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(i) If (4×2 + xy): (3xy – y2) = 12: 5, find (x + 2y): (2x + y). (ii) If y (3x – y): x (4x + y) = 5: 12. Find (x2 + y2): (x + y)2.

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Question no.12  From ML aggarwal book, class10, chapter 7, ratio and proportion

(i) we have given (4×2 + xy): (3xy – y2) = 12: 5, and we have to find (x + 2y): (2x + y).

(ii)also If y (3x – y): x (4x + y) = 5: 12.

Then we have to Find (x2 + y2): (x + y)2.

Ques no. 12

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  1. Solution:

    (i) (4x2 + xy): (3xy – y2) = 12: 5

    We can write it as

    (4x2 + xy)/ (3xy – y2) = 12/ 5

    By cross multiplication

    20x2 + 5xy = 36xy – 12y2

    20x2 + 5xy – 36xy + 12y2 = 0

    20x2 – 31xy + 12y2 = 0

    Now divide the entire equation by y2

    20x2/y2 – 31xy/y2 + 12y2/y2 = 0

    So we get

    20 (x/y)2 – 31 (x/y) + 12 = 0

    20 (x/y)2 – 15(x/y) – 16 (x/y) + 12 = 0

    Taking common terms

    5 (x/y) [4 (x/y) – 3] – 4 [4 (x/y) – 3] = 0

    [4 (x/y) – 3] [5 (x/y) – 4] = 0

     

    Here 4 (x/y) – 3 = 0

    4 (x/y) = 3

    So we get x/y = ¾

    Similarly 5 (x/y) – 4 = 0

    5 (x/y) = 4

    So we get x/y = 4/5

    Now dividing by y

    (x + 2y)/ (2x + y) = (x/y + 2)/ (2 x/y + 1)

    (a) If x/y = 3/4, then

    = (x/y + 2)/ (2 x/y + 1)

    Substituting the values

    = (3/4 + 2)/ (2 × 3/4 + 1)

    By further calculation

    = 11/4/ (3/2 + 1)

    = 11/4/ 5/2

    = 11/4 × 2/5

    = 11/10

    So we get

    (x + 2y): (2x + y) = 11: 10

    (b) If x/y = 4/5 then

    (x + 2y)/ (2x + y) = [x/y + 2]/ [2 x/y + 1]

    Substituting the value of x/y

    = [4/5 + 2]/ [2 × 4/5 + 1]

    So we get

    = 14/5/ [8/5 + 1]

    = 14/5/ 13/5

    = 14/5 × 5/13

    = 14/13

    We get

    (x + 2y)/ (2x + y) = 11/10 or 14/13

    (x + 2y): (2x + y) = 11: 10 or 14: 13

    (ii) y (3x – y): x (4x + y) = 5: 12

    It can be written as

    (3xy – y2)/ (4x2 + xy) = 5/12

    By cross multiplication

    36xy – 12y2 = 20x2 + 5xy

    20x2 + 5xy – 36xy + 12y2 = 0

    20x2 – 31xy + 12y2 = 0

    Divide the entire equation by y2

    20x2/y2 – 31 xy/y2 + 12y2/y2 = 0

    20(x2/y2) – 31 (xy/y2) + 12 = 0

    We can write it as

    20(x2/y2) – 15 (x/y) – 16 (x/y) + 12 = 0

    Taking common terms

    5 (x/y) [4 (x/y) – 3] – 4 [4 (x/y) – 3] = 0

    [4 (x/y) – 3] [5 (x/y) – 4] = 0

     

    Here

    4 (x/y) – 3 = 0

    So we get

    4 (x/y) = 3

    x/y = 3/4

    Similarly

    5 (x/y) – 4 = 0

    So we get

    5 (x/y) = 4

    x/y = 4/5

    (a) x/y = 3/4

    We know that

    (x2 + y2): (x + y)2 = (x2 + y2)/ (x + y)2

    Dividing both numerator and denominator by y2

    = (x2/y2 + y2/y2)/ [1/y2 (x + y)2]

    = (x2/ y2 + 1) (x/y + 1)2

    Substituting the value of x/y

    = [(3/4)2 + 1]/ [3/4 + 1]2

    By further calculation

    = (9/16 + 1)/ (7/4)2

    So we get

    = 25/16/ 49/16

    = 25/16 × 16/49

    = 25/49

    So we get

    (x2 + y2): (x + y)2 = 25: 49

    (b) x/y = 4/5

    We know that

    (x2 + y2): (x + y)2 = (x2 + y2)/ (x + y)2

    Dividing both numerator and denominator by y2

    = (x2/y2 + y2/y2)/ [1/y2 (x + y)2]

    = (x2/ y2 + 1) (x/y + 1)2

    Substituting the value of x/y

    = [(4/5)2 + 1]/ [4/5 + 1]2

    By further calculation

    = (16/25 + 1)/ (9/5)2

    So we get

    = 41/25/ 81/25

    = 41/25 × 25/81

    = 41/81

    So we get

    (x2 + y2): (x + y)2 = 41: 81

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