Question

An engineer deposits $8,000 in year 4,$8,500 in year 5 , and amounts increasing by $500 per year through year 10. At an interest rate of i% per year, the equation to find the present worth in year 0 is:

(a) [8000(P/Ai,7)+500(P/Gi,7)](P/Fi,3)
(b) [8000(P/Ai,7)+500(P/Gi,7)](P/Fi,4)
(c) [8000(P/Ai,6)+500(P/Gi,6)](P/Fi,3)
(d) [8000(P/Ai,6)+500(P/Gi,6)](P/Fi,4)

Answer 1

The correct equation to find the present worth in year 0 is (b) 8000(P/Ai,7)+500(P/Gi,7).

Step-by-step explanation:

Here's an explanation of the different parts of the equation:

The first part, 8000(P/Ai,7), represents the present worth of the $8,000 deposit made in year 4. P/A represents the present worth factor for an annuity, and the "i,7" represents the interest rate and the number of years for which the deposit is made.

The second part, 500(P/Gi,7), represents the present worth of the increasing annual deposits from year 5 to year 10. P/G represents the present worth factor for a growing annuity, and the "i,7" again represents the interest rate and the number of years for which the deposits are made.

The last part, (P/Fi,4), represents the present worth factor for a single future amount, in this case the total present worth of the two parts before.

This equation calculates the present worth of the deposits made in year 4 and year 5 and the increasing annual amounts through year 10, which, when multiplied by the present worth factor for a single future amount, gives us the present worth of the deposits in year 0.