Question

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​

Answer 1

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)

Answer 2

Step-by-step explanation:

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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