A person invested in all 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all the three cases. The money invested at 4% is: Choices: A) 200 B) 600 C) 800 D) 1200 E) None of these
Let the parts be x, y and [2600 – (x + y)].Then,   x×4×1/100=y×6×1/100=[2600−(x+y)]×8×1/100   ∴ y⁄x = 4⁄6 = 2⁄3 or y = 2⁄3x.   So, x×4×1100=[2600−53x]×8100x×4×1100=[2600−53x]×8100   ⇒ 4x = (7800−5x)×8/3⇒52x=(7800×8)   ⇒ x = (7800×8/52)=1200.   ∴ Money invested at 4% = 1200.   Answer 1200
<result>D</result>
A person invested in all 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all the three cases. The money invested at 4% is: