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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. |
Write the structural formula of a cyclic carbon compound sby mentioning its name.2 |
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| 2. |
5. In the given figure, if AB = 14 cm, thenthe value of tan B is:EL-------------D13 cm5 cm(a)ما| در(a) 13 |
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Answer» option A is the right answer option a is the right answer option A is the right answer option A is the right answer A is right.............. A is right answer........... option (a) 4/3 is correct A is the answer for the solution Option A is perfect answer DC=5= AE(AC)^2+(DC)^2= (AD)^2(AC)^2+(5)^2=(13)^2(AC)^2= 169-25AC= √144=12SO ED= 12CMIN ∆ BEDACDE IS A RECTANGLE , AD IS DAIGONAL IT DEVIDE <ADE= <ADC<ADE= 1/2×90= 45°TAN45= BE/CD1= BE/12BE= 12 CMtanB= p/b= DE/BE= 12/12=1 |
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| 3. |
3. (a) The sum of three members is 6. If we multiply the third number by 2 and add the first number to theresult, we get 7. By adding second and third numbers to three times the first number, we get 12. Usedeterminants to find the numbers. |
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| 4. |
L) b bl & Lkl 2 बट b b 13 तौर(el 186. 0002. ४ हाह कह 288 & | 2406. ‘ST topiine |
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Answer» SI = P×R×T/100 = 1000× 5×1/100 = 50A = P + SI = 1000+50=1050 |
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| 5. |
(ay130seconds into min. & sec |
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Answer» 130 secs = 2 mins 10 secs 2 minutes and 10 seconds |
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| 6. |
)130 seconds into min. & se |
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Answer» 1min=60sec1sec=1/60min130sec=130/60min130sec=2min10sec or 2.167sec |
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| 7. |
● Construct a △ABC in which BC=3.6 cm, AB-5 cm and AC=54 cm. Draw theperpendicular bisector of the side BC. |
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| 8. |
A flooring tile has the shape of the pallelogram whose base is 24cm and the corresponding height is 10cm how many such tiles are required to cover the floor of area 1080msquare |
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| 9. |
Complete the following activity.Three coins are tossed.(2)n(S) =Let A be the event of getting at least two tailsP(A) |
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| 10. |
37. The length and the breadth of a rectangular paper are 35cm and 22cm Find thearea of the largest circle which can be cut out of this paper |
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| 11. |
8. Solve for x:7x +14 17-3x4x + 2йН |
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Answer» 1 2 |
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| 12. |
Solve for x7x +14 17-3x |
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| 13. |
A tile measures 10cm x 10cm. How many such tilesare required to cover a wall 4m x 2.5m |
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Answer» Solution :- Dimension of tile = 10*10 sq cm Area of tile = 100 sq cm converting the area in m = 100/(100*100)= 0.01 sq mdimension of wall = 4m * 2.5marea of wall = 10 sq mno. of tiles = (area of wall) / (area of tile) = 10/0.01= 1000no. of tiles required = 1000 please tell me some other answers |
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| 14. |
The floor of a roof is of dimension 4m by 6m, tiles of size 10cm by 12cm have to be lay on it. how many tiles required covered the floor |
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| 15. |
Three different unbiased coins are tossed together, findtheprobabilityofgellngleast two tails (ii) at most two tails. |
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| 16. |
A rectangular sheet of paper 40cm X 22cm is rolled to form a hollowcylinder of height 40cm. What is the radius of the cylinder? |
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| 17. |
26. The centres oftwo cikesa cincle to anodher circleofrads 5 cm and 3 cm are 9cm apart Draw a pair oftangents from the centre of21, lsthegiven fiure MNOPis a trapscium having aroa 24 5 cm2 and MN JOP MN-10cm and OP-4mFend the area ofthe shaded region ifMSP is a qusdrant |
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Answer» 26)STEPS OF CONSTRUCTION: 1.Draw a line segment AB = 9 CM2. Draw a circle with Centre O and radius 5 cm and the circle with Centre B and radius 3 cm.3. Now bisect AB. Let O be the midpoint of AB.4. Take O as Centre and AO as radius and draw a dotted circle which intersects the two given circles at N, Q, M and P.5. Join AN, AQ,BM and BP. These are the required tangents to each circle from the centre of the Other circle. |
