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BACK 1.A certain sum amounts to Rs.7350 in 2 yrs and to Rs.8575 in 3 yrs.Find the sum and rate%? sol: S.I. on Rs.7350 for 1yr =Rs.(8575-7350) =Rs.1225 S.I. on Rs.7350 for 2yrs=Rs.2*1225 =Rs.2450 PTR/100=2450 =>P*2*R/100=2450 Since S.I. on Rs.7350 for 1yr =Rs.(8575-7350) =Rs.1225 Rate R=100*1225/(7350*10 =16 2/3% Since it is C.I ,Let sum be Rs.X Then X[1+(R/100)]2=7350 =>X[1+(50/100)]2=7350 =>X=7350*(36/49) Sum=Rs.5400 2.If the difference between C.I compounded halfyearly and S.I on a sum at 10% per annum for one yr is Rs.25 the sum is sol: p[1+((R/2)/100)]2n-PTR/100=25 P[1+((10/2)/100)]2n-P*1*10/100=25 =>P=Rs.400 3.A man borrowed Rs.800 at 10 % per annum S.I and immediately lent the whole sum at 10% per annum C.I What does he gain at the end of 2yrs? sol: C.I.=Rs.[800[1+(10/100)2]-800]=Rs.168 S.I=Rs.[800*10*2/100]=Rs.160 Gain=C.I-S.I=Rs(168-160) =Rs.8 4.On what sum of money will be S.I for 3 yrs at 8% per annum be half of C.I on Rs.400 for 2 yrs at 10% per annum? sol: C.I on Rs.400 for 2yrs at 10%=Rs.[400*[1+(10/100)]2-400] =Rs.84 Required S.I =1/2*84=42/- New S.I=Rs.42,Time=3yrs Rate=8% Sum=Rs.[100*42/(3*8)] =Rs.175 5.A sum of money placed at C.I doubles itself in 5yrs .It will amount to 8 times itself in------------- sol: p[1+(R/100)]5=2P =>[1+(R/100)]5=2 To become 8 times =>8P p[1+(R/100)]5=2^3P =[1+(R/100)]^(5*3) =[1+(R/100)]^15 n=15years BACK |