Question

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.​

Answer 1

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 2

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%