Discussion Room : Quant


  • Being MBAtious!


    @sumit-agarwal

    Legendre's three-square theorem states that a natural number (say n) can be represented as the sum of three squares of integers if and only if n is not of the form n = 4^a(8b + 7) for integers a and b.



  • Suppose real numbers x and y satisfy x^2 + 9y^2 - 4x + 6y + 4 = 0.what is the maximum value of 4x - 9y?

    Geometric approach

    We know that (x - 2)^2 + (3y + 1)^2 = 1 and we need to maximise 4x - 9y.
    Now, let's replace 3y by Y, so that know equation turns into a circle
    (x - 2)^2 + (Y + 1)^2 = 1 with center at (2, -1) and radius 1.
    And now we need to find the maximum value of 4x - 3Y which will be equal to 4x - 3Y = k (say) and will be achieved when this line is tangential to the circle.

    Doubt - what is the logic behind the point that " for the value of expression to be maximum , that line should be tangential to the circle " ..

    Could anyone help me with understanding this concept please.


  • Being MBAtious!


    @sumit-agarwal
    We are trying to maximize the distance between the line and the centre of the circle. It is also known that the circle lies on the circle so it has to be a tangent.

    Sharing a solution from @kamal_lohia sir

    x^2 + 9y^2 - 4x + 6y + 4 = 0
    i.e. {x^2 - 2(x)^(2) + 2^2} + {(3y)^2 + 2(3y)(1) + 1} = 1
    i.e. (x - 2)^2 + (3y + 1)^2 = 1

    Now we want to maximise 4x - 9y. Let 4x - 9y = k, i.e. x = (k + 9y)/4

    So eliminating x, above equation becomes: (k + 9y - 8)^2 + 16(3y + 1)^2 = 16
    i.e. 225y^2 + {18(k - 8) + 96}y + (k - 8)^2 = 0
    i.e. 225y^2 + (18k - 48)y + (k - 8)^2 = 0

    Now as y is a real number, so discriminant of the above equation should be greater than or equal to zero.
    So, (18k - 48)^2 - 4(225)(k - 8)^2 ≥ 0
    i.e. (18k - 48)^2 - (30k - 240)^2 ≥ 0
    i.e. (3k - 8)^2 - (5k - 40)^2 ≥ 0
    i.e. (-2k + 32)(8k - 48) ≥ 0
    i.e. (k - 6)(k - 16) ≤ 0
    i.e. 6 ≤ k ≤ 16.

    Thus required maximum value of 'k' is 16.



  • A polygon has internal angles of measures 90° and 270° only. If it has18 angles of measure 270°, then what is the number of angles with measure 90°?

    Sol:Let there be n angles of 90°, thn sum of all angles = (90n) + (18 × 270) = (n + 16) 180

    How RHS= (n+16) 180?

    Could anyone explain the part.?


  • Being MBAtious!


    @sumit-agarwal

    Sum of the interior angles of a polygon is: (n - 2) × 180° where n is the number of sides of the polygon.
    now n - sided polygon has n interior angles. here we have 18 angles that measure to 270 and say y angles that measures to 90.
    So n = number of sides = number of interior angles = 18 + y
    sum of all angles = (n - 2) × 180° = (18 + y - 2) * 180° = (y + 16) * 180°

    Detailed solution :
    Let there be x angles of 90°
    Also let there be y angles of 90°
    so we can say n = x + y
    Sum of interior angles = (x + y - 2) × 180°
    we know, sum of interior angles = x * 90° + y * 270°
    x * 90° + y * 270° = (x + y - 2) × 180°
    we know, y = 18
    so x * 90° + 18 * 270° = (x + 16) × 180°
    x * 90° = 18 * 270° - 16 * 180°
    x = (18 * 270° - 16 * 180°)/90°
    = 54 - 32 = 22
    So there are 22 interior angles with measure 90°
    what is the OA ?



