Dear student, the solution is: Let the digits at units and tens places be x and y, respectively ∴ Original number = 10y + x According to the question:10y + x = 4(x + y)⇒ 10y + x = 4x + 4y⇒ 3x − 6y = 0⇒ 3x = 6y⇒ x = 2y (i) Also,10y + x = 2xy⇒ 10y + 2y = 2 2y y [ From (i)]⇒ 12y = 4y2⇒ y = 3 From (i), we get:x = 2 × 3 = 6∴ Original number = 10 × 3 + 6 = 36 Regards

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Dear student,
the solution is:

Let the digits at units and tens places be x and y, respectively . ∴ O riginal number = 10 y + x According to the question: 10 y + x = 4 (x + y) \Rightarrow 10 y + x = 4 x + 4 y \Rightarrow 3 x - 6 y = 0 \Rightarrow 3 x = 6 y \Rightarrow x = 2 y .... (i) Also, 10 y + x = 2 x y \Rightarrow 10 y + 2 y = 2 .2 y . y [From (i)] \Rightarrow 12 y = 4 y^{2} \Rightarrow y = 3 From (i), we get: x = 2 \times 3 = 6 ∴ Original number = 10 \times 3 + 6 = 36

Regards

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