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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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9. Find the number which makes the given expressions true.(a) 2* = 32(b) (-4)' = - 64 |
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Answer» (a) answer is x=5(b)answer is x=3 (a)answer is X =5(b)answer is X=3 (a) answer is x=5(b) answer is x=3 (a) 2(x) =32 >2(5)=32 Hence x=5(b) (-4) x=(-64) >(-4) (3) =(-64) Hence x=3 x = 5 x=3 answer a. answer is x =5b.answer is x = 3 [a] = x=5[b) = x=3 answer (a)= 5answer(b)=3 2^5=32 option (a) is ture dear friend it is answer x=3 dear friend it answer (1) x=5, (2) x=3 (a) answer is X=5(b) answer is X =3 in part a x=5in part b x=3 (a) 2*2*2*2*2=32 and (b) 4*4*4=64 A's answer x=5 and B's answerx=3 (a)answer x=5. (b)answer x=3 (a) x=5 (b) x=3this is the answer x ki value 5 honi hh taki 32 aye x=5 2*2=4 '4*2=8 ;8*2=16;16*2=32 in (a) x=5 and in (b)x=3 x=52 ka 5 bar guna 32 hoga 2*2*2*2*2=32then x=5 for (a) answer is 5 2^5=2*2*2*2*2=32(b) answer is 3 (-4) ^3=-4*-4*-4=64 x=5x=3 respectively (a) answer is x=5(B) answer is x=5 value of x in ques a is 5 and in b is 3 (a) answer is x= 5(b) answer is x= 3 (a) x=5(b) x=3as 2 the power 5 is 32 by putting x=5 both RHS and LHS will be equal. for (b) -4 the power 3 =-64(-4)*(-4)*(-4)=-64 1 answer is 2 ki power 5 (a) x=5because 2 to the power of 5 will be 322*2*2*2*2=32(b) x=3-4*(-4)*(-4)=-64 5 and 3 is the answer 2^5 equals to 32 and -4 ^ 3 = - 64 (a)answer is x=5(b) answer is x=3 a) answer is x=5b) answer is x=3 (a) is true number is x=5(b) is true number is x=3 (a)x=5(b)x=-4. theek hai samajh gaye A = 2 ki power 5B =(-4) ki power 3 x=5as 2 power 5=32-4powers 5 =-64 (a) x= 5 answer( b) x= 3 answer (a) answer x=5(b) answer x=3 answer _a. x=5,b. x=3 (a) answer is x=5(b)answer is x=3ok a.) 5(2^5= 2*2*2*2*2=32)b.) 3 ((-4)^3 = (-4)*(-4)*(-4)=-64 |
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| 2. |
/cof the two parallel sides of a trapezium areectively and its height is 4.2 cm, find its area.PORS is an isosceles traneziumDLrespec |
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Answer» you are very. nice |
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| 3. |
itethevalues of x and y70r |
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Answer» y = 70° .... ( vertically opposite angles) and x +y = 180° ... ( linear pair) so, x = 180-y = 180-70 = 110° |
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| 4. |
Prob.64. Discuss maxima and minima of the funf(x, y)=x^{3}-4 x y+2 y^{2} |
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Answer» Given f(x,y)=x3-4xy+2y2differentiating equation(i),we have fx= df/dx = 3 x2-4y fy= df/dy = -4x+4y and r = fxx= 6 x s= fxy= -4 t=fyy=4 Now for maxima or minima,we must have fx=0,fy=0,we have 3x2-4y=0 (ii) and -4x+4y=0 (iii) Solving equations (ii) and (iii) , we get 3x2-4x=0 or x(3x-4)=0 or x=0, 4/3 Now from equation (iii), we have when x=0 => y=0 and when x=(4/3) => y=(4/3) thus the reuired stationary points are (0,0) and (4/3,4/3) At point (4/3,4/3) r=6 X (4/3) =8 s=-4 t=4 rt-s2=8 X 4 - ( -4)2=32-16=16 > 0 and r > 0 Hence ,f(x,y) has a minima at (4/3,4/3) At point (0,0) r=6 X 0 =0 s=-4 t=4 rt-s2= 0-(-4)2= -16 = -ve and hence there is neither maxima nor minima at (0,0) Minimum value of f(x,y) =[x3- 4xy = 2y2]x = 4/3 , y = 4/3 =(4/3)3-4.(4/3).(4/3)+2.(4/3)2 =(64/27)-(64/9)+(32/9) =(64-192+96)/27 =-(32/27) Ans thank you so much bro |
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| 5. |
Exercise 8C1here.Write the terms of the expressions givena 3a + bf terms given here. |
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Answer» 3a+b= 2 terms4xy-5=2 terms |
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| 6. |
Q.10 Given expressions for the following cases(1) 15 Subtracted from 2x (2) 3 added totwice of q(3) 25 Added to 2x |
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Answer» Expression will bw1) 2x - 15 2) 2q + 3 3) 2x + 25 |
