यदि \(x=\sqrt{13}+\sqrt{5}\) है तो \(x^2-18\) का मान क्या है?
If \(x=\sqrt{13}+\sqrt{5}\), what is the value of \(x^2-18\)?
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A \(\sqrt{65}\)
B (18)
C \(2\sqrt{65}\)
D (65)
Explanation opens after your attempt
Correct Answer
C. \(2\sqrt{65}\)
Step 1
Concept
\(x^2=13+5+2\sqrt{65}=18+2\sqrt{65}\). So \(x^2-18=2\sqrt{65}\).
Step 2
Why this answer is correct
The correct answer is C. \(2\sqrt{65}\). \(x^2=13+5+2\sqrt{65}=18+2\sqrt{65}\). So \(x^2-18=2\sqrt{65}\).
Step 3
Exam Tip
\(x^2=13+5+2\sqrt{65}=18+2\sqrt{65}\) है। इसलिए \(x^2-18=2\sqrt{65}\) है।
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\(\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\)?
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A \(\frac{9+2\sqrt{14}}{5}\)
B \(\frac{9-2\sqrt{14}}{5}\)
C \(9+2\sqrt{14}\)
D (5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{9+2\sqrt{14}}{5}\)
Step 1
Concept
Multiplying by the conjugate gives numerator \(9+2\sqrt{14}\) and denominator (5). So the simplified form is \(\frac{9+2\sqrt{14}}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9+2\sqrt{14}}{5}\). Multiplying by the conjugate gives numerator \(9+2\sqrt{14}\) and denominator (5). So the simplified form is \(\frac{9+2\sqrt{14}}{5}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर अंश \(9+2\sqrt{14}\) और हर (5) मिलता है। इसलिए सरल रूप \(\frac{9+2\sqrt{14}}{5}\) है।
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यदि \(a=6+\sqrt{3}\) और \(b=6-\sqrt{3}\) हैं तो \(a^2-b^2\) का मान क्या है?
If \(a=6+\sqrt{3}\) and \(b=6-\sqrt{3}\), what is the value of \(a^2-b^2\)?
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A \(12\sqrt{3}\)
B (33)
C (72)
D \(24\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
D. \(24\sqrt{3}\)
Step 1
Concept
(a-2 -b-2 =(a-b)(a+b)) where \(a-b=2\sqrt{3}\) and (a+b=12). So the value is \(24\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is D. \(24\sqrt{3}\). (a-2 -b-2 =(a-b)(a+b)) where \(a-b=2\sqrt{3}\) and (a+b=12). So the value is \(24\sqrt{3}\).
Step 3
Exam Tip
(a-2 -b-2 =(a-b)(a+b)) है जहाँ \(a-b=2\sqrt{3}\) और (a+b=12) है। इसलिए मान \(24\sqrt{3}\) है।
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यदि \(p=\frac{1}{\sqrt{19}+4}\) है तो (p) का सरल रूप क्या है?
If \(p=\frac{1}{\sqrt{19}+4}\), what is the simplified form of (p)?
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A \(\frac{\sqrt{19}-4}{3}\)
B \(\sqrt{19}-4\)
C \(\frac{\sqrt{19}+4}{3}\)
D \(4-\sqrt{19}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{\sqrt{19}-4}{3}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (19-16=3). So \(p=\frac{\sqrt{19}-4}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{19}-4}{3}\). Multiplying by the conjugate makes the denominator (19-16=3). So \(p=\frac{\sqrt{19}-4}{3}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (19-16=3) बनता है। इसलिए \(p=\frac{\sqrt{19}-4}{3}\) है।
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यदि \(y=\sqrt{12}+\sqrt{27}\) है तो \(y^2\) का मान क्या है?
If \(y=\sqrt{12}+\sqrt{27}\), what is the value of \(y^2\)?
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A (45)
B (75)
C (39)
D \(12\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so \(y=5\sqrt{3}\). Its square is (75).
Step 2
Why this answer is correct
The correct answer is B. (75). \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so \(y=5\sqrt{3}\). Its square is (75).
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए \(y=5\sqrt{3}\) है। इसका वर्ग (75) है।
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\(\frac{5}{\sqrt{14}-3}\) का परिमेयकृत रूप कौन-सा है?
Which is the rationalised form of \(\frac{5}{\sqrt{14}-3}\)?
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A (5\(\sqrt{14}-3\))
B \(\sqrt{14}+3\)
C (5\(\sqrt{14}+3\))
D (\frac{5\(\sqrt{14}+3\)}{5})
Explanation opens after your attempt
Correct Answer
D. (\frac{5\(\sqrt{14}+3\)}{5})
Step 1
Concept
Multiplying by the conjugate makes the denominator (14-9=5). So the form is (\frac{5\(\sqrt{14}+3\)}{5}), that is \(\sqrt{14}+3\).
Step 2
Why this answer is correct
The correct answer is D. (\frac{5\(\sqrt{14}+3\)}{5}). Multiplying by the conjugate makes the denominator (14-9=5). So the form is (\frac{5\(\sqrt{14}+3\)}{5}), that is \(\sqrt{14}+3\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (14-9=5) बनता है। इसलिए रूप (\frac{5\(\sqrt{14}+3\)}{5}), यानी \(\sqrt{14}+3\), है।
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यदि \(s=\sqrt{108}+\sqrt{300}\) है तो \(\frac{s}{\sqrt{3}}\) का मान क्या है?
