Level 15 • 50/50 questions • 25 seconds per question.
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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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Which is the simplified form of \(\frac{\sqrt{17}+\sqrt{5}}{\sqrt{17}-\sqrt{5}}\)?
A. \(\frac{11+\sqrt{85}}{6}\)
संयुग्मी से गुणा करने पर अंश \(22+2\sqrt{85}\) और हर (12) मिलता है। सरल रूप \(\frac{11+\sqrt{85}}{6}\) है। / Multiplying by the conjugate gives numerator \(22+2\sqrt{85}\) and denominator (12). The simplified form is \(\frac{11+\sqrt{85}}{6}\).
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If \(q=\sqrt{11}+3\), what is the value of \(q^2-6q\)?
A. (2)
\(q^2=20+6\sqrt{11}\) और \(6q=18+6\sqrt{11}\) है। घटाने पर (2) मिलता है। / \(q^2=20+6\sqrt{11}\) and \(6q=18+6\sqrt{11}\). Subtracting gives (2).
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What is the value of (\(\sqrt{11}+\sqrt{7}\)2-\(\sqrt{11}-\sqrt{7}\)2)?
A. \(4\sqrt{77}\)
पहचान ((a+b)2-(a-b)2=4ab) लगाएं। यहाँ मान \(4\sqrt{77}\) है। / Use ((a+b)2-(a-b)2=4ab). Here the value is \(4\sqrt{77}\).
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If \(t=\sqrt{31}+5\), what is the value of \(t+\frac{6}{t}\)?
A. \(2\sqrt{31}\)
\(\frac{6}{\sqrt{31}+5}=\sqrt{31}-5\) है क्योंकि हर (31-25=6) बनता है। इसलिए योग \(2\sqrt{31}\) है। / \(\frac{6}{\sqrt{31}+5}=\sqrt{31}-5\) because the denominator becomes (31-25=6). So the sum is \(2\sqrt{31}\).
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What is the value of \(\frac{\sqrt{300}-\sqrt{108}}{\sqrt{3}}\)?
B. (4)
\(\sqrt{300}=10\sqrt{3}\) और \(\sqrt{108}=6\sqrt{3}\), इसलिए अंश \(4\sqrt{3}\) है। भाग देने पर (4) मिलता है। / \(\sqrt{300}=10\sqrt{3}\) and \(\sqrt{108}=6\sqrt{3}\), so the numerator is \(4\sqrt{3}\). Dividing gives (4).
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A. (16), \(2\sqrt{10}\)
(cd=26-10=16) और \(c-d=2\sqrt{10}\) है। संयुग्मी युग्म में दोनों मान अलग-अलग निकालें। / (cd=26-10=16) and \(c-d=2\sqrt{10}\). Find both values separately in a conjugate pair.
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What is the simplified form of \(\sqrt{847}-\sqrt{363}+\sqrt{147}\)?
B. \(9\sqrt{7}\)
\(\sqrt{847}=11\sqrt{7}\), \(\sqrt{363}=7\sqrt{7}\) और \(\sqrt{147}=3\sqrt{7}\) है। इसलिए परिणाम \(7\sqrt{7}\) नहीं, \(11\sqrt{7}-7\sqrt{7}+3\sqrt{7}=7\sqrt{7}\) है। / \(\sqrt{847}=11\sqrt{7}\), \(\sqrt{363}=7\sqrt{7}\), and \(\sqrt{147}=3\sqrt{7}\). So the result is \(11\sqrt{7}-7\sqrt{7}+3\sqrt{7}=7\sqrt{7}\).
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What is the rationalised form of \(\frac{8}{\sqrt{27}-\sqrt{11}}\)?
C. (\frac{8\(\sqrt{27}+\sqrt{11}\)}{16})
संयुग्मी से गुणा करने पर हर (27-11=16) बनता है। इसलिए रूप (\frac{8\(\sqrt{27}+\sqrt{11}\)}{16}) है। / Multiplying by the conjugate makes the denominator (27-11=16). So the form is (\frac{8\(\sqrt{27}+\sqrt{11}\)}{16}).
