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. |
What is a logic gate? Name the three basic logic gates. |
|
Answer» Logic gatesperformbasic logicalfunctions and are the fundamental building blocks of digital integratedcircuits. Mostlogic gatestake an input of two binary values, and output a single value of a 1 or 0. These gates are the AND, OR, NOT,NAND, NOR, EXOR and EXNOR gates deeply 🤗 |
|
| 2. |
Hence, the Shake ts CtugaExample 35 A pole has to be erected at a point on the boundary of a circular park of diameter13 metres in such a way that the differences of its distances from two diametricallyopposite fixed gates A and B on the boundary is 7 metres. Is it possible to do it so? Ifyes, at what distances from the two gates should the pole be erected? |
|
Answer» Thank you so much but the question is "At what distances from the two gates should the pole be erected? "One is 5 m but what will be the other distance?? |
|
| 3. |
15Use the angle sum property to prove that\angle A C D=\angle C A B+\angle A B C |
| Answer» | |
| 4. |
८ है 1९4३ 2% ४ 9 ६2९4९] 1941:(० [8 0६ 1814 है 241० 24 ४ [2013 0४RoL (¢ [9 006 1ढ1% 216 9189 12॥24 9 24% ४ २०७३५ 0] 24%1- (५० [90 09 9106 फि "08 |
|
Answer» Subhash can copy 50 pages in 10 hours.That means in 40 hours he can copy 50×4=200 pages. Subhash and prakash together can copy 300pages in 40 hours. That means prakash alone can copy 300-200=100 pages in 40 hours. So to copy 30 pages prakash will take 40/100 ×30 = 12 hours. |
|
| 5. |
In a hexagon ABCDEF, side $ A B $ is a parallel to EF and $ \angle B : \angle C : \angle D : \angle E=5 : 4 : 2.1 $Find $ \angle B, \angle C, \angle D $ and $ \angle E $ . |
|
Answer» As AB || FE, FA becomes the tranversal line cutting parallel lines. So angle A + angle F = 180°. sum of other four angles = 720° - 180°= 540°.So 6 x + 4 x + 2 x + 3 x = 15 x = 540°.x = 36°. Angles B C D E are: 216°, 144°, 72°, 108° |
|
| 6. |
In a hexagon ABCDEF, side $A B$ is a parallel to EF and $\angle B : \angle C : \angle D : \angle E=5 : 4 : 2 : 1 .$Find $\angle B, \angle C, \angle D$ and $\angle E$ . |
|
Answer» As AB || FE, FA becomes the tranversal line cutting parallel lines. So angle A + angle F = 180°. sum of other four angles = 720° - 180°= 540°.So 6 x + 4 x + 2 x + 3 x = 15 x = 540°.x = 36°. Angles B C D E are: 216°, 144°, 72°, 108° |
|
| 7. |
\begin { equation } \begin{array}{r}{29} \\ {+\quad 21} \\ \hline\end{array} \end { equation } |
|
Answer» 29 + 21 = 50 50 ...................... |
|
| 8. |
By use of definition of limit, show that u 2x+3)-1 |
| Answer» | |
| 9. |
का खनिकलनन रा +By6X118 |
|
Answer» x^3 + 3x^2 + 6x + 18= x^2(x + 3) + 6(x + 3)= (x^2 + 6)(x + 3 |
|
| 10. |
\left( \begin array c c 2 & 3 \end array \right) \left( \begin array r r 1 & 2 \\ - 3 & 0 \end array \right) \left( \begin array l x \\ 3 \end array \right) = 0 |
| Answer» | |
| 11. |
\left| \begin{array} { r r r } { 1 } & { 2 } & { 9 } \\ { 2 } & { x } & { 0 } \\ { 3 } & { 7 } & { - 6 } \end{array} \right| = 180 |
|