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| 7. |
One side of a parallelogram is 25 m and its corresponding altitude is 10 m. Find the length of theadjacent side of the parallelogram if the height of the altitude to this side is 20 m.10 m25 m |
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| 8. |
3.If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles. |
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Answer» Given:In ∆ABC , AD bisects ∠BAC, & BD= CD To Prove:AB=AC Construction:Produce AD to E such that AD=DE & then join E to C. Proof: In ∆ADB & ∆EDCAD= ED ( by construction)∠ADB= ∠EDC. (vertically opposite angles ( BD= CD (given) ∆ADB congruent ∆EDC (by SAS) Hence, ∠BAD=∠CED......(1) (CPCT) ∠BAD=∠CAD......(2). (given) From eq.1 &2 ∠CED =∠CAD......(3) AB=CE (CPCT).......(4) From eq 3 as proved that ∠CED=∠CAD So we can say CA=CE......(5) [SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL] Hence, from eq 4 & 5 AB = AC HENCE THE ∆ IS ISOSCELES.. |
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| 9. |
If the bisector of an angle of a triangle also bisects the opposite side, provetriangle is isosceles. |
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| 10. |
42. The formula to find the total surface area of a 5-rupee coin is A=A.2trhB.tr2C.2tr2D. 2tr(r+h)43. If n- 100.A 25 and x25 then fd. |
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Answer» surface area of coin=πr^2 |
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| 11. |
7) Find the gencral rule for the perimeter of a rectangle of length l and breadth bilx A pattern of le |
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Answer» Theperimeteris the length of the outline of a shape.To find the perimeter of a rectangle you have to add the lengths of all the four sides. X is in this case the length of the rectanglewhile Y is the width of the rectangle. The perimeter is 2X + 2Y |
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| 12. |
s 1680 m. Find theof a rectangular field is 465 m2 and its length is 31 m. Findhele areal breadth of the fieldcost of fencing it at the rate of15.50 per metre. |
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| 13. |
thout acduof he foloning() 21-23 i 9-941Q.Na "β-) Wnite ithe following as the same of twoConsecudive positive mumbens- |
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Answer» using identity a^2-b^2=(a-b)(a+b)hence 1) 21^2-20^2(21-20)(21+20)=1*41=412)34^2-33^2(34-33)(34+33)=1*67=673)95^2-94^2(95-94)(95+94)=1*189=189 q.no.9 15^2=225=112+11311^2=121=60+6119^2=361=180+181 |
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| 14. |
5. An ant is moving around a few food pieces of different shapes scattered on the floor.would the ant have to take a longer round? RememberFor which food-piececircumference of a circle can be obtained by using the expression c 2tr, where ris the radius of the circle.(b)(c) 81.5 cm-2.8 cm-2.8 cme Tranezium |
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| 15. |
A dr nking glass is in tihe shap oi a rusium of a cone of height14 em The diameters of ite tewo cncuias enda are 16 em and 12 en.Find he capacity of the glass. |
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Answer» please like my answer if you find it useful |
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| 16. |
Wnite the following rational numbers in ascending order1 -2-43 9 33 -2-l |
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| 17. |
2 Se ptemberFUN AT HOMETake some toothpicks and tryshapes you make and write the number of toothpicks used.to make various shapes using them. Draw theLET'S DRAWd square. Using these shap |
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Answer» for square or rectangle, 4triangle 3hexagon 6 |
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| 18. |
The ratio of the sums of m and n terms of an A.P is nr : r. Show that the ratioof m and nth term is (2m 1) (2n 1). |
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| 19. |
