SSC Combined Graduate Level (CGL) Tier-I Exam, 2015 (Second Sitting)

Ans .

4

1. Explanation :

   On;y Conclusion II follows. It was expected that crop would improve after the rains
   

Ans .

4

1. Explanation :

   

Ans .

3

1. Explanation :

   Let the ammount invested in each company be Rs x.
   S.I. = \( \frac{Principal * Rate * Time }{100} \)
   According to the Question,
   \( \frac{x*15*5}{100} - \frac{x*12*4}{100} = 1350 \)
   \( \frac{75x}{100} - \frac{48x}{100} = 1350 \)
   \( \frac{27x}{100} = 1350 \)
   x = \( \frac{1350*100}{27}\) = Rs.5000
   

Ans .

2

1. Explanation :

   According to the Question ,
   Since , Market tax = Rs.165 crores
   Therefore , 33% = Rs. 165 crores
   Therefore, 100-33 = 67% = \( \frac{165*67}{33} \)
   = Rs.335 crores
   

Ans .

1

1. Explanation :

   Since, 100% = Rs.733 crores
   Therefore, 35+10 = 45% = \( \frac{733}{100}*45 \)
   = Rs.329.85 crores
   

Ans .

1

1. Explanation :

   since, 100% = 360°
   Therefore, 1% = \( \frac{360°}{100} \)
   Therefore, 35% = \( \frac{360°}{100}*35 \) = 126°
   

Ans .

1

1. Explanation :

   \(1 + cos ^2 \theta = 3 sin \theta . cos \theta \)
   Dividing both sides by sin \(^2 \theta\)
   \( \frac{1}{sin ^2 \theta} +\frac{cos ^2 \theta}{sin ^2 \theta} = \frac{3 sin \theta . cos \theta}{sin ^2 \theta } \)
   \( cosec ^2 \theta + cot ^2 \theta = 3 cot \theta \)
   \( 1 + cot ^2 \theta + cot ^2 \theta = 3 cot \theta \)
   \( 2 cot ^2 \theta - 3 cot \theta + 1 = 0 \)
   \( 2 cot ^2 \theta - 2 cot \theta - cot \theta + 1 = 0 \)
   \( 2 cot ^ \theta ( cot \theta - 1 )- 1 (cot \theta - 1) = 0\)
   \( (2 cot \theta - 1)(cot \theta - 1) = 0 \)
   \( cot \theta = \frac{1}{2} \)or 1
   

Ans .

3

1. Explanation :

   Average units comsumption in 2012 = \( \frac{600+700+400+300+200}{5} \)
   = \( \frac{2200}{5} \)= 440 units
   Required months => July, August
   

Ans .

3

1. Explanation :

   average units consumption in the year 2013 = \( \frac{550+500+400+350+500}{5} \)
   = \( \frac{2300}{5} \) = 460 units
   

Ans .

4

1. Explanation :

   In the month of November ,
   Differnece = 500 - 200 = 300 units
   In the month of August ,
   Difference = 700-500 = 200 units
   

Ans .

4

1. Explanation :

   Total Consumption in 2012 = 2200 units
   Total consumption in 2013 = 2300 units
   Percentage increase = \( \frac{2300-2200}{2200}*100 \)
   = \( \frac{100}{22} = \frac{50}{11} = 4.5% \)
   

Ans .

3

1. Explanation :

   Let the numbers of 2x & 3x respectively .
   according to the Question,
   \( \frac{2x+8}{3x+8} = \frac{3}{4} \)
   9x+24=8x+32
   9x-8x=32-24=8
   x=8
   Therefore, Sum of number = 2x+3x=5x=5*8=40
   

Ans .

4

1. Explanation :

   Let A,B,C,D,E in Kg. represent their respective weights .
   Then, A+B+C=84*3=252 Kg.
   A+B+C+D=80*4=320 Kg.
   Therefore, D = 320-252 =68Kg.
   E = 68 + 3 = 71 Kg.
   B+C+D+E=79*4=316Kg.
   Now,
   (A+B+C+D)-(B+C+D+E)=320-316
   A-E=4 Kg.
   A = 4+E = 4+71=75 KG.
   

