100 results found for "ordered_operations" in Class 10.
What situation can disordered lines show?
A. भीड़ या बेचैनीCrowd or restlessness
अव्यवस्थित रेखाएं बेचैनी या भीड़ दिखा सकती हैं। परीक्षा में chaotic lines का mood देखें। / Disordered lines can show restlessness or crowd. Exam tip: observe mood of chaotic lines.
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What is the most nuanced difference between peacekeeping and peace enforcement?
A. शांति स्थापना सामान्यतः सहमति और निगरानी पर आधारित होती है जबकि प्रवर्तन बलपूर्वक कदम भी हो सकता हैPeacekeeping is generally based on consent and monitoring while enforcement may include coercive action
शांति स्थापना में पक्षों की सहमति और निष्पक्षता महत्वपूर्ण होती है। परीक्षा में इन शब्दों को एक जैसा न मानें। / Consent and impartiality are important in peacekeeping. Exam tip: do not treat these terms as identical.
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A. क्योंकि वे व्यवहार में विकसित शांति प्रबंधन का साधन बनेBecause they developed in practice as a tool of peace management
शांति स्थापना संयुक्त राष्ट्र की व्यावहारिक भूमिका से विकसित हुई। परीक्षा में इसे चार्टर की शांति भावना से जोड़ें। / Peacekeeping developed from the practical role of the UN. Exam tip: connect it with the peace spirit of the Charter.
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Blue helmets in UN peace operations are associated with what?
A. संयुक्त राष्ट्र शांति सैनिकों सेUnited Nations peacekeepers
नीले हेलमेट संयुक्त राष्ट्र शांति सैनिकों की पहचान हैं। परीक्षा में इन्हें शांति स्थापना से जोड़ें। / Blue helmets identify UN peacekeepers. For exams connect them with peacekeeping.
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What is the main difference between UN peacekeeping operations and military conquest?
A. शांति स्थापना संघर्ष रोकने और स्थिरता में सहायता करती है विजय नहीं चाहतीPeacekeeping helps stop conflict and support stability not conquest
संयुक्त राष्ट्र शांति सैनिक शांति प्रक्रिया और नागरिक सुरक्षा में सहायता करते हैं। परीक्षा में उन्हें नीली टोपी से याद करें। / UN peacekeepers support peace processes and civilian protection. Exam tip: remember them by blue helmets.
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What was the relation of the American capture of Iwo Jima with air operations against Japan?
A. यह लंबी दूरी के बमवर्षकों के लिए सहायक और आपात अड्डा बनाIt became a support and emergency base for long-range bombers
इवो जीमा जापान के निकट वायु युद्ध में रणनीतिक अड्डा था। परीक्षा में द्वीपों का उपयोग सैन्य अड्डों के रूप में समझें। / Iwo Jima was a strategic base for air war near Japan. For exams understand islands as military bases.
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What limitation of airborne operations did Market Garden show?
A. यदि जमीनी सेना जल्दी न पहुंचे तो हवाई सैनिक अलग पड़ सकते हैंAirborne troops can become isolated if ground forces do not arrive quickly
मार्केट गार्डन में पुलों और संकरे मार्गों पर निर्भरता जोखिम बनी। परीक्षा में साहसिक योजना और जमीन की वास्तविकता दोनों पढ़ें। / In Market Garden dependence on bridges and narrow routes became risky. For exams study both bold plan and ground reality.
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What was the connection of capturing Iwo Jima with American air operations?
A. यह जापान के निकट आपात और सहायक वायु अड्डे के रूप में उपयोगी थाIt was useful as an emergency and support air base near Japan
इवो जीमा जापान के निकट वायु अभियानों में सामरिक महत्व रखता था। परीक्षा में द्वीपों को केवल भूमि नहीं बल्कि अड्डों के रूप में देखें। / Iwo Jima had strategic value for air operations near Japan. For exams see islands not only as land but as bases.
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Jallianwala Bagh memorial is linked with firing ordered by which British officer?
A. जनरल डायरGeneral Dyer
जलियांवाला बाग की गोलीबारी जनरल डायर के आदेश से हुई थी। परीक्षा में स्वतंत्रता आंदोलन की घटनाओं को स्थान और व्यक्ति से जोड़ें। / The firing at Jallianwala Bagh was ordered by General Dyer. For exams connect freedom movement events with place and person.
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Who ordered the firing at Jallianwala Bagh?
C. जनरल डायरGeneral Dyer
जलियांवाला बाग में गोली चलाने का आदेश जनरल डायर ने दिया था। इसे रॉलेट एक्ट विरोध की पृष्ठभूमि से जोड़ें। / General Dyer ordered the firing at Jallianwala Bagh. Link it with the background of protests against the Rowlatt Act.
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Who ordered firing at Jallianwala Bagh?
A. जनरल डायरGeneral Dyer
जनरल डायर ने जलियाँवाला बाग में गोली चलाने का आदेश दिया। परीक्षा में इस घटना को राष्ट्रीय आंदोलन के मोड़ से जोड़ें। / General Dyer ordered the firing at Jallianwala Bagh. Exam tip is to link this event with a turning point in the national movement.
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Who ordered firing in the Jallianwala Bagh massacre?
