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the value of root is

 

Posted by Pratham Trikha (Mar 28, 2017 3:48 p.m.) (Question ID: 4571)

If the angle between two radii of a circle is 130° , then find the angle between the tangents at the ends of radii?

Posted by çútê lõõk kìllêr (Mar 26, 2017 10:31 p.m.) (Question ID: 4541)

  • Answers:
  • In the above given figure

    Angle  T +. Angle. O=180 (ptqo cyclic quad.)

    AngleT=180-130

    Angle T=50

     

    Answered by pritish pandey (Mar 28, 2017 12:13 p.m.)
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  • Ans.

    O is the centre of the circle.

    Given, <POQ = 130º

    PT and QT are tangents drawn from external point T to the circle

    <OPT = <OQT = 90º [ Radius is perpendicular to the tangent at point of contact]

    In quadrilateral OPTQ,

    <PTQ + <OPT + <OQT + <POQ = 360º

    => <PTQ + 90º + 90º + 130º = 360º

    => <PTQ = 360º – 310º = 50º

    Thus, the angle between the tangents is 50º.

    Answered by Naveen Sharma (Mar 27, 2017 6:26 a.m.)
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If ratio of the roots of equation px2+qx+q=0 is a:b, prove that √a÷b + √b÷a + √q÷p=0.

Posted by Uzumaki Naruto (Mar 26, 2017 7:34 p.m.) (Question ID: 4533)

  • Answers:
  • Let the roots of the given equation be {tex}\alpha {/tex} and {tex}\beta{/tex}.

    Then, {tex}\alpha {/tex} + {tex}\beta{/tex} = {tex}{{ - q} \over p}{/tex}         and {tex}\alpha .\beta = {q \over p}{/tex}

    Given:     {tex}{\alpha \over \beta } = {a \over b}{/tex}

    Now, {tex}\sqrt {{a \over b}} + \sqrt {{b \over a}} + \sqrt {{q \over p}} = 0{/tex}

    =>          {tex}\sqrt {{\alpha \over \beta }} + \sqrt {{\beta \over \alpha }} + \sqrt {\alpha \beta } = 0{/tex}          [Since, {tex}{\alpha \over \beta } = {a \over b}{/tex} and {tex}\alpha .\beta = {q \over p}{/tex}]

    =>          {tex}{{\sqrt \alpha } \over {\sqrt \beta }} + {{\sqrt \beta } \over {\sqrt \alpha }} + \sqrt {\alpha \beta } = 0{/tex}

    =>          {tex}{{\alpha + \beta + \alpha \beta } \over {\sqrt {\alpha \beta } }} = 0{/tex}

    =>          {tex}\alpha + \beta + \alpha \beta = 0{/tex}

    =>          {tex}{{ - q} \over p} + {q \over p} = 0{/tex}                            [Since, {tex}\alpha + \beta = {{ - q} \over p}{/tex} and {tex}\alpha \beta = {q \over p}{/tex}]

    =>          {tex}-q+q=0{/tex}

    =>          0 = 0

    Hence proved.

    Answered by Rashmi Bajpayee (Mar 27, 2017 12:06 p.m.)
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In quadratic equation ax2  + bx + c =0. a b c are in AP and all are positive .The root of equation p and q which is an integer . Find the value of p+q+pq

Posted by pritish pandey (Mar 25, 2017 7:41 p.m.) (Question ID: 4507)

  • Answers:
  • let a = e-d

        b = e

        c = e+d

    therefore, (e-d)x2 + (e)x +(e+d) =0

    therefore p+q = e-d/e and pq = e+d/e

    p + q + pq = e - d +e +d /e

                  = 2e/e

                  = 2

    Answered by Rohan Jha (Mar 26, 2017 3:23 p.m.)
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A trader bought a number of articles for Rs.900. Five articles were found damaged. He sold 
each of the remaining articles at Rs.80 in the whole transaction. Find the number of articles he 
bought.

Posted by Basu Kailash Bk (Mar 25, 2017 7:18 a.m.) (Question ID: 4497)

  • basu

    plz try to answer my non detail question above

    Posted by SRIKRISHNA N Krish (Mar 25, 2017 11:29 a.m.)
  • don worry fr such questions they wont come any way

     

    Posted by SRIKRISHNA N Krish (Mar 25, 2017 11:28 a.m.)
  • Not easy to understand

     

    Posted by Basu Kailash Bk (Mar 25, 2017 10:25 a.m.)
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Q.ABC is a triangle in which <B = 90°.A circle is drawn with AB as the diameter intersecting the hypotenuse AC at P.Prove that the tangent drawn to the circle at P, bisect BC.

