# ENGECO FINAL

```M1S1: SIMPLE INTEREST
FORMULAS: I = Pin
I = INTEREST
P= PRINCIPAL AMOUNT
i = INTEREST RATE
n = YEAR
QUESTIONS:
1. A MAN DEPOSITED P5,000 IN A BANK AT A RATE OF 10% PER
ANNUM FOR 5 MONTHS AND 25 DAYS. FIND THE ORDINARY
SIMPLE INTEREST.
n = (5ππππ‘βπ  π₯
30πππ¦π
1ππππ‘β
+ 25) = 175days
I = (5,000)(0.10)(175π₯
1π¦πππ
360 πππ¦π
)
2. A MAN DEPOSITED P3,000 IN A BANK AT A RATE OF 10% PER
ANNUM FOR 10 MONTHS AND 25 DAYS. FIND THE ORDINARY
SIMPLE INTEREST.
n = (10ππππ‘βπ  π₯
30πππ¦π
1ππππ‘β
+ 25) = 325days
I = (3,000)(0.10)(325π₯
1π¦πππ
360 πππ¦π
)
3. A MAN DEPOSITED P3,000 IN A BANK AT A RATE OF 15% PER
ANNUM FOR 5 MONTHS AND 17 DAYS. FIND THE ORDINARY
SIMPLE INTEREST.
n = (5ππππ‘βπ  π₯
30πππ¦π
1ππππ‘β
+ 17) = 167days
I = (3,000)(0.15)(167π₯
1π¦πππ
360 πππ¦π
)
4. A MAN DEPOSITED P3,000 IN A BANK AT A RATE OF 15% PER
ANNUM FOR 8 MONTHS AND 9 DAYS. FIND THE
ACCUMULATED AMOUNT.
n = (8ππππ‘βπ  π₯
30πππ¦π
1ππππ‘β
I = (3,000)(0.15)(249π₯
+ 9) = 249days
1π¦πππ
360 πππ¦π
) = 311.25
F = P + I = 3,000 + 311.25
5. A MAN DEPOSITED P5,000 IN A BANK AT A RATE OF 10% PER
ANNUM FOR 10 MONTHS AND 25 DAYS. FIND THE
ACCUMULATED AMOUNT.
n = (10ππππ‘βπ  π₯
30πππ¦π
1ππππ‘β
I = (5,000)(0.10)(325π₯
+ 25) = 325days
1π¦πππ
360 πππ¦π
) = 451.3889
F = P + I = 5,000 + 451.3889
6. A MAN DEPOSITED P5,000 IN A BANK AT A RATE OF 12% PER
ANNUM FROM JANUARY 10,1999 TO NOVEMBER 18, 1999.
FIND THE EXACT SIMPLE INTEREST.
*COMPUTE THE DAYS FROM JAN 10, 99 TO NOV 18, 99*
I = (5,000)(0.12)(
312
) = 512.8767
365
F = 5,000 + 512.8767
7. A MAN DEPOSITED P5,000 IN A BANK AT A RATE OF 10% PER
ANNUM FOR 1 MONTHS AND 25 DAYS. FIND THE ORDINARY
SIMPLE INTEREST.
n = (1ππππ‘βπ  π₯
30πππ¦π
1ππππ‘β
+ 25) = 55days
I = (5,000)(0.10)(55π₯
1π¦πππ
360 πππ¦π
)
8. A MAN DEPOSITED P3,000 IN A BANK AT A RATE OF 15% PER
ANNUM FOR 5 MONTHS AND 4 DAYS. FIND THE ORDINARY
SIMPLE INTEREST.
n = (5ππππ‘βπ  π₯
30πππ¦π
1ππππ‘β
+ 4) = 154days
I = (3,000)(0.15)(154π₯
1π¦πππ
360 πππ¦π
)
9. A MAN DEPOSITED P1,000 IN A BANK AT A RATE OF 12% PER
ANNUM FROM JANUARY 10,1999 TO NOVEMBER 18, 1999.