berate of polishing isて137111 , ( tihe rate oShazlitook a wire of length 44 cm and bent it into the shape of a circle.the radius of that circle. Also find its area. If the same wire is bent into the shapFindasquare, what will be the length of each of its sides? Which figure encloses mo |
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n the shape of a square of side 10-cm. If the wire is relbent into a rectangle of lenigihA wire is in the shap15. m. find its breadth. Which figure encloses more area and by how muckh?n is50 m long 40 m hroad and 10mbi. |
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| 21. |
A flooring tile has the shap of a Parallelogram whosebase is 24cm. and the night is locm. How many suchtiles are required in cover a floor of area 2080 m.The area of rhombus is 240 cm and one of theagonal is 16 cm. Find the other diagonal.trabe9 |
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| 22. |
4... यदि | +1 y—3—)=(3—5.3—l),a}xam ./ ज्ञात कीजिए।- - |
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| 23. |
l breadth are both doubled?What happens to the area of a square when0) its side is doubled?(in) its side is tripled?(im) its side is halved? |
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Answer» the area is doubled tripled and halved |
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| 24. |
A path 1 m wide is built along the border and inside a square garden of side 30 m. Find1 the area of the(ũ) thecost of planting grass in the remaining portion of the garden at the rate of740 per m2. |
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| 25. |
laThe side of an equilateral triangle is 3.5 cm. Find its perimeter.0 |
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| 26. |
(264/33) ×100= |
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Answer» Ans :- (264/33)×100 = 8 × 100 = 800 |
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| 27. |
4:. The inner diameter of a circular well is 3.5cm it is 10m deep find thecost of plastering the C.S.A at rate Rs.40 per m2 |
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| 28. |
6. If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is(a) An - 3 (b) 3n - 4 (c) 4n +3(d) 3n + 4C oming in 51 and bored |
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| 29. |
x/1516äşş8hen the value of x is |
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Answer» Givensqroot (x) /16 = 15/8sqroot (x) = 15*16/8sqroot (x) = 15*2x = (30) ^2x = 900 Value of x = 900 |
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| 30. |
\operatorname { sin } ( 150 ^ { \circ } + x ) + \operatorname { sin } ( 150 ^ { \circ } - x ) = \operatorname { cos } x |
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| 31. |
hen x #5,is continuous nr 1-5when x=5 |
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| 32. |
w henę°+ |
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Answer» when empty 64.27 gwhen 100% filled weight is 150.35 gso weight of water when 100% filled 86.08when 75% it will be 64.56so total weight = 64.27+64.56=128.83 g |
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| 33. |
2. () Letl - Jx2- X d. hen,= | XV |
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| 34. |
e find the area ofs. Find the point on the x-axis which is equidistant from (-2, 5) and (2.-3) HenMax. Marksthe triangle formed by these points.: 80 |
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| 35. |
Pranshasi hen Melthod |
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Answer» 3-5x+2x=6x+53-3x=6x+53-5=6x+3x-2=9xx=-2/9 This much 🙂 This much 🙂 This much 🙂 |
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| 36. |
the 2-hen th |
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Answer» Zeros of polynomial 6x²-7x-3=6x²+2x-9x-3=2x(3x+1)-3(3x+1)=(2x-3)(3x+1)⇒2x-3=0 or ⇒3x+1=0⇒x=3/2 or ⇒ x= - 1/3α=3/2 ,β= - 1/3⇒α+β= -b/a⇒3/2+(-1/3)= - (-3)/6⇒3/2-1/3=1/2⇒7/6 =1/2⇒αβ=c/a⇒3/2(-1/3)= -7/6⇒-1/2= -7/6⇒1/2=7/6 please like the solution 👍 ✔️👍 |
