यदि \(a=6+\sqrt{5}\) और \(b=6-\sqrt{5}\) हैं तो \(a^2-b^2\) का मान क्या है?
If \(a=6+\sqrt{5}\) and \(b=6-\sqrt{5}\), what is the value of \(a^2-b^2\)?
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#irrational-numbers
#hard
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A \(24\sqrt{5}\)
B \(12\sqrt{5}\)
C (31)
D (72)
Explanation opens after your attempt
Correct Answer
A. \(24\sqrt{5}\)
Step 1
Concept
(a-2 -b-2 =(a-b)(a+b)) where \(a-b=2\sqrt{5}\) and (a+b=12). So the value is \(24\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(24\sqrt{5}\). (a-2 -b-2 =(a-b)(a+b)) where \(a-b=2\sqrt{5}\) and (a+b=12). So the value is \(24\sqrt{5}\).
Step 3
Exam Tip
(a-2 -b-2 =(a-b)(a+b)) है जहाँ \(a-b=2\sqrt{5}\) और (a+b=12) है। इसलिए मान \(24\sqrt{5}\) है।
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\(\frac{4+\sqrt{7}}{4-\sqrt{7}}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{4+\sqrt{7}}{4-\sqrt{7}}\)?
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A \(\frac{23+8\sqrt{7}}{9}\)
B \(\frac{23-8\sqrt{7}}{9}\)
C \(9+8\sqrt{7}\)
D \(4+\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{23+8\sqrt{7}}{9}\)
Step 1
Concept
Multiplying by the conjugate gives denominator (16-7=9) and numerator (\(4+\sqrt{7}\)2 =23+8\sqrt{7}). So the correct form is \(\frac{23+8\sqrt{7}}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{23+8\sqrt{7}}{9}\). Multiplying by the conjugate gives denominator (16-7=9) and numerator (\(4+\sqrt{7}\)2 =23+8\sqrt{7}). So the correct form is \(\frac{23+8\sqrt{7}}{9}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (16-7=9) और अंश (\(4+\sqrt{7}\)2 =23+8\sqrt{7}) मिलता है। इसलिए सही रूप \(\frac{23+8\sqrt{7}}{9}\) है।
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यदि \(x=\sqrt{12}+\sqrt{3}\) है तो \(x^2\) का मान क्या है?
If \(x=\sqrt{12}+\sqrt{3}\), what is the value of \(x^2\)?
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A (27)
B \(15+2\sqrt{36}\)
C (9)
D (15)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) so \(x=3\sqrt{3}\). Its square is (27).
Step 2
Why this answer is correct
The correct answer is A. (27). \(\sqrt{12}=2\sqrt{3}\) so \(x=3\sqrt{3}\). Its square is (27).
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) इसलिए \(x=3\sqrt{3}\) है। इसका वर्ग (27) होता है।
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\(\sqrt{147}-\sqrt{75}+\sqrt{27}\) किसके बराबर है?
What is \(\sqrt{147}-\sqrt{75}+\sqrt{27}\) equal to?
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A \(5\sqrt{3}\)
B \(7\sqrt{3}\)
C \(9\sqrt{3}\)
D \(3\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(5\sqrt{3}\)
Step 1
Concept
\(\sqrt{147}=7\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{27}=3\sqrt{3}\). Therefore the result is \(5\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(5\sqrt{3}\). \(\sqrt{147}=7\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{27}=3\sqrt{3}\). Therefore the result is \(5\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{147}=7\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\) है। इसलिए परिणाम \(5\sqrt{3}\) है।
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यदि \(p=\frac{1}{\sqrt{13}+3}\) है तो (p) का सरल रूप कौन-सा है?
If \(p=\frac{1}{\sqrt{13}+3}\), which is the simplified form of (p)?
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A \(\frac{\sqrt{13}-3}{4}\)
B \(\sqrt{13}-3\)
C \(\frac{\sqrt{13}+3}{4}\)
D \(3-\sqrt{13}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{\sqrt{13}-3}{4}\)
Step 1
Concept
Multiplying by the conjugate \(\sqrt{13}-3\) makes the denominator (13-9=4). So \(p=\frac{\sqrt{13}-3}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{13}-3}{4}\). Multiplying by the conjugate \(\sqrt{13}-3\) makes the denominator (13-9=4). So \(p=\frac{\sqrt{13}-3}{4}\).
Step 3
Exam Tip
संयुग्मी \(\sqrt{13}-3\) से गुणा करने पर हर (13-9=4) बनता है। इसलिए \(p=\frac{\sqrt{13}-3}{4}\) है।
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\(\sqrt{9+\sqrt{20}}\times\sqrt{9+\sqrt{20}}\) का मान क्या है?
What is the value of \(\sqrt{9+\sqrt{20}}\times\sqrt{9+\sqrt{20}}\)?
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A \(9+\sqrt{20}\)
B (81+20)
C \(\sqrt{29}\)
D \(9-\sqrt{20}\)
Explanation opens after your attempt
Correct Answer
A. \(9+\sqrt{20}\)
Step 1
Concept
Multiplying the same square root by itself gives the number inside. Therefore the value is \(9+\sqrt{20}\).
Step 2
Why this answer is correct
The correct answer is A. \(9+\sqrt{20}\). Multiplying the same square root by itself gives the number inside. Therefore the value is \(9+\sqrt{20}\).
Step 3
Exam Tip
एक ही वर्गमूल को अपने आप से गुणा करने पर अंदर की संख्या मिलती है। इसलिए मान \(9+\sqrt{20}\) है।
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यदि \(m=\sqrt{48}+\sqrt{192}\) और \(n=12\sqrt{3}\) हैं तो (m-n) का मान क्या है?
