Question 1163118
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Here is a quick and easy way to solve a "mixture" problem like this if a formal algebraic solution is not required.<br>
All $10,000 invested at 1.5% would yield $150 interest; all at 2% would yield $200 interest.<br>
The actual interest, $193, is 43/50 of the way from $150 to $200.  (Picture the three numbers on a number line -- 150 to 200 is a difference of 50; 150 to 193 is a difference of 43.)<br>
That means 43/50 of the total was invested at 2%; so 7/50 of the total was invested at 1.5%.<br>
ANSWER: The amount invested at 1.5% was 7/50 of $10,000, which is $1400.<br>
CHECK:
.015(1400)+.02(8600) = 21+172 = 193<br>