Level 14 • 50/50 questions • 25 seconds per question.
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If \(x=\sqrt{13}+\sqrt{5}\), what is the value of \(x^2-18\)?
C. \(2\sqrt{65}\)
\(x^2=13+5+2\sqrt{65}=18+2\sqrt{65}\) है। इसलिए \(x^2-18=2\sqrt{65}\) है। / \(x^2=13+5+2\sqrt{65}=18+2\sqrt{65}\). So \(x^2-18=2\sqrt{65}\).
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Which is the simplified form of \(\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\)?
A. \(\frac{9+2\sqrt{14}}{5}\)
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। / Multiplying by the conjugate gives numerator \(9+2\sqrt{14}\) and denominator (5). So the simplified form is \(\frac{9+2\sqrt{14}}{5}\).
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If \(a=6+\sqrt{3}\) and \(b=6-\sqrt{3}\), what is the value of \(a^2-b^2\)?
D. \(24\sqrt{3}\)
(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। / (a-2-b-2=(a-b)(a+b)) where \(a-b=2\sqrt{3}\) and (a+b=12). So the value is \(24\sqrt{3}\).
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If \(p=\frac{1}{\sqrt{19}+4}\), what is the simplified form of (p)?
A. \(\frac{\sqrt{19}-4}{3}\)
संयुग्मी से गुणा करने पर हर (19-16=3) बनता है। इसलिए \(p=\frac{\sqrt{19}-4}{3}\) है। / Multiplying by the conjugate makes the denominator (19-16=3). So \(p=\frac{\sqrt{19}-4}{3}\).
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If \(y=\sqrt{12}+\sqrt{27}\), what is the value of \(y^2\)?
B. (75)
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए \(y=5\sqrt{3}\) है। इसका वर्ग (75) है। / \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so \(y=5\sqrt{3}\). Its square is (75).
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Which is the rationalised form of \(\frac{5}{\sqrt{14}-3}\)?
D. (\frac{5\(\sqrt{14}+3\)}{5})
संयुग्मी से गुणा करने पर हर (14-9=5) बनता है। इसलिए रूप (\frac{5\(\sqrt{14}+3\)}{5}), यानी \(\sqrt{14}+3\), है। / Multiplying by the conjugate makes the denominator (14-9=5). So the form is (\frac{5\(\sqrt{14}+3\)}{5}), that is \(\sqrt{14}+3\).
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If \(s=\sqrt{108}+\sqrt{300}\), what is the value of \(\frac{s}{\sqrt{3}}\)?
C. (16)
\(\sqrt{108}=6\sqrt{3}\) और \(\sqrt{300}=10\sqrt{3}\), इसलिए \(s=16\sqrt{3}\) है। भाग देने पर (16) मिलता है। / \(\sqrt{108}=6\sqrt{3}\) and \(\sqrt{300}=10\sqrt{3}\), so \(s=16\sqrt{3}\). Dividing gives (16).
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What is the value of \(\sqrt{8+\sqrt{11}}\times\sqrt{8+\sqrt{11}}\)?
D. \(8+\sqrt{11}\)
एक ही वर्गमूल को अपने आप से गुणा करने पर अंदर की संख्या मिलती है। इसलिए मान \(8+\sqrt{11}\) है। / Multiplying the same square root by itself gives the number inside. Therefore the value is \(8+\sqrt{11}\).
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If \(A=\sqrt{147}+\sqrt{243}\) and \(B=16\sqrt{3}\), which statement is correct?
C. (A=B)
\(\sqrt{147}=7\sqrt{3}\) और \(\sqrt{243}=9\sqrt{3}\), इसलिए \(A=16\sqrt{3}\) है। अतः (A=B) है। / \(\sqrt{147}=7\sqrt{3}\) and \(\sqrt{243}=9\sqrt{3}\), so \(A=16\sqrt{3}\). Hence (A=B).
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Which is the simplified form of \(\frac{\sqrt{13}+\sqrt{3}}{\sqrt{13}-\sqrt{3}}\)?
