Desi Publications

1. If tan A =3/4 then sinA cosA Ans- sina=3/5 cosa =4/5 Tan A=3/4. So it is sinA/cosA=3/4 so sinA=3 and cosA=4 so ans is 12. Where ois the fault 12/25

2. (i)A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60 with the wall, find the height of the wall (ii) An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer. Ans- yes 7.5 and 45 right ans

3. If tan + sec = L , then Sec in terms of L..????

Ans- sec^2 - tan^2 = 1 (sec+tan)(sec-tan) = 1 Add the two eq to get 2sec = L+1/L l^2+1/2l...right ans

4. In a right angled triangel the hypotenuse is four times as long as the perpendicular drawn to it from the opposite vertex .One of the acute angle is???

Ans - Let the triangle be ACB, CP be the perpendicular to AB AC = c, CB=b, CP=h,PB = x and AP = a-x c2 = h2 + (a-x)2

b2 = h2 + x2

on adding, we have 2h2 + x2 + (a-x)2 = a2 ..(1)

also a = 4h and cos B = b/a = x/b

so x = b2/a

on solving (1), we have h2 + x2 -4hx = 0

substituting above gives

b4 + a4/16 -a2.b2 =0 .. (2)

as cos B = b/a on solving eqn. (2) we have , Cos2B = 0.933 or 0.067 Cos B = 0.966 or 0.259 B = 15 or 75 degrees

5. (sin^4 cos^4 +1) cosec^2 = ?? 1 -1 0 2

Ans- 2 is right

6. sin^6 + cos^6 + 3sin^2 cos^2 = ?? 1 -1 0 none of these

Ans- 1. (sin^2@ + cos^2@)^3 = given expression = 1

7. 2sin@ + 15cos^2@ = 7 then cot @ =? 5/4 3/4 1/4

Ans- uadritic hai bhai solve kar lo change cos^@ into sin^2 eq vl b 15sin^2 - 12sin -8 = 0 2sin@ + 15(1-sin^2@) = 7 ........now its quad eq

yesterday glady da told me one thing...it can b done easily by option...cot@=3/4, then sin@=4/5 n cos@=3/5....so again desi method by checking option

8. Sin x + sin. ^2 x=1 Find value of Cos^12 x + 3cos^10 x + 3cos^8 x + cos^6 x?? 1 2 3 0

Ans- (cos^2@ + cos^4@)^3 ...........from given we can deduct sin@ = cos^2@ ...........so ans is 1

9. tan@ + sin@ = m and tan@ -sin@ = n then the value of m^2 - n^2 is a) 2rtmn b) 4mn c) 4rtmn d) 2rt(m/n) Ans- 4rtmn bhai put values of m and n ..... 4rt(tan^2-sin^2) vl give 4sin@tan@

10. Given that sin + 2cos = 1, then 2sin cos=?? Ans- @=30 15/4 haan yaar 15/4

11. if cot @ = 8/15, then wat is the value of rt(1-cos@/1+cos@)??? Ans - cos = 8/hypotenus will come. cot me hypotenus nahi rehta

cot=adj side/opp side, so here 8/15, and u get hyp=rt(8^2+15^2), then u ll get cos=8/hyp now apply this in equ. u ll get answer

b = 8 p = 15 h = 17 Ans will be 3/5....:) welcum shashi bhai

12. SinA= -3/5 and pie

15. if a,b,c r non zero real no. Such dat a+b+c=0 and b^ ca,then d value of a^+b^+c^/b^-ca is Ans- again a typo error if it is b^2 = ca then it would be infinity ans options r 3,2,0,1

16. If ABC is right angled at C, then the value of cos (A+B) is Ans- 0 agar C = 90* then A+B to 90* hoga hi na

17. The angle of elevation of the top of a tower 30 m high, from two points on the level ground on its opposite sides are 45 degrees and 60 degrees. What is the distance between the two points? (1) 30 (2) 51.96 (3) 47.32 (4) 81.96 Ans- tan45=30/x X=30 Tan60 = 30/y Y=10root3 Reqd ans =30+10root3 =30+17.3 =47.3..:)

18. cos^4x/cos^2y + sin^4x/sin^2y = 1 then what is the value of

cos^4y/cos^2x + sin^4y/sin^2x a.4 b.0 c.1/8 d.1

plz expain also Ans - 1 put x=y=45

19. The sides of a triangle are a,b, rt( a^2 + b^2 + ab), the greatest angle is? Ans - angle opposite to rt(a^2 + b^2 +ab ) will be greatest .............i think it may denote as C

20. If cosA-sinA=rt2*sinA,then cosA+sinA=? Options: rt2*sinA 1/2 rt2*cosA 1 Ans- rt2 cosA. (cosA-sinA)^2 + (cosA+sinA)^2 = (root2*sinA)^2 + x^2 Find d value of x, thats ans.

21. the minimum value of cos 2A + cos A is ? ans- cos2A + cosA = (2cos^2A - 1) + cosA = 2(cosA + 1/4)^2 - 9/8 so minimum value=-9/8

2 cos ^2 A -1 + cos A = 2 ( cos ^2 A +1/2* cos A ) - 1..

= 2 ( cos ^2 A + 1/2* cos A + 1/16) - 1 - 2*1/16 ( add and substract 1/16) = 2 ( cos A + 1/4) ^2 - 9/8... so - 9/8 is the minimum value.. Thanks for a detailed solution. Not of 10th stand, but of ssc stand

22. The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:

Ans- iss bar ssc ne hadh hii kar di 10th class mein jo 10-12 min leke bade aaram se karte thai speed nikalne ke height and distance mein woh dal diye :) shadow means base =3 Perpendicular = 1 Tan = 1/3 bhai to angle 60 kaise aaya tan^-1(1/3) aana chahiye na..:p wrong ques i guess 30 degre off course 30 degre

23. if m ( tan A-30) = n tan ( A + 120)....then, (m+n)/ (m-n) = a. 2 cos 2 A, cos 2 A, 2 sin 2A,none of these.. ans- m/n = tan(A+120)/tan(A-30) By C and D m+n/m-n = tan(A+120)+tan(A-30) /tan(A+120) - tan (A-30 ) Lamba ja raha hai...:p Ans- convert tan in to sin / cos .....then apply sin ( A+b ) formulae.. convert tan in to sin / cos .....then apply sin ( A+b ) formulae.. is it 2cos2A..i did the same steps upto shashi and then put A=60 in equations and then in options..only option a satisfies it...

24. if y = 2 sin A / ( 1+cos A +sin A), then ( 1-cos A + sin A ) / ( 1+ sin A ) =? in terms of Y? Ans- multiplying by conjugate ans is y

25.If sin^2 A = cos^3 A,then find the value of ( cot^6 A - cot^ 2 A ) is? 1 0 -1 2 Ans- 1...simply convert cot to cos/sin and put cos^6 = sin ^4...then substract didnt get sin^4@/sin^6@-cos^2@/cos^3@ then?

1/ sin ^2 A - cos ^2 A/ sin ^2 A = ( 1- cos ^2 A)/ sin ^2 A = sin ^2 A/ sn ^2 A

26.If A=sin^4Q+cos^4Q then ( Q-thetha)

1. 0

27. The chord AB of a circle of centre O subtends an angle theeta with the tangent at A to the circle. ang ABO is..? a) theeta b) 90- theeta c) 90+ theeta d) 2(180- theeta) Q2--- The sum of angles in thr four segments exterior to a cyclic quadrilateral is equal to ( in degres)..? a) 450 b) 360 c) 540 d) 720 Q3--- Triangle ABC is an isosceles triangle in which AB = AC. A circle through B touches the side AC at D and intersects the side AB at P. If D is the midpoint of the side AC. Then AP is..? a) AB/4 b) AD^2 c) AP = AD answers....1 . 90 theta 2. 540 3. AB/4

28.find the min. value of 4Cot^2 A+9tan^2 A

my ans is -12 sandes, why not -12? As we can also write d expression as ( A+ B )^2.. here 4Cot^2 A+9tan^2 A = ( 2 cot A +3 tan A )^2 12

to get -12, u need to equate 2 cot A + 3 tan A =0...so 2 cot A = -3 tan A ..or cot ^2 A = -3, which will become complex...so u can not, i think , equate 2 cot A + 3 tan A to zero.. use AM >= GM...one line u can get min value as 12 for any two positinv nos..... arithmetic mean >= geometric mean....let the nos. be a and b then (a+b)/2 >= under root of a*b...here a = 4cot^2A and b = 9tan^2A now use this

29.The area of the traingle formed by the lines 5x+7y=35, 4x+3y=12 and a-axis is a) 160/13sq units b)150/13 sq units c)140/13sq unit d) 10 sq unit Ans- get the intersection point of the two lines by solving them. The Y cordinate would give the ht of the triangle and at y-o for each line would give x get the base. Then area of triangle= 1/2 b*h how do u solve this type of sums? subst. values to all 3 eqns, form triangle or any fig and then apply mensuration formulas? first put y=0 in both eqns for 1st it wud give 7 for other x=3 these pts give intersection pt on x axis and make base the distance is 7-3=4. now solving bothe eqns by subst. or elimn get y=80/13 . this is the perpendicular ht of third pt. Now applying formula = 1/2 base8 ht. gives 160/13

30. First ques here find the value of 4sin50-rt3tan50 Ans- dont think ssc will go to this extant ......we hv to leave this question in exam if it comes Ssc ka koi bharosa nhi, kaise v ques aa sakte h may be 1////right?

31. The area of the triangle formed by the three graphs of the equations x=4, y=3 and 3x+4y=12, is? Ans- 6 shud b the ans the equation 3x+4y itself makes x and y intercept as 4 and 3...so answer ...6*4/2=12 how 12 bro? sorry 3*4/2 =6.... A=b*h/2=4*3/2=6 32. The radius of the circumcircle of the triangle made by x axis, y axis and 4x+3y=12 is? Ans- abc/4 A......3*4*5/4*6 or simply for a right angled triangle its half the hypotneus......so 5/2 = 2.5.... Yes. abc/(4*Area)

33. 4sin50-rt3tan50 =2sin100-rt3sin50 =2cos10-rt3sin50 =2cos(60-50)-rt3sin50 open the bracket of cos n ans is 1 =4sin50cos50-rt3sin50 very gud explanation....i liked it very A good Question.. there is no pt. in lyking or disliking a ques..ssc wont ask us while setting paper I dont understand it and its beyond my ambit.

34. The number of solutions of the equation sin 5 A cos 3 A = sin 6 A cos 2 A in the interval [ 0, pi] Ans- yaar yeh sab engg k questions hai nahi poochenge....do lines ka solution hai y = first line and y = second line intersection ka

35. The greatest number of 5 digits to be added to 8321, so that the sum will be exactly divisible by 15, 20, 24, 27, 32, and 36 is?

Ans- 99679 CGL qustn..

36. sin 2 A + sin 2 B = 1/2, cos 2 A + cos 2 B = 3/2, then cos ^2 ( A -B ) = ? Ans- simple apply the formula of sin ( c+d) and cos ( c+d)...square and add...will get the answer as 5/8....

is there any direct formula for cos ^2 ( A -B ) ? CosA+B*cosA-B =

divide the two eq u vl get tan A+B = 1/3

obtain cos(A+B) from this which vl be 3/rt10 now put this value in above eq u vl get cos A-B = rt10/4

Suare to get the answer

37. if 2cos^2@ + 3sin@ = 0 then sin@= ?

4 2 3 none of the above

Ans-value of sin@ always lies between -1 and 1 so none of the above

ya its none of the above but i wanted to trick you guys by putting this kind of option .....................ans is -1/2

38. Find the value of theeta for maximum value of sin 3 theeta + cos 3 theeta..?

Ans- 15 3q=45 q15

39. If A+B+C=180degree then what is the value of cotA.cotB+cotB.cotC+cotC.cotA

Ans-A= B = C = 60 Ans will be 1....

tanA+tanB+tanC=tanA.tanB.tanC

Cot(a+b+c) ka 40. 5cos@ + 12sin@ = 13 then tan@ = 13/12 12/13 12/5 5/12

Ans- 5cos@+12sin@=13 5+12tan@=13sec@ (divide both side by cos@) 25+144tan^2@+120tan@=169(1+tan^2@) [squaring both] 25tan^2@ - 120tan@ + 144 = 0 (5tan@ - 12)^2=0 tan@=12/5

cos^2 A + sin ^2 A =1...for that 5/13 = cos A and 12/13 = sin A ( same hyp).... so tan A = 12/5

41. Find the value of (cos18-sin18)/ (cos18+sin18) Ans-45-18=27 take tan both d sides n use d formula 4 tan(a+b) Maveric sir's changla(best) way.... Sin45cos18-cos45sin18/cos45cos18+sin45sin18=sin(45-18)/cos(45-18)=sin27/cos27=tan27 In this type que s divide all by cos...here cos 18... which will give, ( 1 - tan 18) / ( 1+ tan 18) = tan ( 45-18) = tan 27 = cot 63 42. If cos(x-y)= -1, find the min +ve value of sinx+siny Ans- 0

x=180 y=0

x-y is 180 x is 360 y is 180 therefore sin360 plus sin180 is zero

ek bar saare trigo functions ke graph dekh lo kaafi help milti hai

43. One more of same type find d value of (cos17+sin17)/ (cos17-sin17) Ans-hey divide by cos 17, u will get ( 1+ tan 17) / ( 1- tan 17) = tan ( 45+17) = tan 62...= cot 28

44. If tan A = ( root 2 - 1), then, sin A cos A is..? Draw a right angled triangle opposite side rt2 -1 adj side 1 find the hyp and solve it.

