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 the sum of intiger and their reciprocal is 145/12. Find the intiger ?

Posted by Pankaj Asery (Jan 18, 2017) (Question ID: 1452)

  • Answers:
  • Solution:

    Let the integer = x

    Then reciprocal = 1/x

    According to Question

    x + 1/x = 145/12

    => (x2+1)/x = 145/12

    => 12x2 + 12 = 145x

    => 12x- 145x + 12 = 0

    => 12x2- 144x - x + 12 = 0

    => 12x(x - 12) - 1(x - 12) = 0

    => (x - 12)(12x - 1) = 0

    => x = 12 and x = 1/12

    as X is integer x can't be 1/12

    Therefore the required integer is 12

    Answered by Naveen Sharma (Jan 19, 2017)
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  • Let the integer be x.

    x + 1/x = 145/12

    12x2 - 145x + 12 = 0

    12x2 - 144x - x + 12 = 0

    12x(x - 12) - 1(x - 12) = 0

    (x - 12)(12x - 1) = 0

    x = 12 and x = 1/12

    Therefore the required integer is 12.

    Answered by Rashmi Bajpayee (Jan 18, 2017)
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how to prove the sum of first terms of ap is 42 10th term to its 30th terms is 1:3 find the first term

Posted by rohit singh (Jan 18, 2017) (Question ID: 1444)

  • i think questions is : The sum of first six terms of an ap is 42 .the ratio of its 10th term is and its 30th term is 1:3. calculate the first and thirteenth term

    Posted by Naveen Sharma (Jan 18, 2017)
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  • Answers:
  • Ans. 

    Let first term of AP is a and comman difference is d.

    as we know 

    an = a + (n-1) dwhere an is nth term

    => a10  = a + (10-1)d

    => a10  = a + 9d

    And a30 = a + (30-1)d

    => a30 = a + 29d

    According to Question

    => a+9da+29d = 13

    => 3a + 27d = a + 29d 

    => 2a = 2d

    => a = d       ............. (1)

    Sum of first 6 terms = 42

    We know 

    Sn = n2× 2a + (n-1)d

    =>S6 = 62 ×2a + (6-1)d

    => 42 = 32a + 5a

    => 42 = 21a

    => a = 2 

    First term is 2. Common difference is 2.

    => a13 = 2 + (13-1)2

    = > a13 = 2 + 24 

    =>  a13 = 26 

    Answered by Naveen Sharma (Jan 18, 2017)
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Which term of AP 121,117,113.................is its first negative term

Posted by Akash A.S (Jan 16, 2017) (Question ID: 1419)

  • Answers:
  • a = 121

    d = -4

    let an= 0

    then

    a= a + (n-1)d

    => 0 = 121 + (n-1)(-4)

    => 0 = 121 - 4n + 4

    => 4n = 125

    => n = 125/4

    => n = 31.25

    as n must be natural number then n= 32

    32nd term is First negative number.

    a32 = 121 + (31)-4

    = 121 - 124 = -3

    Answered by Naveen Sharma (Jan 16, 2017)
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AB is a straight road leading to C,  the foot of a tower, A being at a distance of 120m from C and B being 75m nearer. If the angle of elevation of the tower at B is the double of the angle of elevation of the tower at A,  find the height of the tower. 

Posted by Shruti Sharma (Jan 16, 2017) (Question ID: 1415)

  • Answers:
  • Just consider the image for Reference. 

    Gievn : AB is straigt Road and C is the foot of tower. Let DC is tower of height h, and AC is 120cm and BC is 45 cm as it is 75cm nearer than A.

    Solution : In Triangle BCD 

    tan 60° = h45

    => h = 45 tan 2x  => h = 45 2tanx 1-tan2x   .............. (1) tan 2x = 2tan x1-tan2x

    In Triangle, ACD

    tan x = h120

    => h = 120 tan x       .................... (2)

    From (1) and (2)

    120 tanx = 45×2 tanx 1-tan2x

    => 1-tan2x = 34

    => tan2x = 14

    => tan x = 12

    Put value of tan x in (2)

    we get h = 60 

    so height of tower is 60cm

    Answered by Naveen Sharma (Jan 17, 2017)
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  • Are you sure this is the question? Please check it once.

