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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. |
\frac{5 \sin 17^{\circ}}{\cos 73^{\circ}}+\frac{2 \cos 31^{\circ}}{\sin 59^{\circ}}-\frac{6 \sin 80^{\circ}}{\cos 10^{\circ}} |
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Answer» which book?? can u plz tell |
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| 2. |
List three things that you knowabout some other part of theworld from watching television? |
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Answer» we can see the riverscan listen to their languagecan see the recepiesand many more to comewe can see the historical places |
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| 3. |
1+cos e-sinsin e + sin cose is equal to : |
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Answer» cosA+cos²A/sinA(1+cosA)=cosA/sinA=tanA is the right answer cotA is the right answer cotA is the right answer of the following cotA is the correct answer of this question cot A is the right answer. cot A is the right answer |
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| 4. |
a)₹ P132 25+ 478 95+ 101 50ibtract the follo |
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Answer» 712rs70p is the right answer 712rs 70 paise is the best answer 712 rupya 70p correct answer for this question 712Rs 70p is correct answer. 712 rs 70 paise is right answer please like me 132.25± 478.95± 101.50= 712rs 70p 712rs70paise is your answer. 712.70 rupy Right answer the correct answer is 712rs 70p the answer is 712rupees 70p 712 Ru 70 pe is the answer |
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| 5. |
10150orHow many days of one year |
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Answer» for one year 365 days 365 leap year me 366 There are 365 days in year Their are 365 days in one normal year. and 366 days in leap year One year-365Leap year - 366 |
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| 6. |
llowing : |
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| 7. |
The households in a town are to be sampled in order to estimate the average amount ofassets per household that are readily convertible into cash. The households are stratifiedinto high-rent and a low-rent stratum. A house in the high-rent stratum is thought to haveabout nine times as much assets as one in the low-rent stratum, and S, is expected to beproportional to the square root of the stratum mean. There are 4000 households in thehigh-rent stratum and 20,000 in the low-rent.a) How would you distribute a sample of 1000 households between the two strata?b) If the object is to estimate the difference between assets per household in the twostrata, how should the sample be distributed? |
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| 8. |
5,10-5 125Smplity the llowing0 |
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Answer» 3–5× 10–5× 125/5–7× 6–5= 3–5× 10–5× 53× 57× 6–5 = 3–5× (5 × 2)–5× 53× 57× (3×2)–5 =3–5× 5–5× 2–5× 53× 57× 3–5× 2–5 = 3–5–5× 5–5+3+7× 2–5–5 = 3–10× 55× 2–10 |
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| 9. |
Question 17:Which decimal fraction of 0.3 meter is equalto 6 cm?O 0.18O 0.12O 0.2O 0.02 |
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Answer» 0•2 is correct answer 0•02 is correct answer 0•02 the correct answer 4)0•02 the correct answer 0.3 meter=30 cmand (1/5=0.2)th of 30 cm =6 cmSo, the correct answer is 0.2 decimal fraction. 0.02 is the right answer of the following |
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| 10. |
11. A copper wire, when bent in therorm ot a square encloses a regionhaving area 121 cm2. If the same wire isbent in the form of a circle, then theearea of the region enclosed by the wirewill be|Ď=(A) 154 cm2 (B) 143 cm2(C) 132 cm (D) 121 cm2 |
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| 11. |
Gia) If sin 17, then show that \operatorname{sen} 17-\sin 73 \cdot \frac{x}{y \sqrt{y^{2}-x^{2}}}13. Answer any one question: |
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| 12. |
Virendra owns a plot of rectangular shape. He has fenced it with a wirs of length 759 . Thelength of the plot exceeds the breadth by 5m. Find the length and breadth of the plot |
