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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. |
A P \text { , the sum of } n \text { terms is } \frac { 1 } { 2 } \left( 5 n ^ { 2 } + 3 n \right) . \text { Find its } 40 ^ { \text { th } } |
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| 2. |
(*) - (23) a<\र है -(a ) ० (8) 32(C)-448 (2) 448 |
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| 3. |
prove that :1/secA+tanA - 1/cosA = 1/cosA - 1/secA-tanA |
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| 4. |
448 + = 448 |
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Answer» 1. |
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| 5. |
tan A +secA -1tan A -secA+1(1) |
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Answer» thnx |
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| 6. |
हा. B A+1Prove that: [-————+ \/rec =2cosecAsecA+1 secA-1 |
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Answer» Please hit the like button if this helped you thnx |
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| 7. |
(2) The simple interest on 8000 at certain rate ofinterest for 2 years is 1440. Find the rate of interest. |
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| 8. |
Find the sum of the following series to n terms:1. 2? +4² +63 +8² +... |
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Answer» S=2^2+4^2+6^2...+n^2a1=(2*1)^2a2=(2*2)^2a3=(2*3)^2...an=(2*n/2)^2ak=4k^2 |
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| 9. |
the sum of the following series1+4+7+10+. to 40 terms2 Find the sum |
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| 10. |
How long will it take for 1860 invested at the rate of 9.5% per annum simple interest to amount to€24497 |
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Answer» A=2449,P=1860,R=9.5 p.a.S.I.=A-PS.I.=2449-1860S.I.=589T=S.I.×100÷P×RT=589×100÷1860×9.5T=300.8333 p=1860r=9.5%therefore interest=2449-1860=589so period=(589*100)/(1860*9.5)=3.3yr |
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| 11. |
How long does it take a principal payment to triple at a simple interest rate of 6% ? |
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| 12. |
OB. If the perimeter of the following figure is 47 cm,fintIn fig, APBand AQP are semi-circles, and AO22]the area of the shaded region. | Useπ = |
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Answer» Let OA=OB =r.Therefore, 40= 22/7* r/2 + 22/7*r + r280= 40rr=7So, shaded area= (1/2*22/7*7/2*7/2 + 1/2*22/7*7*7)= 77/4 +77= 77 (1/4+1)= 77* 5/4= 385/4 cm^2= 96.25 cm^2 Like my answer if you find it useful! |
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| 13. |
1. Force on the block by upper spring.(a) Mgb) MaA1(c) Mg Ke(d) Ma K1K1K2K1+ K2 |
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Answer» force balance equation is mg -k1x = k2x=> x = mg/(k1+k2) so, force on block by upper spring is k1x = mgk1/(k1+k2) |
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| 14. |
I. The following table shows the number of students and the time they utilizeddaily for their studies. Find the mean time spent by students for their studies bydirect method. |
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| 15. |
Which term of the series9 l..IS |
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| 16. |
\begin{array} { l } { 129 . \text { Find the wrong term in the following } } \\ { \text { series } } \\ { 2,3,5,8,12,18,23 } \end{array} |
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Answer» The wrong term is 18. Explanation: 2 + 1 = 3 3 + 2 = 5 5 + 3 = 8 8 + 4 = 1212 + 5 = 1717 + 6 = 23 Therefore, the correct terms is 17. |
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| 17. |
ABCDfint value of a |
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Answer» correct answer is x = 140 ek X= 140° is the correct answer of the given question x=140° is the correct answer of the following question |
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| 18. |
\sin 105^{\circ}+\cos 105^{\circ}: |
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| 19. |
105 ^ { \circ } + \operatorname { cos } 105 ^ { \circ } |
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Answer» Sin(60+45) + cos(60+45) =[sin60cos45 +cos60sin45] + [cos60 cos45 - sin60sin45] =cos 60 sin 45 +cos60 cos45 =1/2 * 1/√2 + 1/2 *1/√2 =1/√2 cos45° |
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| 20. |
13cm, 12cm and 5cm are the sides of a triangle Determine whether it is aright angled or not. |
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| 21. |
1f the product of zeros of the pulymnnial m.6-6i.. fint the value ole |
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| 22. |
fint the Lottofactorise the followingon 3+2 +64 3 + 820m3-322 +32 +7 |
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| 23. |
A Fint he 15ch term of the series 1 \frac{1}{2}, 2-\frac{1}{2}, \dots |
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| 24. |
what is the largest gland |
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Answer» the largest gland is liver.... liver is the right answer... |
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| 25. |
6. Find the median in an arranged series of an even number of in terms. |
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Answer» Mid term is nth term and (n+1)th term -> nth term is 2n and (n+1)th term is 2(n+1) median = (2n+2(n+1))/2=2n+1 |
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| 26. |
The difference between the area and the square ofthe radius of a circle is 105 cm2. Find the area ofthe circle. |
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Answer» Square of radius = r^2 Area of circle = π* r^2 A.T.Q π * r^2- r^2= 105 r^2(π - 1 ) = 105 r^2(22/7 - 7 / 7 ) = 105 r^2(15 / 7 ) = 105 r^2= 49 r = 7 cm Circumference = 2 * π* r = 2 * 22 / 7 * 7 = 44 cm Ans: 44 cm |
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| 27. |