If \(s=\sqrt{108}+\sqrt{300}\), what is the value of \(\frac{s}{\sqrt{3}}\)?
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A (12)
B (14)
C (16)
D (18)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{108}=6\sqrt{3}\) and \(\sqrt{300}=10\sqrt{3}\), so \(s=16\sqrt{3}\). Dividing gives (16).
Step 2
Why this answer is correct
The correct answer is C. (16). \(\sqrt{108}=6\sqrt{3}\) and \(\sqrt{300}=10\sqrt{3}\), so \(s=16\sqrt{3}\). Dividing gives (16).
Step 3
Exam Tip
\(\sqrt{108}=6\sqrt{3}\) और \(\sqrt{300}=10\sqrt{3}\), इसलिए \(s=16\sqrt{3}\) है। भाग देने पर (16) मिलता है।
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\(\sqrt{8+\sqrt{11}}\times\sqrt{8+\sqrt{11}}\) का मान क्या है?
What is the value of \(\sqrt{8+\sqrt{11}}\times\sqrt{8+\sqrt{11}}\)?
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A (64+11)
B \(\sqrt{19}\)
C \(8-\sqrt{11}\)
D \(8+\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
D. \(8+\sqrt{11}\)
Step 1
Concept
Multiplying the same square root by itself gives the number inside. Therefore the value is \(8+\sqrt{11}\).
Step 2
Why this answer is correct
The correct answer is D. \(8+\sqrt{11}\). Multiplying the same square root by itself gives the number inside. Therefore the value is \(8+\sqrt{11}\).
Step 3
Exam Tip
एक ही वर्गमूल को अपने आप से गुणा करने पर अंदर की संख्या मिलती है। इसलिए मान \(8+\sqrt{11}\) है।
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यदि \(A=\sqrt{147}+\sqrt{243}\) और \(B=16\sqrt{3}\) हैं तो कौन-सा कथन सही है?
If \(A=\sqrt{147}+\sqrt{243}\) and \(B=16\sqrt{3}\), 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
\(\sqrt{147}=7\sqrt{3}\) and \(\sqrt{243}=9\sqrt{3}\), so \(A=16\sqrt{3}\). Hence (A=B).
Step 2
Why this answer is correct
The correct answer is C. (A=B). \(\sqrt{147}=7\sqrt{3}\) and \(\sqrt{243}=9\sqrt{3}\), so \(A=16\sqrt{3}\). Hence (A=B).
Step 3
Exam Tip
\(\sqrt{147}=7\sqrt{3}\) और \(\sqrt{243}=9\sqrt{3}\), इसलिए \(A=16\sqrt{3}\) है। अतः (A=B) है।
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\(\frac{\sqrt{13}+\sqrt{3}}{\sqrt{13}-\sqrt{3}}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{\sqrt{13}+\sqrt{3}}{\sqrt{13}-\sqrt{3}}\)?
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A \(\frac{16-2\sqrt{39}}{10}\)
B \(16+2\sqrt{39}\)
C \(\frac{8+\sqrt{39}}{5}\)
D (10)
Explanation opens after your attempt
Correct Answer
C. \(\frac{8+\sqrt{39}}{5}\)
Step 1
Concept
Multiplying by the conjugate gives numerator \(16+2\sqrt{39}\) and denominator (10). So the answer is \(\frac{8+\sqrt{39}}{5}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{8+\sqrt{39}}{5}\). Multiplying by the conjugate gives numerator \(16+2\sqrt{39}\) and denominator (10). So the answer is \(\frac{8+\sqrt{39}}{5}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर अंश \(16+2\sqrt{39}\) और हर (10) मिलता है। इसलिए उत्तर \(\frac{8+\sqrt{39}}{5}\) है।
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यदि \(q=\sqrt{7}+3\) है तो \(q^2-6q\) का मान क्या है?
If \(q=\sqrt{7}+3\), what is the value of \(q^2-6q\)?
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A (-2)
B (2)
C \(\sqrt{7}\)
D (16)
Explanation opens after your attempt
Step 1
Concept
\(q^2=16+6\sqrt{7}\) and \(6q=18+6\sqrt{7}\). Subtracting gives (-2).
Step 2
Why this answer is correct
The correct answer is A. (-2). \(q^2=16+6\sqrt{7}\) and \(6q=18+6\sqrt{7}\). Subtracting gives (-2).
Step 3
Exam Tip
\(q^2=16+6\sqrt{7}\) और \(6q=18+6\sqrt{7}\) है। घटाने पर (-2) मिलता है।
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(\(\sqrt{7}+\sqrt{5}\)2 -\(\sqrt{7}-\sqrt{5}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{7}+\sqrt{5}\)2 -\(\sqrt{7}-\sqrt{5}\)2 )?
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A \(4\sqrt{35}\)
B (12)
C \(2\sqrt{35}\)
D (35)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{35}\)
Step 1
Concept
Use ((a+b)2 -(a-b)2 =4ab). Here the value is \(4\sqrt{35}\).
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{35}\). Use ((a+b)2 -(a-b)2 =4ab). Here the value is \(4\sqrt{35}\).
