If A gives Rs. 50 to M, A would then have an amount equal to what M had…

Two quantities with a known SUM (S) and a known DIFFERENCE (D) can always be recovered using the identities: larger value = (S + D) / 2 and smaller value = (S − D) / 2, obtained by adding and subtracting the two defining equations.

1. Let the amount with A be X and the amount with M be Y (both taken as their original, "earlier" amounts).

2. The two amounts together total Rs. 650, so X + Y = 650.

3. After A gives Rs. 50 to M, A is left with X − 50; this must equal what M had earlier, so X − 50 = Y, i.e. X − Y = 50.

4. Add the sum and difference equations: (X + Y) + (X − Y) = 650 + 50, giving 2X = 700, so X = 350.

5. Substitute back into X + Y = 650: Y = 650 − 350 = 300.

6. The required ratio is X : Y = 350 : 300, which simplifies (dividing by 50) to 7 : 6.

Independently verifying: if A starts with Rs. 350 and gives Rs. 50 to M, A is left with Rs. 300 — exactly the Rs. 300 M had earlier, confirming both the sum and transfer conditions are satisfied together.