SOLUTION: An investor has $5000 to invest and decides to invest some of it with company A and the rest with company B. After one year each $1 invested with company A has grown to $1.20, each

Algebra ->  Equations -> SOLUTION: An investor has $5000 to invest and decides to invest some of it with company A and the rest with company B. After one year each $1 invested with company A has grown to $1.20, each      Log On


   

Question 856061: An investor has $5000 to invest and decides to invest some of it with company A and the rest with company B. After one year each $1 invested with company A has grown to $1.20, each dollar invested with company B has grown to $1.05 and the $5000 has grown to $5670. How much of the original $5000 was invested with each company?
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
Say he invested $x in A. So amount invested in B is 5000 - x.
Total earned from A = 1.2 * x  ---> every $ becomes $1.2
Total earned from B = 1.05 * (5000 - x) ----> every $ becomes $1.05

Total = 1.2%2Ax+%2B+1.05%2A%285000+-+x%29+=+5670
i.e. 1.2%2Ax+%2B+5250+-+1.05%2Ax+=+5670
Simplifying:
0.15%2Ax+=+420 or x+=+2800
So, amount invested in A is $2800, and that invested in B is $2200.
Check: 2800 becomes 2800  * 1.2 = 3360
2200 becomes 2200 * 1.05 = 2310
Total = 3360 + 2310 = 5670. Correct!


Hope this is clear :)