B. जनरल डायरGeneral Dyer
जनरल डायर ने जलियांवाला बाग में गोली चलाने का आदेश दिया। परीक्षा में घटना और जिम्मेदार अधिकारी याद रखें। / General Dyer ordered the firing at Jallianwala Bagh. Exam tip: remember the event and the responsible official.
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If (4x+7y=1) and (8x-7y=35), which ordered pair of (x) and (y) is correct?
A. \(x=3,\ y=-\frac{11}{7}\)
जोड़ने पर (12x=36), इसलिए (x=3)। फिर (4x+7y=1) से \(y=-\frac{11}{7}\) मिलता है। / Adding gives (12x=36), so (x=3). Then (4x+7y=1) gives \(y=-\frac{11}{7}\).
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Which ordered pair satisfies (8x-5y=7) and (3x+5y=48)?
A. \(x=5,\ y=\frac{33}{5}\)
दोनों समीकरण जोड़ने पर (11x=55), इसलिए (x=5)। दूसरे समीकरण से \(y=\frac{33}{5}\)। / Adding both equations gives (11x=55), so (x=5). The second equation gives \(y=\frac{33}{5}\).
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Which ordered pair satisfies (7x-3y=22) and (2x+3y=23)?
B. \(x=5,\ y=\frac{13}{3}\)
दोनों समीकरण जोड़ने पर (9x=45), इसलिए (x=5)। दूसरे समीकरण से \(y=\frac{13}{3}\) मिलता है। / Adding both equations gives (9x=45), so (x=5). The second equation gives \(y=\frac{13}{3}\).
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Which ordered pair satisfies (5x+2y=41) and (3x-y=10)?
D. \(x=\frac{61}{11},\ y=\frac{73}{11}\)
दूसरे समीकरण से (y=3x-10) रखें। पहले में रखने पर (11x=61), इसलिए \(x=\frac{61}{11},\ y=\frac{73}{11}\)। / Use (y=3x-10) from the second equation. Substitution gives (11x=61), so \(x=\frac{61}{11},\ y=\frac{73}{11}\).
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Which ordered pair satisfies (3x+2y=19) and (x-y=3)?
D. (x=5,\ y=2)
(x-y=3) से (x=y+3) मिलता है। पहले समीकरण में रखने पर (y=2) और (x=5)। / From (x-y=3), (x=y+3). Substitution in the first equation gives (y=2) and (x=5).
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From (x=2) and (y=x+5), what is ((x,y))?
D. ( (2,7) )
(y=2+5=7), इसलिए हल ( (2,7) ) है। युग्म लिखते समय (x) का मान पहले रखें। / (y=2+5=7), so the solution is ( (2,7) ). Write the value of (x) first in an ordered pair.
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If (x=2y) and (y=3), what is ((x,y))?
D. ( (6,3) )
(x=2(3)=6), इसलिए युग्म ( (6,3) ) है। क्रम (x) पहले और (y) बाद में रखें। / (x=2(3)=6), so the pair is ( (6,3) ). Keep the order as (x) first and (y) second.
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Solve by elimination: (x-y=3) and (x+y=11).
D. ( (7,4) )
दोनों समीकरण जोड़ने पर (2x=14), इसलिए (x=7) और (y=4)। विपरीत चिह्न वाले पद जल्दी कटते हैं। / Adding the equations gives (2x=14), so (x=7) and (y=4). Terms with opposite signs cancel quickly.
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Two lines meet at (A(-1,6)) on a graph. Which pair verifies this?
A. (2x+y=4), (x-y=-7)
((-1,6)) रखने पर (2x+y=4) और (x-y=-7) दोनों सही हैं। प्रतिच्छेद बिंदु दोनों रेखाओं पर होना चाहिए। / Substituting ((-1,6)) makes (2x+y=4) and (x-y=-7) both true. The intersection point must lie on both lines.
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Two lines meet at (A(2,-5)) on a graph. Which pair verifies this?
A. (3x+y=1), (x-y=7)
((2,-5)) रखने पर (3x+y=1) और (x-y=7) दोनों सही हैं। प्रतिच्छेद बिंदु दोनों रेखाओं पर होना चाहिए। / Substituting ((2,-5)) makes (3x+y=1) and (x-y=7) both true. The intersection point must lie on both lines.
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Two lines meet at (A(1,4)) on a graph. Which pair verifies this?
A. (3x+y=7), (x+2y=9)
((1,4)) रखने पर (3x+y=7) और (x+2y=9) दोनों सही हैं। प्रतिच्छेद बिंदु दोनों रेखाओं पर होना चाहिए। / Substituting ((1,4)) makes (3x+y=7) and (x+2y=9) true. The intersection point must lie on both lines.
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If two lines meet at ( (5,1) ), which pair of equations can represent this solution?
B. (x+y=6) और (x-y=4)(x+y=6) and (x-y=4)
( (5,1) ) पर (5+1=6) और (5-1=4)। हल जाँचने के लिए दिए बिंदु को दोनों समीकरणों में रखें। / At ( (5,1) ), (5+1=6) and (5-1=4). To check a solution substitute the point in both equations.
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For the line (5x+y=26), which point is obtained when (y=1)?
A. ( (5,1) )
(5x+1=26) से (5x=25) और (x=5)। इसलिए बिंदु ( (5,1) ) है। / From (5x+1=26), (5x=25) and (x=5). Therefore, the point is ( (5,1) ).
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For the line (x-y=8), which point is obtained when (y=3)?