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Answer» the fencing length is actually perimeter of rectangle which is 750mlet assume that breath be xtherefore x+5therefore 2×(x+x+5)= 7502x+5=750/2=3752x= 370x= 370/2 = 185so breath is 185 andlength is 190 |
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| 13. |
(c) 126,(d) 36(a) 60°,(c) 300(b) 500(d) 40°UTR-(b) |
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Answer» 180 - x = x/2 + 30 3x/2 = 150 x = 100 50 |
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| 14. |
Virendra owns a plot of rectangular shape. He has fenced it with a wire of length 750 m. Thelength of the plot exceeds the breadth by Sm. Find the length and breadth of the plot.. |
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| 15. |
¡The length of a rectangle exceeds the breadth by 8 em. Find the area of the rectangle, if iperimeter is 180 cm. |
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Answer» Let breadth of Reactangle be x Then length of Reactangle = x + 8 Perimeter of Reactangle = 2(length + breadth)= 2(x + x + 8)= 2(2x + 8) 180 = 2(2x + 8)2x = 90-8x = 82/2 = 41 cm Therefore, breadth of Reactangle = 41 cm,Length of the Reactangle = 41+ 8 = 49 cm Area of Reactangle = length * breadth = 41*49 = 2009 cm^2 |
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| 16. |
\left. \begin{array} { l l } { 210 ^ { \circ } \text { and } 300 ^ { \circ } } & { \text { (B) } 24 } \\ { 240 ^ { \circ } \text { and } 300 ^ { \circ } } & { ( D ) 21 } \end{array} \right. |
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Answer» sin theta is negative in 3rd and 4th quadrantthen angle is (180+30)=210° and (360-30)=330° |
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| 17. |
[d] Arun pays Rs5.50 101Rs 15.50 for a burger and Rs8 for a colddrink. If he gives a 50-rupee note to theshopkeeper, how much change will hereceive back?nrs and an ice cream co |
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Answer» for ice cream = Rs 5.50for burger =15.50for cold drinks =Rs 8total =15.50+5.50+8=29but he give Rs50so he receive =50-29=21 |
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| 18. |
(iii)ă(-7) + (-23) |
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Answer» (-7)+(-23)= -7 -23= -30. Please hit the like button if this helped you |
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| 19. |
8. If the polynomial 6x4 + 8x8 - 5x2 + ax + bis exactly divisible by 2x2 - 5. Find the value of a and b.413...2..bol tbot |
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Answer» since the the above polynomial is exactly divisble by 2x^2 -5..it means that the remainder is 0 therefir we can write it as 6x^4 + 8x^3 - 5x^2 + ax + b = (2x^2 - 5)× g(x) where g(x) is a polynomial of 2nd degree simplifying this furthur..we can arrive at 6x^4 + 8x^3 - 5x^2 + ax + b = (√2x -√5)(√2x +√5)× g(x) now r.h.s become 0 when x =√(5/2) and x = -√(5/2) by substituting x =√(5/2) 6(25/4) +8(5√5/2√2) - 5(5/2) + a(√5/√2) +b =0 25 + 20(√5/√2) + a(√5/√2) +b =0 -----(1) by substituting x = -√(5/2) 6(25/4) - 8(5√5/2√2) - 5(5/2) - a(√5/√2) +b =0 25 - 20(√5/√2) - a(√5/√2) +b =0 -----(2) (1) + (2) 50 +2b =0b = - 25 (1) - (2) 40(√5/√2) + 2a(√5/√2) =0a = -20 |
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| 20. |
Find out if the following are Pythagoreantriples.(i) 8, 15, 7(ii) 7, 23, 25 |
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Answer» to find this see that if the largest number square is equal to sum of squares of other two sides15^2=2258^2+7^2=64+49=113hence 225 not equal to 113hence not a Pythagoras triplet |
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| 21. |
Question 17:One side of a rectangle exceeds its other side by 2 cm. If its area is 195cm. determine the sides of the rectangle. |
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Answer» search this on google oneside is 13and second side is 15 one side-13second side-15 one side is 13 cm another side is 15 cm. onesideis 13other 15 two sides are 13, 13 one side be 13 and another one is 15 by solving |
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| 22. |