If \(m=\sqrt{48}+\sqrt{192}\) and \(n=12\sqrt{3}\), what is the value of (m-n)?
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A (0)
B \(\sqrt{3}\)
C \(4\sqrt{3}\)
D \(8\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{192}=8\sqrt{3}\), so \(m=12\sqrt{3}\). Hence (m-n=0).
Step 2
Why this answer is correct
The correct answer is A. (0). \(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{192}=8\sqrt{3}\), so \(m=12\sqrt{3}\). Hence (m-n=0).
Step 3
Exam Tip
\(\sqrt{48}=4\sqrt{3}\) और \(\sqrt{192}=8\sqrt{3}\), इसलिए \(m=12\sqrt{3}\) है। अतः (m-n=0) है।
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\(\frac{\sqrt{5}+2}{\sqrt{5}-2}\) का सरल रूप क्या है?
What 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 \(1+\sqrt{5}\)
D \(5+2\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(9+4\sqrt{5}\)
Step 1
Concept
The conjugate of the denominator 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 conjugate of the denominator 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 \(x=\sqrt{8}+\sqrt{3}\) और \(y=\sqrt{8}-\sqrt{3}\) हैं तो \(\frac{x}{y}\) का मान क्या है?
If \(x=\sqrt{8}+\sqrt{3}\) and \(y=\sqrt{8}-\sqrt{3}\), what is the value of \(\frac{x}{y}\)?
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A \(\frac{11+4\sqrt{6}}{5}\)
B \(11+4\sqrt{6}\)
C \(\frac{11-4\sqrt{6}}{5}\)
D (5)
Explanation opens after your attempt
Correct Answer
A. \(\frac{11+4\sqrt{6}}{5}\)
Step 1
Concept
Multiplying by the denominator conjugate gives denominator (8-3=5). The numerator is (\(\sqrt{8}+\sqrt{3}\)2 =11+4\sqrt{6}).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{11+4\sqrt{6}}{5}\). Multiplying by the denominator conjugate gives denominator (8-3=5). The numerator is (\(\sqrt{8}+\sqrt{3}\)2 =11+4\sqrt{6}).
Step 3
Exam Tip
हर के संयुग्मी से गुणा करने पर हर (8-3=5) होता है। अंश (\(\sqrt{8}+\sqrt{3}\)2 =11+4\sqrt{6}) है।
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यदि \(r=\sqrt{45}+\sqrt{180}\) है तो (r) किसके बराबर है?
If \(r=\sqrt{45}+\sqrt{180}\), what is (r) equal to?
#number-systems
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A \(9\sqrt{5}\)
B \(12\sqrt{5}\)
C \(15\sqrt{5}\)
D \(21\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(9\sqrt{5}\)
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{180}=6\sqrt{5}\). Therefore \(r=9\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(9\sqrt{5}\). \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{180}=6\sqrt{5}\). Therefore \(r=9\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{180}=6\sqrt{5}\) है। इसलिए \(r=9\sqrt{5}\) है।
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\(\sqrt{10+\sqrt{21}}\) के बारे में कौन-सा कथन निश्चित रूप से सही है?
Which statement is definitely true about \(\sqrt{10+\sqrt{21}}\)?
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A यह वास्तविक नहीं है / It is not real
B इसका वर्ग \(10+\sqrt{21}\) है / Its square is \(10+\sqrt{21}\)
C यह \(10+\sqrt{21}\) के बराबर है / It equals \(10+\sqrt{21}\)
D यह परिमेय पूर्णांक है / It is a rational integer
Explanation opens after your attempt
Correct Answer
B. इसका वर्ग \(10+\sqrt{21}\) है / Its square is \(10+\sqrt{21}\)
Step 1
Concept
Since \(10+\sqrt{21}\) is positive, its square root is real. Squaring it gives the inside number.
Step 2
Why this answer is correct
The correct answer is B. इसका वर्ग \(10+\sqrt{21}\) है / Its square is \(10+\sqrt{21}\). Since \(10+\sqrt{21}\) is positive, its square root is real. Squaring it gives the inside number.
Step 3
Exam Tip
क्योंकि \(10+\sqrt{21}\) धनात्मक है, इसका वर्गमूल वास्तविक है। वर्ग करने पर अंदर की संख्या मिलती है।
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यदि \(u=\sqrt{200}-\sqrt{72}\) और \(v=4\sqrt{2}\) हैं तो (u-v) का मान क्या है?
If \(u=\sqrt{200}-\sqrt{72}\) and \(v=4\sqrt{2}\), what is the value of (u-v)?
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#hard
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A \(\sqrt{2}\)
B (0)
C \(2\sqrt{2}\)
D \(6\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{200}=10\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so \(u=4\sqrt{2}\). Hence (u-v=0).
Step 2
Why this answer is correct
The correct answer is B. (0). \(\sqrt{200}=10\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so \(u=4\sqrt{2}\). Hence (u-v=0).
Step 3
Exam Tip
\(\sqrt{200}=10\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\), इसलिए \(u=4\sqrt{2}\) है। अतः (u-v=0) है।
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\(\frac{6}{\sqrt{15}-\sqrt{6}}\) का परिमेयकृत रूप कौन-सा है?
Which is the rationalised form of \(\frac{6}{\sqrt{15}-\sqrt{6}}\)?
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A (\frac{2\(\sqrt{15}+\sqrt{6}\)}{3})
B \(\sqrt{15}+\sqrt{6}\)
C (6\(\sqrt{15}+\sqrt{6}\))
D \(\frac{\sqrt{15}+\sqrt{6}}{6}\)
Explanation opens after your attempt
Correct Answer
A. (\frac{2\(\sqrt{15}+\sqrt{6}\)}{3})
Step 1
Concept
Multiplying by the conjugate makes the denominator (15-6=9). So (\frac{6\(\sqrt{15}+\sqrt{6}\)}{9}=\frac{2\(\sqrt{15}+\sqrt{6}\)}{3}).