B. ( (11,3) )
(x-3=8) से (x=11), इसलिए बिंदु ( (11,3) ) है। घटाव वाले समीकरणों में चिह्न ध्यान से देखें। / From (x-3=8), (x=11), so the point is ( (11,3) ). Watch signs carefully in subtraction equations.
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For (y=4x-2), which point is obtained when (x=3)?
C. ( (3,10) )
(y=4(3)-2=10), इसलिए बिंदु ( (3,10) ) है। गणना के बाद निर्देशांक क्रम न बदलें। / (y=4(3)-2=10), so the point is ( (3,10) ). Do not change the coordinate order after calculation.
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For the line (4x+y=17), which point is obtained when (y=1)?
A. ( (4,1) )
(4x+1=17) से (4x=16) और (x=4)। इसलिए बिंदु ( (4,1) ) है। / From (4x+1=17), (4x=16) and (x=4). Therefore, the point is ( (4,1) ).
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For the line (x-y=5), which point is obtained when (y=2)?
B. ( (7,2) )
(x-2=5) से (x=7), इसलिए बिंदु ( (7,2) ) है। घटाव वाले समीकरण में चिह्न ध्यान से देखें। / From (x-2=5), (x=7), so the point is ( (7,2) ). In subtraction equations, watch the signs carefully.
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For (x-y=3), which point is obtained when (y=1)?
A. ( (4,1) )
(x-1=3), इसलिए (x=4) और बिंदु ( (4,1) ) है। क्रम बदलने से उत्तर गलत हो सकता है। / (x-1=3), so (x=4) and the point is ( (4,1) ). Changing the order can make the answer wrong.
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For (x+y=8), which point is obtained when (x=5)?
A. ( (5,3) )
(5+y=8), इसलिए (y=3) और बिंदु ( (5,3) ) है। बिंदु हमेशा ( (x,y) ) क्रम में लिखें। / (5+y=8), so (y=3) and the point is ( (5,3) ). Always write a point in ( (x,y) ) order.
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Which is the solution of (x+y=10) and (x-y=4)?
A. ( (7,3) )
( (7,3) ) रखने पर (7+3=10) और (7-3=4)। ग्राफ पर यही दोनों रेखाओं का मिलन बिंदु होगा। / Substituting ( (7,3) ) gives (7+3=10) and (7-3=4). On the graph, this is where both lines meet.
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Which point is correct for the equation (x-y=1)?
A. ( (3,2) )
बिंदु ( (3,2) ) में (3-2=1), इसलिए यह रेखा पर है। सरल तरीका है कि हर विकल्प को समीकरण में रखकर जाँचें। / In ( (3,2) ), (3-2=1), so it lies on the line. The easy method is to test each option in the equation.
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On the number line, \( \frac{\sqrt{225}-6}{9} \) is equal to which point?
A. (1)
\( \sqrt{225}=15 \), इसलिए \( \frac{15-6}{9}=1 \)। संचालन का क्रम सही रखें। / \( \sqrt{225}=15 \), so \( \frac{15-6}{9}=1 \). Keep the order of operations correct.
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On the number line, \( \frac{\sqrt{121}-4}{7} \) is equal to which point?
A. (1)
\( \sqrt{121}=11 \), इसलिए \( \frac{11-4}{7}=1 \)। संचालन का क्रम सही रखें। / \( \sqrt{121}=11 \), so \( \frac{11-4}{7}=1 \). Keep the order of operations correct.
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On the number line, \( \frac{\sqrt{64}-3}{5} \) is equal to which point?
A. (1)
\( \sqrt{64}=8 \), इसलिए \( \frac{8-3}{5}=1 \)। संचालन का क्रम सही रखें। / \( \sqrt{64}=8 \), so \( \frac{8-3}{5}=1 \). Keep the order of operations correct.
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B. (1.75)
अंतिम बिंदु (0+2.25-0.5=1.75) है। दाईं ओर जोड़ें और बाईं ओर घटाएं। / The final point is (0+2.25-0.5=1.75). Add for right movement and subtract for left movement.
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B. (2)
अभिव्यक्ति \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\) है। इसलिए (c=1), (r=-8), (s=7), और (c+r+s=0) होता है। / The expression is \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\). Thus (c=1), (r=-8), (s=7), and (c+r+s=0).
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If \(\frac{1}{\sqrt{m}+\sqrt{n}}=\sqrt{m}-\sqrt{n}\) and (m>n>0), what is the value of (m-n)?
A. (1)
दोनों पक्षों को \(\sqrt{m}+\sqrt{n}\) से गुणा करने पर (1=m-n) मिलता है। परीक्षा में संयुग्म गुणनफल सीधे लगाएं। / Multiplying both sides by \(\sqrt{m}+\sqrt{n}\) gives (1=m-n). In exams, apply the conjugate product directly.
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What is the value of \(\frac{\sqrt{363}-2\sqrt{147}+3\sqrt{75}}{\sqrt{3}}\)?
C. (15)
\(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), और \(3\sqrt{75}=15\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) होना चाहिए। / Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
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If (\left\(7^{x}\right\)^{2}\cdot7^{x-1}=16807), what is the value of (x)?