1. 13cm, 12cm and 5cm are the sides of a triangle Determine whether it is aright angled or not. |
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| 28. |
(i) gland which has both exocrine and endocrine components. |
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Answer» Thepancreasgland works both as endocrine and exocrine gland, because it is a glanduar organ in the upper abdomen, but really it serves as two glands in one : a digestive exocrine gland and hormone |
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| 29. |
8.ABCD is a rectangle in which diagonal AC bisects 4 A as well as C. Show that:(i) ABCD is a square (ii) diagonal BD bisects ZB as well as Z D. |
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| 30. |
Evaluateda.CostxtSinu |
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Answer» (1/√2) tan^-1((tan^2x -1)/(√2tanx)) +c Sine, Cosine and Tangent Sine,CosineandTangent(often shortened tosin,cosandtan) are each aratio of sidesof a right angled triangle:  For a given angleθeach ratio stays the sameno matter how big or small the triangle is To calculate them: Divide the length of one side by another side Example: What is the sine of 35°?  Usingthistriangle(lengthsare only to one decimal place): sin(35°)=OppositeHypotenuse=2.84.9=0.57...cos(35°)=AdjacentHypotenuse=4.04.9=0.82...tan(35°)=OppositeAdjacent=2.84.0=0.70... |
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| 31. |
EvaluateLitcost daJV1-Cost |
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| 32. |
Q19. The area of a rectangle is x2 + 12xy + 27 ya and its length is (x+9y). Find the breadth ofthe rectangle.(x+3y) |
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Answer» x+3y is the breadth of rectangle x+3y is the correct answer |
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| 33. |
A rectangle of area 105 cm^2 has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle. |
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| 34. |
13 A rectangle of area 105 cm2 has its length equal to x cm. Write down its breadth interms of x. Given that the perimeter is 44 cm, write down an equation in x and solveit to determine the dimensions of the rectangle. |
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Answer» length x and suppose breadth is yand perimeter is 44so x+y=22so y=22-xarea is 105x(22-x)=105so x^2-22x+105=0si x=15 or x=7so if length is 7 breadth is 15if length is 15 breadth is 7 |
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| 35. |
of a square and a rectangle is same. If a side of the square is 1 5cm and one side of therectangle is 18cm, find the area of the rectangle. |
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| 36. |
16. If x, occurs in the expansion of |x2+, prove that its coeffi- (2n +even terms in the ex |
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| 37. |
10Neerajhas39apples,16 apples found to be rotten. Find the fraction of apples that are goo |
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Answer» (39-16)/39=23/39 is the fraction of apples that are good |
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| 38. |
There are n A.M.'s between 7 and 71 and 5th mean is 27. Find the value of n. |
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Answer» but answer is wrong |
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| 39. |
e adjoining figure, ABCD is aam in which AB is produced to EIn therove that ED bisects BC.s0 |
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| 40. |
EXAMPLE9 ABCD is a parallelogram. AB is produced to E so that BE = AB. Prove that EDbisects BC |
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| 41. |
In the adjoining figure.AD - DCand AB - BC then check whethert BD bisects ZB as well as ZD?(ii) Prove that |
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| 42. |
प्रश्न 10. समीकरण हल कीजिए -[(x+1)/ (x-1 )]-[(x-1)/(x+1)]=5/6,x!=1,-1 |
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| 43. |
,InFig. (i), ABC, ZB 3and CX in descending orderlerzc65°and the bisector of/BAC meets BC in x.Arrange AX, BX65 |
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| 44. |
EvaluateSao da |
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Answer» e^x (x-1) is correct answer |
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| 45. |
Answer the following questions:(a) Five years ago, John is half of Neeraj's age. If John's present age is 10find the present age of Neeraj. |
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Answer» 20 years is the most correct answer the present age of neeraj 15 years old 15 is the right answer the present of neeraj is 15 years present age of neeraj is 15 year olds 15 is the right answer 15 years is correct answer. let five years ago Neeraj's age=xthen five years ago john age=x/2john present age=x/2+5 10=(x/2)+5 10-5=x/2 5×2=xx=10so, present age of neeraj =x+5=10+5=15 years. present age of Neeraj is 15 yrs. Neeraj's present age is 15yrs. neerej's age is 15 years Age of Neeraj is 20 because 10 + 10= 20 |
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| 46. |
6(12. The line drawn through the midpoint of one side of a triangle parallel tanother sidethe 3rd side. |
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Answer» bisects |
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| 47. |
A rectangle- one side 20 other is (x), another rectangle, one side 6 and the other 14,what number makes these similar figures? |
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Answer» 20/6=X/14 for them to be similarX=46.6 |
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| 48. |
\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} |
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| 49. |
4. Simplify:16 X 2n+1(i |
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| 50. |
\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x |
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Answer» thank u deepak mand |
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