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A Man invested 10,000 pesos. A part of it’s yields on interest rate of 10% and the rest at 8%. If he got 920 pesos of interest a year, how much amount was invested at each year?

Sagot :

Let x be the investment in the first year.
     10,000-x be the investment in the succeeding year.

0.10x + 0.08(10,000-x) = 920

0.10x + 800 -0.08x = 920

0.02x = 920 -800

0.02x = 120

x = 6,000

Therefore, the amount invested at the first year with 10% interest is 6,000 pesos. The amount invested the next year with 8% interest is 4,000 pesos.