A. (2)
बाएँ पक्ष \(7^{2x}\cdot7^{x-1}=7^{3x-1}\) है और \(16807=7^{5}\)। इसलिए (3x-1=5) और (x=2)। / The left side is \(7^{2x}\cdot7^{x-1}=7^{3x-1}\), and \(16807=7^{5}\). Hence (3x-1=5), so (x=2).
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What is the value of (\left\(\frac{5}{8}\right\)^{-2}+\left\(\frac{8}{5}\right\)^{-2})?
A. \(\frac{4721}{1600}\)
(\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) और (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64})। योग \(\frac{4096+625}{1600}=\frac{4721}{1600}\) है। / Here (\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) and (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64}). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).
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If \(\sqrt{x}=5\sqrt{2}\), what is the value of \(x^{\frac{3}{2}}\)?
A. \(250\sqrt{2}\)
\(\sqrt{x}=5\sqrt{2}\) से (x=50), और \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें। / From \(\sqrt{x}=5\sqrt{2}\), (x=50), and \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
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What is the simplified form of \(\frac{6b^{-3}+9b^{-3}}{3b^{-5}}\), where \(b\neq0\)?
A. \(5b^{2}\)
ऊपर \(6b^{-3}+9b^{-3}=15b^{-3}\) है। \(\frac{15b^{-3}}{3b^{-5}}=5b^{2}\) मिलता है। / The numerator is \(6b^{-3}+9b^{-3}=15b^{-3}\). Thus \(\frac{15b^{-3}}{3b^{-5}}=5b^{2}\).
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If \(x^{2}-\frac{1}{x^{2}}=60\) and \(x-\frac{1}{x}=6\), what is the value of \(x+\frac{1}{x}\)?
C. (10)
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)) है। इसलिए (60=6\left\(x+\frac{1}{x}\right\)) और मान (10) है। / We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (60=6\left\(x+\frac{1}{x}\right\)), so the value is (10).
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What is the simplified form of (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{-\frac{1}{3}})?
A. \(\frac{4x^{3}y^{4}}{5}\)
(\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{4x^{3}y^{4}}{5}\) मिलता है। / We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).
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If \(s=4+\sqrt{17}\), what is the value of \(s^{2}-\frac{1}{s^{2}}\)?
A. \(16\sqrt{17}\)
\(\frac{1}{s}=\sqrt{17}-4\), इसलिए \(s-\frac{1}{s}=8\) और \(s+\frac{1}{s}=2\sqrt{17}\)। अतः \(s^{2}-\frac{1}{s^{2}}=16\sqrt{17}\)। / Here \(\frac{1}{s}=\sqrt{17}-4\), so \(s-\frac{1}{s}=8\) and \(s+\frac{1}{s}=2\sqrt{17}\). Thus \(s^{2}-\frac{1}{s^{2}}=16\sqrt{17}\).
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What is the simplified form of \(\frac{24^{3}}{2^{6}\cdot3^{2}}\)?
B. (6)
(24^{3}=\(2^{3}\cdot3\)^{3}=2^{9}\cdot3^{3})। भाग देने पर \(2^{3}\cdot3=24\) मिलता है, इसलिए विकल्पों में सही मान नहीं है। / Since (24^{3}=\(2^{3}\cdot3\)^{3}=2^{9}\cdot3^{3}), division leaves \(2^{3}\cdot3=24\), so the correct value is not among the options.
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What is the value of (\left\(\sqrt{29}+\sqrt{20}\right\)\left\(\sqrt{29}-\sqrt{20}\right\)-3^{2})?
A. (0)
संयुग्म गुणनफल (29-20=9) है और \(3^{2}=9\)। इसलिए अंतर (0) है। / The conjugate product is (29-20=9), and \(3^{2}=9\). Hence the difference is (0).
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If \(16^{x}=1024\) and \(32^{y}=1024\), what is the value of (x+y)?
A. \(\frac{9}{2}\)
\(1024=2^{10}\), \(16^{x}=2^{4x}\) से \(x=\frac{5}{2}\), और \(32^{y}=2^{5y}\) से (y=2)। इसलिए योग \(\frac{9}{2}\) है। / Since \(1024=2^{10}\), \(16^{x}=2^{4x}\) gives \(x=\frac{5}{2}\), and \(32^{y}=2^{5y}\) gives (y=2). Hence the sum is \(\frac{9}{2}\).
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What is the simplified form of (\left\(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}\right\)^{-1})?
A. \(9r^{6}s^{-8}\)
अंदर \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\) है। (-1) घात लेने पर \(9r^{6}s^{-8}\) मिलता है। / Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).
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If \(x=\sqrt{11}-\sqrt{6}\), what is the value of \(x^{2}+2\sqrt{66}\)?
C. (17)
\(x^{2}=11+6-2\sqrt{66}=17-2\sqrt{66}\)। इसलिए \(x^{2}+2\sqrt{66}=17\)। / Since \(x^{2}=11+6-2\sqrt{66}=17-2\sqrt{66}\), \(x^{2}+2\sqrt{66}=17\).
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What is the value of \(\frac{5^{-2}+5^{-3}}{5^{-4}}\)?
A. (30)
\(5^{-2}+5^{-3}=\frac{1}{25}+\frac{1}{125}=\frac{6}{125}\) और \(5^{-4}=\frac{1}{625}\)। भाग देने पर (30) मिलता है। / Here \(5^{-2}+5^{-3}=\frac{1}{25}+\frac{1}{125}=\frac{6}{125}\), and \(5^{-4}=\frac{1}{625}\). Division gives (30).
