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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.
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ORIf S, denotes the sum of the first n terms of an AP prove that S30 3 (20 S10)3 (S20 S1o) |
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Answer» Let sum of the first n terms of an A.P, Sn= ( n / 2) [ 2a + ( n -1)d ] Now S30= ( 30 / 2) [ 2a + ( 30 -1)d ] ---------(1) S20- S10= ( 20 / 2) [ 2a + ( 20 -1)d ] - ( 10 / 2) [ 2a + ( 10 -1)d ] S20- S10= 10 [ 2a + ( 20 -1)d ] - 5 [ 2a + ( 10 -1)d ] S20- S10= 10a + 190d - 45d . 3(S20- S10) = 3(10a + 145d ) 3(S20- S10) = 3 x 5(2a + 29d ) 3(S20- S10) = ( 30 / 2) [ 2a + ( 30 -1)d ] --------(2) From (1) and (2) we get S30= 3(S20- S10) |
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| 2. |
Q.4. If S, denotes the sum of the first x terms of an A.P., prove that S30-3 (S20-S10). |
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| 3. |
15. A straight line passes through the points A(2, - 4) and B(5, - 2),Find () the slope of the line AB;(ii) the equation of the line AB;(ii) the value of k, if AB passes through the point P(k + 3, k 4). |
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Answer» thanks |
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| 4. |
8. Find the roots of the equation 5x? - 6x-2=0 by the method of completingExample 8: Find the rootthe square.thenthe chant b5 |
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ind the derivatives of:Example 82+1 3x -I |
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Example 8: Express the following numbers in standard form.(i) 0.000035(ii) 4050000 |
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Ve dryliv] The animals are not scared of Tabaqui.other.2x2242+1=3Give one word for each of the followine anessions: One has(i) An animal that lives on the flesh of other animals iowing expressions : (One has been done for(ii) A disease in which one is scared of water:(iii) One who eats no meat, fish or eggs:(iv) One who writes novels :(v) One who writes poetry:D. Match the words in column-A with their meanings in column-Column-AColumn-B(0) DespiseFrightenedApt(iii) Afraidsuitable*3=3Dislike and having no respect for somebody3=8Grammar2. Underline the participle adjectives in the following sentences :(i) He saw a flying bird.(ii) You should help a wounded man.3. Change the mode of narration:(i) He said, "I am happy".tin She said, "I am a nurse".ch) |
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Answer» despise dislike and having no respect for somebody apt suitableafraid frightened 1 in 22in33in1 it your answer 1in22in33in1your answer |
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| 8. |
Sn denotes the sum of the first n terms of an A.P, prove that S30 = (S20 - S10) |
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| 9. |
24. 1I1 S, denotes the sum of frset term o an Ale powe thatS30 3(S20-So)OR |
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| 10. |
B) A farmer has sheeps and cows in the ratio 8:3.i) How many sheeps has the farmer, if he has 180 cows?ii) Find the ratio of the number of sheeps to the total number ofanimals the farmer has.iii) Find the ratio of the total number of animals with the farmer to thenumber of cows with him. |
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Answer» Let number of sheeps = 8x and number of cows = 3x (1) if number of cows = 180 3x = 180 x = 180/3 = 60 Then, Number of sheeps = 8*60 = 480 (2) Number of sheeps = 480 Total animals = 180 + 480 = 660 Ratio = No. of Sheeps/ Total animals = 480/660 = 8/11 or 8:11 (3) Number of cows = 180 Total animals = 180 + 480 = 660 Ratio = Total animals/ No. of cows = 660/180 = 11/3 or 11: 3 |
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| 11. |
1 If Sn denotes the sum of the first n terms of an AP ,Prove that S30-3(S20 Sio |
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| 12. |
Hence, -65° and y- 110PExample-20. In the adjacent fig, ifOT PR, TOR-0 and 4SPR-S, find x and ySolution: In ATQR, |
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Answer» Where is the figure? |
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| 13. |
1 .If Sn denotes the sum of the first n terms of an AP ,Prove that S30-3(S20 -Sio). |
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| 14. |
24. If S, denotes the sum of first n terms of an A.P, prove that, S30 3(S20 S10) |
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| 15. |
1 .If Sn denotes the sum of the first n terms of an AP ,Prove that S30-3(S20 Sio |
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| 16. |
Example 20 Find the derivative of fix) from the first principle, where flx) is(i) sin x + cos.xI) xsinX |
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Answer» thank you😊 |
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| 17. |
Example 2: A house was purchased for4,00,000 and sold at a gain of 20%. Find theselling price. |
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| 18. |
Example 8 : Find the roots of the equation 5x-6x-20 by the method of completingthe square. |
