(1) S.I. for 2 years =100P×R×T​=1002000×15×2​=Rs 600⇒ Amount = Rs. (2000 + 600) = Rs. 2600Let the additional money be Rs. x. According to the question, C.I=P1[(1+100R​)T−1] 1507=(2600+x)[(1+10020​)2−1] ⇒1507=(2600+x)[(1+51​)2−1]⇒1507=(2600+x)(2536​−1)⇒1507=(2600+x)(2536−25​)⇒1507=(2600+x)(2511​)⇒2600+x=111507×25​=3425 ⇒ x = 3425 – 2600 = Rs. 825