    Answered by Shweta Gulati (Jan 17, 2017)
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Converse of Pythagoras theorem

Posted by Israr Khan (Jan 16, 2017) (Question ID: 1394)

  • Answers:
  • If the square of one side of triangle is equal to the sum of square of other two sides,then the triangle is right triangle.

    Answered by shruti dadhwal (Jan 16, 2017)
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Why volume of cone is 1/3 of cylinder volume

Posted by raman SINGH (Jan 15, 2017) (Question ID: 1391)

  • Answers:
  • Because cone is 1/3 part of a cylinder

    Answered by himani sah (Jan 15, 2017)
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Why volume of cone is1/3 of cylinder volume

Posted by raman SINGH (Jan 15, 2017) (Question ID: 1390)

Find the probability that a non leap year chosen  at random has 

  1. 52 Sundays
  2. 53 Sundays
Posted by Rohit R Mahatungade (Jan 14, 2017) (Question ID: 1367)

  • Answers:
  • 2) A non-leap year has 365 days

    A year has 52 weeks. Hence there will be 52 Sundays for sure.

    52 weeks = 52 x 7 = 364 days .

    365– 364 = 1day extra.

    In a non-leap year there will be 52 Sundays and 1day will be left.

    This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.

    Of these total 7 outcomes, the favourable outcomes are 1.

    Hence the probability of getting 53 sundays = 1 / 7.

    1)
    ∴ probability of getting 52 sundays = 1 - 1/ 7 = 6 / 7.

    Answered by Naveen Sharma (Jan 14, 2017)
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The vertices of a triangle are A(-1,3), B(1,-1), C(5,1). Find the length of the median through the vertex C.

Posted by Arpit Singla (Jan 14, 2017) (Question ID: 1361)

  • Answers:
  • Let the points A(-1, 3), B(1, -1) and C(5, 1) be the vertices of a triangle ABC.

    We know median is the line segment joining the mid point of line to opposite vertex.

    Let D be the mid point of AB. Then the coordinate of D

     x1+x22, y1+y22-1+12, 3-120,1

    Coorditane of D( 0,1)

    Lenght of Median CD= 

    Using distance formula

    x2-x12+y2-y125-02 + (1-1)225 + 0255

    Answered by Naveen Sharma (Jan 14, 2017)
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A train covers a distance of 90kms at a uniform speed. It would have taken 90 minutes less if the speed had been 15km/hr more. Calculate the original duration of the journey.

 

Posted by Arpit Singla (Jan 14, 2017) (Question ID: 1360)

  • Answers:
  • The Questions is it would have taken 30 Minutes less instead of 90 Minutes

    Solution:

    Let “x” be the usual speed of the train

    Let T1 be the time taken to cover the distance 90 km in the speed x km/hr

    Let Tbe the time taken to cover the distance 90 km in the speed (x + 15) km/hr

    Time = Distance/Speed

    T1 = 90x

    T2 = 90x+15

    By using the given condition

    T1 - T2 = 3060

    90x- 90x+15 = 12

    Taking 90 commonly from two fractions

    90 1x - 1x+15 = 12

    x+15 -xxx+15 = 12*90

    15xx+15= 1180

    x2 + 15x  = 2700

     

    x2 + 15x -2700 = 0x2 + 60x - 45x -2700 = 0x (x+60) - 45 (x+60) = 0(x+60) (x-45) = 0x + 60 = 0  or  x -45 = 0x = 45, -60

    Speed can't be negative so,

    Speed is 45km/h

    Answered by Naveen Sharma (Jan 14, 2017)
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3x2-23x-110=0 by using quadratic equation 

Posted by Yash Partap Singh (Jan 12, 2017) (Question ID: 1302)

  • Answers:
  • 3x2-23x-110=0

    3x2-33x+10x-110=0

    3x(x-11)+10(x-11)=0

    (x-11)(3x+10)=0

    x-11=0, 3x+10=0

    x = 11, -10/3

     

    Answered by Naveen Sharma (Jan 12, 2017)
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For what value of  K are the  roots of the  quadratic equation kx(x-2)+6=0

 

Posted by abhishek yadav (Jan 11, 2017) (Question ID: 1273)

  • Answers:
  • For Equal roots D = 0

    kx(x-2) + 6 = 0

    kx2 -2kx + 6 = 0

    comparing with ax2+ bx + c = 0, we get 

    a = k,  b = -2k , c = 6

    D = b2 - 4ac

    Put value of a b c

    (-2k)2 - 4*6*k = 0

    4k2 - 24k = 0

    4k(k-6) =0 

    4k = 0 or k-6=0

    k = 6  

    Answered by Naveen Sharma (Jan 11, 2017)
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Two tangents are drawn from an external point P to the circle such that angle oba 10 then find the value of bpa 

Posted by Shashant Ahuja (Jan 10, 2017) (Question ID: 1267)

The height of a cone is 24 cm.A small cone is cut off at the top by a plane parallel to the base.If the volume of the remaining part be 7/8of the given cone, at what height above the base is the section made?

Posted by Rajeev Kumar (Jan 09, 2017) (Question ID: 1230)

  • Let ABC be a cone of radius R, height 24cm. let this cone be cut by a place A`B` parallel to AB. O` is the center of base of the cut out cone A`B`C`. let h be the height, r be the redius of cone A`B`C`. Cleary ΔA`B`C` ~ Δ AOC

    Posted by Naveen Sharma (Jan 12, 2017)
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  • Answers:
  • Ans.

    So h/24 = r/R

    h = 24r/R ..... (1)

    Volume of Cone ABC =  24πR2/3  = 8πR2      ----- (2)

    Remaining Portion is Know as Frustum as its volume is 7/8 of Cone ABC. 

    So Volume of Smaller Cone A`B`C` = 1/8 of Cone ABC

    Volume of Cone A`B`C`  =  π r2h / 3 

    from (1),   

    π r224r/3R  =  8π r3/R ------ (3)

    Accroding to questions

    8π r3/R  = 8πR2/8

    8π r3/R = πR2

    r3/R3 = 1/8

    r/R = 1/2

    put this value in (1)

    We get h = 24*1/2 = 12cm

    So cone is cut from the height 24-12 = 12 cm

    Answered by Naveen Sharma (Jan 12, 2017)
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The diffrence of the ages of sohrab &his father is 30 years  .if the diffrence of the square of their ages is 1560 then find their ages 

 

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1218)

  • Answers:
  • let age of Sohrab = x 

    then age of his father = x + 30

    Given : (x+30)2 - (x)2 = 1560

    x2 + 900 + 60x - x2 = 1560

    => 60x = 660

    => x = 11

    Age of Sourab = 11 

    age of his father = 41

    Answered by Naveen Sharma (Jan 10, 2017)
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Find the diffrence between the points (a,b) (-a,-b)

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1217)

  • Answers:
  • Difference between (a,b) and (-a,-b)=-a-a2+-b-b2

    =-2a2+-2b2=4a2+4b2

    =4(a2+b2)

    =2a2+b2

    Answered by Neeraj Sharma (Jan 14, 2017)
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The distance between the points (2,-2)&(-1,x) is 5 then one of the value of x is 

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1216)

  • Answers:
  • Given : A (2,-2) and B(-1,x) and Distance b/w these two is 5

    according to distance formula

    (x2-x1)2 + (y2-y1)2 = d2

    => (-1-2)2 + (x+2)2 = 25

    => 9 + x2 + 4 +4x = 25

    => x2 + 4x - 12 = 0

    => x2 + 6x -2x - 12 = 0

    => x (x+6) -2(x+6) = 0 

    => (x+6) (x-2) = 0

    either x + 6 = 0 or x-2=0

    so x = -6, 2

     

    Answered by Naveen Sharma (Jan 10, 2017)
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Show that the points (7,3) (3,0) (0,-4) (4,-1)are vertices of rhombus 

 

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1215)

  • Answers:
  • Let A(7,3) , B(3,0) , C(0,-4) and D(4,-1) represent the points.

    To prove that these points are vertices of a rhombus we have to show that all sides are equal. We may calculate it using distance formula.

    AB= (3-7)2+(0-3)2= 16+9=25= 5 units

    BC=(0-3)2+(-4-0)2=9+16 =25=5 units

    CD=(4-0)2+(-1+4)2=16+9 =25 = 5units

    DA=(7-4)2+(3+1)2=9+16=25=5 units

    The diagonals are not equal. 

