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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. |
ABCD is a square. E, F, Gand H are the mid points of AB, BC, CD and DA respectiveluSuch that AE BF CG DH. Prove that EFGH is a square.9. |
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| 2. |
9. ABCD is a square. E, E, G and H are the mid points of AB, BC, CD and DA respectively.Such that AE BF CG DH. Prove that EFGH is a square |
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| 3. |
ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DArespectively. Such that AE BF CG -DH. Prove that EFGH is a square. |
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| 4. |
In countries like USA and Canada, temperature is measuredin fahrenheit. Whereas in countries like India, it is measuredin celsius. Here is a linear equation that converts fahrenheitto celsius:F=\left(\frac{9}{5}\right) C+32 |
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| 5. |
tana-tan-* दूसरे चतुर्थाश में स्थित है।में स्थित है। |
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| 6. |
If tanA=1/35, tan B=18/17, then tan (A-B) = |
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Answer» tan(A-B) = (tanA-tanB)/(1+tanAtanB) = (1/35-18/17)/(1+1/35*18/17) = -1 so, tan(A-B) = -1 |
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| 7. |
23A pasta recipe requires 2 kg cheese.Approximately, how much pasta can bemade from a 21 kg cheese? |
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| 8. |
YIS T "6рео179) i(= 0um g Itu_ - |
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Answer» Please hit the like button if this helped you. |
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| 9. |
Hint : Simplify LHS and RHS separately\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1 \- \tan A}{1-\cot A^}\right)^{2}=\tan ^{2} A |
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| 10. |
L.H.S.=RHS. कीजिएहरण 4. यदि (1+ (80 4)(1+1आ कै) =2 हो, तो (2+ 2) का मान ज्ञात जिए। |
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| 11. |
4 sec” 30१९ - 121” 45% |
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Answer» hit like if you find it useful |
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| 12. |
Value of sin(90-e) isA) cos(90-e) B) coteC) cosec θ D) co$8Lomentorst events of a random experiment |
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Answer» d oppoatin is the correct sin(90° - θ) = cosθOption (D) is correct |
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| 13. |
e deam466osu3bO CO |
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Answer» Median=l +[{(N/2)-cf}/f]×h =20 + {(50-24)/36}×10 =20 + 260/36 =27.222 |
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| 14. |
f 15 oranges cost 70, what is the cost of 39 oranges? |
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Answer» 15 oranges cost 70 rupees39 oranges cost 182 rupees |
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| 15. |
and H ae the midDA e pectively. Seuch thatposrnh 98,6c, co and bo vepectfvely . Bach thabus a |
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| 16. |
[Hint: Simplify LHS and RHS separately]1 tan A、24) /x) |
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Answer» Like if you find it useful |
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| 17. |
-13. Verify that yis a solution of the differential equation |
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Answer» wrong.... |
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| 18. |
नमस्कार मीT TR S T RY TM |
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| 19. |
x)Hint: Simplify LHS and RHS separately]1 cot A |
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Answer» Like my answer if you find it useful!. |
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| 20. |
Hint : Simplify LHS and RHS separate1 tan'1-tan Atan A |
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Answer» L.H.S1+tan^2A/1+cot^2A=(1+sin^2A/cos2A)/(1+cos^2A/sin^2A)=((cos^2A+sin^2A)/cos^2A)/(sin^2A+cos^2A)/sin^2A=(1/cos^2A)/(1/sin^2A)=1/cos^2A*sin^2A/1=sin^2A/cos^2A=tan^2A M.H.S(1-tanA/1-cotA)^2=(1+tan^2A-2tanA)/(1+cot^2-2cotA)=(sec^2A-2tanA)/(cosec^2A-2cosA/sinA)=(sec^2A-2*sinA/cos A)/(cosec^2A-2cosA/sinA)=(1/cos^2A-2sinA/cosA)/(1/sin^2A-2cosA/sinA)=[(1-2sinAcosA)/cos^2A]/[(1-2cosAsinA)sin^2A]=(1-2sinAcosA)/cos^2A*sin^2A/(1-2sinAcosA)=sin^2A/cos^2A=tan^2A R.H.S=tan^2A Hence L.H.S=M.H.S=R.H.S |
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| 21. |
Hint: Simplify LHS and RHS separately]00 (1+tana). (1-tana tan' A |
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| 22. |
(6 From tre top of a jild 1o wm\}tm %%A/WJ»M, of o oriatickm Sterey, e cant-ord foune| to be 20° and 45 fund [‘H;!Jull 5 |
