Interest problems

1200?

shankysainik wrote:1200?

Right. How it can be solved?

Fokin wrote:Barbara invest $2400 in the national Bank at 5% How much additional money must she invest at 8 % so that total annual income will be equal to 6% of the entire investments ?

a) 3,600
b) 3,000
c) 2,400
d) 1,200
e) 1,000

now let say the additional amount invested be x
so total amount invested =2400+x
total income at the rate of 6 %= 6/100*(2400+x) = 144+3x/50

Barbara invest $2400 in the national Bank at 5%
so income annual=5/100*2400=120
now barbara invest the remaining x at 8 %
so annual income =8/100*x=2x/25
now return on $2400 in the national Bank at 5% +return on x at 8 %= return on totalinvestment(2400+x) at 6%
120+2x/25=144+3x/50
2x/25-3x/50=24
x/50=24
x=1200
so additional amount to be invested is 1200

Lets say the additional income that needs to be invested is p

from the investment of 2400 at 5 % we get = 2400 * 5/100 * 1 = 120

=>120 + p * 8/100 * 1 = 6/100 (p+2400)
=>120 + 8p/100 = 6p/100 + 144
=>2p/100 = 24
=> p= 1200 ( D)

If the Principal amount is X and interest is y%, then total amount at the end of an year = (1 + y/100)*X
Thus, the equation becomes
(1.05*2400) + (1.08 * X) = 1.06(2400 + X)

On simplifying,
.02X = .01 * 2400 --> X =1200 !