt{12}\)) का मान क्या है?
What is the value of (\(\sqrt{31}+\sqrt{12}\)\(\sqrt{31}-\sqrt{12}\))?
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A (43)
B (19)
C \(\sqrt{372}\)
D \(2\sqrt{31}\)
Explanation opens after your attempt
Step 1
Concept
This is conjugate multiplication, so the value is (31-12=19). Identify the \(a^2-b^2\) form.
Step 2
Why this answer is correct
The correct answer is B. (19). This is conjugate multiplication, so the value is (31-12=19). Identify the \(a^2-b^2\) form.
Step 3
Exam Tip
यह संयुग्मी गुणन है इसलिए मान (31-12=19) है। \(a^2-b^2\) रूप पहचानें।
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\(\frac{\sqrt{18}+\sqrt{50}}{\sqrt{8}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{18}+\sqrt{50}}{\sqrt{8}}\)?
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A (2)
B (4)
C (6)
D (8)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\). So the value is (4).
Step 2
Why this answer is correct
The correct answer is B. (4). \(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\). So the value is (4).
Step 3
Exam Tip
\(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\) है। इसलिए मान (4) है।
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यदि \(x=\sqrt{5}+\sqrt{20}+\sqrt{45}\) है, तो \(x^2\) का मान क्या है?
If \(x=\sqrt{5}+\sqrt{20}+\sqrt{45}\), what is the value of \(x^2\)?
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A (150)
B (180)
C (200)
D (225)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so \(x=6\sqrt{5}\). Its square is (180).
Step 2
Why this answer is correct
The correct answer is B. (180). \(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so \(x=6\sqrt{5}\). Its square is (180).
Step 3
Exam Tip
\(\sqrt{20}=2\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\), इसलिए \(x=6\sqrt{5}\) है। इसका वर्ग (180) है।
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\(\frac{5}{\sqrt{17}+\sqrt{8}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{5}{\sqrt{17}+\sqrt{8}}\)?
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A (\frac{5\(\sqrt{17}-\sqrt{8}\)}{9})
B \(\sqrt{17}-\sqrt{8}\)
C \(\frac{\sqrt{17}-\sqrt{8}}{5}\)
D \(5\sqrt{136}\)
Explanation opens after your attempt
Correct Answer
A. (\frac{5\(\sqrt{17}-\sqrt{8}\)}{9})
Step 1
Concept
Multiplying by the conjugate makes the denominator (17-8=9). So the form is (\frac{5\(\sqrt{17}-\sqrt{8}\)}{9}).
Step 2
Why this answer is correct
The correct answer is A. (\frac{5\(\sqrt{17}-\sqrt{8}\)}{9}). Multiplying by the conjugate makes the denominator (17-8=9). So the form is (\frac{5\(\sqrt{17}-\sqrt{8}\)}{9}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (17-8=9) बनता है। इसलिए रूप (\frac{5\(\sqrt{17}-\sqrt{8}\)}{9}) है।
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यदि \(y=\sqrt{14}-\sqrt{7}\) है, तो \(y^2\) का सरल मान कौन-सा है?
If \(y=\sqrt{14}-\sqrt{7}\), which is the simplified value of \(y^2\)?
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A \(21-14\sqrt{2}\)
B (7)
C \(21+14\sqrt{2}\)
D \(\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
A. \(21-14\sqrt{2}\)
Step 1
Concept
(\(\sqrt{14}-\sqrt{7}\)2 =14+7-2\sqrt{98}). Since \(\sqrt{98}=7\sqrt{2}\), the answer is \(21-14\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(21-14\sqrt{2}\). (\(\sqrt{14}-\sqrt{7}\)2 =14+7-2\sqrt{98}). Since \(\sqrt{98}=7\sqrt{2}\), the answer is \(21-14\sqrt{2}\).
Step 3
Exam Tip
(\(\sqrt{14}-\sqrt{7}\)2 =14+7-2\sqrt{98}) है। \(\sqrt{98}=7\sqrt{2}\), इसलिए उत्तर \(21-14\sqrt{2}\) है।
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यदि आयत की लंबाई \(\sqrt{18}+\sqrt{2}\) और चौड़ाई \(\sqrt{18}-\sqrt{2}\) है, तो क्षेत्रफल क्या होगा?
