Mathematics of Investment

MATHEMATICS

of

INVESTMENT

BY: DIGNA PUZON ANTONIO

SET A

(Simple interest)

1. How much is needed to settle a loan of ₱7,500 at 9 2/5 % due in 2 years and 6 months?

Let         P = ₱ 7.500                 r = 9 2/5% or 9. 4% or 0.094                t = 2.5 years

Solution:

Step 1.                I = prt

                          = (7,500) (.094) (2.5)

                          = ₱1762.50

Step 2.                 F= P + I

                          = 7,500 + 1762.50

                          = ₱ 9262.50 (amount needed to settle loan in its due

                                                date)

1. Find the interest on ₱ 9,000 at 10 ½ % from April 12 to November 12, 2016.

Let:          P = ₱ 9,000        

 r = 10.5 % or 0.105  

 t = from April 12 to November 12, 2016

 I=?

        Solution:

        Step 1.      2016                   11                12

                       2016 _                4                12

                                        7                0         (equivalent to 7 mos.)

        Step 2.                I = prt

                                  = (9,000) (0.105) (  )[pic 1]

                                  = ₱551.25      (interest gained within 7 mos.)

1. How long will it take ₱ 12.000 to accumulate to ₱ 17,280 at 11% simple interest rate?

Let:  F = ₱ 17,280                                r = 11% or 0.11

        P = ₱ 12,000                                t =?

Solution:

Step 1. Find interest.                I = F – P

                                          = ₱ 17,280 – ₱ 12,000

                                          = ₱ 5280   (accumulated interest)

Step 2. Find time.                        t =   (derivation of formula)[pic 2]

                                          = [pic 3]

                                          = [pic 4]

= 4 yrs. (time needed to accumulate

                   given amount )

1. What principal will accumulate to ₱ 12,812.50 on the third month if the interest rate is 10 %?

Let:          F = ₱ 12,812.50                r = 10% or 0.10                t = 3/12 or 0.25

Solution:

                                F = P (1 + r t)

                                P =         (derived formula from above)[pic 5]

                                   = [pic 6]

                                   = [pic 7]

                                   = ₱ 12,500  (principal that will accumulate)

1. Compute the payable amount for ₱ 30,000 loan at 10.5% after 90 days.

1. For ordinary interest

Let:         P = ₱ 30,000        r = 10.5% or 0.105                t = [pic 8]

Solution:         [pic 9]

                            = (30,000) (1 + (0.105) ()[pic 10]

                            = (30,000) (1.0625)

                            = ₱ 30,787.50    (payable amount after 90 days)

1. For exact interest

Let:         P = ₱ 30,000        r = 10.5% or 0.105                t =[pic 11]

Solution:                [pic 12]

                            = (30,000) (1 + (0.105) ()[pic 13]

                            = (30,000) (1.025890411)

                            = ₱ 30,776.71     (payable amount after 90 days)

1. Determine how much should be paid for a ₱15,000 loan after 2 years and 6 months at 15.5% simple interest.

Let:         P = ₱15,000                r = 15.5% or 0.155                t = 2.5 yrs

Solution:                [pic 14]

                            = (15,000) (1 + (0.155) (2.5))