Let
x = sum invested at greater interest
y = sum invested at lesser interest
i = interest of
y
4i = interest of
x
F=P(1+rt)
First investment:
F = P37,200, P = x, r = 4i, and t = 3 yrs
372=x(1+12i)
x=37,2001+12i
Second investment:
F = P24,000, P = y, r = i, and t = 10 yrs
240=y(1+10i)
y=24,0001+10i
x+y=50,000
37,2001+12i+24,0001+10i=50,000
3721+12i+2401+10i=500
372(1+10i)+240(1+12i)(1+12i)(1+10i)=500
372(1+10i)+240(1+12i)=500(1+12i)(1+10i)
(372+3,720i)+(240+2,880i)=500(1+22i+120i2)
612+6,600i=500+11,000i+60,000i2
60,000i2+4,400i−112=0
i=0.02 and −0.093
Use i = 0.02 and 4i = 0.08
x=37,2001+12(0.02)=P30,000
y=24,0001+10(0.02)=P20,000
Answer for Part 1: [ C ]
Answer for part 2: [ A ]