Question 1208045: Mr B invested p30000:part at 5% and part at 8%.The total interest on the investment was p2100. How much did she invest at each rate. Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52776) (Show Source):
You can put this solution on YOUR website! .
Mr B invested p30000:part at 5% and part at 8%.
The total interest on the investment was p2100. How much did she invest at each rate.
~~~~~~~~~~~~~~~~~~~~
x invested at 8%;
(30000-x) invested at 5%.
The total annual interest is 0.08x + 0.05(30000-x).
An equation to find x is
0.08x + 0.05(30000-x) = 2100.
Simplify and find x
0.08x + 1500 - 0.05x = 2100
0.08x - 0.05x = 2100 - 1500
0.03x = 600
x = 600/0.03 = 20000.
ANSWER. p20000 invested at 8%. The rest, p30000 - p20000 = p10000 invested at 5%.
CHECK. 0.08*20000 + 0.05*10000 = 2100 for the total annual interest. ! precisely correct !
x = amount invested at 5%
30000-x = amount invested at 8%
The total interest earned was 2100:
ANSWERS:
x = p10000 invested at 5%
30000-x = p20000 invested at 8%
And then a fast and easy informal method for solving any 2-part "mixture" problem like this....
2100 interest on an investment of 30000 is an interest rate of 2100/30000 = .07 = 7%
7% is 2/3 of the way from 5% to 8% (look at the three numbers 5, 7, and 8 on a number line if it helps)
That means 2/3 of the total p30000 was invested at the higher rate
ANSWERS:
2/3 of p30000 = p20000 invested at 8%
1/3 of p30000 = p10000 invested at 5%