1/2rt2 is the right ans

45. Find d max value of 5cos@ +cos(@+60)+3 Ans- At first open the bracket 4 cos,put the value of cos60 n u'l get 13/2*cos@-3rt3/2*sin@ +3 =sqrt of sqre of coef of cos n sin+3 =7+3 =10

46. if MN is 10 meters in length and is tangent to one of the concentric circles at point A. If it is given that the raddi of the both the circles are integers, then what is the radius of the inner circle.

Ans- bhai circle draw karo... with chord MN as tangent to the inner circle... use pytahogorean triplet

Ans- 12

47. tan75-tan30-tan75tan30 is..?

Ans-use tan(A-B) formula where A = 75 and B =30 48. (sin x +sin 3x +sin5x +sin7x) / (cos x +cos 3x +cos5x + cos7x) = ? Ans- put x zero

1) tan x 2) tan 2x 3) tan 4x 4) tan 8x

tan4x

49. cos 55 + cos 65 + cos 175 is..?

Ans- 0

50. Find tan 20 +2tan50..?

Ans- tan(50+20) formula use karlo tan(A+B) wala u vl get the answer Yes n d term tan70tan20= cot20tan20=1 ye concept v use hoga

Yes manish wo type err tha nw edited

tan 70 * tan 20 = 1 ??

I think short cut for this sum is- Do addition & ans is tan70

51. Ram bought 6 mangoes, 8 oranges and 10 bananas for a certain amount of money. Ramesh bought 15 mangoes, 20 oranges and 15 bananas for the twice the amount that hari spent. What percentage of money spent by hari for buying bananas ?

Ans- 6m+8o+10b =x .......multiply 2.5 both side it will be 15m+20o+25b =2.5x where as it is given 15m+20o+15b = 2x................now 10b= 0.5x so hari spent 50%

bhai i think ram ke place pe hari ayega

typo error

hehe.. :-P

CID !!

daya darwazaa tod do :)

52. x^1/3+y^1/3=z^1/3 than x+y+z+27xyz -1 0 1 27

Ans- 0

how?

if a + b + c = 0,,, then a^3+b^3+c^3 = 3abc...

But I think the problem will be -z^(1/3) Then we can go by cubing both sides......

cubing both sides but that process is slightly time taking

Yes thats ok. But z + or -?

+z

hahaha................

Sorry guys mere toh ans nahi aa raha hai :( The problem will be If x^1/3+y^1/3=z^1/3, then the value of (x+y-z)^3 + 27xyz ? time wast type q of ssc

thanks for pointing out error ..+2 for all and who have wasted tym sorry

Cubing both we get, x+y-z= -3*x^1/3*y^1/3*z^1/3 So (x+y-z)^3 + 27xyz= -27xyz+27xyz=0

53. X+Y = 45, Then (1+tanX) (1+tanY)=? 1)1 2)0 3)2 4)-1 Ans- put x =0 and y=45 or x=45 and y=0...answer either 2 or 1/2...jugaadu method hai 54. sin 20 - sin 70 + sin 10 = ?

1) 1 2) 0 3) -1 4) 2 Ans-use sin C + sin D, for sin 50 & sin 10, and convert sin 70 into cos 20.

bhai tumne 70 likha hai 50 nahi

sorry typo error its sin 50 not sin 20 in Ques.

55. in triangle ABC ,angle A,B,C are in A.p then (SinA - SinC)/(CosC-CosA) is equal to

a)Sin B b)Tan B c)Cot B d)Tan(A+B/2) Ans- 2cos (a+c/2) sin (a-c/2) /2sin (a+c/2) sin (a-c/2) =cos(a+c/2)/sin(a+c/2)=cos b/sin b = cot b

AP bola hai na.. so take 30 60 and 90,, u cn still get

56.Cos20cos40cos60cos80=? a. 3/80 b. 2/16 c. 5/16 d.1/16

Ans- Ya its 1/16, aise ques ssc me ayenge to mai to gyi :( shortcut for this tye of que....4 Sin(60-A).SinA.Sin(60+A)=Sin(3A) 57. The length of the portion of the straight line 8x+15y=120 intercepted between the axes is ans- 14unit,15unit,16unit,17unit, plz with some explanation.

Ans-17.. x/15+y/8 =1 rt(15^2+8^2)

58. The no 1,2,3,4,.....1000 are multiplied together.The no of Zero at the end of product will be (kindly explain) Ans-1000! it is ........no of zero would be equal to factor of 5 .......so 1000/5 + 1000/25 +1000/125 +1000/625 = 249 59. If cos theeta - sin theeta = rt2 sin theeta, then cos theeta + sin theeta is..?

Ans-root 2 cos theta

60. tan20 + tan 40+ rt3 tan20tan40 is..?

Ans- rt3

61. (1-1/3)(1-1/4)(1-1/5)----(1-1/25) = ?? Ans- 2/25 62.cot x + cosec x = 5, then cos x= ? a) 5/12 b) 0 c) 1 d) 12/13 Ans- 5 ,12 ,13 triplet ...........always rremember

1+cosx/sinx= 2cos squrx/2/2 sinx/2cosx/2 we get cotx/2=5 and tanx/21/5 and now apply the formula of cos x= 1-tan^2x/2/ 1+ tan^x/2

63.If (sin(x+y))/(sin(x-y)) = (a+b)/(a-b),then a tany is..?

Ans- b tanx apply componendo and div idendo

64. Cos20cos40cos60cos80=? a. 3/80 b. 2/16 c. 5/16 d.1/16

Ans- Cos20Cos40Cos80=1/8 and Cos60=1/2 so....1/16 Cos20cos40cos80 ? 1/8 kaise ??

1/2sin(20)][2sin(20)Cos(20) Cos(40) Cos(80)] [1/4sin(20)][2sin(40)Cos(40) Cos(80)] [1/8sin(20)][2sin(80)Cos(80)] [1/sin(20)](1/8)sin(160) 1/8 as sin(160) = sin(20)

65.tan x = 5/6 & tan y = 1/11, then x+y=?

Ans-use tan(x +y ) = (tan x + tan y)/1-tanx tany and then take inverse x+y=tan( inverse) 1=45

66. 5^3+6^3----10^3 = ??

Ans- yaar calculation karwaeoge......use cubic series expansion frm 1-10 and subtract 1-4 ...formula [n(n+1)/2]^2 67. find the value of sin20sin40sin80?

Ans- rt3/8is right

short cut formula...sinAsin2Asin4A = (1/4)*(sin3A) 68.If sin x + sin y = a and cos x + cos y = b, then tan ( x+y)/2 =? Ans-it is a/b

69.cos ^2 ( (@-$)/2 ) - sin ^2 ( (@+$)/2) is..? a) cos$sin@ b) cos 2@sin$ c) cos@cos$ d) sin@sin$ Ans-c....cos @ cos $ convert cos ^2 in to (1+cos 2A )/2 and sin ^2 in to (1-cos 2A)/2 simply cos(A+B) cos(A-B)

70.The value of rt(2+rt(2+2cos4theeta))..? Ans -2cos theta

71.if 6 multiplied 15 = x10 then value of x=? 1)3 2) 3 3)3 4) 6 Ans-it should be +3 or 3

yahan + and - wala concept nahi aayega so only 3 will be the ans

bhai ye bhi batao k kyu nahi aayega?? we need explanation

simple h yar rt90 is not equal to -rt90

nope.... rt(9) = rt(3) * rt(3)..this is cool... but if we take it as rt(-3) * rt (-3) = rt.3. i * rt 3 .i.. where i = rt(-1).. then LhS not equal RHS

72.if 7x + 3 y + 9=0 and y = kx +7 are two parellel lines, then k =?

Ans- k= --7/3

for perpendicular m1x m2=-1 and for parallel m1=m2

73.Rd sharma's sum in triang ABC,

77. If min hand of a clock is 10cm long,how far the hand wil move in 20min?

Ans- 1/3 * 2* pi* 10

2pie r *120/360 = 20pi/3

78. 1/20 + 1/30 +1/42 +1/56 --------1/132 = ???

Ans- expand as 1/4*5 + 1/5*6 +1/6*7 and so on.....separate the numerator and denominator by lcm and cancel all the terms....the one left wud be the answer

1/4-1/5+1/5-1/6+1/6-1/7+......1/11-1/12 = 1/4-1/12 = 1/6

79. Value of sin36 ?

Ans- 2 sin 18 cos 18....sin 18 = ( rt 5-1)/4 and cos 18 = rt( 10 + 2 rt 5 )/4....I think this kind of sum will not be asked

Just posted ..to highlight the importance of these value....sin36 cos36 sin18 cos18

80. Find d value of tan75

Ans- tan(45+30)= 2+rt3 (rt3+1)/(rt3-1)=2+rt3

81. given the set of numbers, n>1, of which one is 1-1/n and all others are 1. The arithmetic mean of the n numbers.

Ans- answer is 1-1/n^2

82. The Graph of x^2 -4y^2=1 parabola elipse or hyperbole

Ans-hyperbola.... out of sylllabus, i tink..

Yar aise quesn na put kro jo syllabus me nhi h Guys you people are solving sums with so much speed........i am getting inferiority complex... :P

83. If tan A = 1/2, tan B = 1/3, then cos 2A is a) sin B b) sin 2B c) sin 3 B d) cot 3 B Ans- sin 2B

cos2a=1-tan^a / 1+tan^2a = 1-1/4 / 1+1/4 = 3/5 sin 2b = 2/3 / 1+1/9= 2/3 * 9/10 =3/5

84.if (x+1) and (x+2) be the factors of x^3+(a+1)x^2-(b-2)x-6 then the values of a and b will be

Ans- ans is 2 and 8

Sandes bro m also getn a=-1 b=9

there is no option -1 and 9

please then check the ques

thik hai yar its (x-2) Aj k liye bahut trigno ho gyi,kuch eng ka post karen My humble requst to all puys plz give at least 1 discriptiv answr fr each questn.so ppl finding dificulty can undrstand

Those who post question please give answer with explanation yaar .

Puys....fb pe wo feel ni aa rha jo PG par aata tha....PG pr new puys post kiye ja rhe h bt sare senior puys yaha h....apna group 2 parts me divide ho gya h yr....

85. The value of cos 52 + cos 68 + cos 172 is..?

Ans-0

cosa+cosb=2cos(a+b/2)cos(a-b/2) cos52+cos68=2cos60cos8 + cos 172 cos8+cos172 = 2 cos90cos82=0

@godbles 2cos(52+68/2)cos(68-52/2)+cos172=2cos60cos8+cos172=cos8+cos172=2cos(172+8/2)cos(172-8/2)=2cos90cos82=0

86. Base of triangle is 80, n one of base angle is 60,the sum of length of othr two sides is 90.the shortest side is?

Ans- Draw it.. b+c =90 By cosine rule, b^2 = 80^2+c^2-2*80*cos 60 b = 90- c. will get 17. 87. if x+y=90 degree and sinx:siny= root 3:1 then x:y is equal to

Ans- y = 90-x

---> sinx /cosx = root 3 : 1

---> tanx = root 3

x = 60 hence y = 30 ratio 2:1 88. if (x/y)=(z/w) then (xy+zw)^2 is equal to 1.(x^2+z^2)(y^2+w^2) 2.x^2y^2+z^2w^2 3.x^2w^2+y^2z^2 4.(x^2+w^2)(y^2+z^2) Ans-1

take the value of x=1, y=2, z=2 , w=4 now put and solve

x^2w^2 = z^2y^2... thts why... 2 xywz can be written as 2 * x^2w^2 or 2* z^2y^2

89. sin + cos = rt3 , then tan + cot = ??