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If \(p=8-\sqrt{63}\), what is the value of \(\frac{1}{p}-p\)?
A. \(2\sqrt{63}\)
\(\frac{1}{8-\sqrt{63}}=8+\sqrt{63}\), क्योंकि (64-63=1) है। इसलिए \(\frac{1}{p}-p=2\sqrt{63}\)। / Since \(\frac{1}{8-\sqrt{63}}=8+\sqrt{63}\), because (64-63=1). Therefore, \(\frac{1}{p}-p=2\sqrt{63}\).
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Which option is the simplified form of \(\frac{x^{12}-4096}{x^{6}-64}\), where \(x^{6}\neq64\)?
B. \(x^{6}+64\)
(x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\))। समान गुणनखंड कटने पर \(x^{6}+64\) मिलता है। / Since (x^{12}-4096=\(x^{6}\)^{2}-64^{2}=\(x^{6}-64\)\(x^{6}+64\)), cancelling the common factor gives \(x^{6}+64\).
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What is the simplified form of \(\sqrt[3]{343a^{15}b^{12}}\)?
A. \(7a^{5}b^{4}\)
\(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), और \(\sqrt[3]{b^{12}}=b^{4}\)। परीक्षा में घनमूल में घातों को (3) से भाग दें। / We have \(\sqrt[3]{343}=7\), \(\sqrt[3]{a^{15}}=a^{5}\), and \(\sqrt[3]{b^{12}}=b^{4}\). In exams, divide exponents by (3) under a cube root.
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If (\left\(x^{4}y^{-3}\right\)^{k}=x^{16}y^{-12}), what is the value of (k)?
C. (4)
बाएँ पक्ष में घातें (4k) और (-3k) हैं। (4k=16) और (-3k=-12) दोनों से (k=4) मिलता है। / The left side has exponents (4k) and (-3k). Both (4k=16) and (-3k=-12) give (k=4).
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What is the value of (\left\(125^{\frac{2}{3}}\right\)\cdot\left\(25^{-\frac{3}{2}}\right\))?
A. \(\frac{1}{5}\)
(125^{\frac{2}{3}}=(5)^{2}=25) और (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125})। गुणनफल \(\frac{1}{5}\) है। / Here (125^{\frac{2}{3}}=(5)^{2}=25) and (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}). The product is \(\frac{1}{5}\).
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A. \(\frac{x^{2}}{y^{2}z^{2}}\)
अंदर \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), इसलिए उल्टा \(x^{4}y^{-5}z^{-2}\) है। \(\frac{y^{3}}{x^{2}z^{4}}\) से गुणा करने पर \(\frac{x^{2}}{y^{2}z^{6}}\) मिलता है। / Inside, \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), so its reciprocal is \(x^{4}y^{-5}z^{-2}\). Multiplying by \(\frac{y^{3}}{x^{2}z^{4}}\) gives \(\frac{x^{2}}{y^{2}z^{6}}\).
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If \(y=7+4\sqrt{3}\), what is the value of \(y+\frac{1}{y}\)?
A. (14)
\(\frac{1}{7+4\sqrt{3}}=7-4\sqrt{3}\), क्योंकि (49-48=1) है। योग (14) मिलता है। / We have \(\frac{1}{7+4\sqrt{3}}=7-4\sqrt{3}\), because (49-48=1). The sum is (14).
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What is the value of \(\frac{\sqrt{300}+\sqrt{192}-\sqrt{108}}{\sqrt{3}}\)?
C. (12)
\(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), और \(\sqrt{108}=6\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) है। / Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
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If \(\frac{10^{k}\cdot100^{3}}{1000^{2}}=10^{5}\), what is the value of (k)?
C. (5)
\(100^{3}=10^{6}\) और \(1000^{2}=10^{6}\), इसलिए बाएँ पक्ष की घात (k+6-6=k) है। (k=5) मिलता है। / Since \(100^{3}=10^{6}\) and \(1000^{2}=10^{6}\), the exponent on the left is (k+6-6=k). Hence (k=5).
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What is the simplified form of (\frac{\(5x^{-2}\)^{2}\(2x^{4}\)^{2}}{20x^{4}})?
A. (5)
अंश \(25x^{-4}\cdot4x^{8}=100x^{4}\) है। \(\frac{100x^{4}}{20x^{4}}=5\) मिलता है। / The numerator is \(25x^{-4}\cdot4x^{8}=100x^{4}\). Thus \(\frac{100x^{4}}{20x^{4}}=5\).
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If \(3^{a}=81\) and \(9^{b}=729\), what is the value of \(a^{b}-b^{a}\)?
A. \(\frac{37}{8}\)
(a=4) और \(9^{b}=3^{2b}=3^{6}\) से (b=3) है। इसलिए \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), अतः विकल्पों में यह मान नहीं है। / We get (a=4), and \(9^{b}=3^{2b}=3^{6}\) gives (b=3). Thus \(a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17\), which is not among the options.
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Which option gives the correct expansion of (\(4\sqrt{3}-3\sqrt{5}\)^{2})?