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| 18. |
(a) Three circles with centres A,B and C touch each other externally. If AB-5 cm BC-7cmCA-6cm, find the radii of three circles |
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| 19. |
23. Two different dice are thrown together Find the probability that the numbers obtained have(i) even sum, and(ii) even product |
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| 20. |
A train is booked to do the run between two places, 55 km apart, in 1 hr 20 min. If it travels forthe first 30 km at 36 km an hour, at what speed must it travel for the rest of the distance in orderto complete the journey in time? |
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Answer» A----------------------------B55km= ABTotal time= 1 hour 20 minTotal distance travelled= 30kmS1= 36km/hrT1= distance travelled/S1=30/36=5/6 hr=50minremaining time= 40 min = 4/6 hrremaining distance= 25kmrequired speed= 25/(4/6) km/hr=25*6/4 km/hr=150/4 km/hr= 37.5 km/hr |
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| 21. |
Ifittravelsfor16. A train is booked to do the run between two places, 55 km apart, in I hr 20 min.16. A trainthe first 30 km at 36 km an hour, at what speed must it travel for the rest of the distance in orderto complete the journey in time? |
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| 22. |
Gl\/"_—x 2 vi कि |
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| 23. |
The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member ? |
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Answer» Let the sum of nine member (total) =xand the age of old one=zso its average 4 yrs before=(x+z)/10.after 4 yrs let z be replaced by y.so now avg=(x+4*10+y)/10 now (x+z)/10=(x+40+y)/10so after solving it foundz=y+40.so old person is 40yrs older than young one. |
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| 24. |
EXERCISE 2.21. Find the value of the polynomial 5x - 4x2 + 3 at(0) x=0(iii) x=2Find 0) (1) and (2) forcach of the following nolu |
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Answer» The given question is wrong and also the three options are not answer of given polynomial. incomplete question At X=0hence5(0)-4(0)^2+3=3At X=-15(-1)-4(-1)^2+3=-5-4+3=-9+3=-6At X=25(2)-4(2)^2+3=10-16+3=-6+3=-3 |
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| 25. |
(लक s A s4: सोनू दर मोनू की o g 9 43 818 a%‘: P i)5.6 होंगा। उनकी वर्तमान g S R gl | |
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Answer» अनुपात 4x and 5xnow4x+5/5x+5= 5/624x+30= 25x+25x= 5अनुपात= 4*5/5*5= 20/25 |
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| 26. |
if there are 1500 people composed with soldiers,captain.,each group has 15 soldiers and 1 captain ,so how many captain are there |
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Answer» For 15 soldiers there is 1 captian So, for 1500 soldiers there are 1500/15 = 100 captins |
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| 27. |
5 Simplify the following9-5 +43 -(5 - 8-6) - (-3+212. 627)x615) + (-27) X (+14) = 93. 7-16-5+(1-3+ 4)]y (-36) XG5+ (-) = (-30)4.2 +(-36% (-)Find a?5. Match the followingColumn B+36Column AСхо- 5+(-7-3)360 GL5012 (501)nt defined50g. In an election in which 5500 votes wereCasti Gne kandidate got 2 3 of themany votes - did theNotes Lost HowCandidates gott get a |
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| 28. |
90% of 130 |
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Answer» 90% of 130= (90/100)*130= (9/10)*130= 9*13= 117 90% of 130 = 117 |
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| 29. |
TR =1 SRR gL YRR e WY0. यदि ८ तथा 9 द्विघात बहुपद /ए<) न + 2: + 3 का मूल हों , तो हू का पान ©2 ®-2 Ans—(B)3 .4gl |
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| 30. |
कुक? 00gl |
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Answer» (-a-b)*(-x)= -(a+b)*(-x) ax+bx |
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| 31. |
e .(‘ '(“gl (fl |
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| 32. |
In the given figure. AB IICD and a transversal EF cuts them at G and HespectivelyGL and HM are the bisectors of the alternate angles ZAGH andGHD respectively, prove that GL II HM. |
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| 33. |
1. Given f(x)=2x-x? Find (i)f(1)(ii) f(x+1)iii) f(x) + f(1) |
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| 34. |
o] oGl % |