Step 2
Why this answer is correct
The correct answer is A. (\frac{2\(\sqrt{15}+\sqrt{6}\)}{3}). Multiplying by the conjugate makes the denominator (15-6=9). So (\frac{6\(\sqrt{15}+\sqrt{6}\)}{9}=\frac{2\(\sqrt{15}+\sqrt{6}\)}{3}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (15-6=9) बनता है। इसलिए (\frac{6\(\sqrt{15}+\sqrt{6}\)}{9}=\frac{2\(\sqrt{15}+\sqrt{6}\)}{3}) है।
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(\(\sqrt{50}+\sqrt{98}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{50}+\sqrt{98}\)2 )?
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#hard
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A (288)
B (148)
C \(24\sqrt{2}\)
D (196)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{98}=7\sqrt{2}\), so the sum is \(12\sqrt{2}\). Its square is (288).
Step 2
Why this answer is correct
The correct answer is A. (288). \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{98}=7\sqrt{2}\), so the sum is \(12\sqrt{2}\). Its square is (288).
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{98}=7\sqrt{2}\), इसलिए योग \(12\sqrt{2}\) है। इसका वर्ग (288) है।
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यदि \(a=\sqrt{17}+\sqrt{8}\) और \(b=\sqrt{17}-\sqrt{8}\) हैं तो (ab) और (a+b) का सही युग्म कौन-सा है?
If \(a=\sqrt{17}+\sqrt{8}\) and \(b=\sqrt{17}-\sqrt{8}\), which is the correct pair of (ab) and (a+b)?
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#irrational-numbers
#hard
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A (9), \(2\sqrt{17}\)
B (25), \(2\sqrt{8}\)
C \(9\sqrt{136}\), \(2\sqrt{17}\)
D (17), (8)
Explanation opens after your attempt
Correct Answer
A. (9), \(2\sqrt{17}\)
Step 1
Concept
(ab=17-8=9) and \(a+b=2\sqrt{17}\). In a conjugate pair check product and sum separately.
Step 2
Why this answer is correct
The correct answer is A. (9), \(2\sqrt{17}\). (ab=17-8=9) and \(a+b=2\sqrt{17}\). In a conjugate pair check product and sum separately.
Step 3
Exam Tip
(ab=17-8=9) और \(a+b=2\sqrt{17}\) है। संयुग्मी युग्म में गुणन और योग अलग जाँचें।
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(\sqrt{5}\left\(\sqrt{125}-\sqrt{45}\right\)) का मान क्या है?
What is the value of (\sqrt{5}\left\(\sqrt{125}-\sqrt{45}\right\))?
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A (10)
B \(2\sqrt{5}\)
C (20)
D \(\sqrt{80}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so the bracket is \(2\sqrt{5}\). Multiplying by \(\sqrt{5}\) gives (10).
Step 2
Why this answer is correct
The correct answer is A. (10). \(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\), so the bracket is \(2\sqrt{5}\). Multiplying by \(\sqrt{5}\) gives (10).
Step 3
Exam Tip
\(\sqrt{125}=5\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\), इसलिए कोष्ठक \(2\sqrt{5}\) है। \(\sqrt{5}\) से गुणा करने पर (10) मिलता है।
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\(\frac{5}{\sqrt{12}+\sqrt{7}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{5}{\sqrt{12}+\sqrt{7}}\)?
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A \(\sqrt{12}-\sqrt{7}\)
B (5\(\sqrt{12}-\sqrt{7}\))
C \(\frac{\sqrt{12}-\sqrt{7}}{5}\)
D \(5\sqrt{84}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{12}-\sqrt{7}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (12-7=5). So (\frac{5\(\sqrt{12}-\sqrt{7}\)}{5}=\sqrt{12}-\sqrt{7}).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{12}-\sqrt{7}\). Multiplying by the conjugate makes the denominator (12-7=5). So (\frac{5\(\sqrt{12}-\sqrt{7}\)}{5}=\sqrt{12}-\sqrt{7}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (12-7=5) बनता है। इसलिए (\frac{5\(\sqrt{12}-\sqrt{7}\)}{5}=\sqrt{12}-\sqrt{7}) है।
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यदि \(x=4\sqrt{7}-\sqrt{112}\) है तो (x) का मान क्या है?
If \(x=4\sqrt{7}-\sqrt{112}\), what is the value of (x)?
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A (0)
B \(\sqrt{7}\)
C \(8\sqrt{7}\)
D (4)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{112}=4\sqrt{7}\), so \(4\sqrt{7}-4\sqrt{7}=0\). Simplify the radical first.
Step 2
Why this answer is correct
The correct answer is A. (0). \(\sqrt{112}=4\sqrt{7}\), so \(4\sqrt{7}-4\sqrt{7}=0\). Simplify the radical first.
Step 3
Exam Tip
\(\sqrt{112}=4\sqrt{7}\), इसलिए \(4\sqrt{7}-4\sqrt{7}=0\) है। पहले मूल को सरल करें।
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\(\sqrt{6+\sqrt{5}}\times\sqrt{6+\sqrt{5}}\) का मान क्या है?
What is the value of \(\sqrt{6+\sqrt{5}}\times\sqrt{6+\sqrt{5}}\)?
