Amount needed to be saved, A is 10000
Interest rate, r is 0.12% compounded monthly
Time, t is 3 years
Let the amount that needs to be deposited be P
So;
A=== P[[[ 1+rn]nt+\frac{r}{n}]^{nt}+nr]nt
P[[[ 1+0.121200]12×3+\frac{0.12}{1200}]^{12}\times^{3}+12000.12]12×3 === 10000
P[[[ 1+0.0001]36+0.0001]^{36}+0.0001]36 === 10000
1.000136^{36}36 P=== 10000
1.003606P=== 10000
P is 9964.069
Answer is Php9964.06
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