Step 3
Exam Tip
पहचान ((a+b)2 -(a-b)2 =4ab) लगाएं। यहाँ मान \(4\sqrt{35}\) है।
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यदि \(t=\sqrt{23}+4\) है तो \(t+\frac{7}{t}\) का मान क्या है?
If \(t=\sqrt{23}+4\), what is the value of \(t+\frac{7}{t}\)?
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A \(2\sqrt{23}\)
B (8)
C \(\sqrt{23}\)
D \(2\sqrt{23}+8\)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{23}\)
Step 1
Concept
\(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4\) because the denominator becomes (23-16=7). So the sum is \(2\sqrt{23}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{23}\). \(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4\) because the denominator becomes (23-16=7). So the sum is \(2\sqrt{23}\).
Step 3
Exam Tip
\(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4\) है क्योंकि हर (23-16=7) बनता है। इसलिए योग \(2\sqrt{23}\) है।
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\(\frac{\sqrt{98}-\sqrt{50}}{\sqrt{2}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{98}-\sqrt{50}}{\sqrt{2}}\)?
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A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the numerator is \(2\sqrt{2}\). Dividing gives (2).
Step 2
Why this answer is correct
The correct answer is B. (2). \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the numerator is \(2\sqrt{2}\). Dividing gives (2).
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए अंश \(2\sqrt{2}\) है। भाग देने पर (2) मिलता है।
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यदि \(c=\sqrt{21}+\sqrt{5}\) और \(d=\sqrt{21}-\sqrt{5}\) हैं तो (cd) और (c-d) का सही युग्म कौन-सा है?
If \(c=\sqrt{21}+\sqrt{5}\) and \(d=\sqrt{21}-\sqrt{5}\), which is the correct pair of (cd) and (c-d)?
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A (16), \(2\sqrt{5}\)
B (26), \(2\sqrt{21}\)
C \(16\sqrt{105}\), \(2\sqrt{5}\)
D (21), (5)
Explanation opens after your attempt
Correct Answer
A. (16), \(2\sqrt{5}\)
Step 1
Concept
(cd=21-5=16) and \(c-d=2\sqrt{5}\). Find both values separately in a conjugate pair.
Step 2
Why this answer is correct
The correct answer is A. (16), \(2\sqrt{5}\). (cd=21-5=16) and \(c-d=2\sqrt{5}\). Find both values separately in a conjugate pair.
Step 3
Exam Tip
(cd=21-5=16) और \(c-d=2\sqrt{5}\) है। संयुग्मी युग्म में दोनों मान अलग-अलग निकालें।
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\(\sqrt{539}-\sqrt{275}+\sqrt{99}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{539}-\sqrt{275}+\sqrt{99}\)?
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A \(5\sqrt{11}\)
B \(7\sqrt{11}\)
C \(9\sqrt{11}\)
D \(11\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
B. \(7\sqrt{11}\)
Step 1
Concept
\(\sqrt{539}=7\sqrt{11}\), \(\sqrt{275}=5\sqrt{11}\), and \(\sqrt{99}=3\sqrt{11}\). So the result is \(7\sqrt{11}-5\sqrt{11}+3\sqrt{11}=5\sqrt{11}\).
Step 2
Why this answer is correct
The correct answer is B. \(7\sqrt{11}\). \(\sqrt{539}=7\sqrt{11}\), \(\sqrt{275}=5\sqrt{11}\), and \(\sqrt{99}=3\sqrt{11}\). So the result is \(7\sqrt{11}-5\sqrt{11}+3\sqrt{11}=5\sqrt{11}\).
Step 3
Exam Tip
\(\sqrt{539}=7\sqrt{11}\), \(\sqrt{275}=5\sqrt{11}\) और \(\sqrt{99}=3\sqrt{11}\) है। इसलिए परिणाम \(5\sqrt{11}\) नहीं, \(7\sqrt{11}-5\sqrt{11}+3\sqrt{11}=5\sqrt{11}\) है।
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\(\frac{6}{\sqrt{19}-\sqrt{10}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{6}{\sqrt{19}-\sqrt{10}}\)?
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A \(\sqrt{19}+\sqrt{10}\)
B (\frac{2\(\sqrt{19}+\sqrt{10}\)}{3})
C (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9})
D \(6\sqrt{190}\)
Explanation opens after your attempt
Correct Answer
C. (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9})
Step 1
Concept
Multiplying by the conjugate makes the denominator (19-10=9). So the form is (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9}).
Step 2
Why this answer is correct
The correct answer is C. (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9}). Multiplying by the conjugate makes the denominator (19-10=9). So the form is (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (19-10=9) बनता है। इसलिए रूप (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9}) है।
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यदि \(x=\sqrt{17}+\sqrt{6}\) है तो \(x^2-23\) का मान क्या है?
If \(x=\sqrt{17}+\sqrt{6}\), what is the value of \(x^2-23\)?
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A \(\sqrt{102}\)
B (23)
C \(2\sqrt{102}\)
D (102)
Explanation opens after your attempt
Correct Answer
C. \(2\sqrt{102}\)
Step 1
Concept
\(x^2=17+6+2\sqrt{102}=23+2\sqrt{102}\). So \(x^2-23=2\sqrt{102}\).