Ans .

1

1. Explanation :

   The ratio of the areas of two similar triangles is equal to the ratio of squares of any two correspondence sides
   Therefore, \( \frac{area(ΔPQR)}{area(ΔABC)} = \frac{PR^2}{AC^2} \)
   \( \frac{PR^2}{AC^2} = \frac{256}{441} \)
   \( \frac{12^2}{AC^2} = \frac{256}{441} \)
   \( AC = \frac{12*21}{16} \) = 15.75 cm
   

Ans .

4

1. Explanation :

   Expression = \( 3(sin ^4 \theta + cos ^4 \theta)+ 2(sin ^6 \theta + cos ^6 \theta) + 12.sin^2 \theta .cos^2\theta \)
   = \( 3 {(sin ^2 \theta + cos ^2 \theta)^2 - 2.sin ^2 \theta . cos ^2 \theta }+ 2{(sin^2\theta + cos^2\theta)3-3sin^2\theta.cos^2\theta(sin^2\theta+cos^2\theta)+12 sin^2\theta.cos^2\theta} \)
   = \(3 (1 - 2.sin ^2 \theta .cos^2\theta) + 2(1-3.sin^2\theta.cos^2\theta) +12 sin^2\theta.cos^2\theta \)
   = \( 3 - 6 sin^2\theta.cos^2\theta + 2 - 6 .sin^2\theta.cos^2\theta + 12 sin^2\theta.cos^2\theta \)
   = 5
   

Ans .

4

1. Explanation :

   Let the marked price of the camera be Rs.x
   According to the Question
   \( \frac{x*90}{100} = \frac{600*120}{100} \)
   x *90 = 600*120
   x = Rs.800
   

Ans .

2

1. Explanation :

   
   PQ = Tower A = 45 metre
   RS = Tower B = 15 metre
   RS = x metre (let)
   \( \angle PSQ = 60 ; \angle RQS = \theta \)
   From triangle PQS ;
   tan \(\theta = \frac{PQ}{QS}\)
   => \( \sqrt3 = \frac{45}{x} \)
   => \( \sqrt 3 x = 45 \)
   => x = 15 \( \sqrt 3 \)metre
   From Triangle RSQ,
   tan \( \theta = \frac{RS}{QS} = \frac{15}{15\sqrt3} \)
   tan \( \theta = \frac{1}{\sqrt 3} \)
   \( \theta = 30 \)
   Therefore, sin \( \theta = sin 30 = \frac{1}{2} \)
   

Ans .

3

1. Explanation :

   x = 4
   => equation of a line parallel to y-axis
   y = 3
   => Equation of a line parallel to x-axis
   Putting x=0 in the equation
   3x+4y=12
   3*x+4y=12 => y = \( \frac{12}{4} \)=3
   Therefore, Coordinates of the point of intersection on Y-axis =(0,3)
   Again putting y=0 in the equation 3x+4y=12
   3x+4*0=12 => x =\( \frac{12}{3} \)=4
   Therefore, Coordinates of the point of intersction of X-axis = (4,0)
   
   C = 3 units , BC = 4 units
   Therefore, Area of Triangle ABC = \( \frac{1}{2} \)BC*AC
   = \( \frac{1}{2} \)*4*3
   =6 sq.units
   

Ans .

4

1. Explanation :

   
   \( \angle \)BAC = 40°
   \( \angle \)ABC = 65°
   \( \angle \)ACB = 180°-40°-65°=75°
   DE || BC
   Therefore,
   \( \angle \)AED = \( \angle \)ACB = 75°
   Therefore, \( \angle \)CED = 180°-75°=65°
   

Ans .

1

1. Explanation :

   \( x^2+y^2+z^2 = 2(x+z-1) \)
   \( x^2+y^2+z^2=2x+2z-2 \)
   \( x^2-2x+y^2+z^2-2z+2=0 \)
   \( (x-1)^2+y^2+(z-1)^2=0 \)
   Therefore, x-1=0 => x=1
   y=0
   z-1=0=>z=1
   Therefore, \( x^3+y^3+z^3=1+0+1=2 \)
   

Ans .

