SOLUTION: 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.

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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(52777)   (Show Source): You can put this solution on YOUR website!
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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.
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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 !

Solved.



Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!


First a typical formal algebraic solution....

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%
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