Step 2
Why this answer is correct
The correct answer is C. \(2\sqrt{102}\). \(x^2=17+6+2\sqrt{102}=23+2\sqrt{102}\). So \(x^2-23=2\sqrt{102}\).
Step 3
Exam Tip
\(x^2=17+6+2\sqrt{102}=23+2\sqrt{102}\) है। इसलिए \(x^2-23=2\sqrt{102}\) है।
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\(\sqrt{13+\sqrt{42}}\) का वर्ग किसके बराबर है?
What is the square of \(\sqrt{13+\sqrt{42}}\) equal to?
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A \(13-\sqrt{42}\)
B \(\sqrt{55}\)
C (169+42)
D \(13+\sqrt{42}\)
Explanation opens after your attempt
Correct Answer
D. \(13+\sqrt{42}\)
Step 1
Concept
The square of a square root gives the number inside. So (\left\(\sqrt{13+\sqrt{42}}\right\)2 =13+\sqrt{42}).
Step 2
Why this answer is correct
The correct answer is D. \(13+\sqrt{42}\). The square of a square root gives the number inside. So (\left\(\sqrt{13+\sqrt{42}}\right\)2 =13+\sqrt{42}).
Step 3
Exam Tip
वर्गमूल का वर्ग अंदर की संख्या देता है। इसलिए (\left\(\sqrt{13+\sqrt{42}}\right\)2 =13+\sqrt{42}) है।
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यदि एक आयत की लंबाई \(\sqrt{22}+\sqrt{7}\) और चौड़ाई \(\sqrt{22}-\sqrt{7}\) है तो क्षेत्रफल क्या होगा?
If a rectangle has length \(\sqrt{22}+\sqrt{7}\) and breadth \(\sqrt{22}-\sqrt{7}\), what will be its area?
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A (29)
B (15)
C \(\sqrt{154}\)
D \(2\sqrt{22}\)
Explanation opens after your attempt
Step 1
Concept
Area is (\(\sqrt{22}+\sqrt{7}\)\(\sqrt{22}-\sqrt{7}\)=22-7=15). Conjugate dimensions can give a rational area.
Step 2
Why this answer is correct
The correct answer is B. (15). Area is (\(\sqrt{22}+\sqrt{7}\)\(\sqrt{22}-\sqrt{7}\)=22-7=15). Conjugate dimensions can give a rational area.
Step 3
Exam Tip
क्षेत्रफल (\(\sqrt{22}+\sqrt{7}\)\(\sqrt{22}-\sqrt{7}\)=22-7=15) है। संयुग्मी आयामों से परिमेय क्षेत्रफल मिल सकता है।
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(\(\sqrt{80}+\sqrt{45}\)\(\sqrt{80}-\sqrt{45}\)) का मान क्या है?
What is the value of (\(\sqrt{80}+\sqrt{45}\)\(\sqrt{80}-\sqrt{45}\))?
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A (125)
B \(2\sqrt{3600}\)
C (35)
D (-35)
Explanation opens after your attempt
Step 1
Concept
This is the \(a^2-b^2\) form. So the value is (80-45=35).
Step 2
Why this answer is correct
The correct answer is C. (35). This is the \(a^2-b^2\) form. So the value is (80-45=35).
Step 3
Exam Tip
यह \(a^2-b^2\) रूप है। इसलिए मान (80-45=35) है।
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यदि \(y=\sqrt{27}+\sqrt{75}\) है तो \(\frac{y}{\sqrt{3}}\) का मान क्या है?
If \(y=\sqrt{27}+\sqrt{75}\), what is the value of \(\frac{y}{\sqrt{3}}\)?
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A (6)
B (8)
C (10)
D (12)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so \(y=8\sqrt{3}\). Dividing gives (8).
Step 2
Why this answer is correct
The correct answer is B. (8). \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so \(y=8\sqrt{3}\). Dividing gives (8).
Step 3
Exam Tip
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\), इसलिए \(y=8\sqrt{3}\) है। भाग देने पर (8) मिलता है।
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\(\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}\)?
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A \(2+\sqrt{3}\)
B (1)
C \(2-\sqrt{3}\)
D \(\sqrt{3}-2\)
Explanation opens after your attempt
Correct Answer
C. \(2-\sqrt{3}\)
Step 1
Concept
Multiplying by the conjugate gives (\frac{\(\sqrt{6}-\sqrt{2}\)2 }{4}=2-\sqrt{3}). Make the denominator rational.
Step 2
Why this answer is correct
The correct answer is C. \(2-\sqrt{3}\). Multiplying by the conjugate gives (\frac{\(\sqrt{6}-\sqrt{2}\)2 }{4}=2-\sqrt{3}). Make the denominator rational.
Step 3
Exam Tip
संयुग्मी से गुणा करने पर (\frac{\(\sqrt{6}-\sqrt{2}\)2 }{4}=2-\sqrt{3}) मिलता है। हर को परिमेय बनाएं।
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यदि \(m=\sqrt{180}-\sqrt{80}\) और \(n=2\sqrt{5}\) हैं तो (m-n) क्या है?
If \(m=\sqrt{180}-\sqrt{80}\) and \(n=2\sqrt{5}\), what is (m-n)?
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A (0)
B \(\sqrt{5}\)
C \(4\sqrt{5}\)
D \(6\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{180}=6\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\), so \(m=2\sqrt{5}\). Hence (m-n=0).
