An amount of 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and the second, 11% p.a. If the total interest at the end of one year is 9 3⁄4% , then the amount invested in each share was:	Pick one:	A) 52,500; 47,500	B) 62,500; 37,500	C) 72,500: 27,500	D) 82,500; 17,500	E) None of these
Let the sum invested at 9% be x and that invested at 11% be (100000 – x).   Then,   (x×9×1/100)+[(100000−x)×11×1/100]   = (100000×39/4×1/100)   ⇒ 9x+1100000−11x/100=39000/4= 
<gadget id="calculator">39000/4</gadget>
<output>9_750</output>
9750   ⇒ 2x= (1100000 – 975000) = 125000 ⇒ x = 62500.   ∴ Sum invested at 9% = 62500.   Sum invested at 11% = (100000 – 62500) = 37500.   Answer 62,500; 37,500
<result>B</result>
An amount of 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and the second, 11% p.a. If the total interest at the end of one year is 9 3⁄4% , then the amount invested in each share was: