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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. |
ही_ 6, 14, 295. 17 तथा 21 का समान्तर माध्य 18 है 1: का"न i |
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| 2. |
3p-5 > 2p+2p -7 |
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| 3. |
(2p+q+4r+2s):(2p-q+4r-2s)::(2p+q-4r -2s):(2p-q-4r +2s) |
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| 4. |
29. In Fig 10, íź right circular cone of diarneter ran and heightI2 cin tests on the base of a right eincular cvlinder of radius cm.Their bases are in the saree plane and the sylinder is filled withwater upto a height of 12 cm. It the cone is then removed,the height to which water level will fallFiaă10 |
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| 5. |
find p and q such that 2p,2p+q,p+4q,35 are in A.P |
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Answer» Like my answer if you find it useful! |
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| 6. |
35 are in A.P8. Find p and g, such that 2p, 2p +q, p+ 4q, |
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Answer» first term is 2p and no. of terms are 4 common difference should be 2p+q -2p = q so, 35 = a+(4-1)d = a +3d => 35 = 2p +3q but common difference should be 2p+q -2p = q also common difference should be p+4q -2p-q = 3q-p = q => 2q = p so,.. 35 = 2*(2q) +3q=> 7q = 35 , q = 5 aand.. p = 2q = 10 Gud.. thanku for helping me so sweet of u.. |
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| 7. |
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenlyaround it in the shape of a circutar ring of width 4 m to form an embankment. Find thelleightof the embankment.container shaned like a right circular cvlinder having diameter 12 cm and height 15 cm |
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| 8. |
3. Obtain all other zeroes of 3x 6r32x210r 5, iftwo of its zeroes areand |
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Answer» p(x)=3x4+6x3-2x2-10x-5 as we have given that two of its zeroes are root5/3 and -root5/3. so their factors are; (x-root5/3) and (x+root5/3). the product of the factors ; (x-root5/3) (x+root5/3)=x2-5/3 =3x2-5/3=1/3(3x2-5) where k=1/3 now g(x)=3x2-5 dividing p(x) by g(x) we get q(x)=x2+2x+1 by division algorithm; p(x)=g(x)*q(x)+r(x) =(3x2-5)(x2+2x+1) =(3x2-5)(x2+x+x+1) =(3x2-5)(x(x+1)+1(x+1) (3x2-5)(x+1)(x+1) the other two zeroes are -1 and-1 Like my answer if you find it useful! |
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| 9. |
L Solve the following pair of linear equa(i) x+y=5 and 2x-3y = 4 |
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| 10. |
5cos 60°+4cos 2 30°- tan2 45°sin2 30°+ cos 60°20. Evaluate: |
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| 11. |
a aits sa e2) Ajis, 5). B(9, 20) and A-P-B. Find the ratio in which P(11, 15) dividessegment AB. |
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Answer» what about y |
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| 12. |
(30) in ÎÎ BC, show that sin2 2 t. sin2-1 |
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| 13. |
If A and B are (2, -2) and (2, -4), respectively, find the coordinates of P such thatA AB and P lies on the line segment AB. |
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| 14. |
ही »+ xs——l%) 3el* %= (2 cos §)° —3 (2 cos 9)> 8०05? 9-6 ०089 ]=2 (4 cos® 6—3 cos 0) न 2 008 36: |
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| 15. |
62, Prove that: (1 + cot A-cosec A)(1 + tan A +sec A) 2.CRSE 200 |
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Answer» L.H.S = ( 1+ cota-coseca) ( 1+ tana -seca) = (1+cosa/sina-1/sina) (1+sina/cosa-1/cosa) = (sina+cosa-1/sina) (cosa+sina-1/cosa) = ((sina+cosa) - (1)/sina) ((cosa+sina) - (1)/cosa) = ((sina+cosa)raise to power2 - (1)raise to power 2)/sinacosa = sin2a+cos2a+2sinacosa-1/sinacosa = 1+2sinacosa-1/sinacosa (because sin2a+cos2a=1) = 2sinacosa/sinacosa =2 =R.H.S |
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| 16. |
2. Compare the following pairs of ratios(2 marks e(i) V3 55 ' 7Solution:8.7 0.092.9' 0.03Solution: |
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| 17. |
| 1M‘ - el l 4 R(vi > x 3 |
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Answer» 3(1/2) ×14(1/2)=(6+1)/2 × (28+1)/2=(7/2 )× (29/2)=203/2 |
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| 18. |
