Refer figure `V_(max) = 100/2 = 50V.`
`V_(min) = 20/2 = 10 V.`
(i) Percentage modulation `(mu%)`
`= (V_(max) - V_(min))/(V_(max) + V_(min)) xx 100 = (50 - 10)/(50 + 10) xx 100 = 2/3 xx 100 = 66.7%`
(ii) Peak carrier voltage, `V_c = (V_(max) + V_(min))/2 = (50 + 10)/2 = 30 V`.
(iii) Peak information voltage, `V_m = muV_c = 2/3 xx 30 = 20 V`.