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The difference between two positive numbers is 3. If the sum of their squares in 369, then the sum of the numbers is
Let the numbers be $$x$$ and $$y$$
Given : $$(x-y)=3$$ -----------(i) and $$x^2+y^2=369$$ ------------(ii)
Squaring both sides in equation (i), we get : $$x^2+y^2-2xy=9$$
Substituting value from equation (ii), => $$2xy=369-9=360$$
=> $$xy=\frac{360}{2}=180$$
To find : $$(x+y)=z=?$$
Now, we know that $$(x+y)^2=x^2+y^2+2xy$$
=> $$z^2=369+2(180)=729$$
=> $$z=\sqrt{729}=27$$
=> Ans - (C)
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