2

1. Explanation :

   Let the average cost of each book bought (of 64 books) be Rs.x
   According to the question,
   64 * x - 50 (x+1) = 76
   64x-50x-50=76
   14x=76+50=126
   x=\(\frac{126}{14}\)=9
   Therefore, Required average price = 9+1 = Rs.10
   

Ans .

4

1. Explanation :

   \( \angle APB = 110° = \angle CPD \)
   Therefore, \( \angle APD = 180° - 110° = 70° = \angle BPC \)
   

Ans .

2

1. Explanation :

   \(x^2+x=5\)(Given)
   Let,x+3=a
   Therefore, \( \frac{1}{x+3} = \frac{1}{a} \)
   Now,
   a + \(\frac{1}{a}\) = (x+3) + \( \frac{1}{x+3} \)
   = \( \frac{(x+3)^2+1}{x+3} \)
   = \( \frac{x^2+6x+9+1}{x+3} \)
   = \( \frac{x^2+6x+10}{x+3} \)
   = \( \frac{x^2+x+5x+10}{x+3} )
   = \( \frac{5+5x+10}{x+3} \)
   = \( \frac{5x+15}{x+3} = \frac{5(x+3)}{x+3} = 5 \)
   Therefore, \( a^3+\frac{1}{a^3} = (a+\frac{1}{a})^3-3a*\frac{1}{a}(a+\frac{1}{a}) \)
   = \( 5^3-3*5 = 125-15 = 110 \)
   

Ans .

2

1. Explanation :

   In ΔABC,
   \( \angle BAC = 85° \)
   \( \angle BCA = 75° \)
   \( \angle ABC = \) 180° - 85° - 75° = 20°
   The angle subtended by an arc of a circle at the centre is doube the angle sutended by it at any part of the circle .
   Therefore, \( \angle AOZ = \angle ABC = 40° \)
   Therefore, OA=OC=radii
   In ΔOAC,
   \( \angle OAC = \angle OCA \)
   \( \angle OAC + \angle OCA = 180° - 40° = 140° \)
   Therefore, \( \angle OAC = \angle OCA = \frac{140°}{2} = 70° \)
   

Ans .

2

1. Explanation :

   \( sec \theta + tan \theta= 2 + \sqrt 5 \)
   Since, \( sec^2 \theta - tan^2 \theta = 1 \)
   => \( (sec \theta + tan \theta)(sec \theta - tan \theta) = 1 \)
   => \( sec \theta - tan \theta = \frac{1}{\sqrt 5 + 2 } \)
   = \( \frac{1}{\sqrt 5 + 2} * \frac{\sqrt 5 - 2}{\sqrt 5 - 2} = \frac{\sqrt 5 - 2}{5-4} \)
   = \( \sqrt5 - 2 \)
   Therefore, \( sec \theta + tan \theta + sec \theta - tan \theta = 2 + \sqrt 5 + \sqrt 5 - 2 \)
   => 2.\( sec \theta = 2\sqrt 5 \)
   => \( sec \theta = \sqrt 5 \) ........(i)
   Again,
   \( sec \theta + tan \theta - (sec \theta - tan \theta) \)
   = 2 + \( \sqrt 5 - \sqrt 5 + 2 \)
   => 2.\( tan \theta = 4 => tan \theta = 2 \).....(ii)
   Therefore, sin \( \theta = \frac{tan \theta}{sec \theta} = \frac{2}{\sqrt 5} \)
   

Ans .

4

1. Explanation :

   \( \angle QPR = 50° \) Therefore, \( \angle PQR +\angle PRQ \)= 180°-50°=130°
   Therefore, \( \frac{1}{2} \angle PQR + \frac{1}{2} \angle PRQ = 65° \)
   The point of intersection of internal bisectors of angles is in-centre.
   Therefore, \( \angle OQR = \frac{1}{2} \angle PQR \);
   \( \angle ORQ = \frac{1}{2} \angle PRQ\)
   In ΔOQR,
   \( \angle OQR +\angle QOR + \angle ORQ = 180° \)
   => \( \QOR = 180° - 65° = 115° \)
   

Ans .

