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\(5\sqrt{20}=10\sqrt{5}\), \(2\sqrt{125}=10\sqrt{5}\), and \( \sqrt{45}=3\sqrt{5} \). Therefore the result is \(3\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is D. \(3\sqrt{5}\). \(5\sqrt{20}=10\sqrt{5}\), \(2\sqrt{125}=10\sqrt{5}\), and \( \sqrt{45}=3\sqrt{5} \). Therefore the result is \(3\sqrt{5}\).
Step 3
Exam Tip
\(5\sqrt{20}=10\sqrt{5}\), \(2\sqrt{125}=10\sqrt{5}\), और \( \sqrt{45}=3\sqrt{5} \)। इसलिए परिणाम \(3\sqrt{5}\) है।
\(3\sqrt{75}=15\sqrt{3}\), \(4\sqrt{48}=16\sqrt{3}\), and \(2\sqrt{27}=6\sqrt{3}\). Therefore \(25\sqrt{3}\) is correct.
Step 2
Why this answer is correct
The correct answer is B. \(25\sqrt{3}\). \(3\sqrt{75}=15\sqrt{3}\), \(4\sqrt{48}=16\sqrt{3}\), and \(2\sqrt{27}=6\sqrt{3}\). Therefore \(25\sqrt{3}\) is correct.
Step 3
Exam Tip
\(3\sqrt{75}=15\sqrt{3}\), \(4\sqrt{48}=16\sqrt{3}\), और \(2\sqrt{27}=6\sqrt{3}\)। इसलिए \(25\sqrt{3}\) सही है।
The first square is (48) and the second square is (18). The middle term is \(2\times4\sqrt{3}\times3\sqrt{2}=24\sqrt{6}\).
Step 2
Why this answer is correct
The correct answer is A. \(66+24\sqrt{6}\). The first square is (48) and the second square is (18). The middle term is \(2\times4\sqrt{3}\times3\sqrt{2}=24\sqrt{6}\).
Step 3
Exam Tip
पहला वर्ग (48) और दूसरा वर्ग (18) है। मध्य पद \(2\times4\sqrt{3}\times3\sqrt{2}=24\sqrt{6}\) है।
Multiplying by the conjugate gives denominator (48-45=3). So the form is \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \). Multiplying by the conjugate gives denominator (48-45=3). So the form is \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (48-45=3) आता है। इसलिए रूप \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \) है।
Multiplying by the conjugate gives denominator (13-8=5). The numerator also has (5), so the answer is \( \sqrt{13}+\sqrt{8} \).
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{13}+\sqrt{8} \). Multiplying by the conjugate gives denominator (13-8=5). The numerator also has (5), so the answer is \( \sqrt{13}+\sqrt{8} \).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (13-8=5) आता है। अंश में भी (5) है इसलिए उत्तर \( \sqrt{13}+\sqrt{8} \) है।
The value is ( \frac{\(4+\sqrt{7}\)2+\(4-\sqrt{7}\)2}{16-7} ). The numerator is (46) and the denominator is (9).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{46}{9} \). The value is ( \frac{\(4+\sqrt{7}\)2+\(4-\sqrt{7}\)2}{16-7} ). The numerator is (46) and the denominator is (9).
Step 3
Exam Tip
मान ( \frac{\(4+\sqrt{7}\)2+\(4-\sqrt{7}\)2}{16-7} ) होगा। अंश (46) और हर (9) है।
( \(\sqrt{10}-2\)2=10-4\sqrt{10}+4=14-4\sqrt{10} ). The positive principal square root is \( \sqrt{10}-2 \).
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{10}-2 \). ( \(\sqrt{10}-2\)2=10-4\sqrt{10}+4=14-4\sqrt{10} ). The positive principal square root is \( \sqrt{10}-2 \).
Step 3
Exam Tip
( \(\sqrt{10}-2\)2=10-4\sqrt{10}+4=14-4\sqrt{10} )। धनात्मक मुख्य वर्गमूल \( \sqrt{10}-2 \) है।
\( \sqrt{80}\approx8.94 \), so \( -\sqrt{80}\approx-8.94 \). It is greater than ( -9 ) because it is closer to zero.
Step 2
Why this answer is correct
The correct answer is A. \( -\sqrt{80} \). \( \sqrt{80}\approx8.94 \), so \( -\sqrt{80}\approx-8.94 \). It is greater than ( -9 ) because it is closer to zero.
