Question

if compound interest on a certain sum for two year at 3% per annum is 10 1.50 then the simple interest on the same sum at the same rate and for the same time will be​

Answer 1

Answer :

Principal (P) = 1666.7

Simple Interest = 100

$\rule{200}2$

Step-by-step explanation:

Given:-

• Time period (T) = 2 years
• Rate of interest (R) = 3%
• Compound Interest (C.I) = 101.50

Find:-

Principal (P) and Simple Interest.

Solution:-

$\sf{C.I. \:=\:P\:\bigg(1\:+\:\dfrac{R}{100}\bigg)^T\:-P}$

Substitute the known values in above formula. To find the values of Principal (P).

$\implies\:\sf{101.50\:=\:P\:\bigg(1\:+\:\dfrac{3}{100}\bigg)^2\:-\:P}$

$\implies\:\sf{101.50\:=\:P\:\bigg(\dfrac{100\:+\:3}{100}\bigg)^2\:-\:P}$

$\implies\:\sf{101.50\:=\:P\:\bigg(\dfrac{103}{100}\bigg)^2\:-\:P}$

$\implies\:\sf{101.50\:=\:P\:(10.3)^2\:-\:P}$

$\implies\:\sf{101.50\:=\:P(1.0609)\:-\:P}$

$\implies\:\sf{101.50\:=\:1.0609P\:-\:P}$

$\implies\:\sf{101.50\:=\:P(1.0609\:-\:1)}$

$\implies\:\sf{101.50\:=\:0.0606P}$

$\implies\:\sf{P\:=\:\frac{101.50}{0.0606}}$

$\implies\:\sf{P\:=\:1666.67}$

Now,

$\sf{S.I.\:=\:\dfrac{P\:\times\:R\:\times\:T}{100}}$

From above calculations and according to question, we have -

• P = 166.7
• R = 3%
• T = 2 years

Substitute the known values in the above formula

$\implies\:\sf{S.I.\:=\:\dfrac{1666.7\:\times\:3\:\times\:2}{100}}$

$\implies\:\sf{S.I.\:=\:\dfrac{10000}{100}}$

$\implies\:\sf{S.I.\:=\:100}$

Answer 2

Question :-----

• CI for 2 years at 3% = 101.5 Rs.

To Find :----

• SI at same rate and Time ?

Solution :-----

we can solve this problem in 3 ways ..

Let see all methods one by one Now ..

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$\pink{\bold{\underline{\underline{solution(1)}}}}$

Let try it with basic Method First .

Let our Principal = P

Rate = 3 %

Time = 2 years.

so,

using ,

$\huge\red{\boxed{\sf A\:=\:P( 1 + \frac{r}{100})^{t}}}$

we get,

Amount = P(1+3/100)²

→ Amount = P(103/100)²

Now, we know that,

Compound interest = Amount - Principal

101.5 = P(103/100)² - P

taking P common From RHS now,

101.5 = P[(103/100)² - 1]

101.5 = P[(103² - 100²)/100²] [ using (a-b)(a+b) = a²-b² now]

101.5 = P[(103+100)(103-100)/100²]

P = (101.5×100×100)/(203×3)

P = 1666(2/3) Rs.

Now,

our , simple interest for same amount with same rate and time will be :----

SI = P×R×T/100

→ SI = (5000×3×2)/(100×3) = Rs. 100 (Ans) .

so,, we will get simple interest of Rs.100 with same amount and same rate for same years.

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$\red{\bold{\underline{\underline{solution (2)}}}}$

Lets try it with fraction Method now,

Rate = 3% = 3/100

Let 100 = P

than amount after 1 year = 103

so,

100 ------------------- 103

For 2 years

100² ----------------- 103²

10000(P) --------------- 10609(A)

Compound interest = 10609 - 10000 = Rs.609 .

But given compount interest is Rs. 101.5 ,

so,

when ,

Rs.609 ci , our P is = Rs.10000

if 1 Rs. ci, our P is = 10000/609

if 101.5 Rs. ci , our P is = (10000×101.5)/609 = Rs.5000/3

so, our SI will be , with same rate and time =

SI = (5000×3×2)/(100×3) = Rs.100 (Ans)

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$\huge\underline\purple{\mathcal Solution(3)}$

Thats my Favourite ..

Rate is = 3%

Time is = 2 years .

Lets Find successive rate of 3% for 2 years .

Successive rate = 3 + 3 + 3×3/100 = 6.09%

so, we can say that, we get, 6.09% of Principal as CI for 2 years.

And we get, (3×2) = 6% SI for same amount for same year .

so,

6.09% --------------- 101.5

1 % ------------------- (101.5/6.09)

6% ------------------- (101.5×6)/6.09 = Rs.100 (Ans)

we get, our Simple interest for same amount and same time and rate . we dont need to Find Principal also in this method ....

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(Hope it Helps you)