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Answer» Like if you find it useful |
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| 19. |
RepresendD2 on number lime |
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Answer» Place a dot and label it. If you have points on a line marked for 0 and 1 and want to place a dot for√2in correct position relative to them, that would more appropriately be referred to as constructing it. Merely representing it does not have to be so precise |
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| 20. |
3. How many decimetre ate equal to le4. The area of a rectangular field is 600 m2 and its breadth is 15 m. Calculate its length. |
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Answer» area of rectange=l*b 600 =l*15 600/15= l 40=l |
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| 21. |
4. S20 100, d2 s,d a |
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Answer» S20=100n=20d=-2a=?henceSn=n/2(2a+(n-1)d)S20=20/2(2a+19*-2)100=10(2a-38)10+38=2a48=2aa=24 |
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| 22. |
(८ - हेCheck whedhey ४०८८८ VE, ' 2, दि15 o— 2:22 ८८ Vit f-fy il 2PV D2 el |
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Answer» √3 , √12 , √27 ,....√3 , √4×3 , √9×3 ,.......√3 , 2√3, 3√3 ..........are in AP where first term of this AP is √3 and common difference = √3 Hence it is an ap. Like my answer if you find it useful! |
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| 23. |
Solve the differential equationD2- 4D 23- en los an |
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| 24. |
12. A rectangle has length which is 5 cm less than twice its breadth.If the length is decreased by 5 cm and breadth is increased by2 cm, the perimeter of the resulting rectangle will be 74cm.Find the length and breadth of the original rectangle. |
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Answer» Breadth = xLength = 2x - 5Then the length is decreased by 5= 2x - 10The breadth is increased by 2= x + 2Perimeter = 2*breadth + 2*length74 = 2*(x + 2)+ 2*(2x - 10) 74 = 2x + 4 + 4x - 2074 = 6x - 166x = 74 + 166x = 90x = 15 cmBreadth = 15cmLength = 25cm |
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| 25. |
wherefrsExample 20 Find the derivative of fx) from the first principle,(i) sin x +coSx(ii) xsin x |
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| 26. |
1. (i) xsin x2 |
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| 27. |
_ [नह00 कक 90500 [~ पए] नह 095500 * p uts घष्फ 9A01d ST |
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Answer» sinA/secA+tanA-1+ cosA/cosecA+cotA-1 =sinA/(1/cosA+sinA/cosA-1)+cosA/(1/sinA+cosA/sinA-1) =sinA/{(1+sinA-cosA)/cosA}+cosA/{(1+cosA-sinA)/sinA} =sinAcosA/(1+sinA-cosA)+sinAcosA/(1+cosA-sinA) =sinAcosA[(1+cosA-sinA+1+sinA-cosA)/(1+sinA-cosA)(1+cosA-sinA)] =2sinAcosA/(1+sinA-cosA+cosA+sinAcosA-cos²A-sinA-sin²A+sinAcosA) =2sinAcosA/{1+2sinAcosA-(sin²A+cos²A)} =2sinAcosA/(1+2sinAcosA-1) =2sinAcosA/2sinAcosA =1 (Proved) |
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| 28. |
14. Originally a rectangle was twice as long as it is wide. When 4 m were added to its lengthand 3 m subtracted from its width, the resulting rectangle had an area of 600 m2. Find thedimensions of the new rectangle |
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| 29. |
(c) a2 + b2 + c2 + d2 + 2(ac-bd) |
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| 30. |
FQCHOPUSE. . D2 fo 2 |
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Answer» 10x^2-9x-710x^2+5x-14x-75x(2x+1)-7(2x+1)(2x+1)(5x-7) |
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| 31. |
The length and breadth of a square are increased by 40% and 30% respectively. The area of the resulting rectangle exceeds the area of the square by |
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| 32. |
Differentiate xsin , x > 0 w.rt. x. |
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| 33. |
Q.1 Evaluate\lim _{x-2} \frac{x^{5}-32}{x^{3}-8} |
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Answer» a) when x=2 we will get 0/0 formusing L'hospital rulelim x->2 5x*x*x*x/3*x*x=5*16/3*4=5*4/3=20/3 b) when x=0we will get 0/0 form using L'hospital rulelim x->0 cosx/1=cos0/1=1/1=1 |
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| 34. |
The length of a rectangle exceeds its breadth by 7 cm. If the length isdecreased by 4 cm and the breadth is increased by 3 cm, the area of the newrectangle remains the same as the area of the original rectangle. Find thelength and breadth of the original rectangle. |
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| 35. |
a sum gets uts double in 10 years. How much time it will take to get tripple? |
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Answer» There are a lot of calculators out there that can do this in less time A simple tool is the rule of 72. 72/interest rate = time to double. So 72/10 years = about 7.2% annual interest doubles in 10 years. In 30 years it will be triplet of the sum |
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| 36. |