C. \(\frac{8+\sqrt{39}}{5}\)
संयुग्मी से गुणा करने पर अंश \(16+2\sqrt{39}\) और हर (10) मिलता है। इसलिए उत्तर \(\frac{8+\sqrt{39}}{5}\) है। / Multiplying by the conjugate gives numerator \(16+2\sqrt{39}\) and denominator (10). So the answer is \(\frac{8+\sqrt{39}}{5}\).
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If \(q=\sqrt{7}+3\), what is the value of \(q^2-6q\)?
A. (-2)
\(q^2=16+6\sqrt{7}\) और \(6q=18+6\sqrt{7}\) है। घटाने पर (-2) मिलता है। / \(q^2=16+6\sqrt{7}\) and \(6q=18+6\sqrt{7}\). Subtracting gives (-2).
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What is the value of (\(\sqrt{7}+\sqrt{5}\)2-\(\sqrt{7}-\sqrt{5}\)2)?
A. \(4\sqrt{35}\)
पहचान ((a+b)2-(a-b)2=4ab) लगाएं। यहाँ मान \(4\sqrt{35}\) है। / Use ((a+b)2-(a-b)2=4ab). Here the value is \(4\sqrt{35}\).
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If \(t=\sqrt{23}+4\), what is the value of \(t+\frac{7}{t}\)?
A. \(2\sqrt{23}\)
\(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4\) है क्योंकि हर (23-16=7) बनता है। इसलिए योग \(2\sqrt{23}\) है। / \(\frac{7}{\sqrt{23}+4}=\sqrt{23}-4\) because the denominator becomes (23-16=7). So the sum is \(2\sqrt{23}\).
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What is the value of \(\frac{\sqrt{98}-\sqrt{50}}{\sqrt{2}}\)?
B. (2)
\(\sqrt{98}=7\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए अंश \(2\sqrt{2}\) है। भाग देने पर (2) मिलता है। / \(\sqrt{98}=7\sqrt{2}\) and \(\sqrt{50}=5\sqrt{2}\), so the numerator is \(2\sqrt{2}\). Dividing gives (2).
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A. (16), \(2\sqrt{5}\)
(cd=21-5=16) और \(c-d=2\sqrt{5}\) है। संयुग्मी युग्म में दोनों मान अलग-अलग निकालें। / (cd=21-5=16) and \(c-d=2\sqrt{5}\). Find both values separately in a conjugate pair.
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What is the simplified form of \(\sqrt{539}-\sqrt{275}+\sqrt{99}\)?
B. \(7\sqrt{11}\)
\(\sqrt{539}=7\sqrt{11}\), \(\sqrt{275}=5\sqrt{11}\) 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। / \(\sqrt{539}=7\sqrt{11}\), \(\sqrt{275}=5\sqrt{11}\), and \(\sqrt{99}=3\sqrt{11}\). So the result is \(7\sqrt{11}-5\sqrt{11}+3\sqrt{11}=5\sqrt{11}\).
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What is the rationalised form of \(\frac{6}{\sqrt{19}-\sqrt{10}}\)?
C. (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9})
संयुग्मी से गुणा करने पर हर (19-10=9) बनता है। इसलिए रूप (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9}) है। / Multiplying by the conjugate makes the denominator (19-10=9). So the form is (\frac{6\(\sqrt{19}+\sqrt{10}\)}{9}).
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If \(x=\sqrt{17}+\sqrt{6}\), what is the value of \(x^2-23\)?
C. \(2\sqrt{102}\)
\(x^2=17+6+2\sqrt{102}=23+2\sqrt{102}\) है। इसलिए \(x^2-23=2\sqrt{102}\) है। / \(x^2=17+6+2\sqrt{102}=23+2\sqrt{102}\). So \(x^2-23=2\sqrt{102}\).
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What is the square of \(\sqrt{13+\sqrt{42}}\) equal to?
D. \(13+\sqrt{42}\)
वर्गमूल का वर्ग अंदर की संख्या देता है। इसलिए (\left\(\sqrt{13+\sqrt{42}}\right\)2=13+\sqrt{42}) है। / The square of a square root gives the number inside. So (\left\(\sqrt{13+\sqrt{42}}\right\)2=13+\sqrt{42}).