4. Theh term ofan AP is 3-4n find its commondifferene. |
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| 19. |
In Fig. 6.16, ifx +y-+z, then prove that AOBis a line.4.Fig. 6.15.rFig. 6.16 |
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| 20. |
INTEGRALS295Example 1 Write an anti derivative for each of the following functions using themethod of inspection:(i) cos 2x(ii) 3x2 4x3(iii)-. , xor0 |
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| 21. |
EXAMPLEWritethe following in expana1(i) (9x + 2v + z)2(ii)( |
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Answer» hit like if you find it useful |
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| 22. |
The radius of a cylinder is made thrice aslarge as it was. How should the height bechanged so that volume remains the same? |
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| 23. |
If the diameter of cross-session of a wire is decreased by 5%, how much profit will thelength be increased so that volume remains the same. |
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Answer» Diametre be dradius be d/2vol=πr*r*h =πd*d*h/4if d is increased by 5% then d- 5%of dd-5/100*dd-d/2019d/20∴r=2*19d/20=19d/40vol=πd*d*h/4π*19d/40*19d/40*h∴h=400d/361 ( calculate the above one)increase in length=400h/361-l=39l/361%of increase =39h/361/h*100=3900/361% =10.80% |
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| 24. |
34. The radius of a wire is decreased to one third. If volume remains thesame, the length will become(a) 2 times (b) 3 times(c) 6 times(d) 9 times |
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| 25. |
4. In Fig. 4, AB II CD. Find the value of x and y.75600Fig. 4 |
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Answer» Sum of angles in a triangle is 180.= 75 + 60 + angle c = 180 = 135 + angle c = 180= angle c = 45 = Angle x and and B are interior opposite angles in parallel lines, and by property the are equal.Hence, x = 75. = Angle C + angle x + angle y = 180 = 75 + 45 + y = 180 = y = 60. |
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| 26. |
sin2 30 + cos2 30 |
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Answer» sin²(30°) + cos²(30°)We know sin²A + cos² A = 1 So, here A = 30°sin²(30°) + cos²(30°)= 1 thank you |
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| 27. |
I. Check if cos 60° = cos2 30°-sin2 30。 |
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Answer» cos60°=(cos30°)^2- (sin30°)^2 1/2. =(√3/2)^2- (1/2)^2 1/2. = 3/4 - 1/4 = 2/4 =1/2LHS= RHS |
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| 28. |
Show that cos2(45° + θ) + cos2(45°-0)tan (60° + θ) tan (30°-0). |
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Answer» cos^2(45 + theta) + sin^2{90 - (45 -theta)} = tan(60 + theta) cot{90 - (30 - theta)} cos^2(45 + theta) + sin^2(90 - 45 + theta) = tan(60 + theta) cot(90 - 30 + theta) =cos^2(45 + theta) + sin^2(45 + theta) = tan(60 + theta) cot(60 + theta) 1=1 |
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| 29. |
10. 12 2.3n (n +1)n+1 |
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Answer» thanks so much. |
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| 30. |
A and the matrix BExample 7 If Alǐand B1112].then find 2A-B.Example 7 If A =,1 0 2 then find 2A - B. |
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| 31. |
Write the following set in roster form."Set of prime numbers between 1 to 20" |
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Answer» roster form={2,3,5,7,11,13,17,19} thanks angana |
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| 32. |
20°+ 21 400 4 441 4लि "छा. |
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Answer» 400+441/841=841/841= 1 |
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| 33. |
Write the solution set of the equation x^{2}+x-2=0 in roster form. |
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| 34. |
20.21Neth+b)-a-b |
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| 35. |
. Ifx-3y-2z = 4, what is the value of 9xy2-3yz? |
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| 36. |
rea of the.BAD S. ina thetrapezium.28 m15 m90°40 m3. The cross-section of a canal is a trapeziuin shape. If the canal is 15 m wide at thetop and 10 m wide at the bottom and thearea of the cross-section is 125 m2, find thedepth of the canal. In the adjoining 5 cme, are parallel sidesfigure, AB and DC41 cm |
