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. |
Why did various classes and groups of Indiansparticipate in the Civil DisobedienceMovement? |
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Answer» Various classes and different social groups of Indians participated in the Civil Disobedience Movement led by Gandhiji in 1930. All of them joined this movement on account of their own needs, aspirations and limited understanding. In the rural areas, rich farmers and peasant communities such as Patidars (Gujarat) and Jats in Uttar Pradesh were very hard hit by the trade depression and decreasing costs of their commercial crops. They found themselves unable to pay the government’s revenue due to the disappearance of their cash income. For them the fight was a struggle against high revenue. So, the rich peasants participated in the Civil Disobedience Movement and supported the boycott programmes. The poorer peasants were not in favour of Iowering of the revenue demand. Many of them were small tenants who used to cultivate rented land taken from landlords. As the depression continued, cash income dwindled and so the tenants were unable to pay their land-rent. They demanded that their dues of rent should be remitted. The business classes participated in the movement to oppose the colonial polices that restricted business activities. They wanted protection against: imports of foreign goods, and a rupee-sterling foreign exchange ratio that would discourage imports. A few of them attacked colonial control over the Indian Economy and supported the Civil Disobedience Movement. Besides it they supported the movement financially and boycotted the trading of foreign goods. The industrial working classes stayed away from this movement leaving the Nagpur region as industrialists came closer to the congress. Women took part in this movement. They began to see service to the nation as a sacred duty of women. |
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| 2. |
\operatorname { tan } ^ { - 1 } \frac { 1 } { 2 } + \operatorname { tan } ^ { - 1 } \frac { 1 } { 5 } + \operatorname { tan } ^ { - 1 } \frac { 1 } { 8 } = \frac { \pi } { 4 } |
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Answer» Tq |
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| 3. |
\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right)=\frac{\pi}{4} |
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Answer» tan^(-1)((1/2+1/5)/(1-1/10))+tan^(-1)(1/8)=tan^(-1)(7/10 / 9/10)+tan^(-1)(1/8)=tan^(-1)(7/9)+tan^(-1)(1/8)=tan^(-1)((7/9+1/8)/(1-7/72))=tan^(-1)(65/65)=tan^(-1)(1)=pie/4 please answer rest of the questions |
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| 4. |
u two aloys A aud Bhe ratio and 2ama2/ kto |
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Answer» Zinc : Tin A 5x : 2x = 7x B 3y : 4y = 7y => A = 7x = 7 kg. x = 1 kg ∴∴Zinc in Alloy A => 5 kg Tin in Alloy A => 2 kg. => B => 7y = 21 kg. y = 3 kg. Zinc in alloy B =>3×3==9kg Tin in alloy B =>3×4=12kg ∴∴after mix-up the ratio of zinc and tin in new alloy new ratio is ??? |
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| 5. |
4. In an election, 1785436 persons has casted their votes for different parties, but 19209 voteout of them were found invalid. Find the number of valid votes |
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Answer» to find the number of valid votes1785436-19209=1766227 |
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| 6. |
d die sum ot odd integers from 1 to 2001.d the sun of all naturalnumbers lying betwn10 aud 1000 h |
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Answer» The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001. This sequence forms an A.P. Here, first term, a = 1 Common difference, d = 2 Here, a+(n−1)d = 2001 => 1+(n−1)(2) = 2001 => 2n−2 = 2000 => n = 1001 Sn= n/2[2a+(n−1)d] ∴ Sn= 1001/2[2×1+(1001−1)×2] =1001/2[2+1000×2] =1001/2×2002 =1001×1001 =1002001 Hence, the sum of odd numbers from 1 to 2001 is 1002001. |
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| 7. |
ceasing meaning |
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Answer» put an end to a state or an activity |
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| 8. |
\tan ^ { - 1 } \left( \frac { 1 } { 5 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { 7 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { 3 } \right) + \tan ^ { - 1 } \left( \frac { 1 } { 8 } \right) = \frac { \pi } { 4 } |
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Answer» tan^(-1)((1/5+1/7)/(1-1/35))+tan^(-1)((1/3+1/8)/(1-1/24))=tan^(-1)(12/34)+tan^(-1)(11/23)=tan^(-1)(6/17)+tan^(-1)(11/23)=tan^(-1)((6/17+11/23)/(1-66/(17*23))=tan^(-1)(((138+187)/(23×17))/(325/(23+17))=tan^(-1)(325/325)=tan^(-1)(1)=pie/4 |