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B. (15)
क्षेत्रफल (\(\sqrt{22}+\sqrt{7}\)\(\sqrt{22}-\sqrt{7}\)=22-7=15) है। संयुग्मी आयामों से परिमेय क्षेत्रफल मिल सकता है। / Area is (\(\sqrt{22}+\sqrt{7}\)\(\sqrt{22}-\sqrt{7}\)=22-7=15). Conjugate dimensions can give a rational area.
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What is the value of (\(\sqrt{80}+\sqrt{45}\)\(\sqrt{80}-\sqrt{45}\))?
C. (35)
यह \(a^2-b^2\) रूप है। इसलिए मान (80-45=35) है। / This is the \(a^2-b^2\) form. So the value is (80-45=35).
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If \(y=\sqrt{27}+\sqrt{75}\), what is the value of \(\frac{y}{\sqrt{3}}\)?
B. (8)
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\), इसलिए \(y=8\sqrt{3}\) है। भाग देने पर (8) मिलता है। / \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so \(y=8\sqrt{3}\). Dividing gives (8).
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Which is the simplified form of \(\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}\)?
C. \(2-\sqrt{3}\)
संयुग्मी से गुणा करने पर (\frac{\(\sqrt{6}-\sqrt{2}\)2}{4}=2-\sqrt{3}) मिलता है। हर को परिमेय बनाएं। / Multiplying by the conjugate gives (\frac{\(\sqrt{6}-\sqrt{2}\)2}{4}=2-\sqrt{3}). Make the denominator rational.
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If \(m=\sqrt{180}-\sqrt{80}\) and \(n=2\sqrt{5}\), what is (m-n)?
A. (0)
\(\sqrt{180}=6\sqrt{5}\) और \(\sqrt{80}=4\sqrt{5}\), इसलिए \(m=2\sqrt{5}\) है। अतः (m-n=0) है। / \(\sqrt{180}=6\sqrt{5}\) and \(\sqrt{80}=4\sqrt{5}\), so \(m=2\sqrt{5}\). Hence (m-n=0).
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What is the value of (\sqrt{7}\left\(\sqrt{175}-\sqrt{63}\right\))?
B. (14)
\(\sqrt{175}=5\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए कोष्ठक \(2\sqrt{7}\) है। \(\sqrt{7}\) से गुणा करने पर (14) मिलता है। / \(\sqrt{175}=5\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the bracket is \(2\sqrt{7}\). Multiplying by \(\sqrt{7}\) gives (14).
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If \(z=\sqrt{23}-\sqrt{14}\), which is the value of \(z^2\)?
C. \(37-2\sqrt{322}\)
(\(\sqrt{23}-\sqrt{14}\)2=23+14-2\sqrt{322}) है। मध्य पद का चिन्ह ऋणात्मक रखें। / (\(\sqrt{23}-\sqrt{14}\)2=23+14-2\sqrt{322}). Keep the middle term negative.
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What is the rationalised form of \(\frac{3}{\sqrt{17}+\sqrt{8}}\)?
C. \(\frac{\sqrt{17}-\sqrt{8}}{3}\)
संयुग्मी से गुणा करने पर हर (17-8=9) बनता है। इसलिए रूप \(\frac{\sqrt{17}-\sqrt{8}}{3}\) है। / Multiplying by the conjugate makes the denominator (17-8=9). So the form is \(\frac{\sqrt{17}-\sqrt{8}}{3}\).
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If \(x=\sqrt{3}+\sqrt{12}\) and \(y=\sqrt{27}\), what is the value of (x-y)?
A. (0)
\(x=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\) और \(y=3\sqrt{3}\) है। इसलिए (x-y=0) है। / \(x=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\) and \(y=3\sqrt{3}\). Therefore (x-y=0).
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Which number definitely lies between \(\sqrt{18}\) and \(\sqrt{20}\)?
C. \(\sqrt{19}\)
क्योंकि (18<19<20), इसलिए \(\sqrt{19}\) दोनों के बीच होगा। वर्गमूलों में अंदर की संख्या से तुलना करें। / Since (18<19<20), \(\sqrt{19}\) lies between them. Compare square roots using the numbers inside.
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If \(u=\sqrt{28}+\sqrt{63}\) and \(v=5\sqrt{7}\), which statement is correct?
