Question 32387
Assume that the person invests $6,000 at x% rate.
$8,000 is then invested at (2x-1)%
Total interest = x% of 6000 + (2x-1)% of 8000
=(x/100)(6000) + [(2x-1)/100](8000)
=60x + 80(2x-1)
=60x + 160x - 80
=220x - 80
Total earnings from the interest = 1240
=> 220x - 80 = 1240
=> 220x = 1320
=> x = 6
So, $6,000 invested at 6% and $8,000 invested at 11%