Ans- 1

sq both side u will get sin2 + cos2 + 2sin cos = 3 2sin cos = 2,,,,,, sin cos = 1 = sin^2+ cos^2 divide by sincos on both side we get tan+cot=1

off course 1 no doubt

sin^2@ + Cos ^2@ + 2 sin@cos@=3 sin@cos@=1

Me along with Kaltak is suffering from inferiority complex due to level of sum u pagals r doing with speed and accuracy, so math experts who are putting this sums kindly give detail answer.

90. if 1/x+1 + 2/y+2 + 1009/z+1009 =1 then d value of x/x+1 + y/y+2 +z/z+1009

options r 0,2,3,4

Ans- Take x=-1/3 y=2 z=-1

answer is 2 (Y) .. 91. At Daifu university, 40% of all students are both members of a student organization and want to reduce their tuition costs. 20% of those students who want to reduce tuition are not members of the student organization. What percentage of all Daifu students want to reduce tuition? (A) 20% (B) 30% (C) 40% (D) 50% (E) 60% Ans- we need to find out (40+x)

and its given that (40+x) * 20/100 = x x = 10

answer 40+10 = 50

40% common student is 80% of the student who want to reduce tution so all the student who want to reduce tution is 50% of whole student

92. The max and min values of (1+cos 2x) are a) -1 and 1 b) 1 and 2 c) -1/2 and 1/2 d) 0 and 2 Ans- d) 0 and 2 cos x >= -1 cos 2x>= -1 1+cos 2x>= 0 cos x

93. 2sin^2 cos^2 = 2, then find the value of .

Ans- 90..wahi desi method

2sin^2 - Cos^2= 2 ,,,,,, put Cos^2=1-Sin^2 we get, 3sin^2=3 ,,,,,,,sin^2=1 ,,, = pie/2 or pie

94. Length l of a tangent drawn frm a point A to a circle is 4/3 of radius.the shortest distance from A to the circle is?

Ans- 5r/3-r = 2r/3

figure bana pdega .........waise (4r/3)^2+r^2 will give the length of the A to centre .............now minus r frm this

95. there are three points on a circle a,b,c distance of a to b and b to c is 24cm . find the distance between a and c, radius of circle is 20 cm.

Ans- the main point is we have to use sin 2 A... see if u draw the diagram u will get a triangle.. and sin of half tht angle is 4/5 ..but for the other side we need the whole angle tht is sin 2A= 2 sin A cos A

so as per the formula

20 =( 24 * 24 * x ) / ((4 * 24 *24 * 2 *(3/5) * (4/5)) /2) bounce ho gaya bhai

where x is the lenght of the 3rd side

area of isosceles traingle.. (side^2 * sin of the angle )/ 2 you can also use b * rt(4a^2-b^2)/4 96. guys this is a ques frm pg... no 1 cud solve thr....,i solved dis ques ,,,need to verify my answer....so solve it,, A and B run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after A has run 100 meters. They next meet after B has run 150 meters past their first meeting point. Each person runs at a constant speed. What is the length of the track in meters?>>

Ans-Imagine that A and B start at the same place and run in opposite direction. When A and B meet for the first time what is the distance covered by A and B together? Its nothing but the length of the track. So all we need to find out is the Distance covered by A + the distance covered by B

Now they actually start diametrically opposite points, so when they meet for the first time they cover just half of the track and A covers 100 mtrs. So what distance would A have covered if

they ran the full track distance? 2 * 100 = 200

then they again meet after B covers 150 mtrs. Now both start at the same point, so they cover the entire track distance. So B's distance is 150

So total is 200 + 150 = 350.

draw acircle put A n B on the opp.ends of diamtr..now they start moving in opps dirn(towards each othr)..now A covers 100mtr n thus B covers x/2-100mtr(x=circumference)..nowfor 2nd meating B move futher 150mtr to meat A that is A have to cover( x/2-100)+(x/2-50)=x-150...if they takeT tym for 1st meating they wud take 2T for 2nd meating (frm fig).so for A -->100/T=(x-150)/2T.......=>x=350.. 100/x = (x+x+50)/150... if x is distance travelled by b ... and track =2(100+x)... x comes as 75.

97.If sinB = 3sin(2A+B), then the value of 2 tanA + tan(A+B) is? Ans- Taking a=b=0 ans. 0 jugar tech. Jindabad.

98.if tan x + sin x = m and tan x - sin x = n. then (m^2 - n^2) ^2= 1.16mn 2.4mn 3.9mn 4.8mn

Ans-tan x = (m + n)/ 2 ............sin x = (m-n)/2.........(.m^2 - n^2)^2 = 16 tan^2 x sin^2 x..... =16 sin^2x /cos^2x *(1-cos^2x)........= 16*tan^2x - sin^x = 16mn sinx=(m-n)/2and cosx=(m-n)/(m+n) then sq.and add. m+n=2tanx=2tan45=2 m-n=2sinx=2sin45=root2

(m^2 - n^2)^2 {(m+n) (m-n)}^2 (2root2)^2=8

16mn=8 Then its ok.

@Sourav paul wow great method

99.If P, R, T are the area of a parallelogram, a rhombus and a triangle standing on the same base and between the same parallels, which of the following is true. 1. P>R>T 2. T>P>R 3. R= P= T and 4. R= P= 2T

Ans- well parallelogram and rhombus i think have the same area whereas paralellogram has two triangles T so 4

100.sin A+ sin^2 A+ sin^3 A = 1. then, cos ^6A- 4cos ^4A+ 8cos^2 A is?

Ans- Easy one sin A=1 put A= 90 degrees to obtain th answer

Ans- 4.

I hope the above eqs has square,cubes etc Otherwise its wrng on my part as interpretation

can anyone explain?

SinA+SIN^3A=COS^2A SINA(1+SIN^2A)=COS^2A sq both sides n make al d terms in cos.u wil get ans.its lengthy though

101. Two circles with same radius r intersect each other and passed through the centre of the other. then the length of the common chord. a. r b. root3r c root 3/2 r and d root 5 r

Ans- rt3r

Its simple use pythagorus theorm.half of length of chord=rt(r^2-r^2/4) 102. If circles are of two different radii then?...they dont pass through centres but intersect each other. Circles with radii 3cm and 4cm. Find the length of common chord.Distance between their centres is 6cm. then how to solve?

Ans- Ye v vaise hi hoga,yaha distance bw centre diya h.suppose one is x so another wil b 6-x nw use the eqn 3^2-x^2=4^2-(6-x)^2 this eqn wil gv u x Use this x and use pyathagorus theorm again to find length of chord

103.ABC is a triangle . The medians CD and BE intersect each other at O. Then triangle ODE:Triangle ABC. a. 1:3 b. 1:4 c 1:6 d 1:12 answer with explanation please wanna clear concept

Ans- 1:12

explain bro your method don't hesitate it will benefit us.

i think when the medians intersect it divides ABC triangle into 6 equal parts?

Median devides area also. let AF is the median to BC. join DEF. Area of DEF is ABC/4 by the centroid O. DEF further devided to 3 so DEO is ABC/12

wah saab thanx maneesh got it

104.A wheel rotates 3.5 times in one second. What time in seconds does the wheel take to rotate 55 radian of angle? 1. 1.5 b. 2.5 c 3.5 d 4.5

Ans- 2.5

1 sec m vo ghumta hai 7pie radian,to hume nikalna hai ki 55 kitne m ghumega

2pie radian= 360 .........so in 3.5times it make 2pie*3.5 radian angle in center .........so 55 radian will make in time = 55/ 2pie*3.5

2pie radian= 360 .........so in 3.5times it make 2pie*3.5 radian angle in center .........so 55 radian will make in time = 55/ 2pie*3.5

105. A and B are centres of the two circles whose radius are 5 cm and 2 cm respectively. The direct common tangents to the circles meet AB extended at P. The P divides AB in the ratio? a. externally in the ratio 5:2 b. internally in the ratio 2:5 c. internally in the ratio 5:2 d. externally in the ratio 7:2

Ans- externally in the ratio 5:2

106.The radius of circumcircle of triangle made by x- axis, y axis and 4x+3y=12 a. 2 unit b. 2.5 c 3 d 4 easy one.

Ans- 2.5

107. A number is when divided by 5, 6 , 8 and 12 it gives reminder 1 in each case, but when divided by 13 leavs no reminder. Find the smallest such no.

Ans- 481 108. Two equal circles pass through each other's centre .if the radius of each circle is 5cm,what is the length of the common chord? 5, 5rt3, 10rt3, (5rt3)/2 Ans- 5rt3

equation please in pythgoaras

Ok got it 2x2.5root3

109. sin(a+b)=4/5 & sin(a-b)=5/13 then tan2a= a)16/36 b)63/16 c)36/16 d)none of these Ans- 63/16 Tan[(a+b)+(a-b)] 110. the bisector of angle A of triangle ABC cuts BC at D and the circumcircle of the triangle at E.then

A) AB:AC=BD:DC

B) AD:AC=AE:AB

C) AB:AD=AC:AE

D) AB:AD=AE:AC

give explanation too

Ans- A is correct option...its angle bisector theorem which states bisector of any angle devides oposite side devides in the same ratio in which sides forming it r there.

ssc said D

dis.ques.was..also..inmy.set.r.u.representing..dis

111.A & B cab separately do a piece of work in 20 & 15 days. They worked together for 6 days after which B was replaced by C. The work was finished in next 4 days.The number of days in which C alone could do the work? 35, 30, 45, 40

Ans- a+b-6days work= 42/60 remaining work= 3/10 in 4 days total will be in =10*4/3=40/3 days c will comlete it in 3/40-1/20=1/40 or 40 days

112.If x,y satisfy eqn. atan@+bsec@=c then tan(x+y)= ? Ans- are bhaiya tumhari given eqn mein x,y kahan hain jo satisfy karein ??? 113.Easy one for relax of mind

If cosec - cot = 1/2, then cos is equal to?

Ans- Eq wil be (1-cos@)/sin@=1/2 sq both side n replace sin^2@ by cos^2@ u wil get quad.eqn of [email protected] it

114.Find the area of quadrilateral whose vertices taken in order are A(-3,2) B(5,3) C (7,-6) and D (-5,-4) Ans- 75 Yes diavide quadiltrl in 2 two tringl then calculate area using tringl formula..u can use matrix methd as well.

---------formula for area of triangle 1/2{x1(y2-y3)+x2(y3-y1)+x3(y1-y2)} 115. cos24+cos55+cos125+cos300+cos204 =?

Ans- .5

300 = 270 + 30... so 4th quadrant

cos(270+30)=sin30 IV quadrant T-ratio change and cos s positive

APPLY Cos(C+D) 1/2[2cos (125+55/2)cos35+2cos114cos (224-24/2)]+COS 60=1/2 yaar woh cyclic qaudrilateral ka sum se kuch yaad nahi aa raha.. cos 55 + cos 125 = 0.

similarly,,, cos 204 = - cos 24... bacha sirf cos 300 which is 0.5

116.X cot(90+@) + tan(90+@) sin@ + cosec(90+@) = 0, then value of X is?

@=theta

Ans- yes it is sin@ need a lot of practice but i will do it anyway.

118.[cos(90+@) sec(-@) tan(180-@)] / [sec(360-@) sin(180+@) cot(90-@)] =? @=theeta

Ans- Phle jugad k chakar ma 0 ans. Aa gya tha to propr methd lagana pada. yes -1 sec (360-@)= - sec @ 119.If cot (a+b) = 0, then the value of sin (a+2b) ? Ans- cot ( a+b) = cot A cotB - 1 / cota + cot b >> which means.. cota . cot b = 1.. so a and b can take 45 45 respt ..or 30 60.. vice versa :-( Iske toh 3 ans possible ...:-) another one is sin150=1/2

a+ b= 90 so sin(a+2b)= Sin(a+b+b)= Sin(90 +b)= cos b so we have to just find the value of Angle b which can be found from option i think so

so finally iske infinite answer possible hai ........

120.What is the value of cos1 + cos2 + cos3 + ...........+ cos180 ? (1,2,3,......,180 all r in degree)

Please explain the method.

Ans- cos179=-cos1...etc...cos90=0,,,cos180=-1 so ans is -1

cos 90 + cos 180 will only remain so ans must be -1

121.If A, B, C, D are angles of a cyclic quadrilateral , then value of cosA + cosB + cosC + cosD is?

Ans- sinc Cos(A+C)/2= Cos 180/2= cos 90=0 as sum of opposite angle in cyclic quadrilateral is 180

opposite angles are supplementry in cyclic quadrilateral thic concept is used hr..cosA+cos(180-A)=0. Bhai ye to property h jiski mathematics k market ma value=0 h. ok then u better dont use this property in maths..........use sm valuable property instead.

122.What is the value of sin(-420) (cos390) + cos(-660) (sin330) ? Ans- -1

-3/4-1/4=-1

123.What is the value of cos510*cos330 + cos390*cos120 ?

