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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Ex. 6Show that cos 1+sin1sin-11365 |
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Answer» cos¯¹ 12/13 + sin¯¹ 3/5 = sin¯¹ 56/65Taking LHS :sin¯¹ 5/13 + sin¯¹ 3/5 Using the formula, sin¯¹x + sin¯¹y = sin¯¹ ( x√1-y² + y√1-x² )= sin¯¹ ( 5/13 √1-9/25 + 3 √1-25/169)= sin¯¹ ( 5/13 × 4/5 + 3/5 × 12/13)= sin¯¹ (30 + 36 / 65)= sin¯¹ (56/ 65)= RHS Hence proved |
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| 2. |
Ex. 1. Evaluate : -30 !28 ! |
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| 3. |
tan A + tan B+tan C+tanD0100t cos (180 + B) |
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Answer» Since A,B,C and D are the angles of a cyclic quadrilateral, then∠A+∠C=180° and∠B+∠D=180°∴, a.tanA+tanB+tanC+tanD=tan(180°-C)+tan(180°-D)+tanC+tanD=tan{(90°×2)-C}+tan{(90°×2)-D}+tanC+tanD=-tanC-tanD+tanC+tanD=0 (Proved) |
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| 4. |
Example 8: In Fig. 6.38, the sides AB and AC ofAABC are produced to points E and D respectivelyIf bisectors BO and CO of CBE and BCDrespectively meet at point O, then prove thatSolution : Ray BO is the bisector of 2 CBETherefore, < CBO- < CBE- (180-y)= 90°-- (1)Similarly, ray CO is the bisector of ACD.Fig. 6.38Therefore, |
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Answer» its a solved example .. so what is your query say?? I cant understand the solved eg. so for understanding you need a video session .. for this you can join this page and post your query you will get explanatory video this page https://www.facebook.com/Class-6-To-10-Preparation-Cbse-2250343781851299/ Pls specify the answer |
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| 5. |
13.In fig 3, the sides AB and AC of A ABC areproduced to points E and D respectively.If bisectors BO and CO ofZCBE and ZBCD respectivelymeet at point O, then provethat ZBOC 90LBACE.2Fig-314. Factorise the following polynomial using factor theoremx3 13x2 +32x + 20 |
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Answer» 14. |
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| 6. |
3. The sides AB and AC of A ABC are produced to P and Q respectively. The bisectorsof exterior angles at B and C of Δ ABC meet at OProve that BOC = 90°-12 A.int n The bisectors of Z ABCand |
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| 7. |
If AD and BD are bisectors of angleCAB and angleCBA respectively, angleACB =90° find value of angleADB |
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| 8. |
1012262830 |
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Answer» The answer may be 24. 6-8 ==> 30-28 8-10 ==> 28-2610-12 ==> 26-(24)..So answer can be 24. |
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| 9. |
リFind the 101E term of AP. 5, 11,1 |
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| 10. |
1. Find the 25th term of the AP-5, -5/2, 0.5/2...werdenticuations by fartorization |
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| 11. |
(i) 5, 12, 19, 26, |
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Answer» This input string would be 5,12,19,26,33,40,47,54 as it appears to be an Arithmetic Sequence. a2-a1=12-5=7 a3-a2=19-12=7 a4-a3=26-19=7 a5-a4=33-26=7 a6-a5=40-33=7 a7-a6=47-40=7 a8-a7=54-47=7 The difference between every two adjacent members of the series is constant and equal to 7. General Form: an=a1+(n-1)d yes |
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| 12. |
Solve for k.k-24261213 |
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Answer» ∆EFD~∆PQRSo,DE/PQ =EF/QR=DF/PR26/13=24/12=6/k6/k=2k=3 |
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| 13. |
the bisectors ofC andD respectively. ProveIn a quadrilateral ABCD、CO. and DO. arethatCODA+B)13 |
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| 14. |
In ∆ABC, BD and CD are internal bisectors of angleB and angleC respectively.Prove that 180° +y=2x. |
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Answer» in triangle ABCy+<B+<C=180<B+<<C=180-yin triangle BCDx+<B/2 +<C/2 =180x+(<B+<C)/2=180x+(180-y)/2 =1802x+180-y=3602x=180+y |
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| 15. |
( 17 ^ { 2 } - 8 ^ { 2 } ) ^ { \frac { 1 } { 2 } } |
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Answer» (a^2-b^2)=(17^2-8^2)=(a-b)(a+b)=(17-8)(17+8)==(9)(25)=√9*25=3*5=15 |
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| 16. |
/In the quadrilateral ABCD, CO and DO are the bisectors of LC and LD respectively. Prove that |