Inhow much time will a sum of money double if invested at the rate of 8% simple interest perannum? |
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| 37. |
Gautam eats 100 chocolates in half a minute and uGovind eats half of the chocolates in double the time.How many chocolates can they eat together in oneminute? |
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Answer» Gautam can eat 100 chocolates in half a minuteSo, Gautam can eat 200 chocolates in 1 minute. Then, Govind can eat 50 chocolates in 1 minute Together number of chocolates they can eat in one minute = 200 + 50= 250 |
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| 38. |
14. The length of a rectangle is greater than the breadth by 3 cm. If the length isincreased by 9 cm and the breadth is reduced by 5 cm, the area remains the same.Find the dimensions of the rectangle. |
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| 39. |
d22Ify = x + tan x,prove that cos-x-12-2y + 2x=0.de- |
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Answer» Y= x+tanxdy/dx=1+ sec2xd2y/dx2 = 2secx. secxtanx. 2sec2xtanxd2y/dx2 =2tanx/ cos2xCos^2x. d^2y/dx^2- 2tanx=0Cos^2x.d^2y/dx^2 - 2x - 2tanx +2x=0Cos^2x.d^2y/dx^2 - 2y+2x= |
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| 40. |
The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by3 cm, the area of the new rectangle will be 84 cm2 more than that of the given rectangle. Find thelength and breadth of the given rectangle.17. |
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| 41. |
1 (ax-bbxsa, then27 Ify tan(arbr)(1422)(a-b%)2(d) of these |
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| 42. |
14.When V1-47 sin θ and V2-33 sin (e+20°) then (Vi-V2) is given by(a) 19.59 sin (0 700)19.59 sin (e 35°)(c) 19.59 sin (0 350)(d) 19.59 sin (θ + 700) |
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Answer» 19.59 sin (thita - 70degree) |
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| 43. |
\lim _{x \rightarrow 0} \frac{e^{x}-e^{\sin x}}{x-\sin x} |
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| 44. |
9. P and can do a given piece of work in 8 days: Q and R can do the same work in 12 days and Pcomplete it in 6 days. In how many days can P and R finish it? |
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Answer» In one day amount of work done by P and Q = 1/8 In one day amount of work done by Q and R = 1/12 In one day amount of work done by P, Q and R = 1/6 Then, In one day amount of work done R = 1/6 - 1/8= (4-3)/24 = 1/24 In one day amount of work done by P = 1/6 - 1/12= (2-1)/12 = 1/12 Therefore, amount of work done by done P and R in one day = 1/12 + 1/24 = (2 +1)/24= 3/24 = 1/8 P and R can finish work in 8 days |
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| 45. |
\lim _ { x \rightarrow 0 } \frac { e ^ { x } - e ^ { \sin x } } { x - \sin x } |
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Answer» answer 1 ravali No answer will be 0 just check once more sorry answer 1 from book |
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| 46. |
A piece of work is complete by A in 20 days, the same is complete in 10 days by B. in how many days that work can be complete by both of them. |
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Answer» Work done by A in 5 days = 1/20*5 =1/4remaining work=(1-1/4) =3/4now,3/4 work is done by B in 10 dayswhole work will be done by B in 10*4/3= 40/3 daysas 1 day's work of A = 1/201 day's work of B =3/40(A+B)'s 1 day's work= 1/20+3/40 = 1/8so,both finished the work in 8 day's. |
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| 47. |
.6 What is the maximum number of whole cuboids of length = 2 cm, breadthI cm and height-I cm that can be packed into a larger cuboid of length-3 cm, breadth3 cm and heightII cm? |
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Answer» please like my answer if you find it useful |
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| 48. |
Be happyin frontof peoplewhodon'tLike you,9"It kills them" |
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Answer» Please post relevant questions only. Relevant questions ka bhi answers nhi aarha hai to kya post kre |
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| 49. |
here is only one square whose perimeter and area are the saside of this square?The other rectangles whose perimeter and area are the same. Only one of these rectangles has aength and width which are whole numbers. What are the length and width of this rme. What is the length of each |
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Answer» Please hit the like button if this helped you |
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| 50. |
78. If one root of the equation (1 - m) xº + Ix + 1 = 0 isdouble the other and lis real, then what is the greatestvalue of m? |
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Answer» Let the roots be r and 2r So, sum of roots= 3r = -l/(l-m).lProduct of roots= 2r^2 =1/(l-m) Equating them we get: 2l^2- 9l + 9m=0. For l to be real Discriminant, D>=0 So, 81 - 72m >= 0or, m< = 9/8 Therefore, greatest value of m = 9/8 |
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