Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 551-560 of 600 questions.
If \(x=\sqrt{14}+\sqrt{6}\), what is the value of \(x^2-20\)?
B. \(2\sqrt{84}\)
\(x^2=14+6+2\sqrt{84}=20+2\sqrt{84}\) है। इसलिए \(x^2-20=2\sqrt{84}\) है। / \(x^2=14+6+2\sqrt{84}=20+2\sqrt{84}\). So \(x^2-20=2\sqrt{84}\).
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Which is the simplified form of \(\frac{\sqrt{11}+\sqrt{6}}{\sqrt{11}-\sqrt{6}}\)?
A. \(\frac{17+2\sqrt{66}}{5}\)
संयुग्मी से गुणा करने पर अंश \(17+2\sqrt{66}\) और हर (5) मिलता है। हर को परिमेय बनाकर सरल करें। / Multiplying by the conjugate gives numerator \(17+2\sqrt{66}\) and denominator (5). Rationalise the denominator and simplify.
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If \(a=7+\sqrt{2}\) and \(b=7-\sqrt{2}\), what is the value of \(a^2-b^2\)?
B. \(28\sqrt{2}\)
\(a-b=2\sqrt{2}\) और (a+b=14) है। इसलिए \(a^2-b^2=28\sqrt{2}\) है। / \(a-b=2\sqrt{2}\) and (a+b=14). Therefore \(a^2-b^2=28\sqrt{2}\).
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What is the simplified form of \(\sqrt{605}-\sqrt{320}+\sqrt{125}\)?
B. \(8\sqrt{5}\)
\(\sqrt{605}=11\sqrt{5}\), \(\sqrt{320}=8\sqrt{5}\) और \(\sqrt{125}=5\sqrt{5}\) है। इसलिए परिणाम \(8\sqrt{5}\) है। / \(\sqrt{605}=11\sqrt{5}\), \(\sqrt{320}=8\sqrt{5}\), and \(\sqrt{125}=5\sqrt{5}\). Therefore the result is \(8\sqrt{5}\).
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If \(p=\frac{1}{\sqrt{23}+4}\), what is the simplified form of (p)?
A. \(\frac{\sqrt{23}-4}{7}\)
संयुग्मी से गुणा करने पर हर (23-16=7) बनता है। इसलिए \(p=\frac{\sqrt{23}-4}{7}\) है। / Multiplying by the conjugate makes the denominator (23-16=7). So \(p=\frac{\sqrt{23}-4}{7}\).
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If \(y=\sqrt{20}+\sqrt{45}+\sqrt{80}\), what is the value of \(y^2\)?
A. (405)
\(\sqrt{20}=2\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{80}=4\sqrt{5}\) है। इसलिए \(y=9\sqrt{5}\) और \(y^2=405\) है। / \(\sqrt{20}=2\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), and \(\sqrt{80}=4\sqrt{5}\). So \(y=9\sqrt{5}\) and \(y^2=405\).
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Which is the rationalised form of \(\frac{7}{\sqrt{30}-\sqrt{23}}\)?
C. \(\sqrt{30}+\sqrt{23}\)
संयुग्मी से गुणा करने पर हर (30-23=7) बनता है। इसलिए उत्तर \(\sqrt{30}+\sqrt{23}\) है। / Multiplying by the conjugate makes the denominator (30-23=7). So the answer is \(\sqrt{30}+\sqrt{23}\).
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If \(s=\sqrt{242}+\sqrt{128}\), what is the value of \(\frac{s}{\sqrt{2}}\)?
B. (19)
\(\sqrt{242}=11\sqrt{2}\) और \(\sqrt{128}=8\sqrt{2}\), इसलिए \(s=19\sqrt{2}\) है। भाग देने पर (19) मिलता है। / \(\sqrt{242}=11\sqrt{2}\) and \(\sqrt{128}=8\sqrt{2}\), so \(s=19\sqrt{2}\). Dividing gives (19).
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What is the value of \(\sqrt{15+\sqrt{26}}\times\sqrt{15+\sqrt{26}}\)?
C. \(15+\sqrt{26}\)
एक ही वर्गमूल को अपने आप से गुणा करने पर अंदर की संख्या मिलती है। इसलिए मान \(15+\sqrt{26}\) है। / Multiplying the same square root by itself gives the number inside. Therefore the value is \(15+\sqrt{26}\).
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If \(A=\sqrt{300}+\sqrt{108}\) and \(B=16\sqrt{3}\), which statement is correct?
B. (A=B)
\(\sqrt{300}=10\sqrt{3}\) और \(\sqrt{108}=6\sqrt{3}\), इसलिए \(A=16\sqrt{3}\) है। अतः (A=B) है। / \(\sqrt{300}=10\sqrt{3}\) and \(\sqrt{108}=6\sqrt{3}\), so \(A=16\sqrt{3}\). Hence (A=B).
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