Step 2
Why this answer is correct
The correct answer is A. (0). \(\sqrt{180}=6\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\), so \(m=2\sqrt{5}\). Hence (m-n=0).
Step 3
Exam Tip
\(\sqrt{180}=6\sqrt{5}\) और \(\sqrt{80}=4\sqrt{5}\), इसलिए \(m=2\sqrt{5}\) है। अतः (m-n=0) है।
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(\sqrt{7}\left\(\sqrt{175}-\sqrt{63}\right\)) का मान क्या है?
What is the value of (\sqrt{7}\left\(\sqrt{175}-\sqrt{63}\right\))?
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A \(2\sqrt{7}\)
B (14)
C (28)
D \(\sqrt{112}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{175}=5\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the bracket is \(2\sqrt{7}\). Multiplying by \(\sqrt{7}\) gives (14).
Step 2
Why this answer is correct
The correct answer is B. (14). \(\sqrt{175}=5\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the bracket is \(2\sqrt{7}\). Multiplying by \(\sqrt{7}\) gives (14).
Step 3
Exam Tip
\(\sqrt{175}=5\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए कोष्ठक \(2\sqrt{7}\) है। \(\sqrt{7}\) से गुणा करने पर (14) मिलता है।
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यदि \(z=\sqrt{23}-\sqrt{14}\) है तो \(z^2\) का मान कौन-सा है?
If \(z=\sqrt{23}-\sqrt{14}\), which is the value of \(z^2\)?
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A \(37+2\sqrt{322}\)
B (9)
C \(37-2\sqrt{322}\)
D \(\sqrt{9}\)
Explanation opens after your attempt
Correct Answer
C. \(37-2\sqrt{322}\)
Step 1
Concept
(\(\sqrt{23}-\sqrt{14}\)2 =23+14-2\sqrt{322}). Keep the middle term negative.
Step 2
Why this answer is correct
The correct answer is C. \(37-2\sqrt{322}\). (\(\sqrt{23}-\sqrt{14}\)2 =23+14-2\sqrt{322}). Keep the middle term negative.
Step 3
Exam Tip
(\(\sqrt{23}-\sqrt{14}\)2 =23+14-2\sqrt{322}) है। मध्य पद का चिन्ह ऋणात्मक रखें।
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\(\frac{3}{\sqrt{17}+\sqrt{8}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{3}{\sqrt{17}+\sqrt{8}}\)?
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A \(\sqrt{17}-\sqrt{8}\)
B \(\sqrt{17}+\sqrt{8}\)
C \(\frac{\sqrt{17}-\sqrt{8}}{3}\)
D \(3\sqrt{136}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{\sqrt{17}-\sqrt{8}}{3}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (17-8=9). So the form is \(\frac{\sqrt{17}-\sqrt{8}}{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{\sqrt{17}-\sqrt{8}}{3}\). Multiplying by the conjugate makes the denominator (17-8=9). So the form is \(\frac{\sqrt{17}-\sqrt{8}}{3}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (17-8=9) बनता है। इसलिए रूप \(\frac{\sqrt{17}-\sqrt{8}}{3}\) है।
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यदि \(x=\sqrt{3}+\sqrt{12}\) और \(y=\sqrt{27}\) हैं तो (x-y) का मान क्या है?
If \(x=\sqrt{3}+\sqrt{12}\) and \(y=\sqrt{27}\), what is the value of (x-y)?
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A (0)
B \(\sqrt{3}\)
C \(2\sqrt{3}\)
D \(4\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(x=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\) and \(y=3\sqrt{3}\). Therefore (x-y=0).
Step 2
Why this answer is correct
The correct answer is A. (0). \(x=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\) and \(y=3\sqrt{3}\). Therefore (x-y=0).
Step 3
Exam Tip
\(x=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\) और \(y=3\sqrt{3}\) है। इसलिए (x-y=0) है।
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\(\sqrt{18}\) और \(\sqrt{20}\) के बीच कौन-सी संख्या निश्चित रूप से आती है?
Which number definitely lies between \(\sqrt{18}\) and \(\sqrt{20}\)?
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A \(\sqrt{17.5}\)
B (5)
C \(\sqrt{19}\)
D \(\sqrt{21}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{19}\)
Step 1
Concept
Since (18<19<20), \(\sqrt{19}\) lies between them. Compare square roots using the numbers inside.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{19}\). Since (18<19<20), \(\sqrt{19}\) lies between them. Compare square roots using the numbers inside.
Step 3
Exam Tip
क्योंकि (18<19<20), इसलिए \(\sqrt{19}\) दोनों के बीच होगा। वर्गमूलों में अंदर की संख्या से तुलना करें।
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यदि \(u=\sqrt{28}+\sqrt{63}\) और \(v=5\sqrt{7}\) हैं तो कौन-सा कथन सही है?
If \(u=\sqrt{28}+\sqrt{63}\) and \(v=5\sqrt{7}\), 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
\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so \(u=5\sqrt{7}\). Hence (u=v).
Step 2
Why this answer is correct
The correct answer is C. (u=v). \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so \(u=5\sqrt{7}\). Hence (u=v).
Step 3
Exam Tip
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए \(u=5\sqrt{7}\) है। अतः (u=v) है।
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यदि \(x=5+\sqrt{26}\) है तो \(x^2-10x\) का मान क्या है?