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Answer» 3)Area of a trapezium = 1/2 (b + a) h where a and b are parallel sides and h is the altitude or distance between a and b. In a canal, the floor of canal and the top water surface are parallel. Height of trapezium is the depth of canal. 125 meters = 1/2 ( 10+ 15) HeightHeight = 250/25 = 10 meters hit like if you find it useful |
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| 37. |
4. | In Fig. 6.16, ifx + y = w + z, then prove that AOBis a line.4Fig. 6.16 |
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Answer» 360 ke bad kya likha hua hai it's a straight angle |
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| 38. |
1)If the diameter of the cross-section of a wire is decreased by 5%,by what percentage should the length be increased so that the volume remains the same? |
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Answer» V = pi.h.r^2 (cylinder) When diameter is decreased by 5%, radius also is decreased by 5%. So new radius is 19r/20. So V becomes pi.h.(19r/20)^2 = pi.h.(361r^2)/400. But you want to maintain the volume at a constant, so h is to be multiplied by 400/361 to keep it constant. So length is to be incresed by 400/361 - 1 = 39/361 times, i.e. 39 x 100/361 = 10.8033 % Diametre be dradius be d/2vol=πr*r*h =πd*d*h/4if d is increased by 5% then d- 5%of dd-5/100*dd-d/2019d/20∴r=2*19d/20=19d/40vol=πd*d*h/4π*19d/40*19d/40*h∴h=400d/361 ( calculate the above one)increase in length=400h/361-l=39l/361%of increase =39h/361/h*100=3900/361% =10.80% |
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| 39. |
Example 1 Write the solution set of the equation *2+ x-2 = 0 in roster form |
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Answer» S = { x | p(x) = 0 where p(x) = x²+x-2 } (x-1)(x+2)=0 x=1,-2 (x-1)(x+2)x= 1 or x= -2 |
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| 40. |
Write the relation R ((x, x ): x is a primenumber less than 10 in roster form. |
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Answer» Thank you |
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| 41. |
7. Value of sin 30 cos2 30 is |
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Answer» thanks madam 😘😘 |
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| 42. |
x +4 x-7 30 |
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| 43. |
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radiuThe total height of the toy is 15.5 cm. Find the total surface area of the toy |
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Answer» thaxx |
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| 44. |
(20)^-x=1/7 find (20)^2x |
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| 45. |
Example 7 Find the height of a cylinder whose radius is 7 cm and thetotal surface area is 968 cm^2 |
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Answer» 2πrh = 9682*22/7*7*h = 968h =968/44h = 22 cm |
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| 46. |
Find the vector equation of the plane passing through the intersection of the-3k)-7,F.(2i+5j+3k).planes Ĺ .(2 i + 2 j9 and through the point- |
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| 47. |
7(8+ 20)*-21(8 + 20) |
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Answer» 7(a + 2b)^2 - 21(a + 2b)= 7(a + 2b) (a + 2b - 3) |
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| 48. |
8) समीकरण x +7(3+2k)-2x(1+3k)=0 के मूल बराबर हैं।तो k का मान ज्ञात करें। |
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Answer» x^2+ 7(3+2k)-2x(1-3k)=0; x^2+21+14k-(2x+3kx)=0; x^2+21+14k-2x-6kx =0 21 +14k-6kx-2x+x^2 ,, solving for k, added -21; 21-21+14k+6kx-2x+x^2; 14k+6kx-2x+x^2=-21 added -2x, 14k+6kx-2x+2x+x^2=-21+2x added -2x ; 14k+6kx+x^2=-21 +2x added and combine 14k+6kx+x^2-x^2=-21+2x-x^2 14k+6kx+0-0=-21+3x-x^2; combine like the terms 21+14x+-6kx+ - 2x + x^2=0 the solution is could no it is correct answer |
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| 49. |
23. Ifx =-4 is a root of the equation x2 + 2x + 4p-0 , find the values of k for which the equation2+p(1+3k) +7(3 + 2k) 0 has equal roots. |
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| 50. |
Example-7. Locate 2 on number line |
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