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BACK 1.The difference between C.I and S.I. on a certain sum at 10% per annum for 2 yrs is Rs.631.find the sum Sol: MethodI: NOTE: a) For 2 yrs -------->sum=(1002D/R2) b) For 3 yrs -------->sum=(1003D/R2(300+R)) Sum=1002*631/102 =Rs.63100 MethodII: Let the sum be Rs.X,Then C.I.=X[1+(10/100)]2-X S.I=(X*10*2)/100 =X/5 C.I-S.I.=21X/100-X/5 =X/100 X/100=631 X=Rs.63100 2.If C.I on a certain sum for 2 yrs at 12% per annum is Rs.1590. What would be S.I? sol: C.I.=Amount-Principle Let P be X C.I=X[1+(12/100)]2-X =>784X/625-X=1590 =>X=Rs.6250 S.I=(6250*12*2)/100=Rs.1500 3.A sum of money amounts to Rs.6690 after 3 yrs and to Rs.10035b after 6 yrs on C.I .find the sum sol: For 3 yrs, Amount=P[1+(R/100)]3=6690-----------------------(1) For 6 yrs, Amount=P[1+(R/100)]6=10035----------------------(2) (1)/(2)------------[1+(R/100)]3=10035/6690 =3/2 [1+(R/100)]3=3/2-----------------(3) substitue (3) in (1) p*(3/2)=6690 =>p=Rs.4460 sum=Rs.4460 4.A sum of money doubles itself at C.I in 15yrs.In how many yrs will it become 8 times? sol: Compound Interest for 15yrs p[1+(R/100)]15 p[1+(R/100)]15=2P =>p[1+(R/100)]n=8P =>[1+(R/100)]n=8 =>[1+(R/100)]n=23 =>[1+(R/100)]n=[1+(R/100)]15*3 since [1+(R/100)] =2 n=45yrs 5.The amount of Rs.7500 at C.I at 4% per annum for 2yrs is sol: Iyear------------------7500+300(300------Interest on 7500) IIyear ----------------7500+300+12(12------------4% interest on 300) Amount=7500+300+300+12 =Rs.8112 6.The difference between C.I and S.I on a sum of money for 2 yrs at 121/2% per annum is Rs.150.the sum is sol: Sum=1002D/R2=( 1002*150) /(25/2)2=Rs.9600 7.If the S.I on sum of money at 15% per annum for 3yrs is Rs.1200, the C.I on the same sum for the same period at same rate is--------- sol: S.I=1200 P*T*R/100=1200 P*3*5/100=1200 =>P=Rs.8000 C.I for Rs.8000 at 5% for 3 yrs is-------------8000+400 -------------8000+400+20 -------------8000+400+20+20+1 C.I =400+400+20+400+20+20+1 =Rs.1261 BACK |