Answer» it's matrices |
|
| 12. |
ABC is a triangle right angled at C if AB=25 cm and AC =7 cm.find AC |
|
Answer» Using pythagoras theorem (AB)²=(BC)²+(AC)²(25)²=(BC)²+(7)²625=(BC)²+49(BC) ²=625-49(BC)²=576BC=24 C is a right angleso AB^2=AC^2+BC^2BC^2=625-49=576so BC=24 cm |
|
| 13. |
In triangle ABC ,AB=5 ,AC=7 and x>BC>y find value of x and y |
| Answer» | |
| 14. |
A = \left[ \begin{array} { r r r } { 3 } & { - 4 } & { 5 } \\ { - 6 } & { 7 } & { - 8 } \\ { 9 } & { 0 } & { x } \end{array} \right] |
| Answer» | |
| 15. |
An almirah is sold at5,225 after allowing a discount of 5%. Find itsmarked p |
|
Answer» Marked Price = selling price/(1- (discount)/100)MP = 5225/(. 95)=5500 |
|
| 16. |
17. Maximum exterior angle possible for a regularpolygon isa) 120(e) 360(b) 270*(d) 450° |
|
Answer» Answer:a)120° |
|
| 17. |
measure of each exterior angle of regular polygon of 12 side isA.20°B.15°C.30°D.168° |
|
Answer» Sum of all angles of regular polygon is 360Measure of each exterior angle =360/12 =30 thsnks😀😁😂 |
|
| 18. |
( x + y ) p _ { 2 } = 56 \text { and } \quad ( x - y ) p _ { 2 } = 12 , \text { find } x \text { and } y |
| Answer» | |
| 19. |
The volumeofarightcircularcone is 9856 cm. If the diameter of the base is 28 cmfind0) height of the cone(ii curved surface area of the done"(ii) slant height°ftheaofthedone |
| Answer» | |
| 20. |
A semicircular sheet of diameter 28 cm is bent to make a conical shape. Findthe volume of this cone so formed.18. |
| Answer» | |
| 21. |
A serni-circular sheet of metal of diameter 28 cm is henl into an open couical cup.Find the depth and capacity of the cup |
|
Answer» Radius, r = 14 cmCircumference of semi circle = πr= (22/7) x 14 = 44 cmThis becomes circumference of base of cone2πR = 44R = 44 x (7/44)R = 7 cmThe radius of semi circular sheet = slant height of conical cupThat is l = 7 cmWe know that r2+ h2= l2142+ h2= 72196 – 49 = h2h2= 147Therefore, h = 7√3 cmThat is depth of the conical cup is 7Ö3 cmCapacity of cup = (1/3)πr2h = (1/3) x (22/7) x 72x 7√3 = 622.37 cu cm |
|
| 22. |
A semicircular sheet of diameter 28 cm is bent to form an openconical cup. Find the capacity of the cup. (Use sqrt 3 1.732.) |
|
Answer» diameter =d = 28 cmRadius= r = 14 cmCircumference of semi circle = πr= (22/7) x 14 = 44 cm so the circumference of base of cone2πR = 44R = 44 x (7/44)R = 7 cm The radius of semi circular sheet = slant height of conical cupwhich isl = 7 cmAs,r2+ h2= l2 142+ h2= 72 196 – 49 = h2 h2= 147 hence; h = 7√3 cm depth of the conical cup= 7√3 cm Capacity of cup = (1/3)πr2h = (1/3) x (22/7) x 72x 7√3 = 622.37 cu cm approx |
|
| 23. |
10 A semi-circular sheet of metal of diameter 28 cm is bent into an open conical cup. Determine thdepth and capacity of the cup. |
|
Answer» Slant height of the conical cup, 'l' = radius of the semi-circular sheet, R = 14 cmLet radius and height of the conical cup be 'r' and 'h' respectively.Circumference of the base of the cone = Length of arc of the semi-circleOr, 2πr = (1/2)2πROr, 2πr = (1/2)(2π)(14)Or, r = 7 cmNow, we know that l² = h² + r²(14)² = (h)² + (7)²h² = 196 - 49h =√147Height or depth of the conical cup = 12.124 cmNow, capacity of the conical cup = 1/3πr²h= 1/3*22/7*7*7*12.124= 13069.672/21Capacity of the conical cup = 622.365 cu cm |