Posted by Vishnu B Pillai (Mar 23, 2017 3:50 p.m.) (Question ID: 4446)

  • Answers:
  • Given: triangle ABC is a right angled triangle, right angled at B, angle ABC = 90o

               A circle is drawn, taking AB as diameter, intersects AC at P. PQ is a tangent, intersects at Q.

    To prove: BQ = QC

    Construction: Join BP.

    Proof: PQ = BQ       ..........(i)       [Tangents drawn from an external point are equal]

              Therefore, angle PBQ = angle BPQ     [Angles opposite to equal side]

               Also,  angle APB = 90o                        [Angle of semicircle]

               And angle APB + angle BPC = 180o   [Linear pair]

               Now, angle BPC = 180o - 90o = 90o

               Since, angle BPC + angle PBC + angle PCB = 180o   [Angle sum property of a triangle]

               =>      angle PBC + angle PCB = 180o - angle BPC = 180<font size="2">o</font> - 90<font size="2">o </font>= 90<font size="2">o</font><font size="2">     .........(ii)</font>

              Now, angle BPC = 90o

              =>        angle BPQ + angle CPQ = 90o              ..........(iii)

              From eq. (i) and (ii), we get,

              angle PBC + angle PCB = angle BPQ + angle CPQ

              =>        angle PCQ = angle CPQ         [Since angle BPQ = angle PBQ]

              =>        PQ = QC             [In triangle PQC, opposite sides of equal angles]        ..........(iv)

              From, eq. (i) and (iv), we get,

               BQ = QC

              Hence proved.

    Answered by Rashmi Bajpayee (Mar 25, 2017 1:53 p.m.)
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A tarder bought number of article for Rs. 900, five article are found to be damage,  he sold remaining article for Rs. 80.find the number of article. 

Posted by Kishore Kumar (Mar 23, 2017 11:03 a.m.) (Question ID: 4436)

is_/3,_/6_/9_/12  form an a.p. ??

sir I need answer immediately 

Posted by adhyatamjot singh (Mar 23, 2017 8:45 a.m.) (Question ID: 4432)

Hi Sir,

When will you update class 10 syllabus for academic year 2017-2018 and corresponding chapter weightages and clear understanding of the board exam pattern change?.

It will be grateful if you could update it soon as from April many schools would start the syllabus for 2017-2018

Posted by Prasanna Lakshmi (Mar 22, 2017 5:13 p.m.) (Question ID: 4395)

The sum of n term of an ap is zero show that the sum of n term is - am(m+n) /n-1 a being the first term. 

Posted by Ganesh Gupta (Mar 22, 2017 7:01 a.m.) (Question ID: 4373)

One-forth of a herd of a camels was seen in the forest.Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river.Find the total number of camels.

Posted by Rohit Kumar (Mar 21, 2017 7:54 p.m.) (Question ID: 4351)

  • Answers:
  • Ans. Let total number of camel = x2

    Nunber of camel seen in the forest = {tex}x^2\over 4{/tex}

    Number of camel gone to mountain = {tex}2\sqrt {x^2} = 2x {/tex}

    Number of camel seen on  river bank = 15

    According To Question,

    {tex}=>{ x^2\over 4 } + 2x + 15 = x^2{/tex}

    {tex}=> x^2 + 8x + 60 = 4x^2{/tex}

    {tex}=> 3x^2 -8x -60= 0{/tex}

    => 3x2 - 18x + 10x - 60 = 0

    => 3x(x-6) +10(x-6) =0

    => (x-6)(3x+10) = 0

    => x = 6, -10/3

    x cannot be -ve.

    So number of camel = 62 = 36

     

    Answered by Naveen Sharma (Mar 21, 2017 10:25 p.m.)
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What constant must be subtracted or added from equation 4x²+12x+8=0 to solve it by method of completing square. 

Posted by Ayush Meena (Mar 21, 2017 9:35 a.m.) (Question ID: 4306)

  • Answers:
  • ADD 1 AND SUBTRACT 1

    (2X+3)-1=0

    A-B2 FORMULA

    2X+3+1=0    2X+3-1=0

    2X+4=0        2X+2=0

    2X=-4         2X=-2

    X=-2         X=1

    Answered by Gem m (Mar 21, 2017 11:21 a.m.)
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if one root of the quadratic equation is 3+2root5/4.then find the other root

 

Posted by Adi Chauhan (Mar 20, 2017 10:58 p.m.) (Question ID: 4278)

  • Plz send answer. I have my maths exam tomorrow

     

    Posted by Adi Chauhan (Mar 20, 2017 11 p.m.)
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Find four consecutive terms in AP whose sun is 20 and the sum of whose square is 120 ?