C. (u=v)
\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\), इसलिए \(u=5\sqrt{7}\) है। अतः (u=v) है। / \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so \(u=5\sqrt{7}\). Hence (u=v).
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If \(x=5+\sqrt{26}\), what is the value of \(x^2-10x\)?
A. (1)
\(x^2=51+10\sqrt{26}\) और \(10x=50+10\sqrt{26}\) है। घटाने पर (1) मिलता है। / \(x^2=51+10\sqrt{26}\) and \(10x=50+10\sqrt{26}\). Subtracting gives (1).
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What is the simplified form of \(\sqrt{432}-\sqrt{192}+\sqrt{48}\)?
A. \(8\sqrt{3}\)
\(\sqrt{432}=12\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\) है। इसलिए परिणाम \(8\sqrt{3}\) है। / \(\sqrt{432}=12\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\). Therefore the result is \(8\sqrt{3}\).
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What is the value of \(\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{7}-\sqrt{5}}\)?
A. \(\sqrt{7}\)
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। / After rationalising the terms become \(\frac{\sqrt{7}-\sqrt{5}}{2}\) and \(\frac{\sqrt{7}+\sqrt{5}}{2}\). The sum is \(\sqrt{7}\).
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If \(x=\sqrt{12+\sqrt{11}}\), what is the value of \(x^2-12\)?
A. \(\sqrt{11}\)
\(x^2=12+\sqrt{11}\) है। इसलिए \(x^2-12=\sqrt{11}\) होगा। / \(x^2=12+\sqrt{11}\). Therefore \(x^2-12=\sqrt{11}\).
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What is the value of \(\frac{\sqrt{200}+\sqrt{72}}{\sqrt{2}}\)?
C. (16)
\(\sqrt{200}=10\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\), इसलिए अंश \(16\sqrt{2}\) है। भाग देने पर (16) मिलता है। / \(\sqrt{200}=10\sqrt{2}\) and \(\sqrt{72}=6\sqrt{2}\), so the numerator is \(16\sqrt{2}\). Dividing gives (16).
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If \(r=\sqrt{13}+\sqrt{7}\) and \(s=\sqrt{13}-\sqrt{7}\), what is the value of \(r^2-s^2\)?
A. \(4\sqrt{91}\)
(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। / (r-2-s-2=(r-s)(r+s)) where \(r-s=2\sqrt{7}\) and \(r+s=2\sqrt{13}\). So the value is \(4\sqrt{91}\).
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Which is the rationalised form of \(\frac{8}{\sqrt{18}-\sqrt{10}}\)?
A. \(\sqrt{18}+\sqrt{10}\)
संयुग्मी से गुणा करने पर हर (18-10=8) बनता है। इसलिए उत्तर \(\sqrt{18}+\sqrt{10}\) है। / Multiplying by the conjugate makes the denominator (18-10=8). So the answer is \(\sqrt{18}+\sqrt{10}\).
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If \(P=\sqrt{363}+\sqrt{108}-\sqrt{192}\), what is (P) equal to?
B. \(9\sqrt{3}\)
\(\sqrt{363}=11\sqrt{3}\), \(\sqrt{108}=6\sqrt{3}\) और \(\sqrt{192}=8\sqrt{3}\) है। इसलिए \(P=9\sqrt{3}\) है। / \(\sqrt{363}=11\sqrt{3}\), \(\sqrt{108}=6\sqrt{3}\), and \(\sqrt{192}=8\sqrt{3}\). Therefore \(P=9\sqrt{3}\).
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What is the value of (\(\sqrt{44}+\sqrt{99}\)2)?
A. (275)
\(\sqrt{44}=2\sqrt{11}\) और \(\sqrt{99}=3\sqrt{11}\), इसलिए योग \(5\sqrt{11}\) है। इसका वर्ग (275) है। / \(\sqrt{44}=2\sqrt{11}\) and \(\sqrt{99}=3\sqrt{11}\), so the sum is \(5\sqrt{11}\). Its square is (275).
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If \(w=\sqrt{15}+\sqrt{11}\), what is the value of \(w^2-26\)?