If \(x=5+\sqrt{26}\), what is the value of \(x^2-10x\)?
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A (1)
B (-1)
C (26)
D \(2\sqrt{26}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=51+10\sqrt{26}\) and \(10x=50+10\sqrt{26}\). Subtracting gives (1).
Step 2
Why this answer is correct
The correct answer is A. (1). \(x^2=51+10\sqrt{26}\) and \(10x=50+10\sqrt{26}\). Subtracting gives (1).
Step 3
Exam Tip
\(x^2=51+10\sqrt{26}\) और \(10x=50+10\sqrt{26}\) है। घटाने पर (1) मिलता है।
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\(\sqrt{432}-\sqrt{192}+\sqrt{48}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{432}-\sqrt{192}+\sqrt{48}\)?
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A \(8\sqrt{3}\)
B \(10\sqrt{3}\)
C \(12\sqrt{3}\)
D \(\sqrt{288}\)
Explanation opens after your attempt
Correct Answer
A. \(8\sqrt{3}\)
Step 1
Concept
\(\sqrt{432}=12\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\). Therefore the result is \(8\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(8\sqrt{3}\). \(\sqrt{432}=12\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\). Therefore the result is \(8\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{432}=12\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\) है। इसलिए परिणाम \(8\sqrt{3}\) है।
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\(\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{7}-\sqrt{5}}\) का मान क्या है?
What is the value of \(\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{7}-\sqrt{5}}\)?
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A \(\sqrt{7}\)
B \(\sqrt{5}\)
C \(2\sqrt{7}\)
D (2)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{7}\)
Step 1
Concept
After rationalising the terms become \(\frac{\sqrt{7}-\sqrt{5}}{2}\) and \(\frac{\sqrt{7}+\sqrt{5}}{2}\). The sum is \(\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{7}\). After rationalising the terms become \(\frac{\sqrt{7}-\sqrt{5}}{2}\) and \(\frac{\sqrt{7}+\sqrt{5}}{2}\). The sum is \(\sqrt{7}\).
Step 3
Exam Tip
दोनों पदों को परिमेयकृत करने पर \(\frac{\sqrt{7}-\sqrt{5}}{2}\) और \(\frac{\sqrt{7}+\sqrt{5}}{2}\) मिलते हैं। योग \(\sqrt{7}\) है।
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यदि \(x=\sqrt{12+\sqrt{11}}\) है तो \(x^2-12\) का मान क्या है?
If \(x=\sqrt{12+\sqrt{11}}\), what is the value of \(x^2-12\)?
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A \(\sqrt{11}\)
B (11)
C \(\sqrt{23}\)
D (12)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{11}\)
Step 1
Concept
\(x^2=12+\sqrt{11}\). Therefore \(x^2-12=\sqrt{11}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{11}\). \(x^2=12+\sqrt{11}\). Therefore \(x^2-12=\sqrt{11}\).
Step 3
Exam Tip
\(x^2=12+\sqrt{11}\) है। इसलिए \(x^2-12=\sqrt{11}\) होगा।
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\(\frac{\sqrt{200}+\sqrt{72}}{\sqrt{2}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{200}+\sqrt{72}}{\sqrt{2}}\)?
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A (12)
B (14)
C (16)
D (18)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{200}=10\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so the numerator is \(16\sqrt{2}\). Dividing gives (16).
Step 2
Why this answer is correct
The correct answer is C. (16). \(\sqrt{200}=10\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so the numerator is \(16\sqrt{2}\). Dividing gives (16).
Step 3
Exam Tip
\(\sqrt{200}=10\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\), इसलिए अंश \(16\sqrt{2}\) है। भाग देने पर (16) मिलता है।
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यदि \(r=\sqrt{13}+\sqrt{7}\) और \(s=\sqrt{13}-\sqrt{7}\) हैं तो \(r^2-s^2\) का मान क्या है?
If \(r=\sqrt{13}+\sqrt{7}\) and \(s=\sqrt{13}-\sqrt{7}\), what is the value of \(r^2-s^2\)?
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A \(4\sqrt{91}\)
B (20)
C \(2\sqrt{91}\)
D (91)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{91}\)
Step 1
Concept
(r-2 -s-2 =(r-s)(r+s)) where \(r-s=2\sqrt{7}\) and \(r+s=2\sqrt{13}\). So the value is \(4\sqrt{91}\).
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{91}\). (r-2 -s-2 =(r-s)(r+s)) where \(r-s=2\sqrt{7}\) and \(r+s=2\sqrt{13}\). So the value is \(4\sqrt{91}\).
Step 3
Exam Tip
(r-2 -s-2 =(r-s)(r+s)) है जहाँ \(r-s=2\sqrt{7}\) और \(r+s=2\sqrt{13}\) है। इसलिए मान \(4\sqrt{91}\) है।
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\(\frac{8}{\sqrt{18}-\sqrt{10}}\) का परिमेयकृत रूप कौन-सा है?
Which is the rationalised form of \(\frac{8}{\sqrt{18}-\sqrt{10}}\)?