Posted by Adi Chauhan (Mar 20, 2017 5:08 p.m.) (Question ID: 4216)

  • Answers:
  • Ans. Let first term of A.P. is a-2d

    Common difference is d,

    Then four terms are, a-2d, a-d, a and a+d

    According To Ques,

    => a - 2d + a - d + a + a + d = 20

    => 4a - 2d = 20

    => 2a - d = 10

    => d = 2a - 10 ……………(1)

    Also,

    => (a-2d)2 + (a-d)2 + a2 +(a+d)2 = 120

    put value of d from (1)

    => (a - 4a+20)+(a-2a+10)+ a2 +(a+2a-10)2= 120

    => (20-3a)2 + (10-a) + a2 + (3a-10)= 120

    => 400 + 9a2 - 120a + 100+ a2- 20a + a2+ 9a+ 100 - 60a = 120

    => 20a2 - 200a + 600 = 120

    divide by 20

    => a2 - 10a + 30= 6

    => a2 - 10a + 24 = 0

    => a- 6a - 4a + 24 =0

    => a(a-6) -4(a-6) = 0

    => (a-6)(a-4) = 0

    => a = 6, 4

    if a = 6 then d = 2

    and if a = 4 then d = -2

    First A.P = 2, 4, 6, 8,....

    second A.P. = 8, 6, 4, 2,...

     

     

    Answered by Naveen Sharma (Mar 21, 2017 1:34 p.m.)
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Two taps running together can fill a cistern in 6 minutes .if one tap takes 5 minutes more than other to fill it , find the time in which each tap fill the cistern separeately?

Posted by Adi Chauhan (Mar 20, 2017 5:06 p.m.) (Question ID: 4215)

  • Answers:
  • Let the first tap take x minutes to fill the cistern, then second tap will take (x+5) minutes to fill it.

    The first tap will fill the 1/x of the cistern in 1 minute and second tap will fill 1/(x+5).

    Together, they will fill 1/6.

    So, equation becomes

    {tex}{1\over x}{/tex}{tex}{1\over x+5}{/tex}{tex}{1\over 6}{/tex}

    x2+5x= 12x+30

    x2-7x-30 =0

    x2-10x+3x-30=0

    x(x-10)+3(x-10)=0

    x= 10 

    Therefore, first tap will fill in 10 minutes alone and the second tap will take 15 minutes to fill the cistern.

    Answered by Shweta Gulati (Mar 21, 2017 12:45 a.m.)
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Find an ap whose first term is 1 and the sum of first four terms of one third of the sum of next four terms .

Posted by Adi Chauhan (Mar 20, 2017 5:04 p.m.) (Question ID: 4214)

  • Answers:
  • a=1

    S4{tex}{1\over 3}{/tex}[S8-S4]

    4\2 [ 2+3d] = 1\3 [2+7d-2-3d]

    2 (2+3d) = 1/3 (4d)

    6(2+3d) = 4d

    12 +18d = 4d

    12 = -14d

    -6/7 =d 

    So, A.P. becomes 1,1/7,-5/7....

    Answered by Shweta Gulati (Mar 21, 2017 12:51 a.m.)
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Determine the ratio in which the point z(a,-5) divides the line segment joining the points X(2,1) and y(5,-8).also find the value of a?

 

Posted by Adi Chauhan (Mar 20, 2017 5:01 p.m.) (Question ID: 4211)

  • -5 = 5m1 + 2m2 / m1 + m2

    -5m1 - 5m2 = 5m1 + 2m2

    -10m1 = 7m2

    m1:m2 = 7:-10

    The ratio is 7 : -10

    Posted by Saurav Kalita (Mar 20, 2017 9:15 p.m.)
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Chance of winning a game are 60%.if Anil has played the game 20 times ,how many times he can expect to loose?

Posted by Adi Chauhan (Mar 20, 2017 4:59 p.m.) (Question ID: 4210)

  • Answers:
  • Chance of losing the game = 40%

    Total no. of games played = 20

    No. of times lost = 40% of 20 = 8 times

    Answered by Shweta Gulati (Mar 21, 2017 12:53 a.m.)
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The mid point of the line segment joining (2a,4) and (-2,2b) is (1,2a+1).find the value of b?