2

1. Explanation :

   \( \frac{sec \theta + tan \theta}{sec \theta - tan \theta} = 2\frac{51}{79} \)
   \( \frac{158+51}{79} = \frac{209}{79} \)
   \( \frac{sec \theta + tan \theta + sec \theta - tan \theta }{sec \theta + tan \theta - sec \theta + tan \theta } = \frac{209+79}{209-79} \)
   \( \frac{2 sec \theta}{2 tan \theta} = \frac{288}{130} \)
   \( \frac{sec \theta}{tan \theta} = \frac{144}{65} \)
   \( sin \theta \frac{tan \theta}{sec \theta } = \frac{65}{144} \)
   

Ans .

3

1. Explanation :

   Expression = \( \sqrt (\frac{0.324*0.081*4.264}{1.5625*0.0289*72.9*64}) \)
   \( \sqrt (\frac{324*81*4264}{15625*289*729*64}) \)
   \( \sqrt (\frac{18*9*68}{125*17*27*8}) \) = 0.024
   

Ans .

4

1. Explanation :

   Volume of prism = Area of base * height
   => 7200 = \( (\frac{3\sqrt 3}{2})P^2 , 100 \sqrt 3 \)
   7200 = 50*3*3 P²
   P² = \( \frac{7200}{50*3*3} = 16 \)
   P = \sqrt 16 = 4
   

Ans .

3

1. Explanation :

   Single Equivalent discount = \( (10+20-\frac{10*20}{100})% \)
   = (30-2)% = 28%
   Therefore, C>P> of article = 100-28 = Rs.72
   Actual cost price of Article
   = \( \frac{72*110}{100} \)= Rs.79.2
   Therefore,For a profit of 15%
   Required S>P> = \( \frac{79.2*115}{100} \)
   = Rs.91.08
   

Ans .

1

1. Explanation :

   
   AB = 10 cm
   Therefore, AF = FB =5 cm
   CD = 24 sm
   Therefore, CE =DE = 12 cm
   Let OE = x cm
   Therefore, OF = (17-x) cm
   OD = \( \sqrt (OE^2+DE^2) \)
   OD = \( \sqrt (x^2+12^2) \)........(i)
   From ΔOAF,
   OA = \( \sqrt (OF^2+AF^2) \)
   OA = \( \sqrt ((17-x)^2+5^2) \).......(ii)
   Since, OA=OD
   \( \sqrt (x^2+12^2) = \sqrt ((17-x)^2+5^2) \)
   => \( x^2 + 144 = 289 -34x + x^2 + 25 \)
   => 34x = 289 + 25 - 144 = 170
   => x = \( \frac{170}{34} \)= 5
   Therefore, from eqn (i)
   OD = \( \sqrt (x^2+12^2) = \sqrt (5^2+144) = \sqrt 169 = 13 cm. \)
   

Ans .

2

1. Explanation :

   x = z= 225 , y = 226
   Therefore, x+y+z=225+226+225=676
   Therefore, \( x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)[(x-y)^2 + (y-z)^2 + (z-x)^2] \)
   = \( \frac{1}{2}(676)[(225-226)^2 + (226-225)^2 + (225-225)^2] \)
   = \( \frac{1}{2} * 676 * 2 \)
   = 676
   

Ans .

4

1. Explanation :

   Let, a = \( 1 + \frac{1}{10 + \frac{1}{10}} \)
   = \( 1 + \frac{1}{\frac{100+1}{10}} = 1 + \frac{10}{101} \)
   = \( \frac{101+10}{101} = \frac{111}{101} \)
   Again , b = \( 1 - \frac{1}{10 + \frac{1}{10}} \)
   = \( 1 - \frac{1}{\frac{100+1}{10}} = 1 - \frac{10}{101} \)
   = \( \frac{101-10}{101} = \frac{91}{101} \)
   Therefore, Expression = \( (a^2-b^2)/ab \) = [(a+b)(a-b)]/ab
   = \( ( \frac{111}{101} + \frac{91}{101} ) ( \frac{111}{101} - \frac{91}{101}) / (\frac{111}{101}*\frac{91}{101}) \)
   = \( \frac{202}{101}*\frac{20}{101}*\frac{101*101}{111*91} \)
   = ( \frac{4040}{10101} \)
   

Ans .

