Competency Standards: Solving financial problems using mathematical concepts Basic competency: Resolving single interest and compoun...

Competency Standards: Solving financial problems using mathematical concepts

Basic competency: Resolving single interest and compound interest in finance

Indicators:

### 1. Single interest is calculated and used in lending and capital systems

Example:

Wulan borrowed money from cooperatives amounting to Rp. 1.000.000, 00. After a month, Wulan must return the capital and interest amounting to Rp. 1.020.000, 00. Determine the size of flowers and their interest rates?

Interest = Rp. 1020.000, 00 – Rp. 1.000.000, 00 = Rp. 20.000, 00

Interest rate = 20,000, 00/1,000,000, 00 x 100% = 2%

B = Flower

i = interest rate per period

t = number of periods

M = Capital

A capital of Rp. 1,000,000, 00 in the flowers with a single interest rate of 2%/monan. Determine the flower size after 5 months?

M = Rp. 1.000.000, 00

i = 2%/month

t = 5 months

If a deposit money in the Bank then every end of the period, the interest earned is not taken, then the interest will be together capital into a new capital that will be flowering in the next period. The interest earned value will be greater than the interest in the previous period.

Ket:

Mn = Capital in the nth period (final capital)

M = Initial Capital

i = interest rate

n = Period

Capital amounting to Rp. 5.000.000, 00, with 10% compound interest/year. Determine the final capital and interest earned after 6 years?

M = Rp. 5.000.000, 00

i = 10%/year = 0.1/year

n = 6 years

= Rp. 5.000.000, 00 (1.1) ^ 6

= Rp. 5.000.000, 00 x 1.771561

= Rp. 8.857.805, 00

Indicators:

1. The Rente end value is calculated according to its type

Rente is a series of capital or instalments paid or received at any given time period which remains large.

Na = final Value

M = Capital

i = interest rate (%)

t = Period

Every beginning of the year, Niesa kept money in BRI Bank for Rp. 1.000.000, 00. If the bank gives interest 6%/year, determine the Niesa money after saving for 20 years?

M = Rp. 1.000.000, 00

i = 6%/year = 0.06/year

t = 20 years

= (1,000,000, 00 (1 + 0.06) ((〖 1 + 0.06) 〗 ^ 20-1))/0.06

= (1,000,000, 00 + 60,000, 00) (〖 1.06 〗 ^ 20-1)/0.06

= (1,060,000.00 (3,2071-1))/0.06

= (1,060,000.00 (2,2071))/0.06

= 2339563,6005/0.06

= 38,992,726.6757

Na = final Value

M = Capital

i = interest rate (%)

t = Period

Every year end, Diany stores its money in BNI Bank amounting to Rp. 800,000, 00for 25 years. If the bank gives interest 5%/year, specify the amount of Diany deposits?

M = Rp. 800,000, 00

i = 5%/year = 0.05/year

t = 25 years old

= (800,000.00 ((1 + 0.05) ^ 25-1))/0.05

= (800,000.00 (〖 1.05 〗 ^ 25-1))/0.05

= (800,000.00 (3,3863-1))/0.05

= (800,000.00 (2,3863))/0.05

= 1.909.083,9527/0.05

= 38,181,679.0543

M = Capital

i = interest rate (%)

t = Period

Filly will get a scholarship at the beginning of PT UNILEVER each month for Rp. 250,000, 000 for 3 years. If the scholarship will be awarded at the beginning of the first month with interest charged 2%/month, determine the amount of total scholarship that Filly received?

M = Rp. 250,000, 00

i = 2%/month = 0.02/month

t = 3 years = 36 months

= (250,000.00 (1 + 0.02) (1-(1 + 0.02) ^ (-36)))/0.02

= 250,000, 00 (1.02) (1-(1.02) ^ (-36))/0.02

= 250,000, 00 (1.02) (1-0,49022315)/0.02

= 255.000, 00 (0,50977685)/0.02

= 6,499,654.83

Nt = cash Value

M = Capital

i = interest rate (%)

t = Period

At the end of the month, Cinta Damai Foundation received a donation from the World Peace Agency amounting to Rp. 5.000.000, 00 for 3 consecutive years. If the donation will be given at once and subject to interest of 2%/month, determine the total donation received by the Peace Love Foundation?

M = Rp. 5.000.000, 00

i = 2%/month = 0.02/month

t = 3 years = 36 months

= (5,000,000.00 (1-(1 + 0.02) ^ (-36)))/0.02

= (5,000,000.00 (1-(1.02) ^ (-36)))/0.02

= (5,000,000.00 (1-0,49022315))/0.02

= (5,000,000.00 (0,50977685))/0.02

= 127,444,212.5

Indicators:

1. Annuity used in loan system

Annuities are the same amount of payments that are paid every certain period of time, consisting of part of interest and installment.

A = (M. I)/((〖 1-(1 + i) 〗 ^ (-t)))

A = Annuity

M = Capital/Loan

i = interest rate

t = Period

A1 = 1st Installment

A loan of Rp 10 million,-will be repaid with 3 installments with interest rate 12% per year. Determine great annuity.

M = Rp 10 million,-

i = 12% = 0.12

t = 3

= (10 million (0.12))/((1-(1 + 0.12) ^ (-3)))

= 1,200,000/((1-(1.12) ^ (-3)))

= 1,200,000/((1-0,711780))

= 1,200,000/0,28822

= 4,163,483.22

A loan of Rp. 15.000.00, 00 will be paid off for 11 months with a loan interest rate of 2% per month. Specify a large annuity and create a table of its angsurer plans.