If a rectangle has length \(\sqrt{18}+\sqrt{2}\) and breadth \(\sqrt{18}-\sqrt{2}\), what will be its area?
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A (20)
B (16)
C \(18\sqrt{2}\)
D \(2\sqrt{18}\)
Explanation opens after your attempt
Step 1
Concept
Area is (\(\sqrt{18}\)2 -\(\sqrt{2}\)2 =18-2=16). Conjugate dimensions can give rational area.
Step 2
Why this answer is correct
The correct answer is B. (16). Area is (\(\sqrt{18}\)2 -\(\sqrt{2}\)2 =18-2=16). Conjugate dimensions can give rational area.
Step 3
Exam Tip
क्षेत्रफल (\(\sqrt{18}\)2 -\(\sqrt{2}\)2 =18-2=16) है। संयुग्मी आयाम परिमेय क्षेत्रफल दे सकते हैं।
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यदि \(u=3\sqrt{2}+\sqrt{98}-\sqrt{50}\) है, तो \(\frac{u}{\sqrt{2}}\) का मान क्या है?
If \(u=3\sqrt{2}+\sqrt{98}-\sqrt{50}\), what is the value of \(\frac{u}{\sqrt{2}}\)?
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A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so \(u=5\sqrt{2}\). Dividing gives (5).
Step 2
Why this answer is correct
The correct answer is C. (5). \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so \(u=5\sqrt{2}\). Dividing gives (5).
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए \(u=5\sqrt{2}\) है। भाग देने पर (5) मिलता है।
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\(\frac{\sqrt{45}+\sqrt{80}}{\sqrt{5}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{45}+\sqrt{80}}{\sqrt{5}}\)?
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A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\), so the numerator is \(7\sqrt{5}\). Dividing by \(\sqrt{5}\) gives (7).
Step 2
Why this answer is correct
The correct answer is C. (7). \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\), so the numerator is \(7\sqrt{5}\). Dividing by \(\sqrt{5}\) gives (7).
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{80}=4\sqrt{5}\), इसलिए अंश \(7\sqrt{5}\) है। \(\sqrt{5}\) से भाग देने पर (7) मिलता है।
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(\(\sqrt{11}+\sqrt{3}\)2 -\(\sqrt{11}-\sqrt{3}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{11}+\sqrt{3}\)2 -\(\sqrt{11}-\sqrt{3}\)2 )?
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A \(4\sqrt{33}\)
B (14)
C \(2\sqrt{33}\)
D (33)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{33}\)
Step 1
Concept
Use ((a+b)2 -(a-b)2 =4ab). Here the value is \(4\sqrt{33}\).
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{33}\). Use ((a+b)2 -(a-b)2 =4ab). Here the value is \(4\sqrt{33}\).
Step 3
Exam Tip
पहचान ((a+b)2 -(a-b)2 =4ab) लगाएं। यहाँ मान \(4\sqrt{33}\) है।
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यदि \(r=\sqrt{21}+\sqrt{5}\) और \(s=\sqrt{21}-\sqrt{5}\) हैं, तो (rs) का मान क्या है?
If \(r=\sqrt{21}+\sqrt{5}\) and \(s=\sqrt{21}-\sqrt{5}\), what is the value of (rs)?
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A (16)
B (26)
C \(21\sqrt{5}\)
D \(2\sqrt{105}\)
Explanation opens after your attempt
Step 1
Concept
This is conjugate multiplication. Therefore (rs=21-5=16).
Step 2
Why this answer is correct
The correct answer is A. (16). This is conjugate multiplication. Therefore (rs=21-5=16).
Step 3
Exam Tip
यह संयुग्मी गुणन है। इसलिए (rs=21-5=16) होगा।
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\(\sqrt{12+\sqrt{35}}\) का वर्ग किसके बराबर है?
What is the square of \(\sqrt{12+\sqrt{35}}\) equal to?
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A \(12+\sqrt{35}\)
B (144+35)
C \(\sqrt{47}\)
D \(12-\sqrt{35}\)
Explanation opens after your attempt
Correct Answer
A. \(12+\sqrt{35}\)
Step 1
Concept
The square of a square root gives the number inside. So (\left\(\sqrt{12+\sqrt{35}}\right\)2 =12+\sqrt{35}).
Step 2
Why this answer is correct
The correct answer is A. \(12+\sqrt{35}\). The square of a square root gives the number inside. So (\left\(\sqrt{12+\sqrt{35}}\right\)2 =12+\sqrt{35}).