    AC= (7-0)2+(3+4)2=49+49=98=72units

    BD=(4-3)2+(-1-0)2 = 1+1 =2units

    This shows that these points are vertices of rhombus.

    Answered by Shweta Gulati (Jan 10, 2017)
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Find the sum of all 3 digit no. Which leave a remainder 3 when divided by 5 .

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1213)

  • Answers:
  • First 3 Digit Number that leaves remainder 3 when divided by 5 is 103.

    and last will be 998

    A.P. = 103, 108, 113, ..................., 998

    first term a = 103

    Common Difference d  = 5

    nth term = 998

    n = ?

    as we know

    nth term = a + ( n-1)d

    998 = 103 + ( n-1)5

    895/ 5 = n -1

    179 = n -1

    n = 180

     

    now find sum of 180 terms of AP

    S180 = 180/ 2 ( 103 + 998 )

    => S180 = 90 * 1001

    => S180 = 900090

    Answered by Naveen Sharma (Jan 09, 2017)
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If mth term of an A.P. is 1/n and nth term is 1/m .show that its (mn)th term is 1

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1212)

  • Answers:
  • Let a and d are first term and common difference of AP

    Answered by Naveen Sharma (Jan 09, 2017)
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Find k so that  15,k,-1are in A.P. series 

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1211)

  • Answers:
  • As they are in AP

    so 2b = a + c

    2k = 15 - 1 

    2k = 14

    k = 7 

     

    Answered by Naveen Sharma (Jan 09, 2017)
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Solve 4x2-2(a2+b2)x+a2b2=0  find the value of x 

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1210)

  • Answers:
  • a= 4

    b= -2(a2+b2)

    c= a2b2

    Using Discriminant formula,

    D= b2-4ac

    [-2(a2+b2)]2-4 x4xa2b2

    4(a4+b4+2a2b2)-16a2b2

    (2a2-2b2)2

     

    x= (-b±D)/ 2a

    x= [2(a2+b2)+2(a2-b2)]/8 = 4a2/8 = a2/2

    x= [2(a2+b2)-2(a2-b2)]/8 = 4b2/8 = b2/2

    Answered by Shweta Gulati (Jan 10, 2017)
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If ax2+bx+c=0has equal roots then proove that c=b2/4a

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1208)

  • Answers:
  • For equal roots,  discriminant D=0

    i.e. b2-4ac=0

    b2=4ac

    So, c= b2/4a

    Answered by Shweta Gulati (Jan 12, 2017)
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If-5 is root of quadratic equation 2x^2+2px-15=0and the quadratic equation   p(x^2+x)+k=0has equal roots .find the value of k

 

Posted by Kanishk Goyal (Jan 09, 2017) (Question ID: 1202)

  • Thanks for the answer but can u  answers some more questions 

    Posted by Kanishk Goyal (Jan 09, 2017)
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  • Answers:
  • as -5 is root of equation 2x2+2px-15=0

    then it ll satisfy this equation, Put x = -1

    2(-5)2 + 2p(-5) - 15 = 0

    => 50 - 10p -15 = 0

    => 35 = 10p

    p = 3.5

    And it is given that eq. p(x2+x) + k = 0 have equal roots.

    px2+px + k = 0

    3.5x2+3.5x + k = 0

    a = 3.5, b = 3.5, c = k

    so d = 0

    (3.5)2 - 4 (3.5)k = 0

    => 14 k = 12.25

    k = 3.0625

     

     

    Answered by Naveen Sharma (Jan 09, 2017)
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The radius of the cirular ends of a bucket  of height 15cm are 14cm and rcm  (r<14). If the volume of bucket is 5390cm*3 then the value of r.(π=22÷7)

Posted by Himesh kaushik (Jan 07, 2017) (Question ID: 1136)

  • Answers:
  • V = π(r22 - r12)h 

    5390 = 22/7(196-r2)15

    196-r2 = 114.33

    r2= 196-114.33

    r2=81.67

    r = 9.03

    Answered by Devanshu Agarwal (Jan 08, 2017)
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If P(2,4) is equidistant from Q(7,0) and R (X,9), find the value of X. Also find the distance PQ.