Ans- -3 -root3 /4 is it right

Yes

-(rt3+3)/4 124.A circular wire of radius 15 cm is cut and bent so as to lie along the circumference of a loop of radius 120 cm. what is the measure of the angle subtended by it at the centre of the loop?

Ans- 2*pi*r=120 r=60/pi @=l/r @=15/60/pi=pi/4

125.the diagonals of AC and BD of a parallelogram ABCD intersect each other at the point O such that angle DAC=30 degree and angle AOB =70 Degree then angle DBC ???

Ans- 40( alternate angle prop. And v.o. Angles) 126.when a heap of pebbles is arranged into groups of 32 each 10 pebbles are left over .when they are arranged in heaps of 40 each ,18 pebbles are left over and when in groups of 72 each ,50 are left over .the least no. Of pebbles in the heap is

options.. 1450 1440 1418 1412

Ans- 1418 desi

32*44=1408+10..40*35=1400+18..72*19=1368 +50 1418 satisfy all , brother ye 44 35 and 19 kaha se aya

127.At the foot of a mountain the elevation of its summit is 45' after ascending one kilometer towards the mountain upon an incline of 30' the elevation changes to 60' find the height of the mountain.

Ans- (Rt3+1)/2= 1.36km is it?

128.if N is the smallest perfect square which is exactly divisible by 10,12 and 25 then the quotient obtained when N is divided by 25 is

Ans- 36

900/25

129.What is the value of sin10 + sin20 + sin30 + ...........+ sin360 ? (10,20,30,......,360 all r in degree)

Please explain the method.

Ans- 0

sin values are from -1 to +1. so all get cancelled.

Use sin(180+@) =-sin@ So sin190= -sin10 n thus all get canceld

130.A number of light bulbs were purchased to illuminate a gym. However, only of them were needed. The extra 160 light bulbs were returned. 60% percent of their cost, or $96, was reimbursed. How much money was spent on illuminating the gym? (A) 360 (B) 320 (C) 384 (D) 364 (E) 160 Ans- bhai A number of light bulbs were purchased to illuminate a gym. However, only 2/3 of them were needed. The extra 160 light bulbs were returned. 60% percent of their cost, or $96, was reimbursed. How much money was spent on illuminating the gym?

only 2/3 of them were needed....ye line kahan se aa gayi....???

actually i follow gmatclub to waha mene solve kia tha same ques.

2/3 dia hai to bht simple ho gaya 320+64=384

131. if cos@ + sec @= rt 3 then cos^3@ + sec^3@ is equal to : a] -1 b] rt3 c] 0 d] 1 Ans- 0 first que mein take cube of both sides ..u will get the ans

132. max value of 24sin@ + 7 cos@ is : 1] 24 2] 25 3] 7 4] 17 Ans- 25 second que mein there is one direct formula for maximum value of asinx+bcosx=rt(a^2+b^2)=rt(24^2+7^2)=25 cos^3@+sec^3@+3rt=3rt....so ans zero rt((24)^2+(7)^2)=25 133.If sinx + siny +sinz = 0,cosx + cosy + cosz = 0...............then sin2x + sin2y + sin2z = ?

A) -1 B) 0 C) rt3 D) 1 E) NONE Ans- x=90,y=180,z=270...sin 180 + sin 360 + sin 540= 0+0+0=0

x=30, y=150, z=270...they are satisfying both the equations...after putting the values in sin2x + sin2y + sin2z...ans is 0

If x+y+z=180 then sinycosz+coszsiny= sinx ,multiply d given eqn by sinx and put these values.u wil easily get d ans.where i m wrong? Checkn option upto 270 is quite tedious

What vaues are u putting in that??? It's not mentioned that x+y+z = 180....u don't have to check the options till 270....3 values 0f sin or cos should be added up to make 0..this can be done as 1/2+1/2+(-1) or x+(-x)+0....so now from this you can assume the values.....Anyone having shorter method to do this can explain.....

134.A litre of water is evaporated from 6 litres of glucose solution containing 4% glucose. The new concentration of glucose is ?

Ans- yes getting 4.8..5000*x/100=240..so x=4.8

4% means 4g in 1 litre 6l contains 24g now 1l evaporates so (24/5000)*100=x s=4.8

135.There are three diffenrent type of wine each costing Rs.48, Rs.55 and Rs.75 per bag. The bartender wants to make a wine that will cost Rs.67 per bag. In what ratio should he mix all the three ??????????

Ans- ans is 4:1:11.....................you are right

There could be different value too but we always take the lowest one

48 and 75 will be mixed in the ratio 8:19.......and 55 and 75 will be mixed in ratio 2:3 ............so the ratio of 48:55:75 = 8 : 2 : (19+3) =8:2:22 = 4:1:11 ..........this is the way we solve this ..........bt the other possibility just becoz ......... we can write 2:3 above as 8:12 ........so now the three ratio is 8: 8:31 ..........we can also write 2:3 as 6:9 so the three ratio now will be 8 :6:28 = 4 :3:14

mere paas hsortcut hai usse 8:8:31 aya. aise qs mein all three prices likh lo.then jis price ka mixture banana hai wo in price ke uper likh do. likhna is way mein hai ki 48, 55 ,75 likhe hue hain. new mixture 67 rs wala hai to use 55 ke baad uper ki taraf likho. then 75-67=8 fist and second ka ratio hoga. and 67-48 plus 67-55 =31 lat wale ka ratio hoga.ye shortcut hai. aur qs post karo and discuus akrne se ye shortcut clear ho jayega 136. ABCE is an isosceles trapezoid and ACDE is a rectangle. AB= 10, EC= 20, AE= ??

Ans- 10

137.the sum of 4 consecutive two digit number when dived by 10becomes a perfect sqaure .which of the followingcan be one of thoses number?

21 25 41 67 73

Ans- 21+22+23+24= 90 /10 = 9 a perfect square of 3

take four no as n, n+1, n+2, n+3 .........now (4n+6)/10 = k .....4n = 10k-6 ...........as "K" is the perfect square so put value of K = 4, 9, 16......so tht 10K-6 is multiple of 4 ............so ans is 21

138. Three vessels having the volumes in the ratio 1:2:3 full of a mixture of coke and lemon soda. In the first vessels ratio of the coke and the lemon soda is 2:3 , in second 3:7 and in third 1:4 . If the liquid in all three vessels were mixed in a big container , then what is the resulting ratio of the coke and lemon soda ?

Ans- 4:11 asume quntatity 50 100 and 150 calc becomes easy. Coke:Soda=20+30+30/300-20-30-30=80/220=4:11

139.Two circles touch each other internally. The sum of their area is 116 pi . And the distance between their centre is 6 cm. Find the radii of the circles...

Ans 4 & 10

140. A room 5.44m*3.74m is to be paved with square tiles. The least number of tiles required to cover the floor is ?

Ans- 176

find hcf of 5.44 and 3.74 it comes to be 3.4 so 5.44 x 3.74/3.4x3.4

141. (a,b) & (a+3,b+k) are on x=3y-7... K=?? Ans- put a,b into the eq.......a=3b-7

and now (a+3.b+k) in the same eq.,

you will get k=1

u may also say that slope of line would be equal to k/3=1/3..so k=1

142. In circle a chord xy subtends angle 90 with centre.& ox=oy=Radius..if Area of Triangle=32, Area of circle??

Ans- It wud b 64pi. Triangle formed is right angled.let r b the radius of circle.den 1/2 *r*r=32 it gives r=8 hence ar.of circle is 64pi

143.Area of circle inscribed in an eq.triangle is 154 units.find the perimeter of the triangle

Ans- 7=side/2rt3 ..side=14rt3...P=42 rt3

Pi*a*a\2rt3*2rt3=154 so a=14rt3 perimetr=42rt3

144. if sin@ =cos(@-45)where @ and @-45degree are acute angles find the degree measure of @

Ans- @=90-22.5=67.5

135/2 equal to 67.5

cos(90-@)= cos (@-45) equate 90-@= @-45

145. If three circles(equal) of radius 3 cm each touch each other then the area of the shaded portion is:

a)rt3/2*(2-pie) cm^2 b)9/2 *(2rt3 - pie)cm^2 c)9/2*(2rt3 +pie) cm^2 d)2/9 * (2rt3 -pie) cm^2 Ans- Rt3*6*6/4-3pi*3*3*60/360=9/2(2rt3-pi)cm"2 ans. B rt3/4 * (6)^2 - 3pi*(3)^2*60/360=9rt3-9/2pi=9/2(2rt3-pi) 146. {1 - (1/2^2)}*{1 - (1/3^2)}*{1 - (1/4^2)}...........= ?

1 , 2 , (2/pie) , 0.5 , none. Ans- 0.5

15/24~2/pie

147. ABC is a triangle. The internal bisector of the angles angle A, angle B and angle C intersect the circumcirlce at X,Y and Z respectively. If angle A=50 degree, angle CZY=30 degree then angle BYZ will be

a] 35 degree b] 30 degree c]45 degree d]55 degree Ans- a) 35 for this you can find angle

150. In triangle ABC , AD is drawn perpendicular from A on BC. If AD^2=BD.CD then angle BAC is 1] 60 degree 2] 90 degree 3] 30 degree 4]45 degree Ans- IT SHOULD BE A RIGHT ANGLE....BECAUSE GIVEN CONDITION IS ONLY TRUE WHEN TRIANGLE ABC,TRIANGLE ABD N TRIANGLE ADC R SIMILAR SO....ABC MUST BE RIGHT ANGLED AT A

151. The moon's distance from tge earth is 36000 km and its diameter subtends an angle of 31 min at the eye of the observer. Find the diameter of the moon.

Ans- 3470 km

I took 30 min,instead of 31,to ease d calculation

100pi..

152. If x+y+z+w = 17.then find the maximum possible value of (x-1)(y+3)(z-1)(w-2) (x,y,z and w are real nos. greater than 2) a)288 b)232 c)186 d)256 e)300 Ans- 256. Use d concept am>gm.

I dnt think questions on AM,GM will come in SSC

x=5,,y=1,,z=5,,w=6 4*4*4*4=256

how y=1?????????? all values are greater than 2

2*6*4*4 can be never be 256

153. If a^2*b*c^3*c^2 =256 and the value of 2a+b+3c+2d is minimum possible,then find a+2b+3c+4d.

a)16 b)18 c)24 d)20 e)22 Ans- correct ans is 20............for getting a minimum value of 2a+b+3c+2d a=b=c=d= 2...........so a+2b+3c+4d = 20

Oh purely desi question

158. if 3cos@-4sin@=2cos@+sin@ find tan@

Ans- 1/5

Cos@ ya sin@ se divide kr do both side

159. Pawan and Qureshi working together can do a piece of work in 10 days whereas Qureshi and Rohit working together can do the same work in 12 days. All three work together to do job for which they are paid Rs. 300. If Qureshis share is Rs. 140, then what is Pawans share?

(a) Rs. 100 (b) Rs. 60 (c) Rs. 80 (d) Cannot be determined Ans- as in rest of 160 ,,,,,,,,,, pawan is more efficient than rohit so amoumnt of pawan is more thats 100 here isgud option

pay ratio is proportional to work efficiency...

so work efficency is 7:5:3 7=140 then 5=100

160.What would be reminder when 2"256 is divided by 17 and 18? P.s. " means Power

Ans- 2^256 = 16^64 =(-1)^64=1 K....ab to SSC wale CAT ko fod skte h

161.A circle of radius 1 is inscribed in a sector of radius 3.the area of the sector will be??

Ans- bhai diagram bnao..ek arc bnao..aur uske andar ek circle fit kar do...then it would become easier

@mizan 1.5pi ans aaya... Circle ko fit karo..aur 2 radius draw karo perpendiclar to both sides of arc at A& B...abhi beech ka angle AOB=120,so saamne wala angle of aob=60.. So pi*r^2*60/360

162. A trader makes a profit equal to the selling price of 75 articles when he sold 100 articles. What was his % profit ?

Ans- 75X = 100X - Y

Y = 25X.............PROFIT% =300

163.The sum of the third and the ninth term of an A.P is 8. Find the sum of the first eleven terms of the progression.

Ans- 44

11/2(2a+ 10d) Is it in syllabus??

164. Cost of diamonds direcltly propotional to Weight^2 Cost of a Diamond=5184,we broke into 3 pieces in ratio 1:2:3 of their weights.. Loss in cutting=??