2

1. Explanation :

   Let 3Kg of first aloy and 4Kg of seconf alloy be mixed together.
   Therefore, In 3 Kg of Mixture ,
   Tin = 1 Kg.
   Iron = 2 Kg.
   In 4 Kg of mixture ,
   Tin = \( \frac{2}{5}*4 =\frac{8}{15} = 1.6Kg \)
   Iron = \( \frac{3}{5}*4 =\frac{12}{5} = 2.4 Kg. \)
   Therefore, Required Ration = (1+1.6):(2+2.4)=2.6:4.4=13:22
   

Ans .

3

1. Explanation :

   Required mass of lead = 8000 * \( \frac{60}{100} * (1-\frac{3}{400})\)
   = 8000 * \( \frac{60}{100}*\frac{397}{400} \)
   = 4764 Kg.
   

Ans .

3

1. Explanation :

   4a-\(\frac{4}{a}\)+3=0
   On dividing by 4,
   a-\(\frac{1}{a} = -\frac{3}{4} \)
   Therefore,
   \( a^3-\frac{1}{a^3} = (a-\frac{1}{a})^3+3a*\frac{1}{a}*(a-\frac{1}{a}) \)
   = \( (\frac{-3}{4})^3+3*\frac{-3}{4}\)
   = \( (\frac{-27}{64})-\frac{9}{4}\)
   = \( \frac{-171}{64} \)
   \( a^3-\frac{1}{a^3} +3 = \frac{-171}{64} + 3 \)
   = \( \frac{-171+192}{64} = \frac{21}{64} \)
   

Ans .

1

1. Explanation :

   C.P. of cycle = Rs.x(let)
   Therefore, S.P. = \( \frac{110x}{100} \)
   = Rs. \( \frac{11x}{10} \)
   CASE II,
   New C.P. = Rs. \( \frac{9x}{10} \)
   Therefore, \( \frac{11x}{10}+60=\frac{9x}{10}*\frac{125}{100} \)
   =Rs.\( \frac{9x}{8} \)
   => \( \frac{9x}{8}-\frac{11x}{8}=60 \)
   => \( \frac{90x-88x}{80}=60 \)
   => \( \frac{2x}{80}=60 \)
   => \( \frac{x}{40} = 60 => x = 60 * 40 \)
   = Rs.2400
   

Ans .

1

1. Explanation :

   x= \( \frac{\sqrt 5 - \sqrt 3 }{\sqrt 5 + \sqrt 3 } \)
   y = \( \frac{\sqrt 5 + \sqrt 3 }{\sqrt 5 - \sqrt 3 } \)
   Therefore, x+y = \( \frac{\sqrt 5 - \sqrt 3 }{\sqrt 5 + \sqrt 3 } + \frac{\sqrt 5 + \sqrt 3 }{\sqrt 5 - \sqrt 3 } \)
   = \( \frac{(\sqrt 5 - \sqrt 3)^2+(\sqrt 5 + \sqrt 3)^2}{(\sqrt 5 + \sqrt 3)+(\sqrt 5 - \sqrt 3)} \)
   = \( \frac{2((\sqrt 5)^2 - (\sqrt 3)^2)}{5-3} \)
   =5+3=8
   xy =\( \frac{\sqrt 5 - \sqrt 3 }{\sqrt 5 + \sqrt 3 } * \frac{\sqrt 5 + \sqrt 3 }{\sqrt 5 - \sqrt 3 } \) = 1
   Therefore,
   \( \frac{x^2+xy+y^2}{x^2-xy+y^2} = \frac{(x+y)^2-xy}{(x+y)^2-3xy}\)
   =\( \frac{8^2-1}{8^2-3} =\frac{64-1}{64-3} = \frac{63}{61} \)
   

Ans .