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| 17. |
he sides AB and AC Of A ABC are produced to points E & D respectively. If bisectors BoZCBE and4 BCD respectively meet at point O then prove that LBoc-90- ZBAc.2 |
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| 18. |
\operatorname sec 41 ^ \circ \cdot \operatorname sin 49 ^ \circ %2B \operatorname cos 49 ^ \circ \cdot \operatorname cosec 41 ^ \circ - \frac 2 \sqrt 3 |
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Answer» sec 41= 1/cos 41now cos41= sin(90-41)= sin49now cosec 41= 1/sin41nowsin41= cos(90-41)= cos49 now tan 20= cot(90-20)=cot70now after putting all values1+1-(2/√3*√3)2-20thanks |
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| 19. |
Prove thatl. oos (49-A) cos (45°-B) ", sin (450-A) sin (45s-B) = sin (A + B). |
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| 20. |
26. Evaluatesec 41°, sin 49+os49 cosec 41an 20 an 60 tan 70-3(cos 45-sin goe |
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Answer» sec41°sin49°+cos29°cosec61°-2/√3(tan20°tan60°tan70°)]/[3(sin²31°+sin²59°)]=[sec41°sin(90°-41°)+cos29°cosec(90°-29°)-2/√3{tan20°×√3×tan(90°-20°)}]/[3{sin²31°+sin²(90°-31°)]=[sec41°cos41°+cos29°sec29°-2(tan20°cot20°)]/[3(sin²31°+cos²31°]=(1+1-2)/(3×1)=0 |
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| 21. |
4. The HCF of two number is 15 u .5. Find the least 5-digit number which is exactly divisible by 20, 25, 30.aleguind remainders |
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| 22. |
49. यदि 7 sin-03 cos20 = 4, तव tan 0 का मान ज्ञात कीजिए। |
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Answer» 7sina^2+3 cosa^2=7( 1-cosa^2)+3 cosa=7-7 cosa^2+3 cosa= 4cosa=7-4cosa^2; cosa^2=3/4; seca=1/ cosa=4/3, 1+ tana^2=4/3; tana^2=4/3-1=4-3/4=-1/3; tana=-1/V3 |
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| 23. |
1. Find the cube root of each of the following numbers by prime factorisation method(0) 64(D) 512Gin) 10648 (iv) 27000w 15625 (vi) 13824 (vii) 110592(viii) 46656( 175616 ( 91125 |
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Answer» 64= 2*2*2*2*2*2thanks 😊 10648= 2215625=25175616=56110592=48 10648= 2215625=25175616=56110592=48 10648=2215625=25175616=56110592=48 |
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| 24. |
Find the 10th term from the end of the AP5, 10, 15, ..485. |
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Answer» Like if you find it useful |
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| 25. |
Find the 10th term from the last term of the AP: 8, 10, 12,.., 26. |
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Answer» As a= 8d= 210th term = a+9d8+2(9)8+1826Answer |
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| 26. |
411. tan C=12. tan Ď = 4013,sin θ = |
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Answer» 11)tan C=1/3sin c=1/√10cos C=3/√10 |
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| 27. |
11 40(i) 172 |
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Answer» Sum of 11/17 + 40/17 = 3 |
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| 28. |
तो20. यदि +2+3+...........+10 =3025 हो,2-4-6-............-20 का मान ज्ञात कीजिए।(A) 6050(B) 9075(C) 12100(D) 2420) (EN OT |
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Answer» taking 2^3 common we get same expression as given so ans will be 2^3×30258×3025=24200 24200 is the correct answer. C) 12100 is correct answer |
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| 29. |
8000: Find the cube root of 13824 by prime factorisation method. |
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| 30. |
Which term of the AP: 13, 25, 37, 49, ……. Will be 372 more than its 26th term. [Ans: 57th ] |
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| 31. |
en to us(a)24 and 48 (b)20 and 30 (c) 14 and 56 (d) 16 and 64 (e) 15 and 45.jus take4. Find the greatest number which completely divides 144, 189 and 198.hat is the id5. Find the least 5 digit number which on dividing by 4, 12, 20 and 24 leaves a remainder tcolour it!3 in each case.ordine |
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Answer» 9 is the correct answer because h.c.f of 144,189and 198 is 9. |
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| 32. |
find the cube root of each of the following number by prime Factorization method =175616 |
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Answer» 175616 = 2×2×2×2×2×2×2×2×2×7×7×7 cube root of 175616 = 2×2×2×7 = 56 If you find this answer helpful then like it. |