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| 23. |
ix) (\csc A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}Hint: Simplify LHS and RHS separately] |
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| 24. |
1 + sec A sin Asec A C [Hint: Simplify LHS and RHS separa1 - cos At A using the identity cose |
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Answer» (1+secA)/secA = sin²A/(1-cosA)solving LHS(1+secA)/secA=1/secA + 1=cosA+1 soling RHSsin²A/(1-cosA)=(1-cos²A)/(1-cosA)=1+cosA LHS=RHS |
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| 25. |
HOMEWORK 2ndSTD. 5th.Q. 1) Find the value of:a)_2+3+ 16 4 5b2% + 1%c)2-d) _8_33x 11e) 3%+10+649+3Q. 2) Do as directed:a) Solve using BODMAS :i) 19 -- {3 x 2 + 10 + 5 ) = ?ii) ( 84 + (16 x 4) +4 ] -- 19 = ?b) If there are 5000 mangoes in 100 boxes. How many mangoes will be there in 75 boxes?=c) solve :25 x45d) Find the reduced form of1545274566121RAoge) Out of 2000 students, 1225 are boys and remaining are girls. How many more boys are therethan girls ?f) Find the sum of five digit greatest number and 3 digit smallest number?ii) 45 ) 139 iv) 228g) Convert into mixed fraction: i) 169_1117h) What is the sum of prime numbers from 50 to 60? |
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Answer» Q.1 a. 1 17/60 b. 4 13/24 c. 7/18 d. 1/24 e. 13 4/5 or 19/50 f. 1 1/2Q.2 a. 1. 11 2. 89 b. 3,75,000 mangoes c. 1 d. 6/11 e. 775 girls and 450 more boys than girls f. 1,00,009 g. a. 9 16/17 b. 11 1/4 c. 12 7/11 d. 9 21/23 h. 51, 53, 59. 51+53+59= 163 |
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| 26. |
Column - I1-sin-45°-sin2 30° + cos 30°Column -IIa) cos 60°b) 11-cos 45°1- cos2 90°c)1-sin 45d) sin 90°cos 45 |
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| 27. |
more than s 1ith tmand x-1 nu.or |
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| 28. |
ta |खिल ४ + .__-————COtA — 1+ secAcosecC Ae gotA 1- tan Ai |
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Answer» Like my answer if you find it useful! |
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| 29. |
cotA+cosecA -1 1+cos Aसिद्ध करा की, cot A — cosecA + 1 e sinAकि कक |
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Answer» We know that:- 1=cosec^2A-cot^2A 1 =(cosecA+cotA) (cosecA-cotA)……….(1) Now L.H.S. =(cotA+cosecA-1)/(cotA-cosecA+1) On putting the value of 1 from eq.(1) in the Nr. =[(cotA+cosecA)-(cosecA+cotA)(cosecA-cotA)]/(cotA-cosecA+1) =(cosecA+cotA)(1-cosecA+cotA)/(cotA-cosecA+1) =cosecA+ cotA = 1/sinA + cosA/sinA = (1+cosA)/sinA , Proved. Like my answer if you find it useful! |
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| 30. |
e ange ot elevation of the top of a verticai tower from a puint on the ground is 60From another peitt 10 tm vertically above the first, ils angle of elevation is45°. Pini theheight of the tower |
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Answer» sorry the picture is not clear hope u got the answer! Answer: Step-by-step explanation: let height of tower = x + 10m let dist. between base of tower and obs. pt. = y at first angle it makes with ground = 60 degree tan(60) = sqrt{3} = (x + 10)/y . . . . . . (eq 1) next angle = 45 degree pt is 10 m above the ground tan(45)= 1 = x/y :. x=y . . . . . . . . . (eq. 2) substituting value of y in eq. 1 x + 10 = 1.732 x :. x = 13.66 |
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| 31. |
29 Porve that it cota coseedIPe It CorcedteaIt Seco - tanesin1 + Seca +tana"COS |
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Answer» LHS= RHS is the best answer |
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| 32. |
EXERCISE 9A<if 15 oranges cost110, what do 39 oranges cost? |
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Answer» 15 are of 110 so 39 are of(39×110)/15=(3×13×5×22)/(3×5)=13×22=286 |
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| 33. |
EXERCISE 9A1. If 15 oranges cost 110, what do 39 oranges cost? |
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| 34. |
12. For any given A. P., (T27-T20) + (T15-T12)-.........15 12A. dB. 7dC. 3dD. 10d13. If 3 +5+7+9+up to n terms 288, then nA. 12B. 15C. 16D. 17 |
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| 35. |
(iii)1/2cos 2θ cos θ(iv)1/7cos 5θ sin θ |
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| 36. |
1(iv) 1 S |
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Answer» 4/3+9/4=16+27/12=43/12 4/3 + 9/4 = 16 + 27/12 = 43/12 4/3+9/4(16+27)/1243/12 answer 43/12 is the correct answer of the given question 43/12 is the correct answe 43/12 is the best answer of this question 43/12 is the correct answer |