2

1. Explanation :

   Expression = \( 2b^2c^2 + 2c^2a^2 + 2a^2 b^2 - a^4 - b^4 -c^4 \)
   = \( 4b^2c^2 - (2b^2c^2 - 2c^2a^2 - 2a^2b^2 +a^4 +b^4+ c^4) \)
   =\( (2bc)^2 - (a^2-b^2-c^2)^2\)
   =\((2bc+a^2-b^2-c^2)(2bc-a^2+b^2+c^2)\)
   =\((a^2 - (b^2 + c^2 - 2bc ))(b^2+c^2+2bc-a^2)\)
   =\( (a^2 - (b-c)^2 )((b+c)^2-a^2) \)
   = (a-b+c)(a+b-c)(a+b+c)(b+c-a)
   = 0
   

Ans .

4

1. Explanation :

   Rate downstream = (6+1.5)kmph = 7.5 kmph
   Rate upstream = (6-1.5) = 4.5 kmph
   according to the question,
   Time = \( \frac{Distance}{Speed} \)
   Therefore, Required time = \( \frac{22.5}{7.5} + \frac{22.5+4.5}{} \)
   = 3+5=8
   

Ans .

3

1. Explanation :

   Let the C.P. of article be Rs.100 nd the marked price be Rs.x
   Case 1 :
   \(\frac{x*90}{100}=120 \)
   => x = \(\frac{120*100}{90} \)
   = Rs. \( \frac{400}{3} \)
   Case 2 :
   S.P. = \( \frac{x*80}{100} = \frac{4x}{5} \)
   = Rs. \((\frac{4}{5}*\frac{400}{3}) \) = Rs. \( \frac{320}{3} \)
   Profit = Rs. \( (\frac{320}{3}-100) \)
   =Rs. \( \frac{320-300}{3} \)
   =Rs. \( \frac{20}{3} \)
   Therefore, Profit Percent = \( \frac{20}{3} \)% = 6\( \frac{2}{3} \)%
   

Ans .

1

1. Explanation :

   cos 24° + cos 55° + cos 125° + cos 204° + cos 300°
   = cos 24° + cos 55° + cos (180-55)° + cos (180+24)° + cos (360-60)°
   = cos 24° + cos 55° - cos 55° - cos 24° + cos 60°
   = cos 60° = 1\2
   

Ans .

right

1. Explanation :

   \(\angle\)OBC = \(\frac{1}{2}\) \(\angle\)ABC
   \(\angle\)OCB = \(\frac{1}{2}\) \(\angle\) ACB
   From ΔABC.
   \(\angle\)OBC + \(\angle\)OCB + \(\angle\)BOC =180°
   \(\frac{1}{2} \)( \(\angle\)ABC + \(\angle\)ACB) + \(\angle\)BOC = 180°
   \(\frac{1}{2} \)(180° -100°) + \(\angle\) BOC = 180°
   \(\angle\)BOC = 180°-40°=140°
   

Ans .

4

1. Explanation :

   
   In ΔAPB and ΔBCQ ,
   \( \angle \)PAB = \( \angle \)BCQ = 90
   \( \angle \)PBA = \( \angle \)QBC
   By AA - similarity ,
   ΔAPB \( \sim \) ΔBCQ
   Therefore, \( \frac{AB}{BC} = \frac{AP}{QC} \)
   \( \frac{8}{BC} = \frac{6}{3} \)
   BC = 4 cm.
   Therefore,
   PQ = \( \sqrt (AC^2+(r_1+r_2)^2) \)
   = \( \sqrt ((8+4)^2+(6+3)^2\)
   = \( \sqrt (12^2+9^2) \)
   = \( \sqrt (144+81) \)
   =\( \sqrt 225 \)
   =15 cm
   

Ans .

1

1. Explanation :

   Let time taken by A = x days.
   Therefore, Time talen by B = 2x days
   Time take by C = 3x days
   According to the Question ,
   \( \frac{1}{x} + \frac{1}{2x} +\frac{1}{3x} = \frac{1}{6} \)
   x = 11
   Therefore, time taken by C alone = 3x
   = 3 * 11 = 33 days
   

Ans .