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A \(\sqrt{18}+\sqrt{10}\)
B (8\(\sqrt{18}+\sqrt{10}\))
C \(\frac{\sqrt{18}+\sqrt{10}}{8}\)
D \(8\sqrt{180}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{18}+\sqrt{10}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (18-10=8). So the answer is \(\sqrt{18}+\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{18}+\sqrt{10}\). Multiplying by the conjugate makes the denominator (18-10=8). So the answer is \(\sqrt{18}+\sqrt{10}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (18-10=8) बनता है। इसलिए उत्तर \(\sqrt{18}+\sqrt{10}\) है।
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यदि \(P=\sqrt{363}+\sqrt{108}-\sqrt{192}\) है तो (P) किसके बराबर है?
If \(P=\sqrt{363}+\sqrt{108}-\sqrt{192}\), 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
\(\sqrt{363}=11\sqrt{3}\), \(\sqrt{108}=6\sqrt{3}\), and \(\sqrt{192}=8\sqrt{3}\). Therefore \(P=9\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(9\sqrt{3}\). \(\sqrt{363}=11\sqrt{3}\), \(\sqrt{108}=6\sqrt{3}\), and \(\sqrt{192}=8\sqrt{3}\). Therefore \(P=9\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{363}=11\sqrt{3}\), \(\sqrt{108}=6\sqrt{3}\) और \(\sqrt{192}=8\sqrt{3}\) है। इसलिए \(P=9\sqrt{3}\) है।
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(\(\sqrt{44}+\sqrt{99}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{44}+\sqrt{99}\)2 )?
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A (275)
B (143)
C \(25\sqrt{11}\)
D (550)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{44}=2\sqrt{11}\) and \(\sqrt{99}=3\sqrt{11}\), so the sum is \(5\sqrt{11}\). Its square is (275).
Step 2
Why this answer is correct
The correct answer is A. (275). \(\sqrt{44}=2\sqrt{11}\) and \(\sqrt{99}=3\sqrt{11}\), so the sum is \(5\sqrt{11}\). Its square is (275).
Step 3
Exam Tip
\(\sqrt{44}=2\sqrt{11}\) और \(\sqrt{99}=3\sqrt{11}\), इसलिए योग \(5\sqrt{11}\) है। इसका वर्ग (275) है।
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यदि \(w=\sqrt{15}+\sqrt{11}\) है तो \(w^2-26\) का मान क्या है?
If \(w=\sqrt{15}+\sqrt{11}\), what is the value of \(w^2-26\)?
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A \(\sqrt{165}\)
B \(2\sqrt{165}\)
C (26)
D (165)
Explanation opens after your attempt
Correct Answer
B. \(2\sqrt{165}\)
Step 1
Concept
\(w^2=15+11+2\sqrt{165}=26+2\sqrt{165}\). So \(w^2-26=2\sqrt{165}\).
Step 2
Why this answer is correct
The correct answer is B. \(2\sqrt{165}\). \(w^2=15+11+2\sqrt{165}=26+2\sqrt{165}\). So \(w^2-26=2\sqrt{165}\).
Step 3
Exam Tip
\(w^2=15+11+2\sqrt{165}=26+2\sqrt{165}\) है। इसलिए \(w^2-26=2\sqrt{165}\) है।
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(\(\sqrt{23}+\sqrt{7}\)\(\sqrt{23}-\sqrt{7}\)) का मान क्या है?
What is the value of (\(\sqrt{23}+\sqrt{7}\)\(\sqrt{23}-\sqrt{7}\))?
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A (30)
B \(\sqrt{161}\)
C (16)
D \(2\sqrt{23}\)
Explanation opens after your attempt
Step 1
Concept
This is conjugate multiplication so the value is (23-7=16). Identify the \(a^2-b^2\) form.
Step 2
Why this answer is correct
The correct answer is C. (16). This is conjugate multiplication so the value is (23-7=16). Identify the \(a^2-b^2\) form.
Step 3
Exam Tip
यह संयुग्मी गुणन है इसलिए मान (23-7=16) है। \(a^2-b^2\) रूप पहचानें।
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\(\frac{5}{\sqrt{29}+2}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{5}{\sqrt{29}+2}\)?
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A (5\(\sqrt{29}-2\))
B \(\frac{\sqrt{29}-2}{5}\)
C \(\sqrt{29}+2\)
D \(5\sqrt{29}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{\sqrt{29}-2}{5}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (29-4=25). So the form is \(\frac{\sqrt{29}-2}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\sqrt{29}-2}{5}\). Multiplying by the conjugate makes the denominator (29-4=25). So the form is \(\frac{\sqrt{29}-2}{5}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (29-4=25) बनता है। इसलिए रूप \(\frac{\sqrt{29}-2}{5}\) है।
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यदि \(x=\sqrt{24}+\sqrt{6}\) है तो \(x^2\) का मान क्या है?
If \(x=\sqrt{24}+\sqrt{6}\), what is the value of \(x^2\)?
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A (30)
B (36)
C (48)
D (54)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{24}=2\sqrt{6}\) so \(x=3\sqrt{6}\). Its square is (54).
Step 2
Why this answer is correct
The correct answer is D. (54). \(\sqrt{24}=2\sqrt{6}\) so \(x=3\sqrt{6}\). Its square is (54).
Step 3
Exam Tip
\(\sqrt{24}=2\sqrt{6}\) इसलिए \(x=3\sqrt{6}\) है। इसका वर्ग (54) है।
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\(\sqrt{242}-\sqrt{128}+\sqrt{72}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{242}-\sqrt{128}+\sqrt{72}\)?