Step 3
Exam Tip
\( \sqrt{80}\approx8.94 \) है इसलिए \( -\sqrt{80}\approx-8.94 \)। यह ( -9 ) से बड़ा है क्योंकि शून्य के अधिक निकट है।
\( \frac{96}{384}=\frac{1}{4} \) and \(4=2^2\). A decimal terminates when the denominator has only (2) and (5) as prime factors.
Step 2
Why this answer is correct
The correct answer is A. समाप्त दशमलव / Terminating decimal. \( \frac{96}{384}=\frac{1}{4} \) and \(4=2^2\). A decimal terminates when the denominator has only (2) and (5) as prime factors.
Step 3
Exam Tip
\( \frac{96}{384}=\frac{1}{4} \) और \(4=2^2\) है। हर में केवल (2) और (5) होने पर दशमलव समाप्त होता है।
B. असमाप्त आवर्ती दशमलव/Non-terminating repeating decimal
Step 1
Concept
\( \frac{35}{154}=\frac{5}{22} \), and the denominator has (11). Therefore the decimal is non-terminating repeating.
Step 2
Why this answer is correct
The correct answer is B. असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal. \( \frac{35}{154}=\frac{5}{22} \), and the denominator has (11). Therefore the decimal is non-terminating repeating.
Step 3
Exam Tip
\( \frac{35}{154}=\frac{5}{22} \) है और हर में (11) है। इसलिए दशमलव असमाप्त आवर्ती होगा।
B. असमाप्त आवर्ती दशमलव/Non-terminating repeating decimal
Step 1
Concept
\( \frac{132}{360}=\frac{11}{30} \), and the denominator also has (3). Therefore it gives a non-terminating repeating decimal.
Step 2
Why this answer is correct
The correct answer is B. असमाप्त आवर्ती दशमलव / Non-terminating repeating decimal. \( \frac{132}{360}=\frac{11}{30} \), and the denominator also has (3). Therefore it gives a non-terminating repeating decimal.
Step 3
Exam Tip
\( \frac{132}{360}=\frac{11}{30} \) है और हर में (3) भी है। इसलिए यह असमाप्त आवर्ती दशमलव देगा।
A non-terminating and non-repeating decimal is irrational. Do not treat it as rational when there is no fixed repetition.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय वास्तविक संख्या / Irrational real number. A non-terminating and non-repeating decimal is irrational. Do not treat it as rational when there is no fixed repetition.
Step 3
Exam Tip
असमाप्त और अनावर्ती दशमलव अपरिमेय होता है। स्थिर आवृत्ति न होने पर इसे परिमेय न मानें।
The block (81) repeats, so the decimal is repeating. Every repeating decimal is a rational real number.
Step 2
Why this answer is correct
The correct answer is B. परिमेय वास्तविक संख्या / Rational real number. The block (81) repeats, so the decimal is repeating. Every repeating decimal is a rational real number.
Step 3
Exam Tip
(81) बार-बार आ रहा है इसलिए दशमलव आवर्ती है। हर आवर्ती दशमलव परिमेय वास्तविक संख्या होता है।
\( \sqrt{20}=2\sqrt{5} \) and \( \sqrt{45}=3\sqrt{5} \). The total coefficient is (1+2+3=6), so the value is \(6\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{5}\). \( \sqrt{20}=2\sqrt{5} \) and \( \sqrt{45}=3\sqrt{5} \). The total coefficient is (1+2+3=6), so the value is \(6\sqrt{5}\).
Step 3
Exam Tip
\( \sqrt{20}=2\sqrt{5} \) और \( \sqrt{45}=3\sqrt{5} \)। कुल (1+2+3=6) से \(6\sqrt{5}\) मिलता है।
The common denominator is (25-21=4), and the numerator is (10). So the value is \( \frac{10}{4}=\frac{5}{2} \).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{5}{2} \). The common denominator is (25-21=4), and the numerator is (10). So the value is \( \frac{10}{4}=\frac{5}{2} \).
Step 3
Exam Tip
समान हर (25-21=4) और अंश (10) है। इसलिए मान \( \frac{10}{4}=\frac{5}{2} \) है।
Multiply \( \frac{1}{4-\sqrt{7}} \) by the conjugate. The denominator is (16-7=9) and the numerator is \(4+\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{4+\sqrt{7}}{9} \). Multiply \( \frac{1}{4-\sqrt{7}} \) by the conjugate. The denominator is (16-7=9) and the numerator is \(4+\sqrt{7}\).