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A \(6+\sqrt{5}\)
B (36+5)
C \(\sqrt{11}\)
D \(6-\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(6+\sqrt{5}\)
Step 1
Concept
Multiplying the same square root by itself gives the number inside. Therefore the value is \(6+\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(6+\sqrt{5}\). Multiplying the same square root by itself gives the number inside. Therefore the value is \(6+\sqrt{5}\).
Step 3
Exam Tip
एक ही वर्गमूल का अपने आप से गुणन अंदर की संख्या देता है। इसलिए मान \(6+\sqrt{5}\) है।
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\(\sqrt{242}+\sqrt{128}-\sqrt{72}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\sqrt{242}+\sqrt{128}-\sqrt{72}\)?
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A \(13\sqrt{2}\)
B \(11\sqrt{2}\)
C \(15\sqrt{2}\)
D \(17\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(13\sqrt{2}\)
Step 1
Concept
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). Therefore the result is \(13\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(13\sqrt{2}\). \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). Therefore the result is \(13\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\) है। इसलिए परिणाम \(13\sqrt{2}\) है।
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यदि \(t=\sqrt{13}+3\) है तो \(t+\frac{4}{t}\) का मान क्या है?
If \(t=\sqrt{13}+3\), what is the value of \(t+\frac{4}{t}\)?
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A \(2\sqrt{13}\)
B (6)
C \(2\sqrt{13}+6\)
D \(\sqrt{13}\)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{13}\)
Step 1
Concept
\(\frac{4}{\sqrt{13}+3}=\sqrt{13}-3\) because the denominator becomes (13-9=4). Therefore the sum is \(2\sqrt{13}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{13}\). \(\frac{4}{\sqrt{13}+3}=\sqrt{13}-3\) because the denominator becomes (13-9=4). Therefore the sum is \(2\sqrt{13}\).
Step 3
Exam Tip
\(\frac{4}{\sqrt{13}+3}=\sqrt{13}-3\) है क्योंकि हर (13-9=4) बनता है। इसलिए योग \(2\sqrt{13}\) है।
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यदि \(s=\sqrt{75}+\sqrt{192}\) है तो \(\frac{s}{\sqrt{3}}\) का मान क्या है?
If \(s=\sqrt{75}+\sqrt{192}\), what is the value of \(\frac{s}{\sqrt{3}}\)?
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A (13)
B (11)
C (9)
D (15)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{192}=8\sqrt{3}\), so \(s=13\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (13).
Step 2
Why this answer is correct
The correct answer is A. (13). \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{192}=8\sqrt{3}\), so \(s=13\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (13).
Step 3
Exam Tip
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{192}=8\sqrt{3}\), इसलिए \(s=13\sqrt{3}\) है। \(\sqrt{3}\) से भाग देने पर (13) मिलता है।
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\(\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}\)?
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A \(\frac{5-\sqrt{21}}{2}\)
B \(\frac{5+\sqrt{21}}{2}\)
C \(\sqrt{7}-2\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5-\sqrt{21}}{2}\)
Step 1
Concept
Multiplying by the conjugate gives (\frac{\(\sqrt{7}-\sqrt{3}\)2 }{7-3}). So the answer is \(\frac{5-\sqrt{21}}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5-\sqrt{21}}{2}\). Multiplying by the conjugate gives (\frac{\(\sqrt{7}-\sqrt{3}\)2 }{7-3}). So the answer is \(\frac{5-\sqrt{21}}{2}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर (\frac{\(\sqrt{7}-\sqrt{3}\)2 }{7-3}) मिलता है। इसलिए उत्तर \(\frac{5-\sqrt{21}}{2}\) है।
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यदि \(x=\sqrt{5}+\sqrt{20}\) और \(y=\sqrt{45}\) हैं तो (x-y) क्या है?
If \(x=\sqrt{5}+\sqrt{20}\) and \(y=\sqrt{45}\), what is (x-y)?
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A (0)
B \(\sqrt{5}\)
C \(2\sqrt{5}\)
D \(4\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
\(x=\sqrt{5}+2\sqrt{5}=3\sqrt{5}\) and \(y=3\sqrt{5}\). Therefore (x-y=0).
Step 2
Why this answer is correct
The correct answer is A. (0). \(x=\sqrt{5}+2\sqrt{5}=3\sqrt{5}\) and \(y=3\sqrt{5}\). Therefore (x-y=0).
Step 3
Exam Tip
\(x=\sqrt{5}+2\sqrt{5}=3\sqrt{5}\) और \(y=3\sqrt{5}\) है। इसलिए (x-y=0) है।
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(\(\sqrt{19}+4\)2 -\(\sqrt{19}-4\)2 ) का मान क्या है?
What is the value of (\(\sqrt{19}+4\)2 -\(\sqrt{19}-4\)2 )?
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A \(16\sqrt{19}\)
B (23)
C \(8\sqrt{19}\)
D (76)
Explanation opens after your attempt
Correct Answer
A. \(16\sqrt{19}\)
Step 1
Concept
Use ((a+b)2 -(a-b)2 =4ab). Here \(a=\sqrt{19}\) and (b=4), so the value is \(16\sqrt{19}\).
Step 2
Why this answer is correct
The correct answer is A. \(16\sqrt{19}\). Use ((a+b)2 -(a-b)2 =4ab). Here \(a=\sqrt{19}\) and (b=4), so the value is \(16\sqrt{19}\).
Step 3
Exam Tip
पहचान ((a+b)2 -(a-b)2 =4ab) लगाएं। यहाँ \(a=\sqrt{19}\) और (b=4), इसलिए मान \(16\sqrt{19}\) है।
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यदि एक आयत की लंबाई \(5+\sqrt{11}\) और चौड़ाई \(5-\sqrt{11}\) है तो क्षेत्रफल क्या होगा?
If a rectangle has length \(5+\sqrt{11}\) and breadth \(5-\sqrt{11}\), what will be its area?