Posted by Adi Chauhan (Mar 20, 2017 4:53 p.m.) (Question ID: 4207)

  • Answers:
  • Midpoint of two given points are 

    {(x1 + x2)/2, (y1 + y2)/2}

    According to question,

    (2a - 2)/2 = 1

    2a - 2 = 2

    2a = 4

    a = 2

    Again, (4 + 2b)/2 = 2a + 1

    4 + 2b = 4a + 2

    2b = 4 × 2 + 2 - 4

    2b = 8 - 2

    b = 3

    Answered by Rashmi Bajpayee (Mar 22, 2017 6:01 p.m.)
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  • Mid point = (x1+x2)/2 ; (y1+y2)/2

    1 = (2a+1)/2

    2 = 2a+1

    2a = 1

    a =1/2

    2a+ 1 = (4+2b)/2

    4a+2 = 4 +2b

    4a-2 = 2b

    2a-1 = b

    1 - 1 =b

    b=0

    Answered by Shweta Gulati (Mar 21, 2017 1:01 a.m.)
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If one root of the quadratic equation is 3+2root5/4 then what will be the other root

Posted by Adi Chauhan (Mar 20, 2017 4:44 p.m.) (Question ID: 4206)

  • Plz reply 

     

    Posted by Adi Chauhan (Mar 20, 2017 4:51 p.m.)
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Out of a number of Saras birds ,one fourth of the number are moving in lotus plants ,one ninth coupled alongwith one fourth as well as seven times the square root of the number move on a hill . 56 birds remain in vakula trees .what is the total number of birds?

Posted by Adi Chauhan (Mar 20, 2017 4:24 p.m.) (Question ID: 4205)

  • Please send answer for this question urgently.tomorrow is my maths exam

    Posted by Adi Chauhan (Mar 20, 2017 4:27 p.m.)
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  • Answers:
  • Ans. Let the number of birds be x2 Number of birds moving about lotus plant ={tex}x^2\over 4{/tex}

    Number of birds coupled along = {tex}{x^2\over 9} +{ x^2\over 4}{/tex}

    Number of birds moves on hill ={tex}7\sqrt {x^2} = 7x {/tex}

    Number of birds remaining on trees = 56 Now,

    {tex}{x^2\over 4} + {x^2\over 4} +{x^2\over 9} + 7x +56 = x^2{/tex}

    => {tex}=> {11x^2\over 18} + 7x + 56 = x^2{/tex}

    {tex}=> 7x + 56 = {7x^2\over 18}{/tex}

    {tex}=> x + 8 ={x^2\over 18}{/tex}

    => x-18x -144 = 0

    => x2 - 24x +6x -144 = 0

    => x(x-24) +6(x-24) = 0

    => (x-24)(x+6) = 0

    x = 24 or -6

    so Number of birds = 24= 576

    Answered by Naveen Sharma (Mar 21, 2017 10:53 a.m.)
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A man stands at a point X on the bank XY of a river with straight and parallel bank, and observes that the line joining X to a points Z on the opposite Bank makes an angle of 30° with XY. He then goes along the bank a distance of 200 metre to Y and find that the angle ZYX 60°. Find the breadth of the river.

Posted by Rahul Singh (Mar 19, 2017 3:04 p.m.) (Question ID: 4126)

  • Answers:
  • Let the man 200 m to Y till point P.

    Then, tan 30o = ZY / XY

    1/√3 = ZY / (200 + PY)

    ZY = (200 + PY)/√3.            .........(i)

    Again, tan 60o = ZY/PY

    ZY = √3 PY           .......(ii)

    Equatind eq.(i) and (ii), we get,

    (200 + PY)/√3 = √3 PY

    200 + PY = 3PY

    2PY = 200

    PY = 100m

    Therefore width of the river = XY = 200+ 100 = 300 m

    Answered by Rashmi Bajpayee (Mar 19, 2017 4:52 p.m.)
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The perimeter of the ends of the frustum of the cone are 207.24 cm and 169.56 cm .if the height of the frustum be 8 cm,find the whole surface area of the frustum .

Posted by Adi Chauhan (Mar 16, 2017 8:32 p.m.) (Question ID: 3945)

  • Answers:
  • At last, after multiplication the answer is not 7392.52.

    The correct answer is 7592.52 sq.cm

    Answered by Rashmi Bajpayee (Mar 22, 2017 6:20 p.m.)
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  • Ans. Let R and r be the radii of two ends of frustum.