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| 9. |
7. If the point (a, a -2) lies on the graph of the equation3x + 5y = 30. find the value of a. |
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| 10. |
8. Find the equation of straight lines which are perpendicular to the line12x+5y = 17 and at a distance of 2 units from the point (-4,1) |
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| 11. |
\left. \begin{array} { l } { \operatorname { tan } ^ { - 1 } ( \frac { 1 } { 1 + 1.2 } ) + \operatorname { tan } ^ { - 1 } ( \frac { 1 } { 1 + 2.3 } ) + } \\ { + \operatorname { tan } ^ { - 1 } ( \frac { 1 } { 1 + n \cdot ( n + 1 ) } ) = \operatorname { tan } ^ { - 1 } \theta } \end{array} \right. |
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| 12. |
tan^-1(x+1)+tan^-1(x-1)=tan^-1 8/31 |
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| 13. |
9. Candidates of four schools appear in a mathematics test. The data were as follows:No. of Candidates6048Not available40SchoolsAverage Score7580IV50If the average score of the candidates of all the four schools is 66, find the number ofcandidates that appeared from school III. |
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Answer» Thanks teacher your solution is correct.... |
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| 14. |
15. In an election. 4.98,656 votes were found valid, 6768 votes were found invalid and83,865 persons did not cast their votes. How many votes were registered in all? |
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Answer» Total number of votes registered in all=498656+6768+83865=589289 |
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| 15. |
r andy from a lincar pair lf 0, then find the ml984. In the given figure, state if AB is parallel to CD. Give reason for your answer.98㥠|
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Answer» No not parallel as parallel line should have corresponding angle equal considering thay we 98 is not equal 82 . |
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| 16. |
1. In the figure alongside, ZPOR and 00R area lincarpair. f xso find the values of x and y |
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Answer» As x and y are linear pair sox+y = 180and given x-y = 80so we get2x = 260x = 130therefore y is 50 |
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| 17. |
N 4X o © न ि.न्फेउ हक न कि ८5 = 2 Moo - |
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Answer» Like if you find it useful Bit we have to show that tan threta =13 |
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| 18. |
3. The graph of a lincar equation 25y - 10 is the line which meets the y -axis. Find the poind |
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| 19. |
cos-1虱丽すUat 큠cos6 |
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| 20. |
Example 4. A party of 15 persons sit at a round table. Findthe odds against two specified individuals sitting next to eachother. |
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Answer» n(E)=(n-2)!*2!n(s)=(n-1)! as circular arrngmntnwP(E)=(n-2)!*2!/(n-1)!=2/(n-1)P(E)'=(1-(2/n-1))=(n-3)/(n-1)now ATQodd against:P(E)'/P(E)=(n-3)/2=(n-3)/2put n =15hence15-3/2=12/2=6 chances |
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| 21. |
2 x^{2}=\sqrt{1024} x=? |
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| 22. |
(1024) x (81)(243)"X (128)İplify2 |
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Answer» (1024×1024×1024×81×81×81×81)/(243×243×128×128×128×128) = (4×4×4×81×81)/(3×3) = 4×4×4×9×81 = 46656 |
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| 23. |
\lim _{x \rightarrow 2} \frac{x^{10}-1024}{x-2} |
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Answer» thank u very much |
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| 24. |
ar+x- 1805r= 180°ZBAC 2x 72x= 36"Hence ProvedExercise 7A1. Which of the following pairs of triangles are congruent? Alsostate the condition of congruency ineach case:65Date: "Page No.55°60° |
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Answer» i) both triangle are congurent by SAS rule |
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| 25. |
he graph of the lincar equstion x 2y- Atwhich point s) does the graph cut the |
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Answer» At x = 8 it cuts x axis At y = 4 it cuts y axis |
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| 26. |
ŕ¤ŕ¤A Hence Proved. |). 13. Prove that :(1-sin 0 + cos 9)* = 2(1 + cos ) (1 - sin o)e 2 pum |
