SOLUTION: A man saved 3000 in a bank p, whose interest rate was X% per annum and 2000 in another bank Q whose interest rate was Y% per annum. His total interest in one year was 640. If he h

Algebra ->  Finance -> SOLUTION: A man saved 3000 in a bank p, whose interest rate was X% per annum and 2000 in another bank Q whose interest rate was Y% per annum. His total interest in one year was 640. If he h      Log On


   



Question 1149106: A man saved 3000 in a bank p, whose interest rate was X% per annum and 2000 in another bank Q whose interest rate was Y% per annum. His total interest in one year was 640. If he had saved 2000 in P and 3000 in Q for the same period, he would have gained 20 as additional interest. Find
(I) the value of x and Y.
(II) the investment return if the man saves 5000 in bank Q for 2years.

Found 2 solutions by Boreal, Theo:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
interest in first part is 3000*p+2000*q=64000, because I multiplied the interest by 10 to make it an integer
this is 3p+2q=64
the second is 2000*p+3000q=66000
2p+3q=66; multiply top by 2 and bottom by -3
6p+4q=128
-6p-9q=-198
-5q=-70
q=14%
p=12% substituting
5000*0.14*2=$1400 earned at a 14% rate.

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
let x = the interest rate for investment P.
let y = the interest rate for investment Q.

you get:

3000x + 2000y = 640
2000x + 3000y = 660

multiply both sides of the first equation by 2 and multiply both sides of the second equation by 3 to get:
6000x + 4000y = 1280
6000x + 9000y = 1980
subtract the first equation from the second to get:
5000y = 700
solve for y to get:
y = .14
replace y with .14 in the first original equation to get:
3000x + 2000y = 640 becomes 3000x + 2000*.14 = 640 which becomes 3000x + 280 = 640.
subtract 280 from both sides of this equation to get:
3000x = 640 - 280 which becomes 3000x = 360.
solve for x to get x = 360/3000 = .12
you have x = .12 and y = .14
go back to your two original equations and replace x with .12 and y with .14 to get:
3000x + 2000y = 640 becomes 3000*.12 + 2000*.14 = 640 which becomes 360 + 280 = 640 which becomes 640 = 640 which is true.
2000x + 3000y = 660 becomes 2000*.12 + 3000*.14 = 660 which becomes 240 + 420 = 660 which becomes 660 = 660 which is true.
both original equations are true when x = .12 and y = .14.
x% = x * 100 = 12%
y% = y * 100 = 14%.
since they asked for the value of x and y, and not for the value of x% and y%, your solution is x = .12 and y = .14.
if the man has put all 5000 in bank Q, then he would have earned interest of 5000 * .14 = 700.