Step 3
Exam Tip
वर्गमूल का वर्ग अंदर की संख्या देता है। इसलिए (\left\(\sqrt{12+\sqrt{35}}\right\)2 =12+\sqrt{35}) है।
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यदि \(x=5+\sqrt{24}\) है, तो \(x^2-10x\) का मान क्या है?
If \(x=5+\sqrt{24}\), what is the value of \(x^2-10x\)?
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A (-1)
B (1)
C (24)
D \(2\sqrt{24}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=49+10\sqrt{24}\) and \(10x=50+10\sqrt{24}\). Subtracting gives (-1).
Step 2
Why this answer is correct
The correct answer is A. (-1). \(x^2=49+10\sqrt{24}\) and \(10x=50+10\sqrt{24}\). Subtracting gives (-1).
Step 3
Exam Tip
\(x^2=49+10\sqrt{24}\) और \(10x=50+10\sqrt{24}\) है। घटाने पर (-1) मिलता है।
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\(\frac{7}{\sqrt{19}-\sqrt{12}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{7}{\sqrt{19}-\sqrt{12}}\)?
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A \(\sqrt{19}+\sqrt{12}\)
B (7\(\sqrt{19}+\sqrt{12}\))
C \(\frac{\sqrt{19}+\sqrt{12}}{7}\)
D \(7\sqrt{228}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{19}+\sqrt{12}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (19-12=7). So (\frac{7\(\sqrt{19}+\sqrt{12}\)}{7}=\sqrt{19}+\sqrt{12}).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{19}+\sqrt{12}\). Multiplying by the conjugate makes the denominator (19-12=7). So (\frac{7\(\sqrt{19}+\sqrt{12}\)}{7}=\sqrt{19}+\sqrt{12}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (19-12=7) बनता है। इसलिए (\frac{7\(\sqrt{19}+\sqrt{12}\)}{7}=\sqrt{19}+\sqrt{12}) है।
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यदि \(P=\sqrt{363}-\sqrt{147}+\sqrt{75}\) है, तो (P) किसके बराबर है?
If \(P=\sqrt{363}-\sqrt{147}+\sqrt{75}\), what is (P) equal to?
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A \(7\sqrt{3}\)
B \(9\sqrt{3}\)
C \(11\sqrt{3}\)
D \(13\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
B. \(9\sqrt{3}\)
Step 1
Concept
\(\sqrt{363}=11\sqrt{3}\), \(\sqrt{147}=7\sqrt{3}\), and \(\sqrt{75}=5\sqrt{3}\). Therefore \(P=9\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(9\sqrt{3}\). \(\sqrt{363}=11\sqrt{3}\), \(\sqrt{147}=7\sqrt{3}\), and \(\sqrt{75}=5\sqrt{3}\). Therefore \(P=9\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{363}=11\sqrt{3}\), \(\sqrt{147}=7\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\) है। इसलिए \(P=9\sqrt{3}\) है।
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यदि \(t=\sqrt{19}+3\) है, तो \(t+\frac{10}{t}\) का मान क्या है?
If \(t=\sqrt{19}+3\), what is the value of \(t+\frac{10}{t}\)?
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A \(2\sqrt{19}\)
B (6)
C \(2\sqrt{19}+6\)
D \(\sqrt{19}\)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{19}\)
Step 1
Concept
\(\frac{10}{\sqrt{19}+3}=\sqrt{19}-3\) because the denominator becomes (19-9=10). So the sum is \(2\sqrt{19}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{19}\). \(\frac{10}{\sqrt{19}+3}=\sqrt{19}-3\) because the denominator becomes (19-9=10). So the sum is \(2\sqrt{19}\).
Step 3
Exam Tip
\(\frac{10}{\sqrt{19}+3}=\sqrt{19}-3\) है क्योंकि हर (19-9=10) बनता है। इसलिए योग \(2\sqrt{19}\) है।
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(\(\sqrt{32}+\sqrt{50}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{32}+\sqrt{50}\)2 )?
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A (162)
B (98)
C \(18\sqrt{2}\)
D (200)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the sum is \(9\sqrt{2}\). Its square is (162).
Step 2
Why this answer is correct
The correct answer is A. (162). \(\sqrt{32}=4\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the sum is \(9\sqrt{2}\). Its square is (162).