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A (14)
B (25)
C (36)
D \(10\sqrt{11}\)
Explanation opens after your attempt
Step 1
Concept
Area is (\(5+\sqrt{11}\)\(5-\sqrt{11}\)=25-11=14). Multiplying conjugate dimensions can give a rational area.
Step 2
Why this answer is correct
The correct answer is A. (14). Area is (\(5+\sqrt{11}\)\(5-\sqrt{11}\)=25-11=14). Multiplying conjugate dimensions can give a rational area.
Step 3
Exam Tip
क्षेत्रफल (\(5+\sqrt{11}\)\(5-\sqrt{11}\)=25-11=14) है। संयुग्मी आयामों का गुणन परिमेय क्षेत्रफल दे सकता है।
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\(\sqrt{500}-\sqrt{320}+\sqrt{180}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{500}-\sqrt{320}+\sqrt{180}\)?
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A \(12\sqrt{5}\)
B \(10\sqrt{5}\)
C \(8\sqrt{5}\)
D \(\sqrt{360}\)
Explanation opens after your attempt
Correct Answer
A. \(12\sqrt{5}\)
Step 1
Concept
\(\sqrt{500}=10\sqrt{5}\), \(\sqrt{320}=8\sqrt{5}\), and \(\sqrt{180}=6\sqrt{5}\). So the result is \(10\sqrt{5}-8\sqrt{5}+6\sqrt{5}=8\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{5}\). \(\sqrt{500}=10\sqrt{5}\), \(\sqrt{320}=8\sqrt{5}\), and \(\sqrt{180}=6\sqrt{5}\). So the result is \(10\sqrt{5}-8\sqrt{5}+6\sqrt{5}=8\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{500}=10\sqrt{5}\), \(\sqrt{320}=8\sqrt{5}\) और \(\sqrt{180}=6\sqrt{5}\) है। इसलिए परिणाम \(8\sqrt{5}\) नहीं, \(10\sqrt{5}-8\sqrt{5}+6\sqrt{5}=8\sqrt{5}\) है।
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\(\frac{1}{\sqrt{8}+\sqrt{6}}+\frac{1}{\sqrt{8}-\sqrt{6}}\) का मान क्या है?
What is the value of \(\frac{1}{\sqrt{8}+\sqrt{6}}+\frac{1}{\sqrt{8}-\sqrt{6}}\)?
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A \(2\sqrt{8}\)
B \(\sqrt{8}\)
C \(2\sqrt{6}\)
D (7)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{8}\)
Step 1
Concept
The first term becomes \(\frac{\sqrt{8}-\sqrt{6}}{2}\) and the second becomes \(\frac{\sqrt{8}+\sqrt{6}}{2}\). Their sum is \(\sqrt{8}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{8}\). The first term becomes \(\frac{\sqrt{8}-\sqrt{6}}{2}\) and the second becomes \(\frac{\sqrt{8}+\sqrt{6}}{2}\). Their sum is \(\sqrt{8}\).
Step 3
Exam Tip
पहला पद \(\frac{\sqrt{8}-\sqrt{6}}{2}\) और दूसरा \(\frac{\sqrt{8}+\sqrt{6}}{2}\) बनता है। योग \(\sqrt{8}\) है।
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यदि \(z=\sqrt{17}-\sqrt{6}\) है तो \(z^2\) का मान कौन-सा है?
If \(z=\sqrt{17}-\sqrt{6}\), which is the value of \(z^2\)?
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A \(23-2\sqrt{102}\)
B (11)
C \(23+2\sqrt{102}\)
D \(\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
A. \(23-2\sqrt{102}\)
Step 1
Concept
(\(\sqrt{17}-\sqrt{6}\)2 =17+6-2\sqrt{102}). The middle term remains negative.
Step 2
Why this answer is correct
The correct answer is A. \(23-2\sqrt{102}\). (\(\sqrt{17}-\sqrt{6}\)2 =17+6-2\sqrt{102}). The middle term remains negative.
Step 3
Exam Tip
(\(\sqrt{17}-\sqrt{6}\)2 =17+6-2\sqrt{102}) है। मध्य पद का चिन्ह ऋणात्मक रहेगा।
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कौन-सा विकल्प \(\sqrt{98}+\sqrt{162}\) के बराबर है?
Which option is equal to \(\sqrt{98}+\sqrt{162}\)?
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A \(16\sqrt{2}\)
B \(12\sqrt{2}\)
C \(\sqrt{260}\)
D \(20\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(16\sqrt{2}\)
Step 1
Concept
\(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{162}=9\sqrt{2}\). So the sum is \(16\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(16\sqrt{2}\). \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{162}=9\sqrt{2}\). So the sum is \(16\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{162}=9\sqrt{2}\) है। इसलिए योग \(16\sqrt{2}\) है।
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यदि \(q=\sqrt{6}+2\) है तो \(q^2-4q\) का मान क्या है?
If \(q=\sqrt{6}+2\), what is the value of \(q^2-4q\)?
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A (2)
B (6)
C \(\sqrt{6}\)
D (10)
Explanation opens after your attempt
Step 1
Concept
\(q^2=10+4\sqrt{6}\) and \(4q=8+4\sqrt{6}\). Subtracting gives (2).
Step 2
Why this answer is correct
The correct answer is A. (2). \(q^2=10+4\sqrt{6}\) and \(4q=8+4\sqrt{6}\). Subtracting gives (2).
Step 3
Exam Tip
\(q^2=10+4\sqrt{6}\) और \(4q=8+4\sqrt{6}\) है। घटाने पर (2) मिलता है।
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\(\sqrt{14}\) और \(\sqrt{18}\) के बीच कौन-सी संख्या निश्चित रूप से आती है?
