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. |
2/3(4x-1)-(4x-1-3x/2)=x-7/2 |
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Answer» I don't notuydhkjggfghhhhjkjqksjxndkekekdjnnzks |
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| 2. |
1. एक त्रिभुज 4807 में ८८8 न 90९, 48 > 5 सेमी, 807 > 12 सेमी हो तो sin A, cos 4, tan A F HH ज्ञात कीजिए। |
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| 3. |
है. खबर (५13ng) |
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Answer» cos^-1(cos 13pi/6) = cos^-1(cos 2pi + pi/6) = cos^-1(cos pi/6)[cos (2pi + x) = cos x[ = pi/6 As pi/6 £ (0, pi) |
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| 4. |
13n+37x-6=0 |
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Answer» it is middle term splitting so x =2/13 |
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| 5. |
If Îą andbe the zeres of the quadratic poly nomical 3x2 + 5x+7, E al ate |
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| 6. |
16. The ratio between the sum of <nterms of two A.P's is 3n+8 : 7n +15 Find the ratiobetween their 12th terms.(9n 13n + 1 |
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| 7. |
\begin { equation } \begin{array}{l}{\text { Which of the following is an irrational number? }} \\ {\text { (a) } \frac{5}{7}}\end{array} \end { equation } |
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Answer» b) √7 is irrational. Reason : √7 can't be written as p/q form |
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| 8. |
\begin { equation } \begin{array}{l}{\text { Prove the following identity: }} \\ {\sin ^{4} A+\cos ^{4} A=1-2 \sin ^{2} A \cos ^{2} A}\end{array} \end { equation } |
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| 9. |
00Find the lateral swiface greaand tatal surface area of cuboidwhose height, length and breathare 25cm, 20cm and isom |
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| 10. |
2874 \vec HH ^ 3 |
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| 11. |
Evaluate : (HH訊村f TQ)何U)21x:/2971x1297:: |
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Answer» ³√121׳√297 = ³√121×297 = ³√35937 = 33 |
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| 12. |
3 13n17. In an AP the sum of first i teris 15Find the 25th--.term.2 |
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Answer» n term sum = 3n²/2 + 13n/2as we know that nth term = (Sum of nth term ) -( sum of (n-1)th term)method 1so25th term = (sum of 25terms) - ( sum of 24 terms) =(3×(25)²/2 +(13×25)/2) - ( 3×(24)²/2 + (13×24)/2) = 1100 - 1020 =80method 21st term = 8sum of 2 terms = 192nd term = 19 - 8= 11so. difference = 11- 8 = 325th term = (8 + (25 -1)×3) = 80 |
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| 13. |
n When two coins are tossed, what is the probability that two heads are obtained? |
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Answer» Comes under conditional probabaility{H,T} {T,H} {H,H} {T,T} So given atleast one head occurs ~{H,T} {T,H} {H,H} P(E) = n(E)/n(S) = 1/4 |
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| 14. |
A cuboidal block of side 7 cm is surmounted by a hemisphere. What is the greadiameter that the hemisphere can have? Also find the surface area of the solid. |
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| 15. |
The thte Sidas ot a0An almintf ev5x+2 }expAnter+3mn ex5x +3 |
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Answer» s = 24 + 32 + 40 = 96From heron’s formula,A = √s (s-a)(s-b)(s-c)here, a= 24, b=32, c= 40Putting all the values A = √ 96*72*64*56A = √24*4*24*3* 64*4*14A = 24*4*8 √42Answer is 768√42 |
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| 16. |
(4x+5y)2+(4x-5y)2 |
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Answer» (4x+5y)2+(4x-5y)2=8x+10y+8x-10y=16x |
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| 17. |
(4x+5y) (4x+5y) |
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Answer» use identity (a+b)² = a²+b² + 2abso (4x+5y)² = 16x² + 25y² + 40xy |
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| 18. |
How is the law of conservation of matter verified in the experiment of calcium chand sodium sulphate?2. |
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Answer» When calcium chloride and Sodium sulphate react, they form equal amount of sodium chloride and calcium sulphate. Hence the law of conservation of matter is verified |
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| 19. |
Write the complement of each of the following angles.i. 34ii. 89°iii. 25oiv. one-third of a right angle2. Find the supplement of each of the follOL |
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Answer» i) 90-34 = 56 ii) 90-89 = 1 iii) 90-25 = 65 iv) 90-90/3 = 90-60 = 30 1.562.013.754.60these are the complement of the above angles in the question.complement means the sum two angles should be 90 degree |
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| 20. |
11. Sam has coins of denomination 1, 2 and 35. The ratio of the number of these coins with Samis 3:2:4. If the total amount with Sam is 135. How many coins of each denomination doesSam have? |
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| 21. |
Obtained n 9mu than 3 imes the stoxolgi |
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| 22. |
flAcdigidHhimes the |
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| 23. |
lJTTISeVUIIIMestheoher,hin15.Solve the following pair of linear equations by substitution method :\begin{array}{l}{3 x+2 y-7=0} \\ {4 x+y-6=0}\end{array} |
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| 24. |
534 शोधो, : आज +1 नहा 3- |
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| 25. |
56÷57÷34×534+66 |
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Answer» 56÷57÷34×534+66.982÷34×534+66.0288×534+6614.95+6680.95 good |