Ans- let cost = k*w^2. let weight be 1gm, 2gm, 3gm. total weight = 6gm. now total cost = 5184. 5184=k*36. k=144. now cost of each piece = 144*(1)^2 + 144*(2)^2 + 144*(3)^2 = 2016. so loss = 5184 - 2016 = 3186

chupchaap sq root kar aaya 72...ab use 1:2:3 mein baat aaya 12,24,36...teeno k square kar k jod...2016 aaya ghata de 5184 se 165. 3-(3+root5)/4- 1/(3+root5)=?? Ans- bhau....second and third term ko solve karo pehle....denominator same karke....and then subtract it from 3....solve karte raho....3/2 aa jayega 166.In triangle abc,ad-median,ad=(bc)/2,angle bad=30,angle acb=?? Ans- As ad = bc/2. and d is midpoint of bc ad=bd=dc.ADB and ADC are two isosceles triangles formed.

168. A gives B a start of 10m in a 100m race and still beats him by 1.25sec. How long does B takes to complete the 100m race if A runs at the rate of 10m/s ???????? 1-8 2-10 3-16.67 4-12.5

169. A ship is moving at a speed of 30km/hr. To know the depth of the ocean beneath it, it sends a signal which travels at a speed of 200 m/s. The ship receives the signal after it has moved 500 m. The depth of the ocean is

170. AB and CD are two chords of a circle with centre o intersect each other at P. if

175. (x^2- x+1)/ (x^2+x+1)= 2/3, then the value of x+ 1/x is a. 4 b. 5 c. 6 d. 8

Ans- 5 componendo- dividendo ,very easy

176. plz solve it...yar pg pe itne fad ques dalte hai ki solve nhi hota.. 1.cot6. cot42.cot66.cot78=? 2.tan12. tan 24. tan 48 . tan 84=? 3. cos @+cos$=1/3 and sin@+sin$=1/4 then cos(@-$)/2=? 4. if sin@=nsin(@+2$) then tan(@+$)=? 5.sin20 . sin 40 . sin 80 =? 6.if @+$=90 , find the maximum value of sin @ sin $ 7.cos 55+cos65+cos175=? 8.find the max. and min. value of 7cos@+24sin@ 9.value of sin100-sin 10 is positive or negative

Ans- 1,2 type ke ques kaafi tym consuming hai inhe chodne mein hi bhalai hai.(yaa 1 maar do option mein ho agar 1/2 marks ka risk lenaa ho).:) shayad hi ssc aise ques de....trigo 10th ke concept pe de rakhi thi tier1 mein ....pehle thode bacho wale ques puch lete thai jinko dekhte hi desi yaad atta tha....ab thoda traditional de rahe hai..alzebra mein trigo ghussa di coordinate mein mensuration...trigo mein quad..

1.1 2.1 3.5/24 4.q ek baar fir se dekh lo ki n likha hai kya rhs mein ? 5.root3/8 6.1/2 7.0 8.max 25 min -25 9.positive

Ans- 3 multiply both equation, u will get the value of cos ( A-B) i.e, cos A cos B + sin A sin B as -263/288 u can write cos ( A-B ) = 2 cos ^2 ( A-B)/2 -1.... from this equation u can find cos ( A-B )/2 as 5/24 aap first n second eqn mein cosa+cos b vala n sina |+sinb ka formula lagao...dono eqn ko devide karo...u will get cot (a+b)/2=4/3... find cos (a+b)/2=4/5 now put that in eqn 1 u will get ans

177. is type ke ques bhi aa sakte hai kya ssc me......? A group of 10 workers can plough a field in 20 days. This group starts the work and after every 2 days , 2 additional workers join the group. The capacity of each worker is the same. in how many days will the field be ploughed ?

(A) 11 (B) 12 (C) 13 (D) 14 Ans- 13

10*2 + 12* 2 + 14 *2 + 16 *2 + 18*2.. til the count goes to (20 *10 = 200) ok got it ..20+24+28+....+n=200 ...,we have to find the value of n..

178. If Sec - Tan = 4/3, then Sin= ??

Ans- 1 -ve ka kya pangaa hai ??

bhai eqn se solve karoge to smajh jaoge.. dont use desi :-P sina/cos=4/3==>3(1-sina)=4*rt(1-sin^a)..nw taking sqr on both sides we get 9-9sina=16+16sina..so sina=-7/25

179. If sinx+sin^2 x=1 find cos^2 x+cos^4 x

Ans- 1 sinx=cos^2x cos^2x+cos^4x sinx+sin^2x=1

Bwhahahaha issme bhi Desi Method dhoondh nikal hi li= Sinx=x X(x+1)=0 so x=0 Now Cosx=root(1-x^2) Put in eq & put x=0 Then ans=1 (Though its long cut but exam me formula strike nahi hua to yahi kaam aayega)

180. A circular road runs around a circular garden. If the sum of the circumferences of the inner and outer circles is 88 metres and the radius of the inner circle is one-third of the outer circle, then the width of the road is : (A) 4 metres (B) 5 metres (C) 6 metres (D) 7 metres Ans- 7

181. Easy one The radius of a circle is 30 cm. find the length of an arc of this circle, if the length of the chord of the arc is 30 cm.

Ans- 2pi*30*60*360=10pi

chord will form equilateral triangle with centre. very easy

182. In PQR, QR = 10, RP = 11 and PQ = 12, D is the midpoint of PR, DE in drawn parallel to PQ meeting QR in E. EF is drawn parallel to RP meeting PQ in F. What is the length of DF? (a) 11/2 (b) 6 (c) 33/4 (d) 5 Ans- 5

183. Easy one If the arcs of the same length in two circles subtend angles 65 and 110 at the centre, find the ratio of their radii.

Ans- r inv prop q and q in radians

22/13

184. The perimeter of a certain sector of a circle is equal to the length of the arc of a semi circle having the same radius. What is the angle of the sector in degree?

Ans- Length of the arc of a semi circle of radius r= pie*r

As (180/360)*2*pie*r So 2r+(2pier*q/360)=pier q=720/11

185. A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 meters when it has traced out 72 degree at the centre, find the length of the rope.

Ans- 2pi*r*72/360=88 so r=70

186. if x+1/x= root 3 , then the value of x^18 + x^12 + x^6 + 1 is

Ans- x+(1/x)=root3 X^2+1=x*root3 Cubing both x^6+1+3x^2(x^2+1)=x^3*3*root3 x^6+1+3x^2*x*root3=x^3*3*root3 x^6+1=0 x^18+x^12+x^6+1 x^12(x^6+1)+(x^6+1) x^12(0)+(0) 0 ans

187. Find d min value of 3cosx+4sinx+8

Ans- Val is rt(3^2+4^2)+8 =+-5+8 if u hv to find max take + n it wud b 13.for min take - and its 3

-3sinx+4cosx=0 (diffrentiate) tanx=4/3 cosx=3/5 sinx=4/5

3*3/5+4*4/5+8=25/5+8=13 but by desi one..x=0 gives 11

188.

Ans- sin@=cos^2@ cube ( sin@+sin^2@=1) sin^3@+sin^6@+3sin^3@=1 sin^6@+3sin^5@+3sin^4@+sin^3@-1 1-3sin^3@+3sin^5@+3sin^4@-1 1+3sin^3@(sin^2@-1)+3sin^4@-1 1-3sin^4@+3sin^4@-1 =1 - 1 =0 is ans

189. A pole broken by storm of wind & its Top stuck the ground at angle 30 & at dist 20 from foot of pole.Ht of pole before it was broken?? Ssc ans-100root3/3 paramount ans-60root3

Ans- Sorry mera ans to 60/root3 aa raha hai... Hyp=x,Ht=Y.. Sin30=1/2=y/x & cos 30=root3/2=20/x So 40/root3 + 20/root3=60/root3...is my ans correc??

yupss correct 60/rt3 or 20rt3

190. There is a work to be done by three friends A,B,C. Three of them together take 5 hours less than A alone would have taken,one-third that B alone would have taken and two-ninths the time C alone would have taken. How long does the three them take to finish the work? (a) 3 hrs (b) 4 hrs (c) 5 hrs (d) None of these suppose A took "x" hour so A+B+C will take x+5 hour ........so B will take 3(x-5) and c will take 9/2(x-5) hour ...........now we can solve 1/x + 1/3(x-5) + 2/9(x-5) = 1/(x-5)...........so ans is 4h Take A=x+5 ,so A+B+C=x...ye jyada simple hoga But anyway i think such sums should be solved in 2nd round at exams

191. Perimeter of triangle formed by d pts (0,0), (1,0) and (0,1) is Ans- 2+rt2

192. Find d value of (1+tanx+secx)*(1+cotx-cosecx) Ans- 1+sin2x-1/sinx cosx

yes 2 it is..take x=45

Yaar desi lagane ki jarurat nhi h isme,ye bina desi k jyada easy h 193. (sin4A + sin2A)/(1+cos2A+cos4A) = ? a. secA b. sin2A c. tan2A d. cot2A

Ans- traditional method : put sin4A = 2sin2A.cos2A, put 1+cos4A = 2cos^2(2A).....take sin2A common from numrtr. and cos2A common from denom....cancel remaining terms to get tan2A

Desi Rocks A=15

194. AC and BC are 2 equal chords of a circle. BA is produced to any point P and CP, when joined cuts the circle at T. Then (A) CT*TP=AB*CA (B) CT*TP=CA*AB (C) CT*CB=CA*CP (D) CT*CB=CP*CA Ans- I thik PT x PC = PA PB

Bhai ye to option hi ni hai :/

Muskil sawaal..no desi 4 this GEOMETRY ME DESI LAGANA MUSHKIL HI NAHI NAMUMNKIN HAI

195. If tanx=b/a find d value of acos2x+bsin2x

196. If sinx=8/17 find tanx+secx

197. x is a natural number which is a perfect square...then the number x + rt x will end in

a. 0 or 5..b. 0 or 1 or 9....c. 0 or 2 or 6 d. 0 or 4 or 8

Ans- c

i started taking squares....and answer was clear by square of 3....9 + rt9 ends in 2....did u guys use any other method?

Hahaha ....HEIGHT OF DESI METHOD

c ? assuming 4 , then 4+rt4=6

198. 19 ^n - 1 is 1. always divisible by 9 2. always divisible by 20 3. is never divisible by 19 4. only a and c are true

Ans- 4 th option

19 ^n is divisible by 19...so 19^ n -1 will not be divisible by 19..

Put n=1 & solve AGAIN DESI METHOD!

199. if x = 2 ^ (1/3) + 2 ^ ( - 1/3), then 2 x^3 - 6x = ? Ans- 5

cube the given expression...x^3 = 2 + 1/2 + 3x. multiply by 2. 2x^3 = 4 + 1 + 6x. so x^3-6x = 5

try sachislyf's method.. cb. rt.2 ~ 1.25..

x=2^1/3+1/2^1/3 x*2^1/3=2^2/3+1 cube& substract 6x 2x^3-6x=4+1+3.2^2/3(1+2^2/3)-6.2^1/3-6.2^1/3 =5+6.2^1/3+6+6.2^1/3-6.2^1/3-6.2^1/3 =5

200. Rahul bhai...kro..:)

Ans- 22 (3) 23 (3) 24 (4) 25 (2) 26 (1) 27 (2) 28 (2) 201. Which is not the factor of 4^ (6n) - 6 ^ (4n) for any positive intiger n ?

a. 5 b. 25 c. 7 d.none

Ans- 2^4n ( 256^n - 243^n ) .... so apart from 2^n .. the other odd factor is 13 only ... 5 , 25 , 7 doesnt fit the bill .... answer is none

Desi Method- 4^6-6^4=2800 ...which was div by all 3..so none

202. sin^2C = sin^2A + sin^2B. then triangle ABC is right angled, acute, obtuse

Ans- right angled triangel.

C=90 A=B=45

Or Put 30,60,90 Again Desi method! :P

203. Find the length of an arc in terms of pie of a circle of radius 5 cm subtending a central angle measuring 15 degree.

Ans- 15/360*2*pie*5 = 5/12*pie

formula for finding the arc of circle is= @/360 x 2 pie r

204. a ques from pg site..

A balloon of radius x makes an angle y at the eye of an observer and the angle of elevation of its center is z. The height of its center from ground level is

a)x cos(z/2) sec y b)x cos z sec (y/2) c)x sin (z/2) cosec y d)x sin z cosec (y/2) Ans- cat..level..bro..i..'d'..i..think..will..not..attempt..if.comes.in.paper

x cos z secy/2 i had done this question when i am preparing for engineering entrance. so don't give them

205. what is the least multiple of 7,which when divided by2,3,4,5 and 6,leaves the reminders 1,2,3,4,5 respectively?

Ans- 7a=60b-1

119

when the divisor and there remainder have same diff like 1 then u have to take lcm of 2,3,4,5,6 =60 and then multiply it with least dividor given thats 2 .......120 and as remainder difference is 1 then -1 from this 119

206. find the least number which upon being divided by 2,3,4,5,6 leaves in each case a reminder of 1,but when divided by 7 leaves no reminder?