M = Rp. 15.000.000

i = 2%/month = 0.02/month

t = 11 months

Then we determine the large interest in the 1st month namely by multiplying between interest rates with a large loan in the 1st month, ie = 2% x Rp. 15.000.000 = Rp. 300.000. So is next for the Big Flower 2nd month, 3rd month....

For the installment of 1st month obtained by reducing between large annuities with large interest in the 1st month, i.e. = Rp. 1.532.669 – Rp. 300.000 = Rp. 1232.669. So too for the big installment of the next month.

The remainder of the 1st month loan is obtained by decreasing between the big 1st month loan with a large 1st month instalment, i.e. = Rp. 15.000.000 – Rp. 1.232.669 = Rp. 13.767.331. So too for the next big loan remaining.

Large 2nd month loans were acquired from the big remaining 1st month loans. Likewise for large loans the next was gained from the large loan remaining months before.

Pay attention to the table plan.

Interest = 0.02 installments

1 RP 15 million Rp 300.000 0.02 Rp 1,232,669 Rp 13,767,331

2 RP 13,767,331 Rp 275.347 0.02 Rp 1,257,323 Rp 12,510,008

3 RP 12,510,008 Rp 250.200 0.02 Rp 1,282,469 Rp 11,227,539

4 RP 11,227,539 Rp 224.551 0.02 Rp 1,308,118 Rp 9,919,421

5 RP 9,919,421 Rp 198.388 0.02 Rp 1,334,281 Rp 8,585,140

6 RP 8,585,140 Rp 171.703 0.02 Rp 1,360,966 Rp 7,224,174

7 RP 7,224,174 Rp 144.483 0.02 Rp 1,388,186 Rp 5,835,988

8 RP 5,835,988 Rp 116.720 0.02 Rp 1,415,949 Rp 4,420,039

9 RP 4,420,039 Rp 88.401 0.02 Rp 1,444,268 Rp 2,975,771

10 RP 2,975,771 Rp 59.515 0.02 Rp 1,473,154 Rp 1,502,617

11 RP 1,502,617 Rp 30.052 0.02 Rp 1,502,617 Rp (0)

Total Rp 1,859,361 Rp 15 million

Indicators:

1. Depreciation is used in the value of an item

D = (A-S)/n

R = D/A x100%

D =

A =

S =

R =

n =

Interest is the service of lending.

Interest rate = interest/(first loan) x 100%

Settlement:

A single flower is a flower obtained at each end of a certain period that does not affect the amount of borrowed capital. The interest calculation of each period is always calculated based on fixed capital magnitude.

B = i x T x M

Ket:

Example:

Settlement:

Big interest = 2% x 5 x Rp. 1.000.000, 00 = Rp. 100.000, 00

So the amount of interest for 5 months is Rp. 100.000, 00

2. Compound interest is calculated and used in lending systems and capital

Note the illustration below:

M_n = M 〖 (1 + i) 〗 ^ n

Example:

Settlement:

M_6 = Rp. 5.000.000, 00 (1 + 0.1) ^ 6

Interest = Rp. 8.857.805, 00 – Rp. 5.000.000, 00 = Rp. 3.857.805, 00

So the final capital of Rp. 8.857.805, 00 and earn interest of Rp. 3.857.805, 00

Basic competency: Troubleshooting Rente in Finance

A. Rente Pra Numerando Final value

Na = (M (1 + i) 〖 ((1 + i) 〗 ^ T-1))/i

Ket:

Example:

Settlement:

Na = (M (1 + i) 〖 ((1 + i) 〗 ^ T-1))/i

So great savings of Niesa after 20 years is Rp. 38.992.726,6757

B. Rente Post Numerando Final value

Na = (M 〖 ((1 + i) 〗 ^ T-1))/i

Ket:

Example:

Settlement:

Na = (M 〖 ((1 + i) 〗 ^ T-1))/i

So big deposit Diany after 25 years is Rp. 38.181.679 0.05

2. Cash value of Rente calculated according to its type

A. Rente Pra Numerando Cash value

Nt = (M (1 + i) 〖 (1-(1 + i) 〗 ^ (-t)))/i

Ket:

Example:

Settlement:

Nt = (M (1 + i) 〖 (1-(1 + i) 〗 ^ (-t)))/i

So great scholarship that Filly received is Rp. 6,499,654.83

B. Rente Post Numerando Cash value

Nt = (M 〖 (1-(1 + i) 〗 ^ (-t)))/i

Ket:

Example:

Settlement:

Nt = (M 〖 (1-(1 + i) 〗 ^ (-t)))/i

So big donations received peace Love Foundation Rp. 127.444.212,50

Basic competency: Resolving annuity issues in the loan system

A = a_1 x 〖 (1 + i) 〗 ^ t

A_n = A_1 〖 (1 + i) 〗 ^ (T-1)

Ket:

Example:

Settlement:

A = (M. I)/((〖 1-(1 + i) 〗 ^ (-t)))

So big monthly annuality that must be paid Rp. 4.163.483

2. Annuity calculated in loan system

Compiling the Installment plan table

Note the example:

Settlement:

With annuity formula as previous example acquired great annuity is Rp. 1.532.669.

Month to-annuity loan Rp 1,532,669 time loan

Basic competency: Resolving Item Value depreciation

A. Straight line method

With:

B. Fixed percentage method of book value of method balance descending)

R = (1-√ (n&S/A)) x 100%

C. Production Output method

R = (A-S)/Q

D. Method of asset working hours (service hours method)

R = (A-S)/Q

E. Method number of years (sum of the Year's digits method)

2. Depreciation is calculated in the value of an item

Rewritter: Alber

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