    Then

    {tex}=> 2\times 3.14 \times R = 207.24{/tex}

    => R = 33 cm

    {tex}=> 2\times 3.14 \times r = 169.56{/tex}

    => r = 27 cm

    Height of frustum h = 8cm

    Slant Height of Frustum l = {tex}\sqrt {(R-r)^2 + h^2}{/tex}

    {tex}\sqrt {(33-27)^2+(8)^2} = \sqrt {36+64}{/tex}

    => 10

    TSA of frustum = {tex}\pi l(R+r) +\pi R^2 +\pi r^2{/tex}

    => {tex}\pi [l(R+r) + R^2+r^2]{/tex}

    => {tex}3.14[10(33+27) + 1089 + 729]{/tex}

    => 3.14[600+1089+729]

    => 3.14 (2418)

    => 7392.52cm2

     

    Answered by Naveen Sharma (Mar 16, 2017 11:17 p.m.)
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Water is flowing at a rate of 0.7 m/second through circular pipe whose internal diameter is 2 cm into a cylindrical tank,the radius of whose base is 40 cm.determine the increase in the level of water in half an hour

Posted by Adi Chauhan (Mar 16, 2017 8:28 p.m.) (Question ID: 3944)

  • Answers:
  • Ans. Let water Level Rise in Tank = x cm

    Radius of tank = 

    Speed of water = 0.7m/s = 70cm/s

    Diameter of Pipe = 2 cm

    Radius of Pipe = 1cm

    Volume of water flowing In 1 Sec = {tex}{22\over 7}\times 1\times 1\times 70 = 220cm^3{/tex}

    Volume of water flowing in half an hour = 

    {tex}220\times 1800 = 396000cm^3{/tex}

    This Volume = Volume Of Tank

    {tex}=> 396000 = {22\over 7}\times 40\times 40\times x{/tex}

     

    => x = 78.75 cm

    Water Level increase in tank = 78.75cm

    Answered by Naveen Sharma (Mar 17, 2017 8:03 a.m.)
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Four equal circles are described at four corners a square so that each touches two of the others . The shaded area enclosed between the circles is 24/7 Cm square. Find the radius of each circle

Posted by Adi Chauhan (Mar 16, 2017 8:24 p.m.) (Question ID: 3943)

  • Answers:
  • Ans.

    Let ABCD be the square. Let r be the radius of each of the 4 equal circle that are described at each vertices of square ABCD.

    Now, AB = BC = CD = DA = 2r

    We know that each angle of a square is a right angle.

    {tex}=> \angle A= \angle B= \angle C= \angle D = 90{/tex}

    Now, area of a quadrant of the circle having centre at A = {tex}{90\over 360}\pi r^2 = {1\over 4}\pi r^2{/tex}

    area of 4 quadrants = {tex}4\times {1\over 4}\pi r^2 = \pi r^2{/tex}

    Area of Square = {tex}(2r)^2 = 4r^2{/tex}

    Area of Shaded Region = Area of Square - Area of 4 quardants

    {tex}=> {24\over 7} = 4r^2 - \pi r^2 = r^2(4-{22\over 7}){/tex}

    {tex}=> {24\over 7} = r^2\times {6\over 7}{/tex}

    {tex}=> r^2 = 4 => r = 2cm{/tex}

    Answered by Naveen Sharma (Mar 17, 2017 9:10 a.m.)
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A right triangle having sides 15 cm  and 20 cm is made to revolve about it's hypotenuse .find the volume and surface area of the double cone so formed .

Posted by Adi Chauhan (Mar 16, 2017 8:21 p.m.) (Question ID: 3942)

  • Answers:
  • Ans.

    Consider the following right angled triangle ABC is rotated through its hypotenuse AC

     

    BD {tex}\perp{/tex} AC. In this case BD is the radius of the double cone generated.

    Using Pythagoras theorem for {tex}\triangle ABC{/tex} 

    => AC2 = AB2 + BC

    =>  AC2 = 225 + 400 = 625

     => AC = 25 cm

    Let AD = x cm CD = (25 – x) cm

    Using Pythagoras theorem in {tex}\triangle ABD{/tex}

    => AD2 +BD2 = AB2

    => x2 + BD2= 225

    => BD2= 225 – x2 …………(1)

    Using Pythagoras theorem for {tex}\triangle CBD{/tex}

    => BD+ CD= BC2

    => BD2 +(25-x)2= 400

    => BD2 = 400 - (25-x)2 …………(2)

    From (1) & (2)

