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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. |
Factorise:(i)-4 a2 + 4ab-4ca(ii) 2xy +3 + 2y + 3xEa |
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Answer» i) -4a² + 4ab -4ca -4 a (a - b + c) ii) 2xy + 3 + 2y + 3x 2xy + 2y + 3 + 3x 2y ( x + 1) + 3 ( x + 1) ( x + 1) ( 2y + 1) |
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| 2. |
Piove the +costheta-cos(theta)+cos(theta)-cos(theta),=0 |
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Answer» costheta-cos(theta)+cos(theta)-cos(theta)so all will be cancelled out and answer will be zero onlybecause sin (270-theta) = -cos(theta)cos(180+theta)= -cos(theta) |
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| 3. |
Factorise:1. Ď2-16 |
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Answer» LIKE THE SOLUTION |
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| 4. |
Factorise:1. x2 16 |
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| 5. |
Factorise1, x2-16 |
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| 6. |
w H UL I Le number2. The perimeter of a rectangle is 56 cm. The ratioof length and its breadth is 4:3. Find the lengthand breadth of the rectangle. |
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Answer» Perimeter = 2(l + b)2(l + b) = 56l + b = 28l = 28 - bl/b = 4/3(28 - b)/b = 4/3 3(28 - b) = 4b 84 - 3b = 4b 7b = 84 b = 84/7 = 12 cml = 28 - 12 = 16 cm Let length of Reactangle = 4xLet breadth of Reactangle = 3x Given, Perimeter of Reactangle = 56 cm Then,Perimeter of Reactangle= 2(length + breadth) 2(4x + 3x) = 567x = 28x = 28/7 = 4 Therefore,Length = 4*4 = 16 cmBreadth = 3*4 = 12 cm |
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| 7. |
18 cm and 16 cm respectively. What is the curved surface are Ul I II25. If the areas of circular bases of a frustum of a cone are 16 cm? and 36 cm2 respectively arheight of the frustum is 15 cm. What is the volume of the frustum?39. A wnul |
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| 8. |
1) Factorise:-x^2 + 8x +16 |
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Answer» x² + 8x + 16 x² + 2 × x × (4) + 4² (x + 4)² |
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| 9. |
factorise the following p²-6p-16 |
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| 10. |
Factorise thec ollowing expression GYP2 |
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Answer» (6x + 8y)² - 192xy 36x² + 64y² + 96xy - 192xy 36x² + 64y² + 96xy (6x + 8y)² |
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| 11. |
es12.In the given figure, if <BAD-60 and <ADC-105° then determine <DPC3X10-30 |
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Answer» Given--- ∠ ADC = 105° We know,∠ADC +∠ABC = 180° [Opposite angles of the Cyclic Quadrilateral are Supplementary] ∠ABC ( or ∠PBA) = 180° - 105° = 75°∠DAB( or∠PAB) = 60° [Given] Using the Angle Sum Property inΔPAB, ∠PAB + ∠PBA +∠APB = 180° 60° + 75° +∠APB = 180° ∠APB = 180° - 135° ∠APB = 45° Thus, ∠DPC(or ∠APB) is 45°. |
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| 12. |
, Factorise: (x2-3x)2-8(x2-3x)-20. |
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Answer» hit like if you find it useful |
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| 13. |
.Factorise: (x2-3x)2-8(x2-3x) -20. |
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Answer» hit like if you find it useful |
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| 14. |
16. Factorise a2-2ab-c+De M |
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| 15. |
If a + 0 + 2 = 0, Piove uul u+If a + b + c = 0, then find the value of+bccaab |
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| 16. |
UUL3.Write a Pythagorean triplet if one number is 14. |
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Answer» Pythagorean triplets are of the form 2m,m²-1 and m²+1 Given one number,that is 14 14 can not be in the form of m²-1 or m²+1 as we get the following result. m²-1=14m²=13;m=√13 m²+1=14m²=15;m=√15 So,it is of the form 2m2m=14m=7 So,the other Pythagorean triplets are of the form m²-1 and m²+1m²-1=7²-1=49-1=48 m²+1=7²+1=49+1=50 So,the three Pythagorean triplets are 14,48 and 50 |
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| 17. |
A cylinder of fixed capacity 44.8 liters contains helium gas at STP. What is the amount ofheat needed to raise the temperature of the gas in the cylinder by 15°c? Given R8.3211/mol/K. |
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Answer» Like my answer if you find it useful! |
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| 18. |
Factorise 8x+ v64 |
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Answer» If you find this solution helpful, Please give it a 👍 |
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| 19. |
10. Factorise 8x + N64 |
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Answer» √64=88x^3+8y^3=8(x^3+y^3) |
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| 20. |
10. Factorise 8x+ N64 |
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Answer» 8x³ + √64 y³ 8x³ + 8y³ 8(x³ + y³) 8 ( x + y) ( x² + y² - xy) |
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| 21. |
Factorise expression x2-8x + 15 |
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Answer» x^2-(3+5)x+15=0x^2-3x-5x+15= 0x(x-3)-5(x-3)=0(x-5)(x-3)=0x= 3 and 5Answer Please like the solution 👍 ✔️ Wrong Answer |
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| 22. |
Factorise:- x^2 + 8x + 16 |
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Answer» Ans :- x^2 + 8x +16 = x^2 + 2×x×4 + 16 = (x + 4)^2 |
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| 23. |
Factorise 8x3+ V64y |
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| 24. |
10. Factorise 8x3+64y3 |
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| 25. |