Which number definitely lies between \(\sqrt{14}\) and \(\sqrt{18}\)?
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A \(\sqrt{16}\)
B (5)
C \(\sqrt{13}\)
D \(\sqrt{20}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{16}\)
Step 1
Concept
Since (14<16<18), \(\sqrt{16}\) lies between them. For positive square roots compare the numbers inside.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{16}\). Since (14<16<18), \(\sqrt{16}\) lies between them. For positive square roots compare the numbers inside.
Step 3
Exam Tip
क्योंकि (14<16<18), इसलिए \(\sqrt{16}\) दोनों के बीच होगा। धनात्मक वर्गमूलों में अंदर की संख्या से तुलना करें।
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यदि \(A=\sqrt{125}+\sqrt{320}\) और \(B=13\sqrt{5}\) हैं तो कौन-सा कथन सही है?
If \(A=\sqrt{125}+\sqrt{320}\) and \(B=13\sqrt{5}\), 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
\(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{320}=8\sqrt{5}\), so \(A=13\sqrt{5}\). Hence (A=B).
Step 2
Why this answer is correct
The correct answer is A. (A=B). \(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{320}=8\sqrt{5}\), so \(A=13\sqrt{5}\). Hence (A=B).
Step 3
Exam Tip
\(\sqrt{125}=5\sqrt{5}\) और \(\sqrt{320}=8\sqrt{5}\), इसलिए \(A=13\sqrt{5}\) है। अतः (A=B) है।
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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 (4)
B (2)
C (8)
D (12)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{200}=10\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so the numerator is \(4\sqrt{2}\). Dividing gives (4).
Step 2
Why this answer is correct
The correct answer is A. (4). \(\sqrt{200}=10\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so the numerator is \(4\sqrt{2}\). Dividing gives (4).
Step 3
Exam Tip
\(\sqrt{200}=10\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\), इसलिए अंश \(4\sqrt{2}\) है। भाग देने पर (4) मिलता है।
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\(\sqrt{11+\sqrt{30}}\) का वर्ग किसके बराबर है?
What is the square of \(\sqrt{11+\sqrt{30}}\) equal to?
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A \(11+\sqrt{30}\)
B (121+30)
C \(\sqrt{41}\)
D \(11-\sqrt{30}\)
Explanation opens after your attempt
Correct Answer
A. \(11+\sqrt{30}\)
Step 1
Concept
The square of a square root gives the number inside. So (\left\(\sqrt{11+\sqrt{30}}\right\)2 =11+\sqrt{30}).
Step 2
Why this answer is correct
The correct answer is A. \(11+\sqrt{30}\). The square of a square root gives the number inside. So (\left\(\sqrt{11+\sqrt{30}}\right\)2 =11+\sqrt{30}).
Step 3
Exam Tip
वर्गमूल का वर्ग अंदर की संख्या देता है। इसलिए (\left\(\sqrt{11+\sqrt{30}}\right\)2 =11+\sqrt{30}) है।
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यदि \(x=\sqrt{11}+\sqrt{3}\) है तो \(x^2-14\) का मान क्या है?
If \(x=\sqrt{11}+\sqrt{3}\), what is the value of \(x^2-14\)?
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A \(2\sqrt{33}\)
B (14)
C \(\sqrt{33}\)
D (33)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{33}\)
Step 1
Concept
\(x^2=11+3+2\sqrt{33}=14+2\sqrt{33}\). Therefore \(x^2-14=2\sqrt{33}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{33}\). \(x^2=11+3+2\sqrt{33}=14+2\sqrt{33}\). Therefore \(x^2-14=2\sqrt{33}\).
Step 3
Exam Tip
\(x^2=11+3+2\sqrt{33}=14+2\sqrt{33}\) है। इसलिए \(x^2-14=2\sqrt{33}\) है।
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(\(\sqrt{45}+\sqrt{20}\)\(\sqrt{45}-\sqrt{20}\)) का मान क्या है?
What is the value of (\(\sqrt{45}+\sqrt{20}\)\(\sqrt{45}-\sqrt{20}\))?
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A (25)
B (-25)
C (65)
D \(2\sqrt{900}\)
Explanation opens after your attempt
Step 1
Concept
This is the \(a^2-b^2\) form, so the value is (45-20=25). In conjugate multiplication take the difference of squares directly.
Step 2
Why this answer is correct
The correct answer is A. (25). This is the \(a^2-b^2\) form, so the value is (45-20=25). In conjugate multiplication take the difference of squares directly.
Step 3
Exam Tip
यह \(a^2-b^2\) रूप है, इसलिए मान (45-20=25) है। संयुग्मी गुणन में सीधे वर्गों का अंतर लें।
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यदि \(c=\sqrt{10}+\sqrt{3}\) और \(d=\sqrt{10}-\sqrt{3}\) हैं तो (cd) और (c-d) का सही युग्म कौन-सा है?
If \(c=\sqrt{10}+\sqrt{3}\) and \(d=\sqrt{10}-\sqrt{3}\), which is the correct pair of (cd) and (c-d)?
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A (7), \(2\sqrt{3}\)
B (13), \(2\sqrt{10}\)
C \(7\sqrt{30}\), \(2\sqrt{3}\)
D (10), (3)
Explanation opens after your attempt
Correct Answer
A. (7), \(2\sqrt{3}\)
Step 1
Concept
(cd=10-3=7) and \(c-d=2\sqrt{3}\). In conjugate forms find the product and difference separately.
Step 2
Why this answer is correct
The correct answer is A. (7), \(2\sqrt{3}\). (cd=10-3=7) and \(c-d=2\sqrt{3}\). In conjugate forms find the product and difference separately.