Posted by Himesh kaushik (Jan 07, 2017) (Question ID: 1134)

  • Answers:
  • Given: P(2,4) is equidistnat from Q(7,0) and R(X,9).

    then PQ = PR

    Using Distance Formula : = (x2-x1)2 + (y2- y1)2

    PQ = 25 + 16  = 41

    PR = (x-2)2 + (5)2

    PQ = PR 

    Sqauring Both Side, We get 

    x2 + 4 -4x + 25 = 41

    => x- 4x -12 = 0

    => x- 6x +2x -12 = 0

    => x(x-6) +2(x-6) = 0

    => (x-6) (x+2) = 0

    => x = -2 , 6 

    Answered by Naveen Sharma (Jan 12, 2017)
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PQ perpendicular OQ. The tangent to the circle with centre O at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

Posted by Himesh kaushik (Jan 07, 2017) (Question ID: 1133)

  • Answers:
  • Ans.

    Given: AP and AQ are tangents to the circle with centre O, AP ⊥ AQ and AP = AQ = 5 cm

    we know that radius of a circle is perpendicular to the tangent at the point of contact

    ⇒ OP ⊥ AP and OQ ⊥ AQ

    Also sum of all angles of a quadrilateral is 360° ⇒∠O + ∠P + ∠A + ∠Q = 360°

    ⇒∠O + 90° + 90° + 90° = 360°

    ⇒∠O = 360° – 270° = 90°

    Thus ∠O = ∠P = ∠A = ∠Q = 90°

    ⇒ OPAQ is a rectangle but since adjacent sides of OPAQ i.e. AP and AQ are equal.

    Thus OPAQ is a square

    radius = OP = OQ = AP = AQ = 5 cm

    Since diagonals of a square are perpendicular bisector of each other.

    Hence PQ and OA are perpendicular bisectors of each other

    Answered by Naveen Sharma (Jan 12, 2017)
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Solve quadratic equation by factorization 

x-a/ x-b +x -b/ x-a = a/b+ b/a

Posted by Sudhanshu Yadav (Jan 07, 2017) (Question ID: 1126)

  • Answers:
  • x-ax-b+x-bx-a=ab+ba

    let x-ax-b=y x-bx-a=1y

    y+1y=a2+b2ab

    y2+1y=a2+b2ababy2-a2y-b2y-ab=0

    ayby-a-bby-a=0by-aay-b=0

    y=ab or y=ba

    case-1

    if y=abx-ax-b=abbx-ba=ax-babx=ax

    bx-ax=0x(b-a)=0x=0

    Case-2

    if y=bax-ax-b=baax-a2=bx-b2

    ax-bx=a2-b2x(a-b)=(a-b)(a+b)

    x=a+b

    Hence x=0 or a+b

    Answered by Neeraj Sharma (Jan 14, 2017)
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 A 25 FOOT LADDER IS PLACED AGAINST A VERTICAL WALL OF A BUILDING.THE FOOT OF THE LADDER IS 17 FEET FROM THE BASE OF THE BUILDING,IF THE TOP OF THE LADDER SLIPS 4 FEET,THEN THE FOOT OF THE LADDER WILL SLIP.......

Posted by Pawan Goswami (Jan 07, 2017) (Question ID: 1125)

  • Answers:
  • I think it is not 17 feet it is 7 feet.

    Fiirst we need to find at what height ladder is placed. 

    Let BC is ladder and AC is Building with foot at C.

    Now as given AB = 25

    BC = 7 

    Using Pythagorous Theroem 

    (AB2) = (BC)2 + (AC)2

    => 625 = 49 + (AC)2

    => (AC)2 = 576

    => AC = 24

    It slips down 4 feet it means its top is at 20 feet height

    So Now

    AC = 20

    AB = 25 

    then again Using theorem 

    (BC)2 = 625 - 400 

    (BC)2 = 225

    BC = 15 

    it mean it slips 8 feet. 

    Answered by Naveen Sharma (Jan 12, 2017)
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Sir and Madam, In our maths book, we have studied how to construct tangents from any external point without using the center of circle.I have discovered a yet new mode for the same , that is, how to construct tangents by an alternative method...

so, Can I construct the tangents by this method in Exam , If such question is asked ?  

Posted by Lakshmi kumari (Jan 06, 2017) (Question ID: 1115)

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