|
| 24. |
2. Using opposite angles test for parallelogram, prove that every rectangle is a parallelogra |
|
Answer» Firstprovethat ABM ≡ CDMusingthe SAS congruencetest. (alternateanglesare equal): Hence ABCD is aparallelogram, because one pair ofoppositesides are equal and parallel. ... Hence ABCD isrectangle, because it is aparallelogram withone rightangle. |
|
| 25. |
9. The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelograis 60°. Find the angles of the parallelogram. |
| Answer» | |
| 26. |
What is the value of z in the figure shown below?3.6 3.6X1183.8 3.8 |
|
Answer» x=30°(alternate angle) |
|
| 27. |
निम्नलिखित बहुपदों के सम्मुख अंकित मान बहुपद के शून्यक है, सत्यापित कीजिए।(i) p(x) = -1; 3 = 1, -1 (i) p(*) =2x+1; 3=-(iii) p(31) = 4x +5;=(iv) p(x) = 33; 3 = 0(v) p(3) = (x-3)(x+5); x=3, -5(vi) p(3) = ar+b; x =-(vi) P(x) =3x’ -1; 3=-६६(vi) p(५) =3x+2; ४====\ |
|
Answer» i) x^2-1; x=1, -1; (1)^2-1=1-1=0, (-1)^2-1=-1=1; (ii)2x-1; x=-1/2; 2(1/2)-1=1-1=0, (iii)4x+5; x=-5/4; 4(-5/4)+5=-5=5=0; (iv)3x^2; x=0; 3(0)^2=3(0)=0; (v)=( x-3)( x+5); x=3, -5; (3-3)(3+5)=0(8)=0; x=-5; (x-3)(x+5)=(-5-3)(-5+5)=('8)(0)=0;(vi)ax+b; x=-b/a; a(-b/a)+b=a(1/a)=0; (vii); 3x^2-1; x=-1/V3; 1/V3; 3(-1/V3)^2-1=3(1/3)-1=0; 3(1/V3)^2-1=3(1/3)-1=0; p(x)=3x+2; x=-3/2; =3(-3/2)+2=0 |
|
| 28. |
construct a parallelogram ABCD in which AB -5.2 cm, BC 4.7 cmand AC 7.6 cm. |
| Answer» | |
| 29. |
\left(p^{4}+0 p^{3}-2 p^{2}+0 p+9\right) \div(p+3) |
| Answer» | |
| 30. |
| x3 (x+p}} (xp)310. If x, y, z are all different a11. z are all different and yº (y + p)3 (y - p)31 = 0, prove that| 23 (2+ p)3 (2-p)3 |p2(x + y + z) = 3xyz.od koThen operat |
| Answer» | |
| 31. |
- 3 p ^ { 2 } + 3 p q + 3 p x \text { from } 3 p ( - p - a - r ) |
|
Answer» 3p(- p - a - r ) - (-3p²+3pq+3px)-3p² -3pa -3pr + 3p² -3pq -3px-3pa-3pr-3pq-3px-3p(a+r+q+x) thank you |
|
| 32. |
In the adjoining figure, r:y:z 5:4:6. If XOYis a straight line, find the values of x, y and z.zo |
|
Answer» 5k+4k+6k=180;15k=180k=12Then the angles are x=5(k)=5(12°)=60° and y=4(k)=4(12°)=48° and z=6(k)=6(12)=72° I |
|
| 33. |
2 p + \frac { 1 } { p } = y \quad p ^ { 3 } + \frac { 1 } { \gamma p ^ { 3 } } = ? |
|
Answer» 2p+1/p=42p+1=4p2p=1∴p=1/2 p3+1/8p3=(1/2)^3+1/8(1/2)^3=1/8+1/ 8(1/8) =1/8+1=9/8 Bhai Apne to Google Ka ans paste kardiya |
|
| 34. |
4. What should be subtracted from 3x3 8x2+4r 3, so that the resulting expression hasx + 2 as a factor? |
|
Answer» -67 should be subtracted to make the polynomial have a factor (x+2) |
|
| 35. |
A shopkeeper sells an article at 12% loss. If2he sells it for Rs. 92.50 more then he gains,6%.What is the cost price of the article?A. Rs. 510C! Rs. 575B. Rs. 500D. Rs. 600 |
|