If the perimeter of a rectangle be 106 units and the length exceeds the breadth by 7 units, find the area of therectangle.20.OR |
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Answer» Let length be las length exceeds breadth be 7 unitshence b=l-7henceperimeter=2(l+b)106=2(l+l-7)53=2l-760=2ll=30unitshence breadth will be 30-7=23unitshencearea=l*b=23*30=690unit square. *From which book have you taken this question? Please tell us so that we can provide you faster answer.* |
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| 23. |
3900 is to be distributed between A, B andC so that A gets double of C and B getsR 300more than C. Find the share of A, B andC. |
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| 24. |
er's age.19. 3900 is to be distributed between A, B andC so that A gets double of C and B getsR 300more than C. Find the share of A, B and C.s will be |
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| 25. |
DO OO 053BestExcellent |
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Answer» 1/11 is the corrcet answer. yes 1/11 is correct answer 1/11 is the correct answer of the given question 1/11 is the correct answer |
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| 26. |
sinâ0(3) Prove that : S, + cos 0 =sec 0. |
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| 27. |
MY= 2cot-0sec 0-1sec 0 +1.-20 cosec |
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Answer» 1/secA-1 -1/secA+1=secA+1-(secA-1)/(secA-1)(sec+1)= secA+1-secA+1/sec^2-1=2/tan^2A=2×1/tan^2A.= 2cot^2A ggfdjhgjgg fh y it it h eh |
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| 28. |
73322.52V519) Evaluate:15 +3/2 6+/5 |
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Answer» hit like if you find it useful |
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| 29. |
1. Evaluate;(1) 15 +(-8)(iv) (-32) +47(ii) (-16) +9(v) 53+(-26)(iii) (-7)+(-23)(vi) (-48) + (-36) |
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Answer» 7 |
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| 30. |
tan 0 - 001015. Prove that ————— = tan? 0 - cot? 0.sin B cos O |
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Answer» ( tanA - cotA )/( sinAcosA ) = [(sinA/cosA)-(cosA/sinA)]/(sinAcosA) = ( sin²A - cos² A )/ sin²Acos²A = ( sin² A/sin²Acos²A)-(cos²A/sin²Acos²A ) = ( 1/cos² A ) - ( 1/sin² A ) = sec² A - cosec² A ----( 1 ) = ( 1 + tan² A ) - ( 1 + cot² A ) = 1 + tan² A - 1 - cot² A = tan² A - cot² A Hence proved. Hit like if you find it useful |
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| 31. |
Prove that:tan 01âcotBcot 01â-tan 0=1+ sech cosech. |
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Answer» LHS: tanθ/(1 - cotθ) + cotθ/(1 - tanθ) => tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ) => tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1) => 1/(tanθ - 1) { tan²θ - 1/tanθ } => 1/(tanθ - 1) { (tan³θ - 1)/tanθ) [as, a³ - b³ = (a - b)(a² + b² + ab) => {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)} => tanθ + cotθ + 1 => sinθ/cosθ + cosθ/sinθ + 1 => (sin²θ + cos²θ)/sinθ . cosθ + 1 => 1/sinθ . cosθ + 1 => cosecθ . secθ + 1 => 1 + secθ.cosecθ = RHS Hence proved |
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| 32. |
x_3y=3 3x_9y=2 |
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Answer» x - 3y =3; 3x-9y=2; 3( x-3y=3)= 3x -9y=9/ 18y =7; y= 7/18; 3x-9(7/18)=2; 3x-7/2=2; 3x=2+7/2=4+7/2=11/2; x=11/2×3=11/6 upper answer is the right answer |
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| 33. |
tan 0 ,1-cot θcot 0+1-tan θFig. 3Prove that:-1+sec θ cosech1+tan 0 + cot.30, |
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Answer» GIVEN:- tanθ/(1 - cotθ) + cotθ/(1 - tanθ) => tanθ/(1 - 1/tanθ) + (1/tanθ)/(1 - tanθ) => tan²θ/(tanθ - 1) - 1/tanθ(tanθ - 1) => 1/(tanθ - 1) { tan²θ - 1/tanθ } => 1/(tanθ - 1) { (tan³θ - 1)/tanθ) [as, a³ - b³ = (a - b)(a² + b² + ab) => {(tanθ - 1)(tan²θ + 1 + tanθ)}/{(tanθ - 1)(tanθ)} => tanθ + cotθ + 1 => sinθ/cosθ + cosθ/sinθ + 1 => (sin²θ + cos²θ)/sinθ . cosθ + 1 => 1/sinθ . cosθ + 1 => cosecθ . secθ + 1 |
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| 34. |