Step 3
Exam Tip
(cd=10-3=7) और \(c-d=2\sqrt{3}\) है। संयुग्मी रूप में गुणन और अंतर अलग-अलग निकालें।
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\(\frac{\sqrt{11}+\sqrt{2}}{\sqrt{11}-\sqrt{2}}\) का सरल रूप कौन-सा है?
Which is the simplified form of \(\frac{\sqrt{11}+\sqrt{2}}{\sqrt{11}-\sqrt{2}}\)?
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A \(\frac{13+2\sqrt{22}}{9}\)
B \(13+2\sqrt{22}\)
C \(\frac{13-2\sqrt{22}}{9}\)
D (9)
Explanation opens after your attempt
Correct Answer
A. \(\frac{13+2\sqrt{22}}{9}\)
Step 1
Concept
Multiplying by the conjugate gives numerator \(13+2\sqrt{22}\) and denominator (11-2=9). So the simplified form is \(\frac{13+2\sqrt{22}}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{13+2\sqrt{22}}{9}\). Multiplying by the conjugate gives numerator \(13+2\sqrt{22}\) and denominator (11-2=9). So the simplified form is \(\frac{13+2\sqrt{22}}{9}\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर अंश \(13+2\sqrt{22}\) और हर (11-2=9) मिलता है। इसलिए सरल रूप \(\frac{13+2\sqrt{22}}{9}\) है।
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यदि \(x=3+\sqrt{10}\) है तो \(x^2-6x\) का मान क्या है?
If \(x=3+\sqrt{10}\), what is the value of \(x^2-6x\)?
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A (1)
B (-1)
C (10)
D \(2\sqrt{10}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=19+6\sqrt{10}\) and \(6x=18+6\sqrt{10}\). Subtracting gives (1).
Step 2
Why this answer is correct
The correct answer is A. (1). \(x^2=19+6\sqrt{10}\) and \(6x=18+6\sqrt{10}\). Subtracting gives (1).
Step 3
Exam Tip
\(x^2=19+6\sqrt{10}\) और \(6x=18+6\sqrt{10}\) है। घटाने पर (1) मिलता है।
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\(\sqrt{605}-\sqrt{245}+\sqrt{125}\) का सरल रूप क्या है?
What is the simplified form of \(\sqrt{605}-\sqrt{245}+\sqrt{125}\)?
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A \(9\sqrt{5}\)
B \(11\sqrt{5}\)
C \(13\sqrt{5}\)
D \(15\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(11\sqrt{5}\)
Step 1
Concept
\(\sqrt{605}=11\sqrt{5}\), \(\sqrt{245}=7\sqrt{5}\), and \(\sqrt{125}=5\sqrt{5}\). Therefore the result is \(9\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(11\sqrt{5}\). \(\sqrt{605}=11\sqrt{5}\), \(\sqrt{245}=7\sqrt{5}\), and \(\sqrt{125}=5\sqrt{5}\). Therefore the result is \(9\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{605}=11\sqrt{5}\), \(\sqrt{245}=7\sqrt{5}\) और \(\sqrt{125}=5\sqrt{5}\) है। इसलिए परिणाम \(9\sqrt{5}\) है।
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यदि \(r=\frac{1}{\sqrt{17}+4}\) है तो (r) का सरल रूप कौन-सा है?
If \(r=\frac{1}{\sqrt{17}+4}\), which is the simplified form of (r)?
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#irrational-numbers
#hard
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A \(\sqrt{17}-4\)
B \(\frac{\sqrt{17}-4}{1}\)
C \(\sqrt{17}+4\)
D \(4-\sqrt{17}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{17}-4\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (17-16=1). So the simplified form is \(\sqrt{17}-4\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{17}-4\). Multiplying by the conjugate makes the denominator (17-16=1). So the simplified form is \(\sqrt{17}-4\).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (17-16=1) बनता है। इसलिए सरल रूप \(\sqrt{17}-4\) है।
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(\(\sqrt{27}+\sqrt{75}\)2 ) का मान क्या है?
What is the value of (\(\sqrt{27}+\sqrt{75}\)2 )?
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#irrational-numbers
#hard
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A (192)
B (128)
C \(16\sqrt{3}\)
D (300)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so the sum is \(8\sqrt{3}\). Its square is (192).
Step 2
Why this answer is correct
The correct answer is A. (192). \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so the sum is \(8\sqrt{3}\). Its square is (192).
Step 3
Exam Tip
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\), इसलिए योग \(8\sqrt{3}\) है। इसका वर्ग (192) है।
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\(\frac{3}{\sqrt{14}+\sqrt{5}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{3}{\sqrt{14}+\sqrt{5}}\)?
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#hard
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A \(\frac{\sqrt{14}-\sqrt{5}}{3}\)
B \(\sqrt{14}-\sqrt{5}\)
C (3\(\sqrt{14}-\sqrt{5}\))
D \(3\sqrt{70}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{\sqrt{14}-\sqrt{5}}{3}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (14-5=9). So (\frac{3\(\sqrt{14}-\sqrt{5}\)}{9}=\frac{\sqrt{14}-\sqrt{5}}{3}).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{14}-\sqrt{5}}{3}\). Multiplying by the conjugate makes the denominator (14-5=9). So (\frac{3\(\sqrt{14}-\sqrt{5}\)}{9}=\frac{\sqrt{14}-\sqrt{5}}{3}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (14-5=9) बनता है। इसलिए (\frac{3\(\sqrt{14}-\sqrt{5}\)}{9}=\frac{\sqrt{14}-\sqrt{5}}{3}) है।
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\(\frac{\sqrt{18}+\sqrt{72}}{\sqrt{2}}\) का मान क्या है?