Step 3
Exam Tip
\(\sqrt{32}=4\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए योग \(9\sqrt{2}\) है। इसका वर्ग (162) है।
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यदि \(x=\sqrt{10}+\sqrt{6}\) है, तो \(x^2-16\) का मान क्या है?
If \(x=\sqrt{10}+\sqrt{6}\), what is the value of \(x^2-16\)?
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A \(2\sqrt{60}\)
B \(4\sqrt{15}\)
C (16)
D (60)
Explanation opens after your attempt
Correct Answer
B. \(4\sqrt{15}\)
Step 1
Concept
\(x^2=10+6+2\sqrt{60}=16+4\sqrt{15}\). So \(x^2-16=4\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is B. \(4\sqrt{15}\). \(x^2=10+6+2\sqrt{60}=16+4\sqrt{15}\). So \(x^2-16=4\sqrt{15}\).
Step 3
Exam Tip
\(x^2=10+6+2\sqrt{60}=16+4\sqrt{15}\) है। इसलिए \(x^2-16=4\sqrt{15}\) है।
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\(\frac{1}{\sqrt{12}+\sqrt{5}}+\frac{1}{\sqrt{12}-\sqrt{5}}\) का मान क्या है?
What is the value of \(\frac{1}{\sqrt{12}+\sqrt{5}}+\frac{1}{\sqrt{12}-\sqrt{5}}\)?
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A \(\frac{2\sqrt{5}}{7}\)
B \(\frac{\sqrt{12}}{7}\)
C \(\frac{4\sqrt{3}}{7}\)
D \(2\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{4\sqrt{3}}{7}\)
Step 1
Concept
Adding both fractions gives numerator \(2\sqrt{12}\) and denominator (12-5=7). So the value is \(\frac{4\sqrt{3}}{7}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{4\sqrt{3}}{7}\). Adding both fractions gives numerator \(2\sqrt{12}\) and denominator (12-5=7). So the value is \(\frac{4\sqrt{3}}{7}\).
Step 3
Exam Tip
दोनों भिन्नों को जोड़ने पर अंश \(2\sqrt{12}\) और हर (12-5=7) मिलता है। इसलिए मान \(\frac{4\sqrt{3}}{7}\) है।
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यदि \(a=\sqrt{18}+\sqrt{8}\) और \(b=\sqrt{18}-\sqrt{8}\) हैं तो \(a^2-b^2\) का मान क्या है?
If \(a=\sqrt{18}+\sqrt{8}\) and \(b=\sqrt{18}-\sqrt{8}\), what is the value of \(a^2-b^2\)?
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A (24)
B (36)
C (40)
D (48)
Explanation opens after your attempt
Step 1
Concept
(a-2 -b-2 =(a-b)(a+b)), where \(a-b=2\sqrt{8}\) and \(a+b=2\sqrt{18}\). So the value is \(4\sqrt{144}=48\).
Step 2
Why this answer is correct
The correct answer is D. (48). (a-2 -b-2 =(a-b)(a+b)), where \(a-b=2\sqrt{8}\) and \(a+b=2\sqrt{18}\). So the value is \(4\sqrt{144}=48\).
Step 3
Exam Tip
(a-2 -b-2 =(a-b)(a+b)) है जहाँ \(a-b=2\sqrt{8}\) और \(a+b=2\sqrt{18}\) है। इसलिए मान \(4\sqrt{144}=48\) है।
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\(\frac{\sqrt{243}-\sqrt{108}}{\sqrt{3}}\) का सरल मान क्या है?
What is the simplified value of \(\frac{\sqrt{243}-\sqrt{108}}{\sqrt{3}}\)?
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A (2)
B (3)
C (5)
D (7)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{243}=9\sqrt{3}\) and \(\sqrt{108}=6\sqrt{3}\), so the numerator is \(3\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (3).
Step 2
Why this answer is correct
The correct answer is B. (3). \(\sqrt{243}=9\sqrt{3}\) and \(\sqrt{108}=6\sqrt{3}\), so the numerator is \(3\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (3).
Step 3
Exam Tip
\(\sqrt{243}=9\sqrt{3}\) और \(\sqrt{108}=6\sqrt{3}\), इसलिए अंश \(3\sqrt{3}\) है। \(\sqrt{3}\) से भाग देने पर (3) मिलता है।
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