Question

An investment of ​$71,000 was made by a business club. The investment was split into three parts and lasted for one year. The first part of the investment earned​ 8% interest, the second​ 6%, and the third​ 9%. Total interest from the investments was $5490.  The interest from the first investment was 6 times the interest from the second. Find the amounts of the three parts of the investment.

Answer 1

Let

Amount of  first part of the investment =$x

Amount of  second part of the investment =$y

Amount of  third part of the investment =$z

Since investment was of ​$71,000, set up as

x+y+z=71000

Since,Total interest from the investments was $5490, set up as

0.08x +0.06y +0.09z=5490

Since, the interest from the first investment was 6 times the interest from the second

0.08x = 6 (0.06y)

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Solve the system of eqaution as

0.08x = 6 (0.06y) gives x = (6*0.06/0.08)y =4.5 y

put this in x+y+z=71000 to obtain

4.5y +y +z =71000

5.5y +z =71000

z =71000 -5.5y

Substitute z =71000 -5.5y and x=4.5 y in 0.08x +0.06y +0.09z=5490

0.08(4.5y) +0.06y +0.09(71000 -5.5y)=5490

6390 -0.075y =5490

-0.075y = 5490 -6390

-0.075y = -900

y= 900/0.075

y=12000

hence

x= 4.5 *12000 =54000

z=71000 -5.5 *12000 =5000

hence

Amount of  first part of the investment =$54000

Amount of  second part of the investment =$12000

Amount of  third part of the investment =$5000