Given, the difference between simple interest for
2 years and compound interest for 2 years on
the same sum and at the same rate of interest
compounded annually is 120. The difference
between simple interest for 3 years and compound
interest for 3 years on the same sum and at
the same rate of interest is 366. Find the rate
of interest.
Answers
Step-by-step explanation:
Given :-
The difference between simple interest for 2 years and compound interest for 2 years on the same sum and at the same rate of interest compounded annually is 120. The difference between simple interest for 3 years and compound interest for 3 years on the same sum and at the same rate of interest is 366.
To find :-
Find the rate of interest ?
Solution :-
Let the principle be P
Let the rate of interest be R
Time (T) = 2 years and 3 years
The difference between the CI and SI for 2 years = Rs. 120
The difference between the CI and SI for 3 years = Rs. 366
We know that
Simple Interest = PTR/100
Compound Interest = Amount - Principle
=> CI = P[1+(R/100)]^n-P
=> CI = P[{1+(R/100)}^n-1]
Calculating SI for 2 years :-
Simple Interest for 2 years
=>SI = P×2×R/100
=>SI = 2PR/100
=> Simple Interest = PR/100
Calculating CI for 2 years :-
Compound Interest for 2 years
=> CI = P[{1+(R/100)}^n-1]
=> CI = P[{1+(R/100)}²-1]
=> CI = P[{(100+R)²/10000}-1]
Their difference = 120 (Given )
=> P[{(100+R)²/10000}-1]-(2PR/100)=120
=> P[(100+R)²-10000-200PR]/10000=120
=> P[10000+R²+200PR-200PR-10000]/10000=120
=> PR²/10000 = 120
=> P(R/100)² = 120
The difference between CI and SO for 3 years = P(R/100)² = 120------(1)
Calculating SI for 3 years :-
Simple Interest for 3 years
=>SI = P×3×R/100
=> SI = 3PR/100
SI = 3PR/100
Calculating CI for 3 years :-
Compound Interest for 3 years
=>CI = P[{1+(R/100)}^n-1]
=>CI = P[{1+(R/100)}³-1]
=>CI = P[{100+R}/100}³-1]
=>CI = P[{(100+R)³/1000000}-1]
=>CI = P[(100+R)³-1000000)/1000000]
Their difference = 366 (Given )
=>P[(100+R)³-1000000)/1000000]-3PR/100= 366
=> P[(100+R)³-1000000-30000PR]/1000000=366
=> P[R³+30000R+300R²)-30000PR]/1000000 = 366
=> P[R³+30000R+300R²-30000R)]/1000000 =366
=> P[(R/100)³+(3R²/100)] = 366
=> P[(R/100)²{(R/100)+3}] = 366
The difference between CI and SI for 3 years P[(R/100)²{(R/100)+3}] = 36-----(2)
On Substituting the value of P(R/100)²
From (1) in (2) then
=>120× [(R/100)+3] = 366
=> (R/100)+3 = 366/120
=> (R/100)+3 = 61/20
=> (R+300)/100 = 61/20
=> R+300 = (61/20)×100
=> R+300 = 6100/20
=> R+300 = 305
=> R=305-300
=> R = 5
Therefore, R = 5%
Answer:-
Rate of Interest for the given problem is 5%
Used formulae:-
- I = PTR/100
- A =P[1+(R/100)]^n
- A = P+I
Where,
- P = Principle
- T = Time
- R = Rate of Interest
- n = Number of times the interest calculated compoundly.
- A = Amount
- CI = Compound Interest
- SI = Simple Interest
Answer:
5%
Step-by-step explanation:
SI
1. A
2.A
3.A
CI
A
A
A(on principal)+120+120+?=360
6 is what % of 120
6=?/100 x 120
= 5%