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Preeti Dabral 2 years ago
xa+yb=2
bx+ayab=2
bx+ay=2ab........(i)
ax−by=(a2−b2)...(ii)
Multiplying (i) by b and (ii) by a
b2x+bay=2ab2.............(iii)
a2x−bay=a(a2−b2)........(iv)
Adding (iii) and (iv),we get
b2x+a2x=2ab2+a3−ab2
x(b2+a2)=2ab2+a3−ab2
x(b2+a2)=ab2+a3
x(b2+a2)=a(b2+a2)
x=a(b2+a2)(b2+a2)=a
Putting x = a in (i),we get
b×a+ay=2ab
ay=2ab−ab⇒ay=ab or y = b
∴ solution is x = a, y = b
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Preeti Dabral 2 years ago
Let the digits of the required 3 - digit number at hundreds, tens and ones places be a−d, a, a+d respectively
Then their sum = 15
i.e, a−d+a+a+d = 15
⇒ 3a = 15
⇒ a = 5
Required three digit number = 100(a−d)+10a+a+d
= 100a−100d+10a+a+d
= 111a−99d
Number obtained by reversing the digits = 100(a+d)+10a+a−d
= 100a+100d+10a+a−d
= 111a+99d
According to the question,
111a+99d=111a−99d−594
⇒ 594=111a−99d−111a−99d
⇒594=−198d
⇒−594198=d
d=−3
The number is 111a - 99d
111×5−99×−3= 852
555 + 297 = 852
Therefore, Number is 852.
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Preeti Dabral 2 years ago
As per given condition
Area of sector OAPB = 536times the area of circle
∴πr2×θ360=536πr2
∴πr2×x360=536πr2
x360=536 or,
36x=360×5
x=360×536
x=10×5
x=50
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