Given :

Difference between Compound interest and Simple interest for 3 years = 990

Simple interest at same rate of interest and same sum = RS. 2000

To find : Rate of interest

Solution :

Let the same sum i.e Principal be P and and same Rate of interest be R.

Simple interest for 2 years = Rs. 2000

=> PTR / 100 = 2000

=> 2PR / 100 = 2000

=> PR / 100 = 2000 / 2

=> PR × 1 / 100 = 1000

=> Simple interest for 1 years = 1000

Simple interest for 3 years = 3 × 1000 = Rs. 3000

=> PR = 10000

=> P = 100000 / R → EQ( 1 )

Difference between Compound interest and Simple interest for 3 years = 990

=> Compound interest - Simple interest = 990

=> Compound interest for 3 years = 3000 + 990 = Rs. 3990

=> P{ (1 + R/100)^T - 1} = 3990

=> P{ (1 + R/100)³ - 1 } = 3990

From EQ( 1 )

=> ( 100000 / R ) { ( 1 + R/100)³ - 1 } = 3990

=> ( 1 + R/100 )³ - 1 = 399R / 10000

Using identity a³ - b³ = ( a - b )( a² + b² + ab)

=> ( 1 + R/100 - 1 ){ ( 1 + R/100 )² + 1 + 1 + R/100) } = 399R / 10000

=> ( R/100 ) ( 3 + 2R/100 + R²/10000 + R/100 ) = 399R / 10000

=> 3 + R²/10000 + 3R/100 = 399 / 100

=> R²/10000 + 3 - 399 / 100 + 3R/100 = 0

=> R²/10000 + 3 - 399/100 + 3R/100 = 0

=> R² + 30000 - 39900 + 300R = 0

=> R² + 300R - 9900 = 0

=> R² + 300R - 9900 = 0

=> R² + 330R - 30R - 9900 = 0

=> R( R + 330 ) - 30( R + 330 ) = 0

=> ( R - 30 )( R + 330 ) = 0

=> R - 30 = 0 or R + 330 = 0

=> R = 30 or R = - 330

R ≠ 330 SINCE R can't be negative

=> R = 30

Therefore the Rate of interest is 30 %.