Ans- for this Lcm of 2,3,4,5,6 =60 ,, and as 1 is remainder in each case so jst take max. divisor6- remainder(1) = 5 and multiply 60*5 for complete divisibilty and add 1 for remainder 1....301

207. A train travelling with a speed of 60 km/hr catches another train travelling in the same direction and then leaves it 120 m behind in 18 seconds. The speed of the second train is (a) 26 km/hr (b) 35 km/hr (c) 36 km/hr (d) 63 km/hr Ans- use shrt cut,,,,,,,,,,,, as difference of speed = (120/18)(18/5)=24 and hence the other train speed is 60-24=36

208. find the vallue of 1/2sin10*-2sin70*

Ans- arrange them...then use -2sinAsinB=cos(A+B)-cos(A-B) ...finally get cos80/sin10 =1 209. A man covered a certain distance at some speed.Had he moved 3 kmph faster ,he would have taken 40 minutes less.If he would had moved 2 kmph slower,he would taken 40 minutes more.the distance(in km )is. Ans- (3*2*40+40/60)/3-2=8 h then 8*3+8*2=40 210. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of1200 km per hour. How far apart will be the two planes after 3/2hours?

Ans- Solv 100rt549=100rt(9*61)=300rt61 same ans rt (1500sqr+1800sqr)...... 211. If sin 3A + sin 3B + sin 3C = 1, then one angle must be equal to ?

a.90 b. 60 c. 120 d. none

Ans- D. none

sin 180 = sin 360...so try for just 90 and 60 by putting values, it will not satisfy the equation 212. A garden is fenced in the shape of a hexagon with each side 10m in length. A dog is tied with a 14m leash to one of the post at one of the vertices of the garden. What are can the dog cover outside the garden?? 1-527 2-227 3-427 4-327

Ans- area of sector with 14cm radius = 240/360 * 22/7 * 14^2=392/3*pie ............area of sector with 4cm radius = 2 * 60/360 * 22/7 * 4^2 = 16/3*pie............total sum =136*pie = 427.42

213. In a triangle ABC, if Sin A + Sin B + Sin C = 1+rt 2, and cos A + cos b + cos C = rt 2, then the triangle is a. equilateral b. isosceles c. right d. right isosceles

Ans- d.

as one angle is 90 deg and other two are 45 each ,,,,,, then we get above values

214. Easy one find d max value of tanx*tan(90-x) -2sin^2 x Ans- Eqn = tanxcotx- 2sin^2x =cos2x whose max value is 1

215. If a+b=10 a,b>0 find d min value of (a-2)*(b-4) Ans- instead of a=1 and b=9, take a=9 and b=1 then u ll get -21

216. In an obtuse-angled triangle ABC, angle A is the obtuse angle and O is the orthocentre. If angle BOC= 54 degree, then angle BAC is (a) 108 (b) 126 (c) 136 (d) 116 Ans- 126

216. In a square shaped hostel, there are 8 rooms in on each side in every floor.Room No. 101 is the first room facing north in the first floor.The room numbers are continued clockwise.Which direction does Room No.125 face? a)South b)North C)West D)East Ans- simple ... 8*3+1 .... start frm (facing north means situated in south) south..west...north...east.. so as room situated in east ,,it will face west...ans

217. A Candle Burns in 6 Hrs..Another Candle of the same height and width burns in 8 Hrs...After How Many Hours the height of the first candle will be half of the second Candle.

Ans- 2*(6-x)/6 = (8-x)/8.................calculating it we will get X= 4.8 hrs. Kaunse book se sum hai ye??

218. Krishnamurthy earns Rs. 15000 per month and spends 80% of it. Due to pay revision,his monthly income has increased by 20%,but due to price rise, he has to spend 20%more. His new savings are (a) Rs. 3400 (b) Rs. 3000 (c) Rs. 4600 (d) Rs. 4000 Ans- wrong option given by ssc.......may be it was misprinting....

3600

MR.KRISHNMURTHY WORKS AS TA IN CBEC

219. Let a, b, c be distinct such that a + 1/b = b + 1/c = c + 1/a, then abc equals

a) 1 b) 0 c) -1 d) greater than 1 e) Less than -1 Ans- ok if a=b=c=-1 ,ans- -1 a=b=c=1. ans- 1 a=b=c=2 , ans - 8 so a=b=c shld be always there

this is a futile question , time consuming no one is giving this type of question in tier 2, if assume its present leave it. and ankit its clearly given in question u cannot take a=b=c and u r assuming what not to be assumed it. chill it guys the questions will not be of IIT level !!!

220. If a + b + c = 1 and ab + bc + ca =1/ 3then a : b :c is (a) 1 : 2 : 2 (b) 2 : 1 : 2 (c) 1 : 1 : 1 (d) 1 : 2 : 1 Ans- c

221. a^2 + b^2 + c^2 = ac + ab rt3, then the triangle is

1. equilateral 2. isosceles 3. right 4. none

Ans- 3

Bhai cosine law de rakha h decode kar lo...

222. What least number must be subtracted from each of the numbers 17, 17, 34, 42 so that the ratio of first two is the same as the ratio of the next two? (A) 0 (B) 1 (C) 2 (D) 7 Ans- wrng Q

it should be 18,,,,22,,,,34,,,,42 nearest possibilty ,,,,,,,,,,,,, then minus 2 from each will get the same ratio of first two and last two numbers

223. (a+b):(a-b) = 1:5,then (a^2-b^2) : (a^2+b^2) = ? Ans- 5/13

224. sherly started from a fixed point goes 15 m towards north and then after turning to his right goes 15 m . then he goes 10 , 15 and 15 m after turning to his left each tym .hw far is he from starting point ??/

Ans- 10

bhai 15 m north banao phir 15 m east phir 10 m north phir 15 m west phir 15 m south..phir batao answer aa rha h ya nhi

225. Percentage increase in the surface area of cube when each side is increased to 3/2 times the original length

Ans- 125

{(3/2)^2(a^2+b^2+c^2)} - (a^2+b^2+c^) 226. If x2 + y2 2x + 6y + 10 = 0, then the value of (x2 + y2) is (a)4 (b)6 (c)8 (d) 10 Ans- ye circle ka eqn hai centre is (1,-3)...so 1sqr+(-3)sqr=10... 227. If the square of the radii of three concentric circles are in A.P., then the square of the lengths of the tangents from any point to these circles are in (1) G.P. (2) A.P. (3) H.P. (4) None of these Ans- it forms 3 circle with same base but thr sqr of hieghts in ap so hypotan.. is in ap..

228. If the angles A, B , C of a triangle are in AP and sides a,b,c are in GP, then a^2, b^2, c^2 are in...

1. AP, 2. GP, 3. HP, d. Not in Ap or in Gp

Ans- A.P is the correct answer.

gp

one thing when it is given that bsqr =ac ,then why not b^4=a^2*c^2...thus a^2,b^2,c^2 is in gp

have to consider angles also...they are in AP..for your value, i think cos B wil not be equal to 1/2

229. .A three digit number which on being substracted from a another 3 digit number consisting of the same digits n the reverse order gives 594. The minimum possible value of the sum off the digits of the number is ?

Ans- (100z+10y+x)- (100x+10y+z) = 99(Z-x) =594 so z-x =6 ......consider the value of value of z and x so that x become min ....z= 7 , x= 1 ........so no is 107 ...........digit sum is 8..............yup i mistaken earlier

one short trick for this sum is.... 594/99 = 6... so the difference between the face values of the nos in hundreds place and units place is 6... look in the options... u shud get,,,

230. If tanx=3 and x is in third quadrant find sinx

Ans- ASTC yaad rakho

p=3 b=1 h=rt10 and sin will be -veP/H

231. . Two pipes A and B and can fill a cistern in 12 minutes and 15 minutes respectively but a third pipe c can empty the full tank in 6 mintues. A and B are kept open for 5 minutes in the beginning and then c is also opened. In what time is the cistern emptied.

40,,,,,,,,45,,,,,,,50,,,,,55

Ans- tank empty= 1/12+1/15-1/6=60min a & b kept open for 5 min = 45/60 = 3/4 part full remaining 3/4 will empty in = 45 min

232. Find the maximum and minimum values of 6 sinx cos x + 4xcos2x..

Ans- 6sinxcox=3sin2x..ab lgalo wo asinx+bcox wala fromula..max and min value of asinx+bcox is rt(a^2+b^2) and -rt(a^2+b^2) 4 x multiply wala hian ki alphabet wala confuse ho gaya

3 sin 2x + 4 cos 2x so min and max= +-root (3^2+4^2) ye x mat use kiya kare

233. If sinA+sinB = a and cosA+cosB = b, then cos(A+B) (a)a^2+b^2/b^2-a^2 (b)2ab/a^2+b^2 (c)b^2-a^2/a^2+b^2 (d)a^2-b^2/a^2+b^2 Ans- The answer given in the book is (c)...They have solved b^2-a^2 and then a^2+b^2 and then divide and it will be equal to cos(A+B)...but don't know why they have calculated only this and not 2ab or a^2-b^2....FF bhai solution samjhao. Only c option wil give d formula cos(A+B) Cos(A+B)*cos(A-B)=COS^2 A-SIN^2 B= cos^2 B-Sin^2 A Use this formula alongwid cos(a+b) and cos(a-b), rest term would cancel n only cos(a+b) wil remain

If we put A=B=30 then we dont get proper ans..in C we get 11/13 ..& where as cos60=1/2

Whereas if we put A=B=45 then Options C & D comes correct

at A=B=30 ..u will get ans as c option only..as a becomes 1 and b becomes rt 3..put in option c now...it comes 1/2

234. Distance b/w d points (acos 25,0) and (0, acos65) is ? Ans- a rt[(x2-x1)^2+(y2-y1)^2]

235. CGL tier I if X = 2rt6 /(rt3 + rt2) then the value of (x + rt2)/(x-rt2) + (x+ rt3) /(x-rt3) is 1- rt3 2- rt6 3- 2 4- rt2

Ans- 2

Divide first by root 2 and take comp-div again divide by root 3 and take comp-div then add will take hardly 40 secs

X/ rt2 ko comp and div karo to wo (x+rt2)/(x-rt2) hoga na which is the first term X/rt2 ho jayega ........same way . second term X/rt3 ho jayega aisa karo tum dada ka method try karo ..........it is given X=2rt6/(rt3+rt2) nw both side rt3 se divide karo it will be X/rt3 = 2rt2/(rt3+rt2) ...........ab ispe com& div try karo ye tumhara equation ka first term ho jayega .........phir rt2 se divide kar com & div karo wo second term ho jayega .............nw solve

236. A number, when divided successively by 4, 5 and 6, leaves remainders 2, 3 and 4 respectively. The least such number is (a) 50 (b) 53 (c) 19 (d) 214

i think its a wrong ques..asked last yr in tier2

Ans- ans is 94 which is not in option

why not ........214??

214

214/4 -- 53 rem 2 53/5--10 rem 3 10/6 rem 4...

237. time for tough maths..:)

1.A person starts from the origin of the coordinate axis . He travels in this pattern . 1 unit to right , (1/2) units up , (1/4) units right, (1/8) units down , and continues the above pattern . At what point will he ultimately come to stop? (a) (4/3,2/5) (b) (3/2,4/3) (c) (2/5,4/3) (d) (4/5,4/3) Ans- for X axis take term 1 , 1/4, 1/16.............this is in GP with a=1 r= 1/4

so for infinite term its a/(1-r) = 4/3...............for Y axis it is a= 1/2 r= -1/4 ....so its 2/5 ya it should be A (4/3,2/5)......X axis--1/(1-1/4)=4/3.........Y axis----(1/2)/(1+1/4)= 2/5. 238. if a^2+ b^2 =2 and c^2+ d^2=1 then the value of (ad-bc)^2 + (ac+bd)^2 is Ans- desi 2

a=b=c=1 d=0

a=1,b=1,c=1/rt2,d=1/rt2

239. (2a)-(2/a)+3=0 then [(a^3)-(1/a^3)+2]=?? Ans- i think a = -2 and ans is -47/8

240. A man inversts 15680 rs in names of A,B,C in such a way that they get same amount after 2,3,4 yrs respecvly..R=5%..then ratio of amount invested at start??

Ans- (x*2*5)/100 = (y*3*5)/100 = (z*4*5)/100....2x = 3y = 4z.....let it be equal to k. so x = k/2, y= k/3, z= k/4.....so ratio = k/2:k/3:k/4 = 6:4:3

L.C.M OF 2,3 and 4------12.....so ratio--6:4:3

022MM4 waale, Represent karo isse.,,,,1 mark milega

241. Rohan buys 150 articles on which he has to pay Rs.50 on carriage. The articles were marked for sale at Rs.12.50 each. Rohan sells 90 of them at this price and the remaining after allowing a discount of 20% on the marked price. Altogether he finds that he makes a profit of 38% on his outlay. Calculate the amount he pays for each article. (a) Rs.10 (b) Rs.9 (c) Rs.8 (d) Rs.7Jul 15 Ans- SP=90x12.5 + 60x10 =1725... 1725=1.38(CP)..CP=1250..subtract 50(tranporatation)...1200/150=8per piece total SP = (12.50*90) + (80/100)*12.50*60 - 50 = 1675......CP of each article = 100/138 * 1675 * 1/150...m i right?

242. A lotus is seen 5 cm above the water level of a lake. With the onset of the wind it sinks in the water 10 cm away from its place. How deep is the water in that place? 1) 5 cm 2) 5Sqrt5 cm 3) 7.5 cm 4)10 cm Ans- x^2 + 10^2 = (x+5)^2 assume a lotus that is in water upto x cm and above water 5 cm. when it sinks a right aangled triangle is formed with its dimensions as x cm, (x+5) cm, 10 cm. using pythogorous again we get (x+5)^2 = x^2 + 10^2 or 25 + 10 x = 100 or x = 7.5 cm....courtsey orkut community

243. ==Desi== X^2+Y^2+2X+1=0 Than X^239+Y^478=?? 0 -1 1 NONE OF THESE

Ans- Eqn can b written as (x+1)^2+y^2 thus x=-1 and y=0 244. In a right angled isoceles triangle, the ratio of the circumradius and inradius is ?

Ans- let sides be x, so hypotenuse is rt2x. circumradius of right angles tr. is hypo/2. so R = x/rt2. Area = inradius * s. so we get r = x/2+rt2. ratio = (x/rt2)/(x/2+rt2). solve to get rt2 + 1. this is the method i used.....did u use any short-cut

245. the least positive integer that should be subtracted from 3011*3012 so that the difference is a perfect square..how to solve ths type of ques..

Ans- its like a(a+1)===>a^2+a..so if we subtract a from it..it will bcom prfct square..means 3011

246. Values of a & b for which (8x^3-ax^2+54x+b) is perfec cube?? Ans- for the given sum to be a cube , it should be in the form ( c-d )^3

==> c = 2x....and d = cube root b

3 c^2 d = ax^2 and 3 c d^2 = 54 x

12 x^2 d = a x^2 ==>> 12 d = a

3 * 2x * d^2 = 54 x ==>> d^2 = 9, d =3...

a = 12* 3 = 36.........b = 27...

247. P,Q,R are employed to do a work for rs 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. wage of Q in rupees is..a. 2850 b. 3750 c. 2750 d. 1000

Ans- p & Q did 19/23 work. so R did 4/23 work. R & Q did 8/23 work. so Q did 4/23 work. so work ratio is P:Q:R = 15/23:4/23:4/23....:) 4/23*5750 = 1000

248. HEY WHAT IS THE ANSWER TO THE QUES RT 6 * RT 15 = X RT 10, THEN X = ?

ACCORDING TO SSC KEY...IS IT 3 OR +/- 3..

Ans- 3

249. If one side of the triangle is double the other, and the angles on opposite sides differ by 60 degree, then the triangle is ?

1. equi....2. obtuse 3. right 4. acute..

Right

2

Ans- again 30 60 90

250. A pipe of 2 inch diameter fills the water tank in one hour. If the diameter of the pipe is 4 inch in what time will the pipe fill the same tank? (1) 10 minutes (2) 15 minutes (3) 30 minutes (4) 45 minutes Ans- r^2 = k/t.............1=k/60........k=60

now,when r=2.......(r)^2 =k/t.......4=60/t.......t=15min

251. In climbing a round pole of 80 metres height, a monkey climbs 5 metres in a minute and slips 2 metres in the alternate minute. To get to the top of the pole, the monkey would take : (A) 51 minutes (B) 54 minutes (C) 58 minutes (D) 61 minutes Ans- after 50 mins---75metres in next 1 min +5 metres..........so in 51th min. he would be on top.

in 2 min - 3 m 2*25 min i.e, 50 min - 75 m 1 min + 50 min = 5 m + 75 m so in 51 min 80 m

252. In right handed Triangle,AD is perpendicular to Hypotenuse BC.if AC=2AB...Then BD=??

1.bc/2 2.bc/3 3.bc/5

Ans- OK fresh expalnation first find out BC...........then use the similar triangle proprty to find out the BD =AB/rt5.........(1)...............AC^2 =BC*CD..................DC=4/rt5*AB...............BC=BD+CD...........BC=rt5*AB or AB =bc/rt5...............NOW PUTTING THS IN (1) BC/rt5*(1/rt5) = BD .........BD =BC/5 253. P, Q and S are moving on a circular stadium of circumference 2100 m. When P completes one round, Q is still 700 m behind. When S completes one round, Q is 300 m ahead of him. How far from the starting point three of them will meet for the first time ?

1) 168 2) 2568 3) 2100 4) None of these Ans- when P covers 2100 Q covers 1400........so P:Q = 3:2...........when S cover 2100 Q cover 2400 so S:Q = 7:8..........so ratio of P:Q:S= 12:8:7.................so they will meet the starting point in 168th round ..............so total distance covered till then is 168*2100 ..............so this will be the point 168 m away from the starting point ..................ANS is right 168

254.Two pipes can separately fill a tank in 20hr & 30hr respectively. Both pipes are opened to fill the tank bt when the tank is 1/3 full a leak develops in the tank through which 1/3 of water supplied by both the pipes leak out. What is the total time taken to fill the tank?

Ans- Without leak its 12, bt after 4hr (1/3rd) its only 2/3 efficient so 4+12=16. since leak develop after 4hr so 8hr's work finished in 12hr due to efficiency decreases by 1/3rd. Hence total time taken by pipes is 4+12, instead of 4+8.

255. A constant distance is covered by a man at 40kmph. The person rides back the same distance at 30kmph. Find his avg speed during whole journey. Ans- constant dist avg speed -- 2ab/a+b , constant time avg speed--a+b/2 whre a and b r speed

using 2xy/(x+y)................2*40*30/(40+30)=240/7 kmph 256. Two pipes A and B fill a tank in 15 hours and 20 hours respectively while a third pipe c can empty the full tank in 25 hours. All the three pipes are opened in the beginning, after 10 hours, c is closed. In how much time, will the tank be full ?

10,,,,,,,,,12,,,,,,,,,14,,,,,,16

Ans- 12

10 +2 ..

257. A man makes a proft of 20% on d sale by selling 20 articls for Re1. The numbr of articles he bought by Re 1 is???

Ans- 20*120/100--rule of frac.

20 articles for 1....means SP of 1 article = 5 paise...

on selling it by 5 paise, he makes a profit of 20 %....so CP of 1 article = 5 * 100 /120...= 500/120...

so articles bought for 1 rupee = 100/500/120 = 100*120/500 =24

258. If m and n are natural numbers such that 2^m 2^n = 960, what is the value of m ? (a) 10 (b) 12 (c) 16 (d) Cannot be determined Ans- m=10, n=6

259. 2p +1/p=4 find the value of p^3 + 1/8p^3 same as the ssc paper

Ans- (2p + 1/p) = 4 .....multiply both side by 1/2 it will be p+ 1/2p = 2.......now (p+1/2p)^3 = 2^3.........p^3+1/8P^3 + 3*p*1/2p(p+1/2p)=8..........p^3+1/2p^3 + 3/2*2 = 8 so p^3+1/2p^3 = 5 (2)'3-3(2) ANS IS 2

260. Three grades of milk are 1%, 2%, 3% fat by volume. If x gallons of the 1% grade, y gallons of the 2% grade, and z gallons of the 3% grade are mixed to give x+y+z gallons of a 1.5% grade, what is x in terms of y and z?

1) y + 3z 2) y + z/4 3) 2y + 3z 4) 3y + 2 5) 3y + 4.5z Ans- x/100+y*2/100 + z*3/100 =3/2*(x+y+z)/100).............2x+4y+6z =3x+3y+3z.........y+3z = x..:) x+2Y+3Z=1.5(X+2Y+3Z)

X=Y+3Z

261. Ram started his journey at 9.00 a.m. at 8 km/hour. Hamid started from the same spot in the same direction at 9.30 a.m. at 10 km/hour. Hamid overtakes Ram at : (A) 11.00 a.m. (B) 12.30 p.m. (C) 12.00 noon (D) 11.30 a.m Ans- 4/2=2 9.30+2=11.30 D

262. (X^4)+(1/x^4)=119 then x-(1/x)=? Ans- My desi method- Trace X^4 =119 3^4=81 4^4=256 So x must be between 3 & 4 = 3.5 Put x=3.5 in reqd Q...ans is approx 3....& thats correct ans

X^4 + {1/(X^4)} = 119........{X^+1/X^2}^2 = 121..........X^2+1/X^2 =11 X^2+1/X^2 - 2 = 11-2............{X-(1/X)}^2 = 3^2 SO,X-(1/X) = 3 263. X=(root3)/2 then [root(1+x)+root(1-x)]=?? Ans- let [rt(1+x) + rt(1-x)] = y squaring both sides....1+x + 1-x + 2 rt[(1+x)(1-x)] = 2 + 2rt(1-x^2) substituting value of x = 2+ 2rt(1-3/4) = 2+ 2rt(1/4) = 2+2*1/2 = 2+1 = 3 so y^2 = 3. i.e. y=rt3

Note down My theorem- ( one`s attepmts & score in CGL tier2 maths paper) will be directional propotional to (No of sums in which one successfully able to apply DESI METHODS)

264. Greatest no of 5 digits to b added to 8321,so that sum will b exactly divisible by 15, 20 ,24, 27 ,32 ,&36??

Ans- 99679+8321=108000 108000 is divisible by all digits

dekhte hi samajh agaya 108 hi to dekhna tha start with biggest no opt b) ye 15 se hi divide nahi ho raha, other nos se check akrne ki need nahi then check for second biggest opt a) ye bhi 15 se divide nahi ho raha then check for c) easily 15 se divide, 20se bhi hoga ek look dek ke hi pata lag rha hai. 24 se bhi divide, 27 and 32 se bhi divide.calcualtion practice ho to 1 min ka qs hai.

265. (X^4)+(1/x^4)=119 then x-(1/x)=? Ans- 3

266. Remainder when (17^37+ 29^37) is divided by 23 is=?? Ans- 0

in a^n+b^n if n is odd and postive integer then it will have a+b as one factor, so 17+29=46 will be one factor.

267. (3+Root6)/(5*root3-2*root12-root32+root50)=? Ans- Rt3(rt3+rt2)/(5rt3-4rt3-4rt2+5rt2)=rt3 268. Sunil has 4/5th of the number of berries that Rahul has. If Sunil sells the berries at 2/3rd the price per kg at which Rahul sells, and Rahul has 20% profit, what is the profit or loss percentage of Sunil? (a) 10% profit (b) 20% loss (c) 20% profit (d) 10% loss (e) neither profit nor loss Ans- 20% loss

269. Mr.Khanna earns 1/3rd of his total income from his salary while 1/5th of the rest by working for an office on weekends. He earns of the remaining from royalty payment as the author of a best seller he had written some time back and the remaining amount from investments in stocks. If he earns Rs.1200 by working on the weekends, what is the interest he gets from the investments? (a) 2400 (b) 1200 (c) 3300 (d) 2000 Ans- na re.. 1/3 + (2/3 * 1/5) = 1/3 + 2/15 = 7/15... bacha 8/15..which he divides into equal half so 4/15 and 4/15.. now its given 2/15 = 1200 so 4/15 = 2400

Goru bhai gaur frmaye.assume erning 300 so salry 100..weeknd=200*1/5=40 rest 160 soby invstmnt 80..w/i=1/2 so invst=2400

270. Three spherical balls of 1 cm, 2 cm and 3 cm are melted to form a single spherical ball. In the process, the loss of material is 25%. The radius of the new ball is (a) 6 cm (b) 5 cm (c) 3 cm (d) 2 cm Ans- 75/100* (1^3+ 2^3 + 3^3)= 27..aftr tht take cube root of 27 271. if the length of rectangle is increased by 10 percent and breadth decreased by 10% what will be change in area?

Ans- 1% decrease

hooori baba.........1% decrease

272. a very easy question. 1. the copper wire of length 36 m and diameter 2 mm is melted to form a sphere. the radius of the sphere in cm?

