Latest self Aptitude Question SOLUTION: x/y=6/5 find value of (x^2+y^2)/(x^2-y^2)

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(#M40021018) SELF question numerical Keep an EYE

x/y=6/5 find value of
(x^2+y^2)/(x^2-y^2)

kumar (10 years ago)

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usha devi  (10 years ago)

x/y=6/5
while taking as proportions x=6k and y=5k
thus (x^2+y^2)=((6k)^2+(5k)^2)
=>(36(k^2))+(25(k^2))
=>61(k^2)
similarly (x^2-y^2)=((6k)^2-(5k)^2)
=>(36(k^2))-(25(k^2))
=>11(k^2)
finally (x^2+y^2)/(x^2-y^2)=61(k^2)/11(k^2)=61/11

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Vaibhav Mishra  (10 years ago)

using componendo and dividendo rule,
x/y = 6/5
(x+y)/(x-y)=(6+5)/(6-5)
(x+y)/(x-y)=11,here (x-y)=(x+y)/11
Now we have
(x^2+y^2)/(x^2-y^2)=(x^2+y^2)/(x+y)(x-y)
x^2+y^2=(x+y)^2-2xy
now,(x+y)^2-2xy/(x+y)(x-y)
11((x+y)^2-2xy)/(x+y)^2, here we substitute the value of (x-y)=(x+y)/11
now take (x+y)^2 common,
we have,
11(1-2xy/(x+y)^2)
11(1-2xy/(x^2+y^2+2xy))
11(1-2y/x-2x/y-1)
we have x/y=6/5,y/x=5/6 , now
11(-2*5/6-2*6/5)
(-44.73) Ans

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Soumya Sharma  (10 years ago)

ans: 61/11
solution: (x2+y2)/(x2-y2)= ((x-y)^2-2xy)/(x2-y2)
= (x-y)^2/(x2-y2) - 2xy/(x2-y2)
= (x-y)^2/((x+y)(x-y)) - 2xy/((x+y)(x-y))
= (x-y)/(x+y) - 2xy/(xy(1+y/x)(x/y-1)
= (x-y)/x+y) - 2/(1+y/x)(x/y-1) ------- eqn1
we know x/y=6/5,,aplly componendo and dividend
x-y/x+y=1/11
put in eqn the value of x-y/x+y, x/y and y/x

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