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Answer» 15 multiplied by 3 and then x= 45/9 i.e = 9 |
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| 35. |
If f(x) be defined on [-2, 2) and is given by f(x) =C S-1x-1, -25% 0, 0<x<2and g(x) = f(x) + f(x)]. Find g(x). |
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| 36. |
14) A function f is defined as follows:f(x) = 5-x for 0 SX4Find the value of x such thatf(x) = 3 and f(x) = 5 |
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Answer» f(x)=5-x; f(3)=5-3=2; f(5)=5-5=0 |
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| 37. |
If f : R--> ( 0 , 2 ) defined byXx-Xf(x)=-e+ 1isinvertible find f-1(x) |
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| 38. |
2x+3y = 12 ; 2x+3y=62x+3y = 12 |
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| 39. |
8.Find x t y+z90130 |
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Answer» z = 180° - 30° = 150° ( linear pair) x = 180° - 90° = 90° ( linear pair) y = 90° + 30° = 120 ( exterior angle theorem) x + y + z 150° + 90° + 120° 270° + 90° 360° |
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| 40. |
= (x - )-8, find the values of ((x+5) 6 (x- |
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| 41. |
ple-16. Rationalise the denominator of -4+55 |
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Answer» 1/4+√51×(4-√5)/(4+√5)(4-√5)4-√5/(4)^2-(√5)^24+√5/16-54+√5/11 1/4+V5 x4- V5/4-V5=4-V5/16-5 = 4-V5/11 |
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| 42. |
x^-3/4=1/8 find x? |
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Answer» x^-3/4 = 1/8 (1/x)^3/4 = 1/8 1/x = (1/8)^4/3 1/x = (1/2)^3*4/3 1/x = (1/2)^4 1/x = 1/16 x = 16 Therefore, value of x = 16 |
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| 43. |
Find the equation of the plane which contains the line of intersectionof the planes ř.(it2J + 3k)-4-0, ř.(2ī +j-k)+5-0 and isperpendicular to the plane r.(5i +3j-6k)+8-06 |
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| 44. |
Solve the following simultaneous equations by cross-multupilationiol35. 3x + 4y = 1 0 and 2x-2y = 26 2x +3y 11 and 2x-4y-24 |
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Answer» 2x+3y = 11 2x-4y = -24 subtraction second equation from first equation 2x+3y-2x+4y = 11+24 7y = 35 so y = 5 2x+3(5) = 11 2x+15 = 11 2x = -4 so x = -2 |
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| 45. |
15. Three lines x +2y +3-0; x +2y-70 and2x-y-4=0 form the three sides of twosquares. The equation to the fourth side of eachsquare is1) 2x -y +14 0 and 2x-y+602) 2x-y+14 = 0 and 2x-y-6-03) 2x-y-14 0 and 2x-y-602x-y-14 0 and 2x-y+6 0 |
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Answer» 2 answers·Mathematics Best Answer Let's get them in point slope form... y=(-x-3)/2, y=(7-x)/2, y=2x+4 notice that the first two have the same slope of -1/2 this means that they are parallel and are one pair of opposite sides... thus the missing line must have the slope of the 3rd line above...which is 2x. y=2x+b and must meet lines one and two above.. Now let's see how far apart dy is between the opposite sides... y=(-x-3)/2, y=(7-x)/2 we can set both x=0 zero to see their separation in height... -3/2 and 7/2 so dy is (7--3)/2=5 this means that our parallel sides must have a dx of 5.... our missing line y=2x+b must differ in x by 5 from y=2x+4 when the y values are equal, we'll just set them equal to each to each other... 2x+b=2x+4 let x1=0 and x2=5 b=10+4=14 So now we know that the missing line is: y=2x+14 so our line is y=2x+14 |
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| 46. |
-)EXERCISE 5 (A)Factorise by taking out the common factors:13.abla + -)-bele-4-b) + ca(+1. 2(2x - 5y) (3x + 4y) - 6(2x - 5)(x - y)2.4. 2x(a - b) + 3y(5a - 5b) + 43(2b - 2a)r-2-) - (2y2 - 31-) + 2x(1522 - 104 |
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| 47. |
solve for x and y 8/2x-3y+21/2x+3y=11 , 5/2x-3y+7/2x+3y=6 |
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| 48. |
√(2x +√(2x-4))= 4. in this equation the value of x is(2) 4(4) 66.(3) 5 |
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Answer» √(2x+√(2x+4)) = 4 by doing trial and error lets take x = 6 √(2×6+√(2×6+4)) = √(12+√16) = √(12+4) = √16 = 4 |
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| 49. |
and Learnhat are the missing numbers?17 x= 17(b)x 30 |
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Answer» 17_=17=x=17/17=17÷17=1 |
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| 50. |
((-5)/17)*(b*(y*(51/((-60))))) |
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Answer» -5/17*51/-60=1/12*3=1/4 |
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