Rheecaidn
Answered

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A man invests some amount of money at 6% interest rate per annum and some amount at 9% interest rate per annum. If he puts P12,000.00 more in the second investment than in the first investment and his perceived annual income from both investments is P4,830.00, how much must he invest at each rate?

Sagot :

Mlcparra16idn
So the formula is 
[tex]Interest= Principal*rate*time[/tex]
(Principal being the money he invested, time would be in years)

We let the principal of the first first investment be [tex]x[/tex]
this would make the principal of the second investment be [tex]x+12000[/tex]

The interest of the first investment would be
[tex]x* \frac{6}{100} *1= \frac{6x}{100} [/tex] 
 
The interest of the second investment would then be
[tex](x+12000)* \frac{9}{100} *1= \frac{9x}{100} +120*9= \frac{9x}{100} +1080[/tex]

The sum of these two should be 4,830
[tex] \frac{6x}{100} +\frac{9x}{100} +1080=4830[/tex]

We combine like terms and subtract 1080 from both sides
[tex] \frac{15x}{100} =4830-1080=3750[/tex]

Multiply both sides by [tex] \frac{100}{15} [/tex]
[tex]x=3750* \frac{100}{15} =25000[/tex]

Therefore he invested P25,000 in the first investment and
P25,000+P12,000=P37,000 in the second investment.



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