A. \(93-24\sqrt{15}\)
(\(4\sqrt{3}\)^{2}=48), (\(3\sqrt{5}\)^{2}=45), और मध्य पद \(24\sqrt{15}\) है। इसलिए विस्तार \(93-24\sqrt{15}\) है। / Here (\(4\sqrt{3}\)^{2}=48), (\(3\sqrt{5}\)^{2}=45), and the middle term is \(24\sqrt{15}\). Therefore, the expansion is \(93-24\sqrt{15}\).
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What is the value of (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}})?
A. \(\frac{343}{125}\)
(\left\(\frac{25}{49}\right\)^{\frac{1}{2}}=\frac{5}{7}), इसलिए (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}}=\left\(\frac{5}{7}\right\)^{-3}=\frac{343}{125})। परीक्षा में पहले वर्गमूल निकालें। / Since (\left\(\frac{25}{49}\right\)^{\frac{1}{2}}=\frac{5}{7}), (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}}=\left\(\frac{5}{7}\right\)^{-3}=\frac{343}{125}). In exams, take the square root first.
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If \(x^{5}=3\), what is the value of \(x^{15}+x^{10}\)?
B. (36)
(x^{15}=\(x^{5}\)^{3}=27) और (x^{10}=\(x^{5}\)^{2}=9)। इसलिए योग (36) है। / Here (x^{15}=\(x^{5}\)^{3}=27) and (x^{10}=\(x^{5}\)^{2}=9). Therefore, the sum is (36).
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What is the value of \(\frac{1}{\sqrt{26}-5}+\frac{1}{\sqrt{26}+5}\)?
A. \(2\sqrt{26}\)
हरों का गुणनफल (26-25=1) है और अंश (\(\sqrt{26}+5\)+\(\sqrt{26}-5\)=2\sqrt{26}) है। परीक्षा में संयुग्म भिन्नों को साथ जोड़ें। / The product of denominators is (26-25=1), and the numerator is (\(\sqrt{26}+5\)+\(\sqrt{26}-5\)=2\sqrt{26}). In exams, add conjugate fractions together.
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What is the simplified form of \(\frac{x^{10}-1024}{x^{5}-32}\), where \(x^{5}\neq32\)?
B. \(x^{5}+32\)
(x^{10}-1024=\(x^{5}\)^{2}-32^{2}=\(x^{5}-32\)\(x^{5}+32\))। समान गुणनखंड कटने पर \(x^{5}+32\) बचता है। / We use (x^{10}-1024=\(x^{5}\)^{2}-32^{2}=\(x^{5}-32\)\(x^{5}+32\)). Cancelling the common factor leaves \(x^{5}+32\).
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If \(r=\sqrt{21}+\sqrt{14}\), what is the value of \(r^{2}-14\sqrt{6}\)?
C. (35)
\(r^{2}=21+14+2\sqrt{294}=35+14\sqrt{6}\)। इसलिए \(r^{2}-14\sqrt{6}=35\)। / Since \(r^{2}=21+14+2\sqrt{294}=35+14\sqrt{6}\), \(r^{2}-14\sqrt{6}=35\).
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What is the value of (\left\(25^{\frac{3}{2}}\right\)\cdot\left\(125^{-\frac{2}{3}}\right\))?
B. (5)
(25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) और (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2})। गुणनफल (5) है। / Here (25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) and (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2}). The product is (5).
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If \(4^{x}+4^{x+1}+4^{x+2}=336\), what is the value of (x)?
B. (2)
सामान्य पद \(4^{x}\) लेने पर (4^{x}(1+4+16)=336) मिलता है। इसलिए \(21\cdot4^{x}=336\), \(4^{x}=16\), और (x=2)। / Factoring \(4^{x}\), we get (4^{x}(1+4+16)=336). Thus \(21\cdot4^{x}=336\), \(4^{x}=16\), and (x=2).
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A. (1)
अंदर \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), इसका वर्ग \(16x^{-16}y^{12}\) है। फिर \(\frac{x^{16}}{16y^{12}}\) से गुणा करने पर (1) मिलता है। / Inside, \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), and its square is \(16x^{-16}y^{12}\). Multiplying by \(\frac{x^{16}}{16y^{12}}\) gives (1).
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If \(A=19+6\sqrt{10}\), what is the simplified form of \(\sqrt{A}\)?
A. \(3+\sqrt{10}\)
क्योंकि (\(3+\sqrt{10}\)^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), इसलिए \(\sqrt{A}=3+\sqrt{10}\)। परीक्षा में पूर्ण वर्ग करणी पहचानें। / Because (\(3+\sqrt{10}\)^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), \(\sqrt{A}=3+\sqrt{10}\). In exams, identify perfect-square surd forms.
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A. \(\frac{x^{2}+y^{2}}{x^{2}y^{2}}\)
मान लें \(A=x^{-2}\) और \(B=y^{-2}\), तो \(\frac{A^{2}-B^{2}}{A-B}=A+B\)। इसलिए उत्तर \(x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}\) है। / Let \(A=x^{-2}\) and \(B=y^{-2}\). Then \(\frac{A^{2}-B^{2}}{A-B}=A+B\), so the answer is \(x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}\).
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If \(3^{x}\cdot27^{x-1}=243\), what is the value of (x)?
B. (2)
\(27^{x-1}=3^{3x-3}\), इसलिए कुल घात (x+3x-3=4x-3) है। \(243=3^{5}\), इसलिए (4x-3=5) और (x=2)। / Since \(27^{x-1}=3^{3x-3}\), the total exponent is (x+3x-3=4x-3). Since \(243=3^{5}\), (4x-3=5), so (x=2).