Step 3
Exam Tip
\( \frac{1}{4-\sqrt{7}} \) को संयुग्मी से गुणा करें। हर (16-7=9) और अंश \(4+\sqrt{7}\) होगा।
\(a^2=7+2\sqrt{10}\) and \( \frac{1}{a^2} \) is not simply \(7-2\sqrt{10}\) because (a\(\sqrt{5}-\sqrt{2}\)=3). Check options carefully using reciprocal rules.
Step 2
Why this answer is correct
The correct answer is A. (98). \(a^2=7+2\sqrt{10}\) and \( \frac{1}{a^2} \) is not simply \(7-2\sqrt{10}\) because (a\(\sqrt{5}-\sqrt{2}\)=3). Check options carefully using reciprocal rules.
Step 3
Exam Tip
\(a^2=7+2\sqrt{10}\) और \( \frac{1}{a^2}=7-2\sqrt{10} \) नहीं है क्योंकि (a\(\sqrt{5}-\sqrt{2}\)=3)। सही मान सीधे (a-2+\frac{1}{a-2}=\frac{\(a^2\)2+1}{a-2}) से कठिन है इसलिए विकल्प जाँचें।
\(x-3=\sqrt{8}=2\sqrt{2}\), so \( \frac{1}{x-3}=\frac{\sqrt{2}}{4} \). The total should be \(3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(3+\frac{5\sqrt{2}}{2}\). \(x-3=\sqrt{8}=2\sqrt{2}\), so \( \frac{1}{x-3}=\frac{\sqrt{2}}{4} \). The total should be \(3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}\).
Step 3
Exam Tip
\(x-3=\sqrt{8}=2\sqrt{2}\) इसलिए \( \frac{1}{x-3}=\frac{\sqrt{2}}{4} \)। कुल \(3+2\sqrt{2}+\frac{\sqrt{2}}{4}=3+\frac{9\sqrt{2}}{4}\) होना चाहिए।
\( \sqrt{27}>4 \) because (27>16). So \(4-\sqrt{27}\) is negative and the absolute value is \( \sqrt{27}-4 \).
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{27}-4 \). \( \sqrt{27}>4 \) because (27>16). So \(4-\sqrt{27}\) is negative and the absolute value is \( \sqrt{27}-4 \).
Step 3
Exam Tip
\( \sqrt{27}>4 \) क्योंकि (27>16) है। इसलिए \(4-\sqrt{27}\) ऋणात्मक है और निरपेक्ष मान \( \sqrt{27}-4 \) होगा।
The common denominator is (13-4=9), and the numerator is \(2\sqrt{13}\). Therefore the value is \( \frac{2\sqrt{13}}{9} \).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{2\sqrt{13}}{9} \). The common denominator is (13-4=9), and the numerator is \(2\sqrt{13}\). Therefore the value is \( \frac{2\sqrt{13}}{9} \).
Step 3
Exam Tip
समान हर (13-4=9) और अंश \(2\sqrt{13}\) है। इसलिए मान \( \frac{2\sqrt{13}}{9} \) है।
\( \sqrt{13} \) and (2+3=5) are not equal. Therefore square root cannot be distributed directly over addition.
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{4+9}\neq\sqrt{4}+\sqrt{9} \). \( \sqrt{13} \) and (2+3=5) are not equal. Therefore square root cannot be distributed directly over addition.
Step 3
Exam Tip
\( \sqrt{13} \) और (2+3=5) बराबर नहीं हैं। इसलिए वर्गमूल को जोड़ के अंदर सीधे नहीं बाँटते।
\( \sqrt{36}+\sqrt{49}=6+7=13 \), and \( \sqrt{85}<10 \). Therefore the first value is greater.
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{36}+\sqrt{49}>\sqrt{85} \). \( \sqrt{36}+\sqrt{49}=6+7=13 \), and \( \sqrt{85}<10 \). Therefore the first value is greater.
Step 3
Exam Tip
\( \sqrt{36}+\sqrt{49}=6+7=13 \) और \( \sqrt{85}<10 \) है। इसलिए पहला मान बड़ा है।
\( \sqrt{3}\times\sqrt{27}=\sqrt{81}=9 \), which is rational. The other products do not become square roots of perfect squares.
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{3}\times\sqrt{27} \). \( \sqrt{3}\times\sqrt{27}=\sqrt{81}=9 \), which is rational. The other products do not become square roots of perfect squares.
Step 3
Exam Tip
\( \sqrt{3}\times\sqrt{27}=\sqrt{81}=9 \) परिमेय है। बाकी गुणनफल पूर्ण वर्ग के वर्गमूल में नहीं बदलते।