B. \(2\sqrt{165}\)
\(w^2=15+11+2\sqrt{165}=26+2\sqrt{165}\) है। इसलिए \(w^2-26=2\sqrt{165}\) है। / \(w^2=15+11+2\sqrt{165}=26+2\sqrt{165}\). So \(w^2-26=2\sqrt{165}\).
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What is the value of (\(\sqrt{23}+\sqrt{7}\)\(\sqrt{23}-\sqrt{7}\))?
C. (16)
यह संयुग्मी गुणन है इसलिए मान (23-7=16) है। \(a^2-b^2\) रूप पहचानें। / This is conjugate multiplication so the value is (23-7=16). Identify the \(a^2-b^2\) form.
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What is the rationalised form of \(\frac{5}{\sqrt{29}+2}\)?
B. \(\frac{\sqrt{29}-2}{5}\)
संयुग्मी से गुणा करने पर हर (29-4=25) बनता है। इसलिए रूप \(\frac{\sqrt{29}-2}{5}\) है। / Multiplying by the conjugate makes the denominator (29-4=25). So the form is \(\frac{\sqrt{29}-2}{5}\).
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If \(x=\sqrt{24}+\sqrt{6}\), what is the value of \(x^2\)?
D. (54)
\(\sqrt{24}=2\sqrt{6}\) इसलिए \(x=3\sqrt{6}\) है। इसका वर्ग (54) है। / \(\sqrt{24}=2\sqrt{6}\) so \(x=3\sqrt{6}\). Its square is (54).
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What is the simplified form of \(\sqrt{242}-\sqrt{128}+\sqrt{72}\)?
C. \(9\sqrt{2}\)
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\) और \(\sqrt{72}=6\sqrt{2}\) है। इसलिए परिणाम \(9\sqrt{2}\) है। / \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). Therefore the result is \(9\sqrt{2}\).
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If \(q=\sqrt{10}+3\), what is the value of \(q^2-6q\)?
A. (1)
\(q^2=19+6\sqrt{10}\) और \(6q=18+6\sqrt{10}\) है। घटाने पर (1) मिलता है। / \(q^2=19+6\sqrt{10}\) and \(6q=18+6\sqrt{10}\). Subtracting gives (1).
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What is the value of \(\frac{\sqrt{27}+\sqrt{12}}{\sqrt{3}}\)?
B. (5)
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\) है। \(\sqrt{3}\) से भाग देने पर (5) मिलता है। / \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (5).
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If \(z=\sqrt{31}-\sqrt{22}\), which is the value of \(z^2\)?
D. \(53-2\sqrt{682}\)
(\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}) है। मध्य पद का चिन्ह ऋणात्मक रखें। / (\(\sqrt{31}-\sqrt{22}\)2=31+22-2\sqrt{682}). Keep the middle term negative.
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What is the value of (\(\sqrt{29}+5\)2-\(\sqrt{29}-5\)2)?
B. \(20\sqrt{29}\)
पहचान ((a+b)2-(a-b)2=4ab) लगाएं। यहाँ \(a=\sqrt{29}\) और (b=5), इसलिए मान \(20\sqrt{29}\) है। / Use ((a+b)2-(a-b)2=4ab). Here \(a=\sqrt{29}\) and (b=5), so the value is \(20\sqrt{29}\).
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If \(P=\sqrt{363}+\sqrt{147}-\sqrt{75}\), what is (P) equal to?
C. \(13\sqrt{3}\)
\(\sqrt{363}=11\sqrt{3}\), \(\sqrt{147}=7\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\) है। इसलिए \(P=13\sqrt{3}\) है। / \(\sqrt{363}=11\sqrt{3}\), \(\sqrt{147}=7\sqrt{3}\), and \(\sqrt{75}=5\sqrt{3}\). Therefore \(P=13\sqrt{3}\).
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What is the rationalised form of \(\frac{9}{\sqrt{26}+\sqrt{17}}\)?
A. \(\sqrt{26}-\sqrt{17}\)
संयुग्मी से गुणा करने पर हर (26-17=9) बनता है। इसलिए उत्तर \(\sqrt{26}-\sqrt{17}\) है। / Multiplying by the conjugate makes the denominator (26-17=9). So the answer is \(\sqrt{26}-\sqrt{17}\).
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