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What is the simplified form of \(\sqrt{242}-\sqrt{128}+\sqrt{98}-\sqrt{72}\)?
C. \(4\sqrt{2}\)
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), और \(\sqrt{72}=6\sqrt{2}\)। कुल \(4\sqrt{2}\) मिलता है। / We have \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). The total is \(4\sqrt{2}\).
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What is the simplified value of \(\frac{13^{4}\cdot169^{-1}}{2197^{-1}}\)?
B. \(13^{5}\)
\(169^{-1}=13^{-2}\) और \(2197^{-1}=13^{-3}\), इसलिए \(\frac{13^{4}\cdot13^{-2}}{13^{-3}}=13^{5}\)। परीक्षा में ऋणात्मक घात से भाग करते समय घात जुड़ती है। / Here \(169^{-1}=13^{-2}\) and \(2197^{-1}=13^{-3}\), so \(\frac{13^{4}\cdot13^{-2}}{13^{-3}}=13^{5}\). In exams, division by a negative power adds the exponent.
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If \(x+\frac{1}{x}=7\), what is the value of \(x^{2}+\frac{1}{x^{2}}\)?
B. (47)
(\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2) होता है। इसलिए \(49=x^{2}+\frac{1}{x^{2}}+2\) और मान (47) है। / We use (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).
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What is the value of (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}})?
A. \(\frac{64}{27}\)
(\left\(\frac{81}{256}\right\)^{\frac{1}{4}}=\frac{3}{4}), इसलिए (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27})। परीक्षा में पहले चौथा मूल निकालें। / Since (\left\(\frac{81}{256}\right\)^{\frac{1}{4}}=\frac{3}{4}), (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27}). In exams, take the fourth root first.
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What is the simplified form of (\left\(\frac{p^{-5}q^{4}}{p^{-1}q^{-2}}\right\)^{-2})?
A. \(p^{8}q^{-12}\)
अंदर (p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^{6}) है। (-2) घात देने पर \(p^{8}q^{-12}\) मिलता है। / Inside, (p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^{6}). Raising to (-2) gives \(p^{8}q^{-12}\).
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If \(6^{x+1}-6^{x}=900\), what is the value of (x)?
B. (3)
\(6^{x+1}-6^{x}=6\cdot6^{x}-6^{x}=5\cdot6^{x}=900\), इसलिए \(6^{x}=180\) नहीं बनता। इसलिए दिए विकल्पों में कोई भी सही नहीं है। / Here \(6^{x+1}-6^{x}=5\cdot6^{x}=900\), so \(6^{x}=180\), which is not a listed integral power. Therefore none of the listed options is correct.
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What is the value of \(\frac{1}{5-\sqrt{24}}-\frac{1}{5+\sqrt{24}}\)?
A. \(2\sqrt{24}\)
हरों का गुणनफल (25-24=1) है और अंश \(2\sqrt{24}\) बनता है। परीक्षा में संयुग्म हरों का गुणनफल पहले निकालें। / The product of the denominators is (25-24=1), and the numerator becomes \(2\sqrt{24}\). In exams, first find the product of conjugate denominators.
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C. \(\frac{8\sqrt{34}}{9}\)
(u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}) और (uv=9) है। इसलिए मान \(\frac{8\sqrt{34}}{9}\) है। / Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}), and (uv=9). Hence the value is \(\frac{8\sqrt{34}}{9}\).
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What is the simplified value of \(\frac{5^{9}\cdot25^{-2}\cdot125}{5^{4}}\)?
C. \(5^{4}\)
\(25^{-2}=5^{-4}\) और \(125=5^{3}\), इसलिए कुल घात (9-4+3-4=4) है। परीक्षा में सभी पदों को समान आधार में बदलें। / Since \(25^{-2}=5^{-4}\) and \(125=5^{3}\), the total exponent is (9-4+3-4=4). In exams, convert all terms to the same base.
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If \(a\neq0\) and \(\frac{a^{3p-2}\cdot a^{p+5}}{a^{2p-1}}=a^{10}\), what is the value of (p)?
B. (3)
कुल घात ((3p-2)+(p+5)-(2p-1)=2p+4) है। (2p+4=10) से (p=3) मिलता है। / The total exponent is ((3p-2)+(p+5)-(2p-1)=2p+4). From (2p+4=10), we get (p=3).
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A. \(\frac{x}{4}\)
\(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), इसलिए व्युत्क्रम \(\frac{x^{5}}{4}\) है और \(x^{-4}\) से गुणा करने पर \(\frac{x}{4}\) मिलता है। परीक्षा में पहले कोष्ठक को सरल करें। / Here \(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), so its reciprocal is \(\frac{x^{5}}{4}\), and multiplying by \(x^{-4}\) gives \(\frac{x}{4}\). In exams, simplify the bracket first.
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If \(x=\sqrt{2}+\sqrt{5}\), what is the value of \(x^{3}-7x\)?
A. \(10\sqrt{2}+4\sqrt{5}\)
\(x^{2}=7+2\sqrt{10}\), इसलिए \(x^{3}=17\sqrt{2}+11\sqrt{5}\) और \(x^{3}-7x=10\sqrt{2}+4\sqrt{5}\)। परीक्षा में पहले \(x^{2}\) निकालकर फिर (x) से गुणा करें। / Here \(x^{2}=7+2\sqrt{10}\), so \(x^{3}=17\sqrt{2}+11\sqrt{5}\) and \(x^{3}-7x=10\sqrt{2}+4\sqrt{5}\). In exams, first find \(x^{2}\) and then multiply by (x).