4

1. Explanation :

   Part of tank filled by pipes A and B in 1 minute
   = \( \frac{1}{30} + \frac{1}{45} = \frac{3+2}{90} = \frac{1}{18} \)part
   Part of tank filled by pipes A and B in 12 minute
   = \( \frac{12}{18} = \frac{2}{3} \)part
   Remaining Part = 1 - \(\frac{2}{3} = \frac{1}{3}\)
   When pipe C is opened ,
   {art of tank by all three pipes = \( \frac{1}{30}+\frac{1}{45}-\frac{1}{36} \)
   = \( \frac{6+4-5}{180} = \frac{5}{180} = \frac{1}{36} \)
   Therefore, Time taken in filling \(\frac{1}{3} \) part = \( \frac{1}{3} *36\) = 12 minutes
   Therefore, Total time = 12+12 = 24 minutes
   

Ans .

4

1. Explanation :

   
   let the radius of swimming pool be r metre .
   Therefore, OB = (r+4) metre
   According to the Question,
   \(\pi * OB^2 - \pi * OA^2\) = \( \frac{11}{25} \pi * OA^2 \)
   => \( (r+4)^2 - r^2 = \frac{11}{25} r^2\)
   => \( r^2 + 8r + 16 - r^2 = \frac{11}{25} r^2 \)
   => \( 8r + 16 = \frac{11}{25}r^2\)
   => \( 11r^2 - 200r - 400 = 0 \)
   r = 20 metre because r \( \neq -\frac{20}{11} \)metre
   

Ans .

2

1. Explanation :

   Let C complete the work in x days.
   Therefore, B's day's work = \( \frac{1}{20}-\frac{1}{x} \)
   and A's day's work = \( \frac{2-3}{60}+\frac{1}{x} = \frac{1}{x} - \frac{1}{60}\)
   According tp the question,
   \( 5(\frac{1}{x}-\frac{1}{60})+15(\frac{1}{20}-\frac{1}{x})+\frac{18}{x}=1 \)
   \( \frac{5}{x} - \frac{1}{12} +\frac{15}{20} -\frac{15}{x} +\frac{18}{x} = 1 \)
   \( \frac{5}{x} - \frac{15}{x} +\frac{18}{x} = 1 + \frac{1}{12} - \frac{3}{4} \)
   => \( (\frac{5-15+18}{x}) =(\frac{12+1-9}{12}) \)
   => \( \frac{8}{x} = \frac{1}{3} \)
   => x=8*3=24 days
   

Ans .

1

1. Explanation :

   \( x+\frac{1}{x}=1 \)
   => \( x^2+1=x => x^2-x+1=0 \)
   = \( \frac{2}{x^2-x+2} = \frac{2}{x^2-x+1+1} =\frac{2}{0+1} = 2\)
   

Ans .

2

1. Explanation :

   tan A +cot A = 2
   tan A + 1/tan A = 2
   \( \frac{tan^2 A+1}{tan A} = 2 \)
   \( tan^2A\)+1=2tanA
   \( tan^2A\)-2tanA+1=0
   (tanA-1)²=0
   tan A - 1 = 0
   Tan A = 1
   Cot A = 1
   \(tan^10 A + cot^10 A\) = 1+1 = 2
   

Ans .

3

1. Explanation :

   Here distance is constant .
   Therefore, Speed is inversly proportional to time
   Therefore, Ratio of the speeds of A and B = \( \frac{\frac{7}{2}}{4} \)= 7:8
   Therefore, A's speed = 7x kmph
   B's speed = 8x kmph
   therefore AB = 28x kmph
   Let both trains cross each other after t hours from & a.m.
   According to the Question,
   7x(t+2)+8x*t=28x
   t = \(\frac{14}{15}\)hours
   = \((\frac{14}{15}*60)\) minutes
   56 minutes
   Therefore, Required time = 7;56 a.m.
   

Ans .

4

1. Explanation :

   Radius of cylindrical vessel = r cm (let)
   Volume of conical piece of iron = \(\frac{1}{3} \pi * R^2 \)* h
   = \( \frac{1}{3} \pi *14*14*30 \)cu.cm
   Volume of raised water = \(\pi *r^2\)*6.4 cu.cm
   Therefore, \(\pi *r^2\)*6.4 = \( \frac{1}{3} \pi *14*14*30 \)
   \( r^2 = \frac{14^2*10^2}{8^2} \)
   \( r = \frac{14*10}{8} \)
   \( 2r = 2*\frac{14*10}{8} \)
   diameter = 35 cm