Math investment problem?

A(12/100)+B(8/100)=5,280

A+B=50,000

12A+8B=528,000

12A+12B=600,000

4B=72000

B=P18,000 with 8% rate

A=P32,000 with 12% rate

Let, the amount was invested at an instrument that earns 12% per year be P x

Income from this for one year = x * 12 * 1/ 100 = 12 x /100

the rest is 50000 - x

Income from this for one year = (50000 - x) * 8 * 1/ 100 = 8 *(50000 - x) / 100

So, 12 x /100 + 8 *(50000 - x) / 100 = 5280

12x + 400000 - 8x = 528000

4x = 528000 - 400000 = 12800

x = 12800 / 4 = 32000

Investment in 1st instrument = P 32, 000 and other one P 18, 000

Let the ammount invested at 12% = A

Let the ammount invested at 8% = B

We know then that A + B = 50000

and that 1.12*A + 1.08*B = 55280, this is ammount A plus interest and ammount B plus interest and equals the total.

We now solve these simultaneous equations.

Multiply the first equation by 1.08 to give

1.08A + 1.08B = 54000.

Subtract this from the second equation:

1.12A + 1.08B = 55280

1.08A + 1.08B = 54000

--------------------------------

0.04A = 1280

so A = 1280/0.04 = 32000

So Mr S invested P32,000 at 12% and the remainder (50,000-32,000 = 18,000) at 8%.

;-)

a million. Jason invested a entire of $20,000 for 365 days. component to which became into invested at an annual activity fee of 6%, something at an annual fee of 10%. If the return on the ten% investment is $4 hundred greater effective than the return on the 6% investment, how plenty became into invested at each and each fee? 6% fee account significant = p fee = 0.06 activity = i = 0.06p 10% fee account significant = 20000 - p fee = 0.10 activity = i + 4 hundred = 0.06p + 4 hundred on the ten% fee account, enable activity = significant * fee: 0.06p + 4 hundred = (20000 - p) * (0.10) ---> 0.06p + 4 hundred = 2000 - 0.10p ---> 0.16p = 1600 ---> p = $10,000 in view that p = $10,000, all of us be attentive to that the 6% fee account (p) had $10,000 invested in it, and the ten% fee account (20000 - p) additionally had $10,000 invested in it. --------------------------------------... 2. The ration of the quantity of money invested at 6% annual fee to the quantity invested at 9% is 3:5. The return on the 9% investment is 216 greater effective than two times the return on the 6% investment. How plenty is invested at each and each fee? 6% fee account significant = p fee = 0.06 activity = i = 0.06p 9% fee account significant = 5/3 p fee = 0.09 activity = 2i + 216 = 0.12p + 216 on the 9% fee account, enable activity = significant * fee: 0.12p + 216 = (5/3 p) * (0.09) ---> 0.12p + 216 = 0.15p ---> 0.03p = 216 ---> p = $7200 in view that p = $7200, all of us be attentive to that the 6% fee account (p) had $7200 invested in it, and the 9% fee account (5/3 p) had $12,000 invested in it. --------------------------------------...