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If \(x=\sqrt{21}+\sqrt{8}\), what is the value of \(x^2-29\)?
C. \(2\sqrt{168}\)
\(x^2=21+8+2\sqrt{168}=29+2\sqrt{168}\) है। इसलिए \(x^2-29=2\sqrt{168}\) है। / \(x^2=21+8+2\sqrt{168}=29+2\sqrt{168}\). So \(x^2-29=2\sqrt{168}\).
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What is the square of \(\sqrt{16+\sqrt{45}}\) equal to?
D. \(16+\sqrt{45}\)
वर्गमूल का वर्ग अंदर की संख्या देता है। इसलिए (\left\(\sqrt{16+\sqrt{45}}\right\)2=16+\sqrt{45}) है। / The square of a square root gives the number inside. So (\left\(\sqrt{16+\sqrt{45}}\right\)2=16+\sqrt{45}).
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B. (19)
क्षेत्रफल (\(\sqrt{30}+\sqrt{11}\)\(\sqrt{30}-\sqrt{11}\)=30-11=19) है। संयुग्मी आयामों से परिमेय क्षेत्रफल मिलता है। / Area is (\(\sqrt{30}+\sqrt{11}\)\(\sqrt{30}-\sqrt{11}\)=30-11=19). Conjugate dimensions give rational area.
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What is the value of (\(\sqrt{112}+\sqrt{63}\)\(\sqrt{112}-\sqrt{63}\))?
C. (49)
यह \(a^2-b^2\) रूप है। इसलिए मान (112-63=49) है। / This is the \(a^2-b^2\) form. So the value is (112-63=49).
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If \(y=\sqrt{75}+\sqrt{147}\), what is the value of \(\frac{y}{\sqrt{3}}\)?
B. (12)
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{147}=7\sqrt{3}\), इसलिए \(y=12\sqrt{3}\) है। भाग देने पर (12) मिलता है। / \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{147}=7\sqrt{3}\), so \(y=12\sqrt{3}\). Dividing gives (12).
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Which is the simplified form of \(\frac{\sqrt{10}-\sqrt{6}}{\sqrt{10}+\sqrt{6}}\)?
C. \(4-\sqrt{15}\)
संयुग्मी से गुणा करने पर (\frac{\(\sqrt{10}-\sqrt{6}\)2}{4}=4-\sqrt{15}) मिलता है। हर को परिमेय बनाएं। / Multiplying by the conjugate gives (\frac{\(\sqrt{10}-\sqrt{6}\)2}{4}=4-\sqrt{15}). Make the denominator rational.
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If \(m=\sqrt{245}-\sqrt{125}\) and \(n=2\sqrt{5}\), what is (m-n)?
A. (0)
\(\sqrt{245}=7\sqrt{5}\) और \(\sqrt{125}=5\sqrt{5}\), इसलिए \(m=2\sqrt{5}\) है। अतः (m-n=0) है। / \(\sqrt{245}=7\sqrt{5}\) and \(\sqrt{125}=5\sqrt{5}\), so \(m=2\sqrt{5}\). Hence (m-n=0).
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What is the value of (\sqrt{11}\left\(\sqrt{539}-\sqrt{275}\right\))?
B. (22)
\(\sqrt{539}=7\sqrt{11}\) और \(\sqrt{275}=5\sqrt{11}\), इसलिए कोष्ठक \(2\sqrt{11}\) है। \(\sqrt{11}\) से गुणा करने पर (22) मिलता है। / \(\sqrt{539}=7\sqrt{11}\) and \(\sqrt{275}=5\sqrt{11}\), so the bracket is \(2\sqrt{11}\). Multiplying by \(\sqrt{11}\) gives (22).
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If \(z=\sqrt{29}-\sqrt{18}\), which is the value of \(z^2\)?
C. \(47-2\sqrt{522}\)
(\(\sqrt{29}-\sqrt{18}\)2=29+18-2\sqrt{522}) है। मध्य पद का चिन्ह ऋणात्मक रखें। / (\(\sqrt{29}-\sqrt{18}\)2=29+18-2\sqrt{522}). Keep the middle term negative.
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What is the rationalised form of \(\frac{5}{\sqrt{26}+\sqrt{17}}\)?
C. (\frac{5\(\sqrt{26}-\sqrt{17}\)}{9})
संयुग्मी से गुणा करने पर हर (26-17=9) बनता है। इसलिए रूप (\frac{5\(\sqrt{26}-\sqrt{17}\)}{9}) है। / Multiplying by the conjugate makes the denominator (26-17=9). So the form is (\frac{5\(\sqrt{26}-\sqrt{17}\)}{9}).
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If \(x=\sqrt{5}+\sqrt{45}\) and \(y=\sqrt{80}\), what is the value of (x-y)?
A. (0)
\(x=\sqrt{5}+3\sqrt{5}=4\sqrt{5}\) और \(y=4\sqrt{5}\) है। इसलिए (x-y=0) है। / \(x=\sqrt{5}+3\sqrt{5}=4\sqrt{5}\) and \(y=4\sqrt{5}\). Therefore (x-y=0).
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Which number definitely lies between \(\sqrt{26}\) and \(\sqrt{30}\)?
C. \(\sqrt{28}\)
क्योंकि (26<28<30), इसलिए \(\sqrt{28}\) दोनों के बीच होगा। वर्गमूलों में अंदर की संख्या से तुलना करें। / Since (26<28<30), \(\sqrt{28}\) lies between them. Compare square roots using the numbers inside.
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If \(u=\sqrt{112}+\sqrt{175}\) and \(v=9\sqrt{7}\), which statement is correct?
C. (u=v)
\(\sqrt{112}=4\sqrt{7}\) और \(\sqrt{175}=5\sqrt{7}\), इसलिए \(u=9\sqrt{7}\) है। अतः (u=v) है। / \(\sqrt{112}=4\sqrt{7}\) and \(\sqrt{175}=5\sqrt{7}\), so \(u=9\sqrt{7}\). Hence (u=v).
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What is the value of (\(\sqrt{23}+5\)2-\(\sqrt{23}-5\)2)?
C. \(20\sqrt{23}\)
पहचान ((a+b)2-(a-b)2=4ab) लगाएं। यहाँ \(a=\sqrt{23}\) और (b=5), इसलिए मान \(20\sqrt{23}\) है। / Use ((a+b)2-(a-b)2=4ab). Here \(a=\sqrt{23}\) and (b=5), so the value is \(20\sqrt{23}\).
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If \(x=6+\sqrt{35}\), what is the value of \(x^2-12x\)?
A. (-1)
\(x^2=71+12\sqrt{35}\) और \(12x=72+12\sqrt{35}\) है। घटाने पर (-1) मिलता है। / \(x^2=71+12\sqrt{35}\) and \(12x=72+12\sqrt{35}\). Subtracting gives (-1).
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What is the simplified form of \(\sqrt{588}-\sqrt{300}+\sqrt{192}\)?
C. \(12\sqrt{3}\)
\(\sqrt{588}=14\sqrt{3}\), \(\sqrt{300}=10\sqrt{3}\) और \(\sqrt{192}=8\sqrt{3}\) है। इसलिए परिणाम \(12\sqrt{3}\) है। / \(\sqrt{588}=14\sqrt{3}\), \(\sqrt{300}=10\sqrt{3}\), and \(\sqrt{192}=8\sqrt{3}\). Therefore the result is \(12\sqrt{3}\).
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What is the value of \(\frac{1}{\sqrt{10}+\sqrt{6}}+\frac{1}{\sqrt{10}-\sqrt{6}}\)?
A. \(\frac{\sqrt{10}}{2}\)
दोनों पदों को जोड़ने पर अंश \(2\sqrt{10}\) और हर (10-6=4) मिलता है। इसलिए मान \(\frac{\sqrt{10}}{2}\) है। / Adding the two terms gives numerator \(2\sqrt{10}\) and denominator (10-6=4). So the value is \(\frac{\sqrt{10}}{2}\).
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If \(x=\sqrt{17+\sqrt{13}}\), what is the value of \(x^2-17\)?
A. \(\sqrt{13}\)
\(x^2=17+\sqrt{13}\) है। इसलिए \(x^2-17=\sqrt{13}\) होगा। / \(x^2=17+\sqrt{13}\). Therefore \(x^2-17=\sqrt{13}\).
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What is the value of \(\frac{\sqrt{392}+\sqrt{200}}{\sqrt{2}}\)?
C. (24)
\(\sqrt{392}=14\sqrt{2}\) और \(\sqrt{200}=10\sqrt{2}\), इसलिए अंश \(24\sqrt{2}\) है। भाग देने पर (24) मिलता है। / \(\sqrt{392}=14\sqrt{2}\) and \(\sqrt{200}=10\sqrt{2}\), so the numerator is \(24\sqrt{2}\). Dividing gives (24).
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If \(r=\sqrt{17}+\sqrt{11}\) and \(s=\sqrt{17}-\sqrt{11}\), what is the value of \(r^2-s^2\)?
A. \(4\sqrt{187}\)
(r-2-s-2=(r-s)(r+s)) है जहाँ \(r-s=2\sqrt{11}\) और \(r+s=2\sqrt{17}\) है। इसलिए मान \(4\sqrt{187}\) है। / (r-2-s-2=(r-s)(r+s)), where \(r-s=2\sqrt{11}\) and \(r+s=2\sqrt{17}\). So the value is \(4\sqrt{187}\).
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Which is the rationalised form of \(\frac{10}{\sqrt{35}-5}\)?
A. \(\sqrt{35}+5\)
संयुग्मी से गुणा करने पर हर (35-25=10) बनता है। इसलिए उत्तर \(\sqrt{35}+5\) है। / Multiplying by the conjugate makes the denominator (35-25=10). So the answer is \(\sqrt{35}+5\).
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If \(P=\sqrt{500}+\sqrt{180}-\sqrt{320}\), what is (P) equal to?
B. \(8\sqrt{5}\)
\(\sqrt{500}=10\sqrt{5}\), \(\sqrt{180}=6\sqrt{5}\) और \(\sqrt{320}=8\sqrt{5}\) है। इसलिए \(P=8\sqrt{5}\) है। / \(\sqrt{500}=10\sqrt{5}\), \(\sqrt{180}=6\sqrt{5}\), and \(\sqrt{320}=8\sqrt{5}\). Therefore \(P=8\sqrt{5}\).
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What is the value of (\(\sqrt{125}+\sqrt{320}\)2)?
A. (845)
\(\sqrt{125}=5\sqrt{5}\) और \(\sqrt{320}=8\sqrt{5}\), इसलिए योग \(13\sqrt{5}\) है। इसका वर्ग (845) है। / \(\sqrt{125}=5\sqrt{5}\) and \(\sqrt{320}=8\sqrt{5}\), so the sum is \(13\sqrt{5}\). Its square is (845).
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If \(w=\sqrt{18}+\sqrt{13}\), what is the value of \(w^2-31\)?
B. \(2\sqrt{234}\)
\(w^2=18+13+2\sqrt{234}=31+2\sqrt{234}\) है। इसलिए \(w^2-31=2\sqrt{234}\) है। / \(w^2=18+13+2\sqrt{234}=31+2\sqrt{234}\). So \(w^2-31=2\sqrt{234}\).