  • @zabeer

    Thanks a lot:)



  • What is the greatest common divisor of the 2010 digit and 2005 digit numbers below?
    333.........333(2010 times)
    777........777(2005 times)

    Let A = 333..33 and B = 777..77, then A/3 and B/7 both contains the digit 1 only (2010 times and 2005
    times respectively). As HCF(2010, 2005) = 5 that means HCF(A/3, B/7) = 11111.
    Now A is also divisible by 7 as 2010 = 6 × 335 (Remember a number formed by repeating same digit 6 times is divisible by 7). So HCF(A, B) = 77777

    Here, As HCF(2010, 2005) = 5 that means HCF(A/3, B/7) = 11111. ???

    Could anyone explain the solution?



  • In ∆ABC (not drawn to scale), the altitude from A, the angle bisector of ∠BAC, and the
    median from A to the midpoint of BC divide ∠BAC into four equal angles. What is the measure in degrees of angle ∠BAC?

    Here the solution says apply sine rule
    What I could get is AB sin3A/4 = AC sin A/4
    But nt able to establish relation between AB and AC. How to find the angle BAC then?


  • Being MBAtious!


    @sumit-agarwal
    A = 3333 ...... (2010 times)
    B = 7777 ...... (2005 times)

    A = 3 * 1111 ..... (2010 times)
    B = 7 * 1111 ..... (2005 times)

    HCF(11111.... (m times), 1111 .... (n times)) = 1111.. (HCF(m,n) times)

    A is divisible by 7. B is not divisible by 3
    HCF(A, B) = 11111 * 7 = 77777



  • 30,000 is divided into three equal installments.The remaining sum at rate 4 % is added to the intallment amounts.Find the value of each installment.



  • ABCD is a square with side length 2 cm. It is divided into five rectangles of equal areas, as shown in the figure. The perimeter of the rectangle BEFG is

    0_1516899509808_0f9deca4-db08-4d50-a9b0-8d16877b2efc-image.png

    (a) 51/16
    (b) 36/11
    (c) 58/15
    (d) 47/13



  • Can anyone explain please.


  • Being MBAtious!


    @vinaycat2017

    Side length = 2 cm.
    So area of square = 4 cm^2
    Given square is divided into 5 rectangles of equal areas. So each rectangle has an area of 4/5 cm^2.
    Proceed with this understanding (i.e, if we know one side of rectangle = 2 then other side should be 2/5 to get an area of 4/5) and we get EB = 4/3 cm and BG = 3/5 cm. Figure as below

    0_1517200748962_ac81beed-447d-4d5a-8429-731d8711bd1b-image.png

    Perimeter = 58/15 cm (does it match with the OA?)





  • @0_1516904528318_QUES.png



  • @zabeer
    Sir one more
    0_1516904775903_how.png



  • @vinaycat2017

    58/15.
    One side comes out to be 4/3 and another one comes out to be 3/5



  • @vinaycat2017

    One way could be
    You take two triplets:-
    9,12,15
    5,12,13
    a=12, b=5, c=9
    Area comes out to be 84...by heron's.
    If u put these values in options...only third option would give 84 as answer.


  • Being MBAtious!


    @vinaycat2017
    Tan(A + B) = (Tan A + Tan B)/(1 - TanA * TanB)
    Tan70 = (Tan50 + Tan20)/(1 - Tan50 * Tan20)
    Tan70 - Tan70 * Tan50 * Tan20 = Tan50 + Tan20
    We know, Tan70 * Tan20 = 1
    So, Tan70 - Tan50 = Tan50 + Tan20
    Tan70 = Tan20 + 2Tan50.



  • A swimming pool is of length 50 m. A and B enter a 300 m race starting simultaneously at one end of the pool at speeds of 3 m/s and 5 m/s. How many times will they meet while travelling in opposite directions before B completes the race?
    a) Twice
    b) Thrice
    c) Once
    d) 5 times

    In every meet their combined distance travelled is 100 m.

    Let there be "a" meets.

    =>[(100a) * 5/(5+3)]=300
    Which gives a=4.8

    Though it cant be decimal.

    But Where am I proceeding wrong with this approach????could anyone help me?

    I guess the problem lies when they might be travelling in same direction ..


 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.