What is the value of \(\frac{\sqrt{18}+\sqrt{72}}{\sqrt{2}}\)?
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#hard
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A (6)
B (9)
C (12)
D (15)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\) so the numerator is \(9\sqrt{2}\). Dividing by \(\sqrt{2}\) gives (9).
Step 2
Why this answer is correct
The correct answer is B. (9). \(\sqrt{18}=3\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\) so the numerator is \(9\sqrt{2}\). Dividing by \(\sqrt{2}\) gives (9).
Step 3
Exam Tip
\(\sqrt{18}=3\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\) है इसलिए अंश \(9\sqrt{2}\) है। \(\sqrt{2}\) से भाग देने पर (9) मिलता है।
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यदि \(a=\sqrt{12}+\sqrt{75}\) और \(b=\sqrt{48}\) हैं तो (a-b) का सरल रूप क्या है?
If \(a=\sqrt{12}+\sqrt{75}\) and \(b=\sqrt{48}\), what is the simplified form of (a-b)?
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#hard
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A \(\sqrt{3}\)
B \(2\sqrt{3}\)
C \(3\sqrt{3}\)
D \(7\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
C. \(3\sqrt{3}\)
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\). Therefore \(a-b=3\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is C. \(3\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\). Therefore \(a-b=3\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\) है। इसलिए \(a-b=3\sqrt{3}\) है।
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यदि \(x=\sqrt{7}+\sqrt{3}\) है तो \(x^2-10\) का मान क्या है?
If \(x=\sqrt{7}+\sqrt{3}\), what is the value of \(x^2-10\)?
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#hard
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A \(\sqrt{21}\)
B (10)
C (21)
D \(2\sqrt{21}\)
Explanation opens after your attempt
Correct Answer
D. \(2\sqrt{21}\)
Step 1
Concept
\(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\). Therefore \(x^2-10=2\sqrt{21}\).
Step 2
Why this answer is correct
The correct answer is D. \(2\sqrt{21}\). \(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\). Therefore \(x^2-10=2\sqrt{21}\).
Step 3
Exam Tip
\(x^2=7+3+2\sqrt{21}=10+2\sqrt{21}\) है। इसलिए \(x^2-10=2\sqrt{21}\) है।
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\(\frac{4}{\sqrt{13}-\sqrt{5}}\) का परिमेयकृत रूप क्या है?
What is the rationalised form of \(\frac{4}{\sqrt{13}-\sqrt{5}}\)?
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#irrational-numbers
#hard
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A \(\sqrt{13}-\sqrt{5}\)
B \(\frac{\sqrt{13}+\sqrt{5}}{2}\)
C \(\sqrt{13}+\sqrt{5}\)
D \(4\sqrt{65}\)
Explanation opens after your attempt
Correct Answer
B. \(\frac{\sqrt{13}+\sqrt{5}}{2}\)
Step 1
Concept
Multiplying by the conjugate makes the denominator (13-5=8). So (\frac{4\(\sqrt{13}+\sqrt{5}\)}{8}=\frac{\sqrt{13}+\sqrt{5}}{2}).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{\sqrt{13}+\sqrt{5}}{2}\). Multiplying by the conjugate makes the denominator (13-5=8). So (\frac{4\(\sqrt{13}+\sqrt{5}\)}{8}=\frac{\sqrt{13}+\sqrt{5}}{2}).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (13-5=8) बनता है। इसलिए (\frac{4\(\sqrt{13}+\sqrt{5}\)}{8}=\frac{\sqrt{13}+\sqrt{5}}{2}) है।
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यदि एक आयत की लंबाई \(\sqrt{11}+\sqrt{6}\) और चौड़ाई \(\sqrt{11}-\sqrt{6}\) है तो क्षेत्रफल क्या होगा?
If a rectangle has length \(\sqrt{11}+\sqrt{6}\) and breadth \(\sqrt{11}-\sqrt{6}\), what will be its area?
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#irrational-numbers
#hard
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A (17)
B (5)
C \(\sqrt{66}\)
D \(2\sqrt{11}\)
Explanation opens after your attempt
Step 1
Concept
Area is (\(\sqrt{11}+\sqrt{6}\)\(\sqrt{11}-\sqrt{6}\)=11-6=5). Conjugate dimensions can give a rational area.
Step 2
Why this answer is correct
The correct answer is B. (5). Area is (\(\sqrt{11}+\sqrt{6}\)\(\sqrt{11}-\sqrt{6}\)=11-6=5). Conjugate dimensions can give a rational area.
Step 3
Exam Tip
क्षेत्रफल (\(\sqrt{11}+\sqrt{6}\)\(\sqrt{11}-\sqrt{6}\)=11-6=5) है। संयुग्मी आयामों से परिमेय क्षेत्रफल मिल सकता है।
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किस विकल्प में दो अपरिमेय संख्याओं का गुणनफल परिमेय है?
In which option is the product of two irrational numbers rational?
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A \(\sqrt{2}\times\sqrt{3}\)
B \(\sqrt{5}\times\sqrt{7}\)
C \(\sqrt{10}\times\sqrt{2}\)
D \(\sqrt{12}\times\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
D. \(\sqrt{12}\times\sqrt{3}\)
Step 1
Concept
\(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6\), which is rational. Check whether multiplication creates a perfect square.
Step 2
Why this answer is correct
The correct answer is D. \(\sqrt{12}\times\sqrt{3}\). \(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6\), which is rational. Check whether multiplication creates a perfect square.
Step 3
Exam Tip
\(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6\) है जो परिमेय है। विकल्पों में गुणन के बाद पूर्ण वर्ग बनने की जाँच करें।
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