iple /l 5 Gin 6 — 1 1% Prove that ना- using the identity sinB+cosO—1 secH— tan 0ec? 0 =1+ tan? 0. |
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Answer» LHS = (sinθ - cosθ + 1)/(sinθ + cosθ - 1) dividing by cosθ both Numerator and denominator = (sinθ/cosθ - cosθ/cosθ + 1/cosθ)/(sinθ/cosθ + cosθ/cosθ - 1/cosθ)= (tanθ + secθ - 1)/(tanθ - secθ + 1) Multiply (tanθ - secθ) with both Numerator and denominator = (tanθ + secθ - 1)(tanθ - secθ)/(tanθ - secθ + 1)(tanθ - secθ)= {(tan²θ - sec²θ) - (tanθ - secθ)}/(tanθ - secθ + 1)(tanθ - secθ)= (-1 - tanθ + secθ)/(tanθ - secθ + 1)(tanθ - secθ) [ ∵ sec²x - tan²x = 1 ]= -1/(tanθ - secθ) = 1/(secθ - tanθ) = RHS please like the solution |
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| 35. |
If1312then sec 0 -tan?0sec 0 + tan |
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| 36. |
0. What journal entries would be recorded for the following transactions on thedissolution of a firm after various assets (other than cash) on the third partyliabilities have been transferred to Reliasation account.1. Arti took over the Stock worth Rs. 80,000 at Rs. 68,000.2· There was unrecorded Bike of Rs. 40,000 which was taken over By Mr. Karim.3. The firm paid Rs. 40,000 as compensation to employees.4. Sundry creditors amounting to Rs. 36,000 were settled at a discountof 15%.Loss on realisation Rs. 42,000 was to be distributed between Arti and Karimin the ratio of 3:4.5. |
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Answer» give me the realisation account of this question❓ |
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| 37. |
2^x=3^y=6^zthen prove that z=xy/x+y |
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| 38. |
\left. \begin{array} { l } { x - 3 y + z = 2 } \\ { 3 x + y + z = 6 } \\ { 5 x + y + 3 z = 3 } \end{array} \right. |
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| 39. |
x - 3 y - 3 = 03 x - 9 y - 2 = 0 |
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| 40. |
: 2210. If the equation of a line 4B are3-LL2-3 2 6z+2,find the direction cosines of a lineparallel to AB |
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| 41. |
log (1+cos x) dx |
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Answer» thnx |
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| 42. |
उन कक दर 5110: |
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| 43. |
6) ह 7०7 7117700 ९002 “रे 35५, ट ९02. X5y OMOO पल 4ित/ ०-00 दि ol दी erannum Priya lemo o Sasme & uan[2 H'g_ ot (57 -pey QNI g m.Fing 3-10 ०. |
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| 44. |
B कोजिए 5110 - 00509 + 1 थे 1" sinO+cosO-1 secH-tan0 |
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Answer» hit like if you find it useful |
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| 45. |
Gदि 3 o थ्ल्ल्ः Sl ek cos' x, - ——lA2 V2 |
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| 46. |
SECTION- D (4xa) Subtract 3pq(p-q)from 2pq(ptb) Simplify g(a2 I , 1ă |
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Answer» If you like the solution, Please give it a 👍 |
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| 47. |
WW223. Find the domain and range of f(x) = log cos1 +where [:denotes the greatest integer function. |
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Answer» mein to 6th mein hame nahi pata |
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| 48. |
85. Let f(x) = min{x + [x]. - x - -x]}, where [ ] denotesthe greatest integer function. Then f(x) dx isequal to |
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| 49. |
Find the angle between two non-parallel lines whose direction cosines satisfy theequations 3l+m+5n =0 and 6mn-2nl+5lm = 0. |
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| 50. |
g A2) Annales School |
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Answer» Ans :- The Annales school is a group of historians associated with a style of historiography developed by French historians in the 20th century to stress long-term social history. The third generation stressed history from the point of view of mentalities, or mentalités. PLEASE LIKE THE ANSWER this much only |
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