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A. \(\frac{17}{4}\)
अभिव्यक्ति \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=;2x^{-5}y^{7}\) है। इसलिए (c+r+s=2-5+7=4) है। / The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence (c+r+s=2-5+7=4).
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What is the value of \(\frac{\sqrt{192}-2\sqrt{48}+3\sqrt{12}}{\sqrt{3}}\)?
C. (12)
\(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), और \(3\sqrt{12}=6\sqrt{3}\)। अंश \(6\sqrt{3}\) है, इसलिए मान (6) है। / Here \(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), and \(3\sqrt{12}=6\sqrt{3}\). The numerator is \(6\sqrt{3}\), so the value is (6).
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If (\left\(5^{x}\right\)^{2}\cdot5^{x-2}=3125), what is the value of (x)?
B. \(\frac{7}{3}\)
बाएँ पक्ष \(5^{2x}\cdot5^{x-2}=5^{3x-2}\) है और \(3125=5^{5}\)। इसलिए (3x-2=5) और \(x=\frac{7}{3}\)। / The left side is \(5^{2x}\cdot5^{x-2}=5^{3x-2}\), and \(3125=5^{5}\). Hence (3x-2=5), so \(x=\frac{7}{3}\).
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What is the value of (\left\(\frac{4}{7}\right\)^{-2}+\left\(\frac{7}{4}\right\)^{-2})?
A. \(\frac{2657}{784}\)
(\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) और (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49})। योग \(\frac{2401+256}{784}=\frac{2657}{784}\) है। / Here (\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) and (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49}). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).
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If \(\sqrt{x}=4\sqrt{3}\), what is the value of \(x^{\frac{3}{2}}\)?
A. \(192\sqrt{3}\)
\(\sqrt{x}=4\sqrt{3}\) से (x=48), और \(x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें। / From \(\sqrt{x}=4\sqrt{3}\), (x=48), and \(x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
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What is the simplified form of \(\frac{4b^{-2}+6b^{-2}}{5b^{-3}}\), where \(b\neq0\)?
A. (2b)
ऊपर \(4b^{-2}+6b^{-2}=10b^{-2}\) है। \(\frac{10b^{-2}}{5b^{-3}}=2b\) मिलता है। / The numerator is \(4b^{-2}+6b^{-2}=10b^{-2}\). Thus \(\frac{10b^{-2}}{5b^{-3}}=2b\).
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If \(x^{2}-\frac{1}{x^{2}}=40\) and \(x-\frac{1}{x}=5\), what is the value of \(x+\frac{1}{x}\)?
C. (8)
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)) है। इसलिए (40=5\left\(x+\frac{1}{x}\right\)), और मान (8) है। / We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (40=5\left\(x+\frac{1}{x}\right\)), so the value is (8).
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What is the simplified form of (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{-\frac{1}{3}})?
A. \(\frac{3x^{2}y^{3}}{4}\)
(\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{3x^{2}y^{3}}{4}\) मिलता है। / We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).
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If \(s=3+\sqrt{10}\), what is the value of \(s^{2}-\frac{1}{s^{2}}\)?
A. \(12\sqrt{10}\)
\(\frac{1}{s}=\sqrt{10}-3\), इसलिए \(s-\frac{1}{s}=6\) और \(s+\frac{1}{s}=2\sqrt{10}\)। अतः \(s^{2}-\frac{1}{s^{2}}=12\sqrt{10}\)। / Here \(\frac{1}{s}=\sqrt{10}-3\), so \(s-\frac{1}{s}=6\) and \(s+\frac{1}{s}=2\sqrt{10}\). Thus \(s^{2}-\frac{1}{s^{2}}=12\sqrt{10}\).
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What is the simplified form of \(\frac{18^{3}}{2^{2}\cdot3^{5}}\)?
B. (6)
(18^{3}=\(2\cdot3^{2}\)^{3}=2^{3}\cdot3^{6})। भाग देने पर \(2^{1}\cdot3^{1}=6\) मिलता है। / Since (18^{3}=\(2\cdot3^{2}\)^{3}=2^{3}\cdot3^{6}), division leaves \(2^{1}\cdot3^{1}=6\).
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What is the value of (\left\(\sqrt{17}+\sqrt{8}\right\)\left\(\sqrt{17}-\sqrt{8}\right\)-\sqrt{81})?
A. (0)
संयुग्म गुणनफल (17-8=9) है और \(\sqrt{81}=9\)। इसलिए अंतर (0) है। / The conjugate product is (17-8=9), and \(\sqrt{81}=9\). Hence the difference is (0).
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If \(9^{x}=729\) and \(27^{y}=729\), what is the value of (x+y)?
A. \(\frac{7}{2}\)
\(729=3^{6}\), \(9^{x}=3^{2x}\) से (x=3), और \(27^{y}=3^{3y}\) से (y=2)। इसलिए (x+y=5)। / Since \(729=3^{6}\), \(9^{x}=3^{2x}\) gives (x=3), and \(27^{y}=3^{3y}\) gives (y=2). Therefore, (x+y=5).
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