Solution

(Y² - 540Y + 72000 = 0) Y² - 540Y + 72000 = 0 Y² - (300+240)Y + 72000 = 0 Y² - 300Y - 240Y + 72000 = 0 Y(Y- 300) - 240(Y- 300) = 0 (Y - 300) (Y - 240) = 0 Y = 300, 240 It is assumed that the monthly salary of Rohit is 1600 times of the difference between the both of the roots of ‘Y’. monthly salary of Rohit = 1600x(300-240) = 1600x60 = 96000 Rohit spent (R-4)% of his monthly income on food. Out of the remaining (R+7)% was spent on travelling. (1/6) of the remaining was spent on education. (2R–1)% of the remaining was spent on house rent and after that the remaining amount was saved by him. If the annual savings of Rohit is Rs. 402480. 96000 of [100-(R-4)]% of [100-(R+7)]% of [1-(1/6)] of [100-(2R–1)]% = 402480/12 96000 x [100-R+4]% x [100-R-7]% x (5/6) x [100-2R+1]% = 33540 0.096 x [100-R+4] x [100-R-7] x (5/6) x [100-2R+1] = 33540 0.016 x [104-R] x [93-R] x 5 x [101-2R] = 33540 0.08 x [104-R] x [93-R] x [101-2R] = 33540 [104-R] x [93-R] x [101-2R] = 419250 (−R+18) (2R ² − 459R + 30979) = 0 So R = 18 Monthly expenditure on travelling = 96000 of [100-(R-4)]% of (R+7)% Put the value of ‘R’ in the above equation. = 96000 of [100-(18-4)]% of (18+7)% = 96000 of [100-14]% of 25% = 96000 of 86% of 25% = 96000 x (86/100) x (25/100) = Rs. 20640