1 m of same clor? 245름. Sanjubuys 262 m of same2. Rama bought 181 m of cloth for 245 . Sanju8should Sanju pay the shopkeeper? |
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| 26. |
10. Factorise 8x3+ N64y |
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Answer» √64=88x^3+8y^3=8(x^3+y^3)=8(X+y)(x^2-xy+y^2) |
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| 27. |
10 Factorise 8x3+ N64 |
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Answer» 8x³ + √64y³ 8x³ + 8 y³ 8( x³ + y³) 8 ( x + y) ( x² + y² - xy) |
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| 28. |
10. Factorise 8x3+ V64y |
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Answer» 8x³ + 8y³ (2x)³ + (2y)³ x³+y³ = (x+y)(x²-xy+y²) (2x+2y)(4x²-2xy+4y²) Like my answer if you find it useful! |
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| 29. |
factorise 8x3+27y3+36x2y+54xy2 |
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Answer» (2x+3y)^3 how give full solution (2x)^3+(3y)^3+3(2x)(3y)(2x)+3(2x)(3y)(3y)=(2x+3y)^3 |
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| 30. |
Factorise:1. x2 +8x +16 |
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| 31. |
\cos \frac{\pi}{11} \cdot \cos \frac{2 \pi}{11} \cdot \cos \frac{3 \pi}{11} \cdot \cos \frac{4 \pi}{11} \cdot \cos \frac{5 \pi}{11}=\frac{1}{32} |
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| 32. |
uding 15,8,2,3 only ene |
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| 33. |
The reagent used for the distinction of 1º,2º and 3 amines is |
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Answer» Aqueoussodium hydroxidesolution andbenzenesulfonyl chloride. |
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| 34. |
Factorise 8x3+ 64y |
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Answer» 8x*x*x + √64 y*y*y = 8x*x*x + 8y*y*y using formula a*a*a + b*b*b = (a+b)(a*a - ab + b*b) we can write following = 8(x+y)(x*x - xy + y*y) If you find this answer helpful then like it. |
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| 35. |
Factorise 8x*+ 64y |
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Answer» hit like if you find it useful |
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| 36. |
(a) 27x3 + 108xy + 144xy +64y b) |
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Answer» (3x+4y)^3(a+b)^3=a^+b^3+3ab(a+b)(3x+4y)=27x^3+64y^3+3*3x*4y(3x+4y)=27x^3+64y^3+36xy(3x+4y)=27x^3+64y^3+108x^y+144xy^2 SO the answer is (3x+4y)^3 Like my answer if you find it useful! |
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| 37. |
As shown in figure 1.57, two poles ofheight 8 m and 4 m are perpendicularto the ground. If the length of shadowof smaller pole due to sunlight is Q6 m then how long will be the shadowof the bigger pole at the same time ?R B4Fig. 1.57 |
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| 38. |
drawn has a rumber which is multiple of 3 or s.that the ticketQ 11) Tweother, ind the value of a?z)rwasi sme fare (40-af and (120+2a). If both the angles are supplementary of one |
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Answer» both angles are supplementary therefore their sum must be 180°40 - a + 120 + 2a = 180°a = 180° - 160°a = 20° |
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| 39. |
(3/2)^3*((2/11)^4*(11/3)^2) |
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Answer» [2×2×2×2/11×11×11×11]×[11×11/3×3]×[3×3×3/2×2×2](2/121)×3=6/121 IS the right answer No this is wrong 2/121 is correct answer 6/121 is correct answer of following |
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| 40. |
. Factorise. ()x 64y |
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| 41. |
1)x +64y5) 24a' + 81 |
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| 42. |
Two sides AB and BC and median AMof one triangle ABC are respectivelyequal to sides PQ and QR and medianPN of A PQR (see Fig. 7.40). Show that:B4(ii)IlΔ ABC-A POR |
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| 43. |
x^3+64y^3 |
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Answer» We will use formula of a^3+b^3= (a+b)(a^2-ab+b^2)(x+4y)(x^2-4xy+16y^2)will be the factors |
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| 44. |
- 21(5p. (1)x+64y(5) 240' + 81baga (Factors of a |
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| 45. |
he diagonals of a rectangle ABCD intersect at O. If BOC-44, find<OAD.BCD is a rhombus with ABC-560, Determine CAD.C A50°44°56°A DFig. Q.7Fig. Q. 4Fig. Q.5Fig. Q.6 |
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| 46. |
ABCDis a rhombus with LABC- 56. Determine ZCAD.C A50°144A56A DFig, O. 4Fig. Q.5Fig. Q.6 |
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Answer» BOA=triangle 56+y+y=180=62A=62CAD=62 |
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| 47. |
If the zeroes of the polynomial x3-3x2 + x + 1 are a-b, a, a + b, find a and b. |
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| 48. |
Name two cylindrical objects. |
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Answer» Two cylindrical objects are Roller and Pipe. A test tubeA steel pipe gas cylinder and rod Gas cylinder and pipe. gas cylinder and pilar gas cylinder and rod |
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| 49. |
\frac { \operatorname { cos } 45 ^ { \circ } } { \operatorname { sec } 30 ^ { \circ } + \operatorname { cosec } 30 ^ { \circ } } |
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Answer» The value of this equation is 0.22. |
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| 50. |
\begin{array} { l } { \text { 5. The degree of the polynomial } } \\ { 5 x ^ { 3 } - 2 x ^ { 2 } y ^ { 2 } + x ^ { 2 } + 9 y ^ { 2 } \text { in } x \text { and } y \text { is } } \end{array} |
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