Answer» Let cost price be CP If article sold at loss of 12 1/2 %SP1 = CP(1 - 25/200)SP1 = 7CP/8 If article sold for Rs 92.50 more 7CP/8 + 92.50 = CP(1 + 6/100)53CP/50 - 7CP/8 = 92.50(212 - 175)CP/200 = 92.5037CP = 18500CP = 185000/37 = 500 (B) is correct option and |
|
| 36. |
A shopkeeper sells 6 eggs for30. What would be the cost of 10 eggs? |
|
Answer» cost of 6 eggs=30cost of 1 egg=30/6=5cost of 10 eggs=5×10=50 |
|
| 37. |
drilateral ABCD, the bisector of ZC andLD intersect at OProve that 2COD =-(ZA + 2B) |
|
Answer» Angle COD= 180°- 1/2angle C-1/2 angle D=180°-1/2(angle C+D)----------(1) Angle (A+B+C+D)=360° AngleC+D= 360°- AngleA- angle B--------------(2) Substituting (2) in (1) Angle COD= 180°-1/2(360°- angle A- Angle B) Angle COD= 180°-180°+1/2 Angle A + 1/2 Angle B= 1/2(Angle A+ Angle B) thanks |
|
| 38. |
3.In a quadrilateral ABCD, the bisector of ZC andZD intersect at OProve that COD A +4B) |
|
Answer» 1 |
|
| 39. |
Write the given set in roaster form 1)seven basic sounds of sargam |
|
Answer» Solution : Let set A is seven basic sounds of a sargam . List form of set A : i ) In Carnatic music : A = { Sa , Ri , Ga,Ma,Pa,Dha,Ni } ii ) In Hindustani classical : A = { Sa, Re, Ga, Ma, Pa, Dha, Ni } |
|
| 40. |
M. Prove that <МОС--nLe.21. The bisectors of ZB and ZC of an isosceles AABC with AB- AC intersecteach other at a point O. Show that the exterior angle adjacent to LABCis equal to ZBOC.I in line through P, parallel to BA |
|
Answer» If you find this solution helpful, Please like it. thankhs but ext angle ABD +angle B=180 |
|
| 41. |
10. In triangle ABC, LA-4B-15° and 2B-ZC-30°, find ZA.4B and ZC. |
|
Answer» 1st 2nd |
|
| 42. |
EXERCISE 6.3In Fig. 6.39, sides QP and RQ of A PQR are produced to points S and T respectively.If SPR 135° and Z PQT 1100, find 4 PRQIn Fig. 6.40, LX = 62°, XYZ = 54°. IfYO and ZO are the bisectors of< XYZ andLXZY respectively of A XYZ, find ZOZY and 2 YOZ.3. In Fig. 6.41, ifAB | DE, BAC 35° and ZCDE 53°, find L DCE.P0135B.▽35062°110530RYFig. 6.39Fig. 6.40Fig. 6.414. In Fig. 6.42, if lines PQ and RS intersect at oint T urhth |
| Answer» | |
| 43. |
17. OB and OC bisect ZB and ZC respectiely. If ZB=ZC, can you say that triangle OBC is isosceles.Justify you answer stating the axiom used. |
| Answer» | |
| 44. |
BE ACcinDE BIn the given figure, DBL BC, DEL AB and AC丄BC Prove thatA. |
| Answer» | |
| 45. |
zo m |
| Answer» | |
| 46. |
I.Distance of the centre of mass of a solid uniforvertex is zo. If the radius of its base is R and its height is h, thenZo is equal to(a) n(2015 Main)3h4Sh4R3/h8R |
|
Answer» option A , |
|
| 47. |
22. In the given figure, ABL BC, GFä¸BC andDE LAC. Prove that AADE ~ ÎGCF. |
| Answer» | |
| 48. |
(1) Write the symbolic statements in words for |
|
Answer» Representing the symbol statement given we have:- 4/3 is an element of set Q. |
|
| 49. |
(1) Write the symbolic statements in words for EQ |
|
Answer» Statement :4/3 belongs to set of rational number |
|
| 50. |
4(1) Write the symbolic statements in words for eQ |
|
Answer» Statement : 4/3 belongs to set of rational numbers |
|