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If \(x=\sqrt{22}+\sqrt{10}\), what is the value of \(x^2-32\)?
A. \(2\sqrt{220}\)
\(x^2=22+10+2\sqrt{220}=32+2\sqrt{220}\) है। इसलिए \(x^2-32=2\sqrt{220}\) है। / \(x^2=22+10+2\sqrt{220}=32+2\sqrt{220}\). So \(x^2-32=2\sqrt{220}\).
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Which is the simplified form of \(\frac{\sqrt{31}+\sqrt{19}}{\sqrt{31}-\sqrt{19}}\)?
A. \(\frac{25+\sqrt{589}}{6}\)
संयुग्मी से गुणा करने पर अंश \(50+2\sqrt{589}\) और हर (12) मिलता है। सरल रूप \(\frac{25+\sqrt{589}}{6}\) है। / Multiplying by the conjugate gives numerator \(50+2\sqrt{589}\) and denominator (12). The simplified form is \(\frac{25+\sqrt{589}}{6}\).
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If \(a=8+\sqrt{7}\) and \(b=8-\sqrt{7}\), what is the value of \(a^2-b^2\)?
B. \(32\sqrt{7}\)
\(a-b=2\sqrt{7}\) और (a+b=16) है। इसलिए \(a^2-b^2=32\sqrt{7}\) है। / \(a-b=2\sqrt{7}\) and (a+b=16). Therefore \(a^2-b^2=32\sqrt{7}\).
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What is the simplified form of \(\sqrt{726}-\sqrt{486}+\sqrt{150}\)?
A. \(7\sqrt{6}\)
\(\sqrt{726}=11\sqrt{6}\), \(\sqrt{486}=9\sqrt{6}\) और \(\sqrt{150}=5\sqrt{6}\) है। इसलिए परिणाम \(7\sqrt{6}\) है। / \(\sqrt{726}=11\sqrt{6}\), \(\sqrt{486}=9\sqrt{6}\), and \(\sqrt{150}=5\sqrt{6}\). Therefore the result is \(7\sqrt{6}\).
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Which is the rationalised form of \(\frac{12}{\sqrt{43}-\sqrt{31}}\)?
C. \(\sqrt{43}+\sqrt{31}\)
संयुग्मी से गुणा करने पर हर (43-31=12) बनता है। इसलिए उत्तर \(\sqrt{43}+\sqrt{31}\) है। / Multiplying by the conjugate makes the denominator (43-31=12). So the answer is \(\sqrt{43}+\sqrt{31}\).
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If \(q=\sqrt{15}+4\), what is the value of \(q^2-8q\)?
A. (-1)
\(q^2=31+8\sqrt{15}\) और \(8q=32+8\sqrt{15}\) है। घटाने पर (-1) मिलता है। / \(q^2=31+8\sqrt{15}\) and \(8q=32+8\sqrt{15}\). Subtracting gives (-1).
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B. (24)
क्षेत्रफल (\(\sqrt{41}+\sqrt{17}\)\(\sqrt{41}-\sqrt{17}\)=41-17=24) है। संयुग्मी आयामों से परिमेय क्षेत्रफल मिलता है। / Area is (\(\sqrt{41}+\sqrt{17}\)\(\sqrt{41}-\sqrt{17}\)=41-17=24). Conjugate dimensions give rational area.
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In which option is the quotient of two irrational numbers rational?
A. \(\frac{\sqrt{75}}{\sqrt{3}}\)
\(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\) है जो परिमेय है। भाग के बाद पूर्ण वर्ग बनने की जाँच करें। / \(\frac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\), which is rational. Check whether division forms a perfect square.
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This level is designed for 50 active questions. Currently 50 questions are available for the selected class and difficulty.
Yes, the timer uses 25 seconds per question for Expert difficulty and shows the total remaining time on the page.
Yes, every question has its own SEO-friendly page with answer, explanation and related practice links.
Use Google login or mobile OTP. Admin can block users and control login providers.