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| 37. |
15.Give possible expressions for the length and breadth of each of the followingrectangles, in which their areas are given:Area: 25a'-35a+ 12Area: 35y2 + 13y-12 |
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| 38. |
Hene each of the |
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| 39. |
15. Give possible expressions for the length and breadth of each of the followingrectangles, in which their areas are givenArea: 35y + 13-12Area: 250-350+12 |
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| 40. |
solid cube is cut into two cuboids of equal volumes. Findthe ratio of the total surface area of the given cube and thatof one of the cuboids3.A |
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| 41. |
A path I m wide is built along the border and inside a square garden of side 30mla0) the area of the path(ii) the cost of planting grass in the remaining portion of the garden at the5.40 per m |
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| 42. |
A path I m wide is builtalong the border and inside a square garden of side 30m. Find:0)(1)5.the area of the paththe cost of planting grass in the remaining portion of the garden at the rate of740 per m2. |
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| 43. |
5. A path 1 m wide is built along the border and inside a square garden of side 3n6) the area of the path(ii) the cost of planting grass in the remaining portion of the gardenat the:40 per m2 |
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| 44. |
5. A path I m wide is built along the border and inside a square garden of side 30m Fa0)(i)the area of the paththe cost of planting grass in the remaining portion of the garden at the raeriR 40 per m2 |
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| 45. |
solid cubeofis cut into tuo cuboids of equal volumes. Find the ratiothe total surface area of the given cube and that one of the cuboids.Ans. 3: 2] |
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Answer» Like if you find it useful |
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| 46. |
t Asolid cube is cut into two cuboids of equal volumes. Find the ratioof the total surface area of the given cube and that one of the cuboids.Ans. 3: 2 |
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Answer» Like if you find it useful |
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| 47. |
15. If two straight lines intersect each other then prove that the ray opprtieto the bisector of one of the angles so formed bisects the veráopposite angle.16. Prove that the bisectors of two adjacent supplementary anglesindalea right angle |
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Answer» 16)Solution:-Given,∠ DAB + EBA = 180°. CA and CB are bisectors of∠DAB∠ EBA respectively.∴∠ DAC +∠ CAB = 1/2 (∠DAB).....(1)⇒∠ EBC +∠ CBA = 1/2 (∠ EBA)....(2)⇒∠ DAB +∠ EBA = 180°⇒ 2 (∠CAB) + 2 (∠ CBA) = 180° [using (1) and (2)]⇒∠ CAB +∠ CBA = 90°InΔ ABC,∠ CAB +∠ CBA +∠ ABC = 180° (Angle Sum property)⇒ 90° +∠ ABC = 180°⇒∠ ABC = 180° - 90°⇒∠ ABC = 90°So, the bisectors of the two adjacentsupplementary angles include a right angle.Hence proved. |
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| 48. |
%Curvedsurfaceareaofarightcircular cylinderis4.4 mi. fthe radcylinder is 0.7 m, find its height.ius of the base of the |
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| 49. |
17. Cost of ploughing at the rate of Rs. 10 per m? is Rs. 1,540. Findte ius of the circular feld 122 |
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Answer» Area = 1540/10 = 154 m² Area of Circle = πr² => πr² = 154 => r² = (154 * 7)/(22)=> = 49=> r = 7m |
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| 50. |
-6*sqrt(35)*sqrt(7)/14 %2B (2*sqrt(30))/sqrt(6) %2B sqrt(55)/((3*sqrt(11))) |
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Answer» is it write answer |
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