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| 37. |
994 3. tan? (cosec?0 - 1) | ARM-(1) tan? (ii) cota e(iii) cosec? o(iv) 1 |
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Answer» Answer:(iv)1 Explanation |
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| 38. |
(b) The remainder, when Grt+ 1) is divided by x+1will be :e Tm45ii)-1 (ii)2(iv) 1 |
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Answer» hit like if you find it useful |
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| 39. |
(iv)1 + secAsec A.sin' A1 -cos A[Hint : Simplify LHS and RHS separatelyA1moet |
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Answer» LHS1+secA/secA1+1/cosA /1/cosA= solve rhs sin^2A 1+secA/secA= sin^2A/1-cosALHS1+secA/secA= 1+1/cosA/1/cosA=cosA+1/cosA / 1/cosA= cosA+1RHSsin^2A/1-cosA1-cos^2A/1-cosA= (1+cosA)(1-cosA)/(1-cosA)= 1+ cosALHS=RHSproved |
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| 40. |
Kerosene oil in a jar:______ |
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Answer» it will be in ml in a case of jar and it may be litre as well |
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| 41. |
f If 2 5 7-|-1 321B--1 41fans. (i)[-3, 1] (ir 1313.17]}1 15 16](n) 6|4-7.17 |
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| 42. |
On decreasing the price of fans by 30%, the sale is increased by 20%. What is the effect on money(increase/decrease %) receipt by the shopkeeper ? |
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Answer» Thank you so much |
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| 43. |
15 The area of a square park whose perimeter is 48 m.(b) 150 sq m(b) 400 sq m(c)144 sq m. (d) 130 sq m.16. How many hours are there in 7500 minutes ?(a)125 hours (b) 200 hours(c) 144 hours (d) 140 hours.d in 18 days are17. A factory produces 5625 toy cars in 25 days.Numbers of cars will be manufacture(a) 255 (b) 225 (c) 425 (d) 14018. find the missing numbers- rule : the total of each line is 27 use numbers from 5 to 1613 910(a) 10,5 (b) 11,5 (c) 7,5(d) 11,78 17 1219、2/5 of 60 =(a) 30 (b) 11 (c) 17(d) 2420. 1 lakh =ten thousands(a) 25 (b) 10 (c) 40 (d) 30 |
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Answer» 11,5 is the rjght answer bbbbbbvbbbbbbbbbbbbbbbb |
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| 44. |
Make a number that has :a) 7 crore 80 lakhs, 90 thousand, 0 hundred, 4 tens, 5 ones.Lund Abundrod otone Dones |
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Answer» 7,80,90,045 is the answer you are looking for 7,80,90,045 is the answer 7,80,90,045 is the correct answer of this question 78090045 is the right answer 7,80,90,045is the answer 7809045 is your answer 7,80,90,045 is the right answer 7,80,90,045 is your answer 7,80,90,045 is your answer of the following question. 78090045 is correct answer |
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| 45. |
root 72 + root 800 - root 18 |
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Answer» (6+20-9 )root2= 23.root 2 √72+√800-√18= √36×2+√400×2-√9×2= 6√2 + 20√2-3√2= 6+20-3(√2)= 26- 3(√2)= 23√2The answer is 23√2... 6√2 + 20√2 - 3√2 = 23√2 thx gyzzz |
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| 46. |
Determine by substitution if:(6) 7is the root of 3x- 21(ii) 2 is the root of 4x-1 7(ii)i-5 is the root ofx-1 6(iv)-1 is the root ofx -1 0(v) 3 is the root of 2x +53xc +vi) 5 is the root of (x-3) (x-2) |
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Answer» thanks |
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| 47. |
3. Fllu6 months.4. Find the time when simple interest on3.3 lakhs at 65% per annum i?75075 |
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| 48. |
6. The diameter of the wheel of a car is 70 cm. How many revolutions will it make to travel1.65 km? |
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| 49. |
A shopkeeper buys 5 fans for 4050. He spends on the transportation. He sells them at again of 15%.Find the selling price of a fan |
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Answer» Cost of 5 fans =4050 M9ney spent on transport =50 So total cost prise = 4050+50 =4100 Selling price =? Gain %=15% S.p.=100+gain %/100*C. P. S. P. =100+15/100 115/100*4100 115*41 =rs4715 Cost of each fan=4715/5 =rs943 |
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| 50. |
Ashopkeeper buys 5 fans for? 4,050. He spends? 50 on the transportation He sells them at a gain of 15%.Find the selling price of a fan. |
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Answer» Firstly, 4050+50=4100Then find 15%of 410015/100*4100=615Then add 615 to 4100 to find the selling price,4100+615=4715. The selling price is Rs. 4715 |
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