Ans- pie r^2h= n x 4/3 pie r^3

aab toh geometry,trigo,alzebra ki itni aadat hogayi hai ki aise ques dhudne padte hai paper mein..:) , seriously..! karne se jyadda dhudne mein tym lagta hai..:) 273. If sinx=1/rt 10 and siny=1/rt5 find x+y

Ans- sin(x+y) nikal lo..usse aa jayega bt thn hw v vill get the value of cos x and cos

cos^2 x = 1- sin^2x

cos x = 3/root 10 and cos y= 2/ root 5 by pythogoras theorm

274. Aj k liye bas easy quesn so dat evry1 can do it if 35cm long line consist of x no. of pts. then no of pts in a line of length 15cm?

Ans- Infinity is d right ans

i guess it is based on the definition 'a line consists of infinite no. of points' correct me if my theory is wrong :-) Rssj is playing with our mathematical emotions Ye to clas 5 ka quesn h.hw can u count no of pts on a given line

275. ABC is an equilateral triangle of side 4 cm. If R, r, and h are the circum radius, inradius and altitude, resp. then ( R+r)/h = ?

a. 4 b. 2 c. 1 d. 3

Ans-1

R=a/rt3 , r= a/2 rt3, altitute is 4^2=2^2=x^2 (x= altitude) put values by pythagoras theorem x=2 rt3

276. Two teams participating in a competition had to take a test in a given time. Team B chose the easier test with 300 questions, and team A the difficult test with 10% less questions. Team A completed the test 3 hours before schedule while team B completed it 6 hours before schedule. If team B answered 7 questions more than team A per hour, how many questions did team A answer per hour ? (a) 15 (b) 18 (c) 21 (d) 24 Ans- 18

Group mein total 100+ members h par answers dene k wale Sirf 15-20 log...aur sab kaha h yaar.??..Jo log jawab de rhe h wo toh phod k aye h aur aage b phodenge..nthn new for them..:) 277. Triangle ABC has Bc = 1 and AC =2, then the maximum possible value of angle A?

Ans- 30

i think when triangle is isoceles i.e third side is 2 angle will be greatest

278. Sin x sin (60-x) sin (60+x ) = 1/4 *sin 3 X

===>> cos x cos ( 60 -x) cos ( 60 + x) = 1/4 * cos 3 X

====>> tan x tan (60-x) ( 60 + x) = tan 3X

This seems right,....whats ur opinion..please post....

Ans- this is the way how to create sortcuts

279. The bisector of angle A of triangle ABC cuts BC at D and the circumcircle of the triangle at E then 1- AB:AC = BD:DC 2-AD:AC = AE:AB 3-AB:AD = AC:AE 4-AB:AD = AE:AC

Ans- 1 & 4 both are possible...

280. A traveler walks a certain distance. Had he gone half a kilometer an hour faster, he would have walked it in4\5th of the time, and had he gone half a km an hour slower, he would have traveled5\2 hours longer. What is the distance? (a) 10 km (b)15 km (c) 30 km (d) Data insufficient. Ans- Since the traveler takes4\5 th of the time he makes his speed 5/4 of original speed. He increases his speed by 1/4. 1/4 = 1/2 km/hr. speed = 2km/hr. In second case he goes 1/2 km/hr slower. 3/4 of original speed he takes 4/3 of time. 1/3 of time = 5/2 hrs. time = 15/2hrs distance = 2 x 15/2= 15 km

281. Find tan 15* tan 30* tan 60?

Ans- 2-3

somebody given yesterday that tan A * tan 2A * tan 4A = tan 3 a.

have to change formula....right?

282. A triangle has sides 6,7, and 8. The line through its incentre parellel to the shortest side is drawn to meet the other two sides at P and Q. Find PQ? a. 25/7, b. 30, c. 40/7, d. 30/7

Ans 30/7

draw the figure.........

A/S = r.....A = 1/2*base*h

find the ratio of r and h....

u will find 2 similar triangles, APQ and ABC....find PQ by equating 2 sides...

( height of triangle APQ will be h-r) 283. If A lies in the third quadrant and 3tanA -4 = 0 then, 5sin2A+3sinA+4cosA is..?

Ans- 0, yep zero sin and cos are negative in thir d quadrant

only tan and cot are positive in third quadrant.

284. ABC is an acute angled triangle with circumcentre O, orthocentre H. If AO = AH, then angle A = ?

Ans- It means O n H are the same point.. Angle A+ Angle BOC= 180 & Angle A= 1/2 Angle BOC Dono se angle 60 degree aa jayega only in case of equilateral triangle orthocentre, circumcentre, incentre & centroid will b same so A= 60

285. If in a triangle sin A cos B = ( rt 2 -1 ) / rt 2....and sin B cos A = 1/ rt 2, then the triangle is..

1. equilateral....2. isosceles...3. right ...4. right angled isosceles

Ans- right angled very- very easy apply Sin(A+B) 286. If A, B, C are angles of a triangle such that angle A is obtuse, then tan B tan C will be less than... a. 1/ rt3 ...b. rt3/2...c. 1....d.none

Ans- 1

287. Easy one A 100 ml flask contains 30% acid solution. What quantity of the solution should be replaced with 12% acid solution so that the resultant solution contains 21% acid? (a) 50 ml (b) 44.44 ml (c) 33.33 ml (d) 64 m Ans- Let x ml of the solution be replaced with 12% acid solution. Therefore, 30/100*(100-X) + 12/100 *X = 21/100 X=50

288. Find d value of cot(45+x)*cot(45-x) Ans- 1

289. If secx+tanx= k then secx-tanx =?

Ans- SECX-TANX)(SECX+TANX) = K(SECX-TANX).........SECX-TANX =1/K 289. In a km race, A gives B a start of 20 seconds and beats him by 40m. However, when he gives B a start of 25 seconds they finish in a dead heat. What is As speed in m/sec?

Ans-10

let A in t secs complete 1000m and B in t+20 sec 960m... eqn1 also B in t+25 sec completes 1000 m... eqn2... use them to find ans

290. Find d Value of k for which (x+2) is a factor of (x+1)^7 +(3x+k)^3 Ans- put x=-2 ... 5^7 +(k-6)^3=0 (k-6)^3=- 5^7 (k-6)=(-5)^(7/3) k=(-5)^7/3+6 =25.(-5)^1/3+6 291. In an examination, A attempted 80% of questions with 50% accuracy and B attempted 60% of questions with 80% accuracy. The negative marking is 25% for wrong answer, there is no negative marks for un-attempted questions.

What is the difference of marks obtained by A and B?

Ans- it can be 1, 2 or anything

hmm... the marks of each question is not given..

Thx god cgl12 ka paper aap nhi bna rahe ho

292. 2 cm of rain has fallen on a square km of land.Assuming that 50% of raindrops have been collected and contained in a pool having area of base 1000m^2,by what level would the water level in pool have inbcreased. Options:10m,8m,15m,17m

Ans- 10m

293. The chord RS of length 8 cm , of a circle with center C,cuts one of the diameter PQ in a point T such that CT=TQ, If RT = 6,then the diameter of the circle is (a) 14 cm (b) 8 cm (c) 16 cm (d) None of these

PS-Same concept as last ques from maneesh

Ans- Let radius of the circle be r PT = 3r/2 and TQ = r/2. (3r/2) (r/2) = 6 x 2 = 12. 3r2/4 = 12 r2 = 16. r = + 4. Hence the diameter of the circle is 8 cm. But it is impossible to draw a chord of length 8 cm (other than the diameter) in a circle of diameter 8 cm. The circle must be imaginary.

arre yaar ye imaginary circle kya hota hai.

i don't know bhai!!

294. What must be added to 1/x to make it x.

Ans- x^2-1/x

295. a^2+b^2+2b+4a+5=0 than a-b/a+b 3 -3 1/3 -1/3

Ans- a=-2,b=-1..so ans 1/3

296. If a+b+c = 0, then a^2/bc + b^2/ca + c^2/ab = ?

Ans- 3

go desi a=2 b=c=-1

If a+b+c=0 then a^3+ b^3 + c^3 = 3abc

2976. If x = (16^3 + 17^3 + 18^3 + 19^3 ), then x divided by 7 leaves a remainder of options:1,3,7,9,none of these

Ans- 0

(16+17+18+19)/7=10 so rem=0 my take 298. 3^x - 3^(x-1) = 18, then the value of x^x=? Ans- 3^3 - 3^2 = 18.....so x= 3.......and the required ans= 27

299. (a-1)^2+(b+2)^2+(c+1)^2=0 than 2a-3b+7c = ?? Easy ques but very impt. concept wrt to Tier 2

Ans- 1, whats the concept yaar...how do u people do in split second?!

sum of square is 0 hence all of them are zero

Put a=1,B=-2,c-1...

300. In an increasing arithmetic progression, the product of the 5th term and the 6th term is 300. When the 9th term of this a.p. Is divided by the 5th term, the quotient is 5 and the remainder is 4. What is the first term of the a.p.? (a) 12 (b) 40 (c) 16 (d) 5 Ans- (a+4d)(a+5d) = 300..................(1)

a+8d = a+4d)*5 + 4

a = -(1+3d)

putting it in (1).

(d-1)(2d-1) = 300

d = 13

a = -40

301. What must b added to x^3-3x^2-12x+19 so dat result is exactly divisible by x^2+x-6

Ans- Na,ans is in algebric term, rd sharma clas 9th sum

x^ + 3x - 1 :-( Its 2x+5

ha,, -2x -5 is the remainder .... yes tht shud be added to it.. didnt know wht to do with it.. :-(,,

302. AB and CD are two parallel chords drawn on two opposite sides of their parallel diameter such that AB = 6cm, CD = 8cm.If the radius of the circle is 5cm, the distance between the chords, in cm, is

Ans- I am also getting 7 but it is not in the options...ye question bhi galat tha exam mein..have to represent.

whats the formula for this?

303. If x = 2+rt3, the value of (x^6 + x^4 + x^2 + 1)/x^3 is Ans- Given eqn can b written as x^3+1/x^3 +x+1/x x=2+rt3, 1/x =2-rt3 put it in d eqn

@rssj cgl..pehli baar kisi ladki ka maths itna acha dekha hai... Hehe mom ne kaha tha maths aaye na aaye roti banana aana chahiye,maine advice ulta le liya :) 304. Let C and K be constants. If x^2 + Kx + 5 factors into (x + 1)(x + C), the value of K is (A) 0 (B) 5 (C) 6 (D) 8 (E) none Ans- X^2+6x+5=(x+1)(x+5 on x= -1 k=6 yahi method na? short explanation bhi de diya karo ans ke saath bhai logo.

305. If sin@ + sin^2@ + sin^3@ = 1, then cos^6@ - 4cos^4@ + 8cos^2@is equal to

Ans- sin@ + sin^2@ + sin^3@ = 1

sin@ + sin^3@ = 1 - sin^2@

sin@(1 + sin^2@) = cos^2@ Sq both sides (1-cos^2@)(2-cos^2@)^2 = cos^4@ expand kar lo u vl get the answer

306. If secx + cosx = 3, than tan^2x-sin^2x is:

Ans- 5

sqaure karo,,, scene solve

307The ratio of the angles in TringABC is 2 : 3 : 4. Which one of the following triangles is similar to ABC ? (A) TringDEF has angles in the ratio 4 : 3 : 2. (B) TringPQR has angles in the ratio 1 : 2 : 3. (C)Tring LMN has angles in the ratio 1 : 1 : 1. (D)Tring STW has sides in the ratio 1 : 1 : 1. (E)Tring XYZ has sides in the ratio 4 : 3 : 2. Ans- (A) TringDEF has angles in the ratio 4 : 3 : 2. 308. Max n min value of 5cosx + 3cos(x+pi/3) + 3 Solutions any1??

Ans- Open d bracket of cos usin cos(a+b) u wil get 13/2 cos@-rt3/2sin@+3 Lagta h padhai shuru karni padegi..ppl here r very intelligent..!

= +- sqrt (sq of 13/2+sq 3rt3/2)+3 =+-7+3 thus max 10 min -4

Term vl b 13/2 cos@-3rt3/2sin@+3

Ha wo typo ho gya,bt dat doesnt matter solve krne ka idea to mil gya

309. If sec A= (x+1/4x) then sec A+ tan A ?? Ans- 2x or 1/2x

why 1/2x

sir u can write tan x = x - 1/4x or 1/4x -x

310. A and B are centres of the two circles whose radii are 5 cn and 2 cm respectively.The direct common tangents to the circle meet AB extended at P.Then P divides AB in.. 1.externally in ratio 5:2 2.internally ratio 2:5 3.internally 5:2 4.externally 7:2

Ans- 1?

the points of intersection of direct common tangents and indirect common tangents to two circles divide the line segment joining the two centres externally and internally rsepectively in the ratio of their rad

311. Two spheres of radii 6 cm and 1 cm are inscribed in a right circular cone. The bigger sphere touches the smaller one and also the base of the cone. What is the height of the cone? (a) 14 cm (b) 1