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| 33. |
270(A) 0°(B) 450(C) 30s |
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Answer» hiii |
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| 34. |
Find the cube root of each of the following numbers by the prime factorisation method 64? |
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Answer» 64 = 2×2×2×2×2×2 Cuberoot(64) = Cuberoot(2×2×2×2×2×2) = 2×2 = 4 If you find this answer helpful then like it. |
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| 35. |
37. The 17th term of an A.P. is 5 more than twice its gth term. If the 11th term of the A.P. isCBSE 2012]43, find the nth term.hor of all three disit natural numbers which are divisible by 9,CBSE 2013] |
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| 36. |
8. For any positive integer n, prove that n3-n is divisible by 6. |
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| 37. |
{22÷2×11+3}-{40÷10+3-5 |
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Answer» (22/22+3)-(40/10+3-5)=(1+3)-(4+3-5)=4-(7-5)=4-2=2 |
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| 38. |
37. The sum of first m terms of an AP is (4m2-m). If its nth term is 107, findCBSE 201]the value of n. Also, find the 21st term of this AP. |
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| 39. |
Write the nth term of the AP: 5, 9, 13, 17 |
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Answer» AP is 5, 9, 13, 17..... a = 5 d = 9 - 5 = 4 a(n) = a + (n -1) d = 5 + ( n -1) 4 = 5 + 4n - 4 = 4n +1 |
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| 40. |
N|enतत्वNien |
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Answer» (2/3)*(2/3)*(2/3) = (2*2*2)/(3*3*3)= 8/27 Please hit the like button |
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| 41. |
If 10 times of 10th term is equal to 20 times of 20th term of an A. P. find its 30tterm. |
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Answer» Let x be first term and a be the common difference10th term = x+9a20th term=x+19a30th term=x+29agiven10*10th term=20*20th term10(x+9a)=20(x+19a)10x+90a=20x+380a20x-10x+380a-90a=010x+290a=010(x+29a)=0x+29a=030th term=0. |
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| 42. |
Subtract the following using column method:(i) -X° +7xy - 3y from 3x' + 8 xy - 2y |
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| 43. |
N, EN420 |
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Answer» Apologies we are currently taking questions in English language only |
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| 44. |
If 97+79 is divisible by 2". Then find the greatest value of n, where n EN |
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Answer» We have79+97=−(1−8)9+(1+8)779+97=−(1−8)9+(1+8)7 ⇒−(1−9C1.81+9C2.82−......−9C9.89)+(1+7C1.81+7C2.82+....+7C7.87)⇒−(1−9C1.81+9C2.82−......−9C9.89)+(1+7C1.81+7C2.82+....+7C7.87) ⇒16×8+64[−(9C2−........−9C9.87)(7C2+.....+7C7.85)]⇒16×8+64[−(9C2−........−9C9.87)(7C2+.....+7C7.85)] ⇒64⇒64(an integer) |
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| 45. |
Subtract by column method(a)3x + 2y from 2x + 3y |
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Answer» 2x + 3y - (3x + 2y)2x + 3y - 3x - 2y(2x - 3x) + (3y - 2y)y - x |
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| 46. |
Find themethod 2cubeof as bycolumn |
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Answer» Here is your answer =(25)^3=25x25x25=625x 25=15625. (25)³=25×25×25=15625 (25)³= 25×25×25 =625×25 =15625is the best answer 15625 is the write answer 15625. is the correct answer 25^3=25×25×25 625×25=15625 (25)*3=25×25×25=625×25=15625 |
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| 47. |
an A.P. a = 2, d = 8 if sum of n terms is 90en value of n will be: |
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Answer» This is the right one* |
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| 48. |
Find the sum of 17 terms of the AP, 5, 9, 13, 17, ..... |
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| 49. |
8for any positive integer nen les divisible by 6.prove that |
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Answer» take n =to any no. since take it 1so, n3-n=divisible by 6(1)3-1=factor of 6solve 1-1=0therefore n3-n=divisible by 6you can try this for several no. like karna mat bhulna |
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| 50. |
zeroIf 10 times of 10th term is equal to 20 times of 20th term of an A.P. find its 30s 37term. |
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