A sum of money invested at 5 per cent simple interest amounts to 1,380 in 2 1/2 years at at the rate of 6% r year find the sum
Answers
Step-by-step explanation:
answer :- Rs. 12000
Let the sum (Principle) = Rs. 100
C.I. of 1st year = Rs. \frac{100\times 5\times 1}{100} = Rs. 5$$
And, amount of 1st year = Rs. 100 + Rs. 5=Rs. 105$$
⇒ The principke for 2nd year = Rs. 105
C.I. of 2nd year =Rs.
100
105×5×1
=Rs.5.25
And, amount of 2nd year =Rs.105+Rs.5.25=Rs.110.25
⇒ The principle for 3rd year =Rs.110.25
C.I. of 3rd year =Rs.
100
110×5×1
=Rs.5.5125
Difference between C.L of 1st year and C.L of 3rd year =Rs.5.5125−Rs.5=Rs.0.5125
Now, when the difference of interest =Rs.0.5125,sum=Rs.100
And, when the difference of interest =Rs.61.50,sum=Rs.
0.5125
100 ×61.50=
Rs.12,000 (Ans)
Step-by-step explanation:
It is given that
It is given that Simple Interest (SI) = 1380
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) Here
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750Rate of interest (R)= 12% p.a or 6% half- yearly
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750Rate of interest (R)= 12% p.a or 6% half- yearlyPeriod (n) = 1 year - 2 half years
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750Rate of interest (R)= 12% p.a or 6% half- yearlyPeriod (n) = 1 year - 2 half yearsSo we get
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750Rate of interest (R)= 12% p.a or 6% half- yearlyPeriod (n) = 1 year - 2 half yearsSo we getAmount (A)= P(1 + R/100)
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750Rate of interest (R)= 12% p.a or 6% half- yearlyPeriod (n) = 1 year - 2 half yearsSo we getAmount (A)= P(1 + R/100)Substituting the values
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750Rate of interest (R)= 12% p.a or 6% half- yearlyPeriod (n) = 1 year - 2 half yearsSo we getAmount (A)= P(1 + R/100)Substituting the values =5750(1+6/100)2
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750Rate of interest (R)= 12% p.a or 6% half- yearlyPeriod (n) = 1 year - 2 half yearsSo we getAmount (A)= P(1 + R/100)Substituting the values =5750(1+6/100)2By further calculation
It is given that Simple Interest (SI) = 1380Rate of interest (R) = 12% p.a.Period(T)= 2 years (i) We know that Sum(P) = (SI×100)/(R×T)Substituting the values =(1380×100)/(12×2)= 5750(ii) HerePrincipal (P)= 5750Rate of interest (R)= 12% p.a or 6% half- yearlyPeriod (n) = 1 year - 2 half yearsSo we getAmount (A)= P(1 + R/100)Substituting the values =5750(1+6/100)2By further calculation= 5750×(53/50)2