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A \(7\sqrt{2}\)
B \(8\sqrt{2}\)
C \(9\sqrt{2}\)
D \(11\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
C. \(9\sqrt{2}\)
Step 1
Concept
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). Therefore the result is \(9\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(9\sqrt{2}\). \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). Therefore the result is \(9\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\) है। इसलिए परिणाम \(9\sqrt{2}\) है।
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यदि \(q=\sqrt{10}+3\) है तो \(q^2-6q\) का मान क्या है?
If \(q=\sqrt{10}+3\), what is the value of \(q^2-6q\)?
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A (1)
B (-1)
C (10)
D \(2\sqrt{10}\)
Explanation opens after your attempt
Step 1
Concept
\(q^2=19+6\sqrt{10}\) and \(6q=18+6\sqrt{10}\). Subtracting gives (1).
Step 2
Why this answer is correct
The correct answer is A. (1). \(q^2=19+6\sqrt{10}\) and \(6q=18+6\sqrt{10}\). Subtracting gives (1).
Step 3
Exam Tip
\(q^2=19+6\sqrt{10}\) और \(6q=18+6\sqrt{10}\) है। घटाने पर (1) मिलता है।
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\(\frac{\sqrt{27}+\sqrt{12}}{\sqrt{3}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{27}+\sqrt{12}}{\sqrt{3}}\)?
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A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (5).
Step 2
Why this answer is correct
The correct answer is B. (5). \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (5).
Step 3
Exam Tip
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\) है। \(\sqrt{3}\) से भाग देने पर (5) मिलता है।
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यदि \(z=\sqrt{31}-\sqrt{22}\) है तो \(z^2\) का मान कौन-सा है?
If \(z=\sqrt{31}-\sqrt{22}\), which is the value of \(z^2\)?
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A (9)
B \(53+2\sqrt{682}\)
C \(\sqrt{9}\)
D \(53-2\sqrt{682}\)
Explanation opens after your attempt
Correct Answer
D. \(53-2\sqrt{682}\)
Step 1
Concept
(\(\sqrt{31}-\sqrt{22}\)2 =31+22-2\sqrt{682}). Keep the middle term negative.
Step 2
Why this answer is correct
The correct answer is D. \(53-2\sqrt{682}\). (\(\sqrt{31}-\sqrt{22}\)2 =31+22-2\sqrt{682}). Keep the middle term negative.
Step 3
Exam Tip
(\(\sqrt{31}-\sqrt{22}\)2 =31+22-2\sqrt{682}) है। मध्य पद का चिन्ह ऋणात्मक रखें।
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(\(\sqrt{29}+5\)2 -\(\sqrt{29}-5\)2 ) का मान क्या है?
What is the value of (\(\sqrt{29}+5\)2 -\(\sqrt{29}-5\)2 )?
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A \(10\sqrt{29}\)
B \(20\sqrt{29}\)
C (34)
D (116)
Explanation opens after your attempt
Correct Answer
B. \(20\sqrt{29}\)
Step 1
Concept
Use ((a+b)2 -(a-b)2 =4ab). Here \(a=\sqrt{29}\) and (b=5), so the value is \(20\sqrt{29}\).
Step 2
Why this answer is correct
The correct answer is B. \(20\sqrt{29}\). Use ((a+b)2 -(a-b)2 =4ab). Here \(a=\sqrt{29}\) and (b=5), so the value is \(20\sqrt{29}\).
Step 3
Exam Tip
पहचान ((a+b)2 -(a-b)2 =4ab) लगाएं। यहाँ \(a=\sqrt{29}\) और (b=5), इसलिए मान \(20\sqrt{29}\) 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 \(9\sqrt{3}\)
B \(11\sqrt{3}\)
C \(13\sqrt{3}\)
D \(15\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
C. \(13\sqrt{3}\)
Step 1
Concept
\(\sqrt{363}=11\sqrt{3}\), \(\sqrt{147}=7\sqrt{3}\), and \(\sqrt{75}=5\sqrt{3}\). Therefore \(P=13\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(13\sqrt{3}\). \(\sqrt{363}=11\sqrt{3}\), \(\sqrt{147}=7\sqrt{3}\), and \(\sqrt{75}=5\sqrt{3}\). Therefore \(P=13\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{363}=11\sqrt{3}\), \(\sqrt{147}=7\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\) है। इसलिए \(P=13\sqrt{3}\) है।
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\(\frac{9}{\sqrt{26}+\sqrt{17}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{9}{\sqrt{26}+\sqrt{17}}\)?
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A \(\sqrt{26}-\sqrt{17}\)
B (9\(\sqrt{26}-\sqrt{17}\))
C \(\frac{\sqrt{26}-\sqrt{17}}{9}\)
D \(9\sqrt{442}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{26}-\sqrt{17}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (26-17=9). So the answer is \(\sqrt{26}-\sqrt{17}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{26}-\sqrt{17}\). Multiplying by the conjugate makes the denominator (26-17=9). So the answer is \(\sqrt{26}-\sqrt{17}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (26-17=9) बनता है। इसलिए उत्तर \(\sqrt{26}-\sqrt{17}\) है।
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