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Answer» Consider the LHS:(1-sinA+cosA)^2= [(1-sinA) + cosA]^2= (1-sinA)^2+ cos^2A+ 2(1-sinA)cosA= 1 + sin^2A− 2sinA+ cos^2A+ 2(1-sinA)cosA= 1 + (sin^2A+ cos^2A)− 2sinA+ 2(1-sinA)cosA= 1 + 1− 2sinA+ 2(1-sinA)cosA [Since,sin^2A+ cos^2A =1]= 2− 2sinA+ 2(1-sinA)cosA= 2(1− sinA) + 2(1-sinA)cosA= 2(1− sinA)(1 +cosA)= RHS |
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| 27. |
Check: 1" term of LHS =43x-7 -1- 31-21 - 3131x4--52-13- 31x4 31-13 79 -39 + 7940Hence, LHS 31 93 93 93Again, RHS = -7%Therefore, LHS = RHS31 x 69Exercise 14.1Q1.(h) 7ySolve the following equations and verify the result.(a) 2(0.35y - 1) + 2.3y = 8 (b) 2x - 45 = 15 (c) 3a -(e) 3(5a - 2) + 5(2a + 4) = 4 (6) 9x + 5 = 4(x - 2) + 8(g) 15ly - 4) - 2ly - 9) + 5(y + 6) = 0уRam and Mohan are brothers. Ram's present age is three yeMohan. If the sum of their present age is 27, find their ages.Sum of three consecutive multiples of 8 is 96. Find the multipQ2.Q3.Mathematics for Class 8 - CBSE |
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Answer» SORRY I COULD NOT ANSWER THIS QUESTION NEXT TIME I WILL SURELY ANSWERING YOU Answer: let Mohan's present age be x Then Ram's age = 2x+3 According to the question: x+2x+3=27 3x=27-3 3x=24 x=24/3 x=6 Therefore , Ram's age=x=6 years Mohan's age= 2x+3 = 2(6)+3 = 15 years |
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| 28. |
Hence Proved.ample 9. In the given figure, AD is perpendicular to BC and EFBC, if < EABFAC,show that triangles ABD and ACD are congruent.Also, find the values of x and y if AB-2r+3, ACBD,and DC =y+1.3y+1, |
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Answer» Where is the figure? |
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| 29. |
2. ABC is an isosceles triangle in which ABAC. Using RHS property, prove that <B-2C.[Hint: Draw AD丄BC. Prove ΔΑΙΒ兰AADC by RHS congruence] |
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Answer» thanks |
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| 30. |
Q-> proved theetProved-2 |
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Answer» correct is answer is the =2 2 is the right answer 2 is the right answer |
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| 31. |
Proved +uat |
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| 32. |
[Hint: Simplify LHS and RHS separately] |
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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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| 33. |
tihe followin4 cm4 cm4 cm2 cm |
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Answer» If you find this solution helpful, Please like it. |
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| 34. |
l tihe argles so obtained.3. Using your protractor, drawan angle of measure 108°. With this angle as given,draw an angle of 540 |
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| 35. |
15. Draw the graph of the linear equation x +2y-8. At which poin(6) does the graph cut tihex-axis and y-axis. |
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Answer» Given is graph of x + 2y = 8 x - axis cut at 8 y - axis cut at 4 tqu |
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| 36. |
Vertfy the Tndentrty f046 -0+206+bgeome tri colly b ag 2nits, 6ihg d2 unië Units | |
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Answer» LHS = (a-b)² = (2-6)² = -4² = 16 RHS = a²-2ab+b² = 2²-2(2)(6)+6² = 4-24+36 = 16since LHS = RHSverified. that is not ( a-b)^2.that is (a+b)^2 |
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| 37. |
If a, b, c, d are in G.P., prove that:(i) (a2- b2), (b2 - c2), (c2 -d2) are in G.P.2a. b' b2 + c2 , c2+d2 are in GP |
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| 38. |
25. If a, b, c and d are in GP. show that(аг + b2 + c*) (b? + c2 + d2)-(ab + bc + cd), |
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Answer» a,b,c,dare in G.P. Therefore, bc=ad… (1) b2=ac… (2) c2=bd… (3) It has to be proved that, (a²+b²+c²) (b²+c²+d²) = (ab+bc–cd)² R.H.S. = (ab+bc+cd)² = (ab+ad+cd)²[Using (1)] = [ab+d(a+c)]² =a²b²+ 2abd(a+c) +d²(a+c)² =a²b²+2a²bd+ 2acbd+d²(a²+ 2ac+c2) =a²b²+ 2a²c²+ 2b²c²+d²a²+ 2d²b²+d²c²[Using (1) and (2)] =a²b²+a²c²+a²c²+b²c²+b²c²+d²a²+d²b²+d²b²+d²c² =a²b²+a²c²+a²d²+b2×b2+b²c²+b²d²+c²b²+c²×c²+c²d² [Using (2) and (3) and rearranging terms] =a²(b²+c²+d²) +b²(b²+c²+d²) +c²(b²+c²+d²) = (a²+b²+c²) (b²+c²+d²) = L.H.S. ∴ L.H.S. = R.H.S. |
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| 39. |
B-6.If a, b, c, d are in G.P., prove that:(i)(a2-b?), (b?-c2), (c2-day are in GP.a" + b2 , b2 + c2 , c2+d2 are in GP. |
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| 40. |
tanA=n/m than proved the value of sinA and secA |
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| 41. |
Axioms or postulates are the assumptions which are obvious universal truths. Tproved. |
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Answer» Axiomsandpostulatesare the basicassumptionsunderlying a given body of deductive knowledge. They are accepted without demonstration. All other assertions (theorems, if we are talking about mathematics) must be proven with the aid of these basicassumptions. |
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| 42. |
Which is greater in each of the following:-5-46 3-3 2--3(iii)3' 2v) 4'4 |
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| 43. |
-+मा बि,4-Provedthat,+ 1 हल40:1044-a9-021 |
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| 44. |
Write any four benefits of Micro-propogation |
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Answer» Ans :- 1. Requires relatively small space for growing.2. Pathogen free plants can be raised and maintained economically.3.Small size of propagules and their ability to proliferate in the soil free environment facilitate their storage on a large scale.4. Stocks of germplasm can be stored and maintained for many years. Micropropagation has a number of advantages over traditional plant propagation techniques: The main advantage of micropropagation is the production of many plants that are clones of each other. Micropropagation can be used to produce disease-free plants. It can have an extraordinarily highfecundityrate, producing thousands ofpropaguleswhile conventional techniques might only produce a fraction of this number. It is the only viable method of regeneratinggenetically modifiedcells or cells afterprotoplastfusion. |
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| 45. |
5.Why is Business considered as an Economic Activity? |
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| 46. |
ofAttempt any four questions from this section,)Question 5.2(a) If-cos θ +ysin θ = 1 and a sin θ-b1, then prove that _ + 3-2cos θrth30 The |
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Answer» Thanxx.. |
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| 47. |
a bus has 50 seats How much such buses are required to take 400 children for picnic? |
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Answer» Solution:Number of Seats in a bus = 50Number of children for picnic = 400Number of buses = ? Number of buses = 400/50Number of buses = 40/5Number of Number of Buses = 400/50 = 40/5 = 8 So,Number of buses = 8 8 is the correct answer. number of childrens = 400number of seats = 50number of buses = ? according to question no. of buses = 400/50 no. of buses = 8 400/50=8 8 buses are required to take 400 children for picnic. |
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| 48. |
Find the angle between the vectorsa i+j-k and b-i-j+k |
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| 49. |
uwhat is Cicumjevance o Civale? |
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Answer» In geometry, the circumference of a circle is the distance around it. That is, the circumference would be the length of the circle if it were opened up and straightened out to a line segment. Since a circle is the edge of a disk, circumference is a special case of perimeter. thnxs what is formula of circumference of circle? Thecircumference formulais used to calculate the distance around a circle.Circumferenceformulas: C = 2πr or C = πd. r is the radius and d is the diameter. |
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| 50. |
Pumpe timberuers show that the square of any positive odd integer inof the form 4q4i an and ag for some integer &Set |
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Answer» Let positive integer a be the any positive integer. Then, b = 4 . By division algorithm we know here 0 ≤ r < 4 , So r = 0, 1, 2, 3. When r = 0 a = 4m Squaring both side , we get a² = ( 4m )² a² = 4 ( 4m²) a² = 4q , where q = 4m² When r = 1 a = 4m + 1 squaring both side , we get a² = ( 4m + 1)² a² = 16m² + 1 + 8m a² = 4 ( 4m² + 2m ) + 1 a² = 4q + 1 , where q = 4m² + 2m When r = 2 a = 4m + 2 Squaring both hand side , we get a² = ( 4m + 2 )² a² = 16m² + 4 + 16m a² = 4 ( 4m² + 4m + 1 ) a² = 4q , Where q = 4m² + 4m + 1 When r = 3 a = 4m + 3 Squaring both hand side , we get a² = ( 4m + 3)² a² = 16m² + 9 + 24m a² = 16m² + 24m + 8 + 1 a² = 4 ( 4m² + 6m + 2) + 1 a² = 4q + 1 , where q = 4m² + 6m + 2 Hence ,Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer. |
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