    => 225 - x2= 400 - 625 - x+ 50x

    => 50x = 450

    => x = 9 cm

    AD = 9, CD = 16

    From (1) BD = 12

    Surface area of the double cone formed = L.S.A of upper cone + L.S.A of the lower cone

    {tex} => \pi \times BD \times AB + \pi \times BD \times BC{/tex}

    {tex} => \pi \times BD \times (AB + BC){/tex}

    {tex}=> \pi \times 12(20+15){/tex}

    {tex}=> \pi \times 12\times 35{/tex}

    {tex}=>420 \pi {/tex}cm2

    Answered by Naveen Sharma (Mar 17, 2017 8:31 a.m.)
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In an ap , the sum of first n terms is 3n2/2 + 13n/2 . Find the 25th  term.

Posted by Adi Chauhan (Mar 16, 2017 7 p.m.) (Question ID: 3932)

  • Thank u 

    Posted by Adi Chauhan (Mar 16, 2017 7:26 p.m.)
  • Please send answer for this question.

    Posted by Adi Chauhan (Mar 16, 2017 7:06 p.m.)
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  • Answers:
  • Sn = 3n2/2 + 13n/2

    S1 = a = 3/2 + 13/2 = 8

    S2 = a + a2 = 12/2 + 26/2 = 19

    a2 = 19 - 8 = 11

    Then,.       d = 11 -8 = 3

    Now, a25 = a +(25 - 1)d

    = 8 + 24 × 3

    = 80

    The 25th term is 80

    Answered by Rashmi Bajpayee (Mar 16, 2017 7:24 p.m.)
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P and Q are centres of circles of radii 9 cm and 2 cm respectively . PQ is 17 cm . R is the centre of the circles of radius cm which touches the above circle externally . Given that angle PRQ is 90 degree .write an equation in and solve it.

Posted by Adi Chauhan (Mar 16, 2017 6:55 p.m.) (Question ID: 3931)

  • Please send solution for this question. I have maths exam tommorow

    Posted by Adi Chauhan (Mar 16, 2017 7:11 p.m.)
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  • Answers:
  • Ans. If two circles touch each other externally, then sum of their radii is equal to the distance between their centres.

    RP = x+9

    RQ= x+2

    PQ = 17

    Given, angle PRQ= 90, So In Triangle PRQ => RP2 + RQ2 = PQ2   …(using Pythagoras)

    => (x+9)2 + (x+2)2= (17)2

    => x2 + 18x + 81 + x+ 4x + 4 = 289

    => 2x+ 22x -204 =0

    => x2 + 11x -102 = 0

    On Solving We get

    x = -17 and 6

    So radius of  circle is 6cm.

     

    Answered by Naveen Sharma (Mar 16, 2017 8 p.m.)
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ABC is an isosceles triangle with AB=AC. D is the midpoint of AC. Taking BD as diameter a circle is drawn which meets AB at E. Prove that AE=1/4AC

Posted by Anurag Joardar (Mar 16, 2017 2:22 a.m.) (Question ID: 3891)

  • Answers:
  • Ans.

    Given: ABC is isosceles triangle. AB = AC. D is mid-point of AC.

    To prove : {tex}AE = {1\over 4}AC{/tex}

    Proof:

    AB = AC …………(1)   (Given)

    AD = AC/2   ……………(3) (d is midpoint)

    As AD is tangent to the triangle and AEB is secant then,

    {tex}=> AE \times AB = AD^2{/tex}

    {tex}=> AE \times AC = ({AC\over 2})^2{/tex}

    {tex}=> AE \times AC = {AC^2\over 4}{/tex}

    {tex}=> AE = {AC\over 4}{/tex}

    Hence Proved.

    Answered by Naveen Sharma (Mar 16, 2017 7:29 a.m.)
    Thanks (3)
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In a family of 3children the probability of having at least one boy is :

Posted by Puttaswamy R M (Mar 15, 2017 11:56 p.m.) (Question ID: 3889)

  • <hr/>

    Aarohi mehra, Can u please show me the method of how u have solved? ¿

     

    Posted by Puttaswamy R M (Mar 16, 2017 12:11 a.m.)
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  • Answers:
  • Ans. Three children can be as follow:

    BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG.

    No of total cases = 8

    No of favorable cases = 7

    Probability = 7/8

    Answered by Naveen Sharma (Mar 16, 2017 7:06 a.m.)
    Thanks (1)
    • 7/8
    Answered by Aarohi Mehra (Mar 16, 2017 12:08 a.m.)
    Thanks (2)
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