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| 26. |
n vs 6 more thanA two digit numberhait of the original nunber Find the difterence of tve digas ofane given number49.ภsirven imes the pum of its dgits The rrumber formed by reversrtE the d(h) 3 |
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| 27. |
in yeathan the quaxeage ushen Nisha gerssAaha's ageimes |
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Answer» Let Nisha's age is x years Aisha's age = (x² + 2) years Now, A/C to question , when Nisha grows to her mother present age .means Nisha's grows = Aisha's present age - Nisha's present age = x² + 2 - x Then, this time Aisha's age = Aisha's present age + growing time = x² + 2 + x² + 2 - x = 2x² - x + 4 Now, question said , Aisha's age = 10× times of Nisha's age -1 2x² - x + 4 = 10x -1 2x² - 11x + 5 = 0 2x² -10x - x + 5 = 0x = 5 and 1/2 x ≠ 1/2 because age doesn't in moths Hence , Nisha's present age = 5 years Aisha's present age = (5)² + 2 = 27 years |
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| 28. |
2y2-5y+10=0 |
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Answer» 2y^2-5y+10=02y^2-5y=-10y^2=-10/2+5yy=√5+√5yy=5√y y=5√y is a right answer d=55 is the correct answer h 2y^2-5y+10=02y^2-5y=-10 y=5 is the right answer of this question y=5√y is correct answer Y=5√y is a correct answer |
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| 29. |
Prove that in a triangle, other than an equilateral triangle, angle opposite the longest sideis grea2ter thanof a right angle. |
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| 30. |
(i) 3y2-20 = 160-2y2 |
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Answer» 3y² - 20 = 160 - 2y² 3y² + 2y² = 160 + 20 5y² = 180 y² = 180/5 y² = 36 y = √(36) y = + or - 6 |
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| 31. |
9. Verily UThe solution of 3x - 5 = 7 is x = 4. |
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Answer» Put in X=4hence3(4)-5=12-5=7hence proved it's a solution. |
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| 32. |
5. Verily that |
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Answer» thanks |
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| 33. |
- 462 cm24 A conical tent is 10 m high and the radius of its base is 24 m.Find(0 slant height of the tent.(ü) cost of the canvas required to make the tent, if the cost of1 m2 canvas is 70, |
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| 34. |
TUR+w 35,23,87cm 26,3", 4", 534ATLARyemek yemn=1 |
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| 35. |
tur at the rate of 10% ?aS. In how many years the sum of money 3 imes8) 10C) 15D) 25A) 20 |
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Answer» send clear image answer is 20 |
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| 36. |
z. onethe triangle is right angle2 If each angle of a triangle is less than the suthat the triangle is acute angled.two, shor than the sum of the other tur |
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| 37. |
6. A conical tent is 10 m high and the radius of its base is 24 m. Find theslant height of the tent. If the cost of 1 m2 canvas is R 70, find the cost ofcanvas required to make the tent. |
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| 38. |
4.A conical tent is 10 m high and the radius of its base is 24 m. Find(i) slant height of the tent.(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70. |
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| 39. |
() TadituA conical tent is 10 m high and the radius of its base is 24 m. Find(i) slant height of the tent.4.(i) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is 70. |
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| 40. |
9, 2x2 + 2y2-x=0 |
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| 41. |
7、The sum of the radii of two circles is 7 cm, and the difference of theircircumferences is 8 cm. Find the circumferences of the circles |
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| 42. |
11^(-3)*11^(4*x)=11*121^4 |
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Answer» Given:11⁴x × 11-³= 11× 121⁴ 11⁴x × 11-³= 11× (11²)⁴ 11⁴x × 11-³= 11× 11^8 11⁴x=11^9/11-³11⁴x=11^(9+3)11⁴x=11^12since base are equal4x=12x=12/4 =3 |
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| 43. |
\frac { 3 } { 2 } + \frac { - 8 } { 11 } + \frac { - 4 } { 11 } + \frac { 5 } { 22 } |
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| 44. |
The sum of the radii of two circles is 7 cm, and the difference ot theicircumferences is 8 cm. Find the circumferences of the circles.resnectivelv |
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| 45. |
7. The sum of the radii of two circles is 7 cm, and the difference of theircircumferences is 8 cm. Find the circumferences of the circles |
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| 46. |
find the circumferences of the circle whose diameters are: |
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| 47. |
,Find the values ofk, if 2x2tur ht khas equal roots? |
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Answer» equation has two equal rootsso b^2-4ac=0k^2-4(2)(k)=0k^2-8k=0k(k-8)=0so k=0 or k=8 |
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| 48. |
(3)/By using slopes, if the following points arecollinear, find the value ofk. |
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| 49. |
what value ofk-x+3 (K-1)-0 has difference of noots equal so1 |
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| 50. |
4. A conical tent is 10 m high and the radius of its base is 24 m. Findpu(i) slant height of the tent.(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is? 70. |
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