Question 1184710
A man invested  part of P20,000 at 18% and the rest at 16%. The annual income from 16% investment was P620 less than three times the annual income from 18% investment. How much did he invest at 18%?
<pre>Let amount invested at 18% be E
Then amount invested at 16% = 20,000 - E
Income from 18% and 16% investments: .18E, and .16(20,000 - E), respectively
Based on what's given, we get: .16(20,000 - E) = 3(.18E) - 620
3,200 - .16E = .54E - 620
3,200 + 620 = .16E + .54E
3,820 = .7E
Amount invested at 18%, or {{{highlight_green(matrix(1,9, E, "=", "3,820"/.7, "=", "P5,457.14", "(to", 2, decimal, "places)"))}}}