FIND THE EXACT SIMPLE INTEREST.
*COMPUTE THE DAYS FROM JAN10, 99 TO NOV 18, 99*
I = (1,000)(0.12)(
312
) = 102.5753
365
F = 1,000 + 102.5753
M2S1: TYPES OF INTEREST RATES
FORMULA:
i = INTEREST RATE
i=
π
π
j = (i) (n)
j = NOMINAL RATE
n = METHOD OF COMPOUNDING
METHOD OF COMPOUNDING
ANNUALY
SEMI – ANNUALLY
QUARTERLY
BI – MONTHLY
MONTHLY
WEEKLY
NUMBER OF PERIOD PER YEAR
1
2
4
6
12
52
QUESTIONS:
1. AN INTEREST RATE OF 10% COMPOUNDED BI-MONTHLY IS
GIVEN, COMPUTE FOR THE INTEREST RATE PER PERIOD.
0.10
π₯ 100
6
π¨π΅πΊπΎπ¬πΉ: = 1.67%
2. AN EFFECTIVE RATE OF INTEREST WHICH IS 7%, IS
EQUIVALENT TO WHAT PERCENT IF COMPOUNDED
CONTINUOUSLY?
x = 0.07
0.07 = π π₯ − 1 *SHIFT SOLVE*
3. AN EFFECTIVE RATE OF INTEREST WHICH IS 12.75%, IS
EQUIVALENT TO WHAT PERCENT IF COMPOUNDED
CONTINUOUSLY?
x = 0.01275
0.01275= π π₯ − 1 *SHIFT SOLVE*
4. THE INTEREST RATE 10%. FIND THE EFFECTIVE RATE OF
INTEREST IF IT IS PAID SEVEN TIMES A YEAR.
(1 + 0.10)7 − 1
5. AN EFFECTIVE RATE OF INTEREST WHICH IS 10%, IS
EQUIVALENT TO WHAT PERCENT IF COMPOUNDED
CONTINUOUSLY?
x = 0.10
0.10= π π₯ − 1 *SHIFT SOLVE*
6. THE INETEREST RATE IS 10%, FIND THE EFFECTIVE RATE OF
INTEREST IF IT IS PAID FOUR TIMES A YEAR.
(1 + 0.10)4 − 1
7. AN EFFECTIVE RATE OF INTEREST WHICH IS 13%, IS
EQUIVALENT TO WHAT PERCENT IF COMPOUNDED
CONTINUOUSLY?
x = 0.13
0.13 = π π₯ − 1 *SHIFT SOLVE*
8. A FINANCING CHARGES 1.5% EVERY TWO MONTHS ON A
LOAN. FIND THE EQUIVALENT EFFECTIVE RATE OF INTEREST.
(1 + 0.015)12/2 − 1 ANSWER: = 9.34*
9. AN EFFECTIVE RATE OF INTEREST WHICH IS 8%, IS
EQUIVALENT TO WHAT PERCENT IF COMPOUNDED
CONTINUOUSLY?
x = 0.08
0.08 = π π₯ − 1 *SHIFT SOLVE*
10.
THE INETEREST RATE IS 10%, FIND THE EFFECTIVE RATE
OF INTEREST IF IT IS PAID THRICE TIMES A YEAR.
(1 + 0.10)3 − 1
M3S1: EQUIVALENT COMPOUNDING
FORMULAS:
π
π ππ (1 + )
π
ππ(1 + π)
π1 π1
π2 π2
(1 + ) = (1 + )
π1
π2
QUESTIONS:
1. WHAT IS THE EQUIVALENT RATE OF 10% COMPOUNDED
SEMI-ANNUALLY TO COMPOUNDED MONTHLY?
0.10 2
π₯ 12
(1 +
) = (1 + )
2
12
2. CONVERT 16% COMPOUNDED MONTHLY TO EQUIVALENT
NOMINAL RATE WHICH COMPOUNDED DAILY.
12
365
0.16
π₯
(1 +
)
= (1 +
)
12
365
3. A 15% COMPOUNDED SEMI-ANNUALLY HAS AN EQUIVALENT
RATE OF 4.9395%, HOW MANY TIMES DOES IT PAY EVERY
YEAR?
0.15 2
(1 +
) = (1 + 0.049395)π₯
2
4. WHAT IS THE EQUIVALENT RATE OF 10% COMPOUNDED
SEMI-ANNUALY TO COMPOUNDED WEEKLY?
0.10 2
π₯ 52
(1 +
) = (1 + )
2
52
M3S2: COMPOUND INTEREST
FORMULA:
j −yn
P = F (1 + )
n
j yn
P = F (1 + ) − P
n
π ππ¦
π = (1 + )
π
QUESTIONS:
1. IF THE ACCUMULATE AMOUNT IS P9,500 AT 5%
COMPOUNDED MONTHLY. CALCULATE THE PRINCIPAL
AMOUNT AFTER 2 YEARS.
0.05 −2(12)
P = 9,500 (1 +
)
12
2. HOW LONG IN YEARS WILL A CERTAIN SUM OF MONEY TO
DOUBLES ITS AMOUNT WHEN DEPOSITED AT A RATE OF 8%
COMPOUNDED BI-MONTHLY?
0.08 6π¦
2 = (1 +
)
6
3. IF THE ACCUMULATE AMOUNT IS P5,500 AT 9%
COMPOUNDED MONTHLY. CALCULATE THE PRINCIPAL
AMOUNT AFTER 2 YEARS.
0.09 −2(12)
P = 5,500 (1 +
)
12
4. HOW LONG IN YEARS WILL A CERTAIN SUM OF MONEY TO
TRIPLES ITS AMOUNT WHEN DEPOSITED AT A RATE OF 10%
COMPOUNDED MONTHLY?
0.10 12π¦
3 = (1 +
)
12
5. HOW LONG IN YEARS WILL A CERTAIN SUM OF MONEY TO
DOUBLES ITS AMOUNT WHEN DEPOSITED AT A RATE OF 6%
COMPOUNDED BI-MONTHLY?
0.06 6π¦
2 = (1 +
)
6
6. HOW LONG IN YEARS WILL A CERTAIN SUM OF MONEY TO
TRIPLES ITS AMOUNT WHEN DEPOSITED AT A RATE OF 9%
COMPOUNDED MONTHLY?
0.09 12π¦
3 = (1 +
)
12
7. HOW LONG IN YEARS WILL A CERTAIN SUM OF MONEY TO
TRIPLES ITS AMOUNT WHEN DEPOSITED AT A RATE OF 7%
COMPOUNDED MONTHLY?
0.07 12π¦
3 = (1 +
)
12
8. IF THE ACCUMULATED AMOUNT IS P11,500 AT 4%
COMPOUNDED ANNUALY. CALCULATE THE PRINCIPAL
AMOUNT AFTER 7 YEARS.
0.04 −7(1)
P = 11,500 (1 +
)
1
9. HOW LONG IN YEARS WILL A CERTAIN SUM OF MONEY TO
TRIPLES ITS AMOUNT WHEN DEPOSITED AT A RATE OF 2%
COMPOUNDED MONTHLY?
0.02 12π¦
3 = (1 +
)
12
10. HOW LONG IN YEARS WILL A CERTAIN SUM OF MONEY
TO TRIPLES ITS AMOUNT WHEN DEPOSITED AT A RATE OF 3%
COMPOUNDED MONTHLY?
0.03 12π¦
3 = (1 +
)
12
M4S1: TIME VALUE OF MONEY
FORMULA:
F = P(1 + i)yn
P = F(1 + i)−yn
QUESTIONS:
1. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P55,000 AT THE END
OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 55,000(1 + 0.10)9(1) = 129687.123
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
2. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P35,000 AT THE END
OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 35,000(1 + 0.10)9(1) = 82528.16919
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
3. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P30,000 AT THE END
OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 30,000(1 + 0.10)9(1) = 70738.43073
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
4. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P40,000 AT THE END
OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 40,000(1 + 0.10)9(1) = 94317.90764
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
5. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P45,000 AT THE END
OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 45,000(1 + 0.10)9(1) = 106107.6461
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
6. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P20,000 AT THE END
OF 1ST YEARS, P20,000 AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 20,000(1 + 0.10)9(1) = 47158.95382
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
7. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P60,000 AT THE END
OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 60,000(1 + 0.10)9(1) = 141476.8615
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
8. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P15,000 AT THE END
OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 15,000(1 + 0.10)9(1) = 35369.21537
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
9. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD IN
INSTALLMENT BASIS. FIRST PAYMENT IS P50,000 AT THE END
OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS, P30,000 AT
THE END OF 6TH YEAR AND P8,000 AT THE END OF 8TH YEAR.
HOW MUCH IS THE TOTAL PAYMENT OF ENGR. SALVADOR
AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 50,000(1 + 0.10)9(1) = 117897.3846
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
M4S2: PRINCIPLE OF DISCOUNTING
FORMULA:
d=
i
1+i
i=
d
1−d
1. A 7% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE INTEREST
RATE.
i=
0.07
1−0.07
2. A 2% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE INTEREST
RATE.
i=
0.02
1−0.02
3. A 8% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE INTEREST
RATE.
i=
0.08
1−0.08
4. A 1% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE INTEREST
RATE.
i=
0.01
1−0.01
5. A 4% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE INTEREST
RATE.
i=
0.04
1−0.04
6. A 10% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE
INTEREST RATE.
i=
0.10
1−0.10
7. A 5% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE INTEREST
RATE.
i=
0.05
1−0.05
8. A 6% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE INTEREST
RATE.
i=
0.06
1−0.06
9. A 9% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE INTEREST
RATE.
i=
0.09
1−0.09
10. A 3% DISCOUNT RATE IS GIVEN. COMPUTE FOR THE
INTEREST RATE.
i=
0.03
1−0.03
11. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD
IN INSTALLMENT BASIS. FIRST PAYMENT IS P60,000 AT THE
END OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS,
P30,000 AT THE END OF 6TH YEAR AND P8,000 AT THE END OF
8TH YEAR. HOW MUCH IS THE TOTAL PAYMENT OF ENGR.
SALVADOR AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 60,000(1 + 0.10)9(1) = 141476.8615
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
12. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD
IN INSTALLMENT BASIS. FIRST PAYMENT IS P35,000 AT THE
END OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS,
P30,000 AT THE END OF 6TH YEAR AND P8,000 AT THE END OF
8TH YEAR. HOW MUCH IS THE TOTAL PAYMENT OF ENGR.
SALVADOR AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 35,000(1 + 0.10)9(1) = 82528.16919
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
13. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD
IN INSTALLMENT BASIS. FIRST PAYMENT IS P20,000 AT THE
END OF 1ST YEARS, P20,000 AT THE END OF 3RD YEARS,
P30,000 AT THE END OF 6TH YEAR AND P8,000 AT THE END OF
8TH YEAR. HOW MUCH IS THE TOTAL PAYMENT OF ENGR.
SALVADOR AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 20,000(1 + 0.10)9(1) = 47158.95382
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
14. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD
IN INSTALLMENT BASIS. FIRST PAYMENT IS P30,000 AT THE
END OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS,
P30,000 AT THE END OF 6TH YEAR AND P8,000 AT THE END OF
8TH YEAR. HOW MUCH IS THE TOTAL PAYMENT OF ENGR.
SALVADOR AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 30,000(1 + 0.10)9(1) = 70738.43073
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
15. ENGR. SALVADOR HAS TO PAY HER BILL IN CREDIT CARD
IN INSTALLMENT BASIS. FIRST PAYMENT IS P15,000 AT THE
END OF 1ST YEARS, P20,000AT THE END OF 3RD YEARS,
P30,000 AT THE END OF 6TH YEAR AND P8,000 AT THE END OF
8TH YEAR. HOW MUCH IS THE TOTAL PAYMENT OF ENGR.
SALVADOR AFTER 10 YEARS AT A RATE OF 10% PER ANNUM?
*n = CALCULATE CURRENT YEAR TO FOCAL DATE*
F1 = 15,000(1 + 0.10)9(1) = 35369.21537
F3 = 20,000(1 + 0.10)7(1) = 38974.342
F6 = 30,000(1 + 0.10)4(1) = 43923
F8 = 8,000(1 + 0.10)2(1) = 9680
M5S1: ORDINARY ANNUITY
FORMULA:
FUTURE WORTH:
j yn
(1 + ) − 1
n
A[
]
j
n
PRESENT WORTH:
j −yn
1 − (1 + )
n
A[
]
j
n
QUESTIONS:
1. HOW MUCH DO EIGHT P64,055 QUARTERLY PAYMENTS
AMOUNT AT PRESENT, IF THE INTEREST RATE IS 2
COMPOUNDED QUARTERLY?
64055 [
0.02 −8
)
4
0.02
4
1−(1+
]
2. IF MONEY IS WORTH 9.94% ANNUALLY, WHAT ANNUAL
PAYMENT IS REQUIRED TO RAISE P15,043 AFTER 1
YEAR?15043
15043 = x [
(1+0.0994)1 −1
0.0994
]
3. HOW MUCH MONEY MUST YOU INVEST TODAY IN ORDER TO
WITHDARW P15,864 ANNUALLY FOR 4 YEARS IF THE
INTEREST RATE IS 4%.57584.6578
15864 [
1−(1+0.04)−4
0.04
]
4. A P72,179 DEBT IS TO BE PAID OFF IN EIGHT EQUAL
QUARTERLY PAYMENTS AT 7% COMPOUNDED QUARTERLY.
WHAT SHOULD BE THE AMOUNT OF EACH PAYMENT?
72179 = x [
0.07 −8
)
4
0.07
4
1−(1+
]
5. A MAN SECURED A P7,901 LOAN AT 3% INTEREST RATE. FIND
THE ANNUAL AMORTIZATION TO EXTINGUISH IT IN 1 YEAR.
7901 = x [
1−(1+0.03)−1
0.03
]
6. WHAT IS THE ACCUMULATED AMOUNT OF A 7 YEAR
ANNUITY PAYING P1,520 AT THE END OF EACH YEAR, WITH
INTEREST RATE OF 2% COMPOUNDED ANNUALLY?
1520 [
(1+0.02)7 −1
0.02
]
7. A GASOLINE ENGINE IS AVAILABLE ON INSTALLMENT BASIS
WITH A DOWN PAYMENT OF P11,100 AND P18,070 AT THE
END OF EACH MONTH FOR ONE YEAR. WHAT IS THE CASH
PRICE OF THE ENGINE IF THE INTEREST IS SET AT 6.4%
COMPOUNDED MONTHLY?
18070 [
0.064 −12
)
12
0.064
12
1−(1+
] + 11100
8. MONEY BORROWED TODAY IS TO BE PAID IN 6 EQUAL
PAYMENTS AT THE END OF EACH QUARTER. IF THE INTEREST
IS 2% COMPOUNDED QUARTERLY, HOW MUCH WAS INIATLLY
BORROWED IF QUARTERLY PAYMENT OF P27,928?
27928 [
0.02 −6
)
4
0.02
4
1−(1+
]
9. AT 2% INTEREST RATE, HOW MUCH SHOULD YOU INVEST
TODAY TO BE ABLE TO WITHDARW P1,250 ANNUALLY FOR 5
YEARS?
1250 [
1−(1+0.02)−5
0.02
]
10. THE PRESIDENT OF A GROWING FIRM WISHES TO GIVE
EACH OF 13 EPLOYEES A HOLIDAY BONUS. HOW MUCH IS
NEEDED TO INVEST MONTHLY FOR A YEAR AT 4% NOMINAL
INTEREST RATE, COMPOUNDED MONTHLY, SO THAT EACH
1000(13) = x [
0.04 12
) −1
12
0.04
12
(1+
]
M5S2: DUE ANNUITY
FORMULA:
FUTURE WORTH:
j yn
(1 + ) − 1
π
n
A[
] (1 + )
j
π
n
PRESENT WORTH:
j −yn
1 − (1 + )
π
n
A[
] (1 + )
j
π
n
QUESTIONS:
1. A GASOLINE ENGINE IS AVAILABLE ON INSTALLMENT BASIS
WITH A DOWN PAYMENT OF P10,688 AND P2,328 AT THE
END OF EACH MONTH FOR ONE YEAR. WHAT IS THE CASH
PRICE OF THE ENGINE IF THE INTEREST IS SET AT 9%
COMPOUNDED MONTHLY IF THE PAYMENT ARE DONE EVERY
BEGINNING OF THE MONTH?
0.09 −12
1 − (1 +
)
0.09
12
2328 [
] + 10686 (1 +
)
0.09
12
12
2. HOW MUCH DO TEN P6,945 QUARTERLY PAYMENTS
AMOUNT AT PRESENT, IF THE INTEREST RATE IS 3
COMPOUNDED QUARTERLY IF THE PAYMENT ARE DONE
EVERY BEGINNING OF THE MONTH?
6945 [
0.03 −10
)
4
0.03
4
1−(1+
] (1 +
0.03
4
)
3. A P20,854 DEBT IS TO BE PAID OFF IN EIGHT EQUAL
QUARTERLY PAYMENTS AT 4% COMPOUNDED QUARTERLY.
WHAT SHOULD BE THE AMOUNT OF EACH PAYMENT IF THE
PAYMENT ARE DONE EVERY BEGINNING OF THE MONTH?
20854 = x [
0.04 −8
)
4
0.04
4
1−(1+
] (1 +
0.04
4
)
4. HOW MUCH MONEY MUST YOU INVEST TODAY IN ORDER TO
WITHDARW P12,156 ANNUALLY FOR 4 YEARS IF THE
INTEREST RATE IS 4.23% IF THE PAYMENT ARE DONE EVERY
BEGINNING OF THE MONTH?
12156 [
1−(1+0.0423)−4
0.0423
] (1 + 0.0423)
5. MONEY BORROWED TODAY IS TO BE PAID IN 9 EQUAL
PAYMENTS AT THE END OF EACH QUARTER. IF THE INTEREST
IS 2% COMPOUNDED QUARTERLY, HOW MUCH WAS INIATLLY
BORROWED IF QUARTERLY PAYMENT OF P20,160 IF THE
PAYMENT ARE DONE EVERY BEGINNING OF THE MONTH?
20160 [
0.02 −9
)
4
0.02
4
1−(1+
] (1 +
0.02
4
)
6. AT 4% INTEREST RATE, HOW MUCH SHOULD YOU INVEST
TODAY TO BE ABLE TO WITHDARW P4,325 ANNUALLY FOR 5
YEARS IF THE PAYMENT ARE DONE EVERY BEGINNING OF THE
MONTH?
4325 [
1−(1+0.04)−5
0.04
] (1 + 0.04)
7. A MAN SECURED A P9,556 LOAN AT 2% INTEREST RATE. FIND
THE ANNUAL AMORTIZATION TO EXTINGUISH IT IN 3 YEAR IF
THE PAYMENT ARE DONE EVERY BEGINNING OF THE
MONTH?
9556 = π₯ [
1−(1+0.02)−3
0.02
] (1 + 0.02)
8. THE PRESIDENT OF A GROWING FIRM WISHES TO GIVE EACH
OF 15 EPLOYEES A HOLIDAY BONUS. HOW MUCH IS NEEDED
TO INVEST MONTHLY FOR A YEAR AT 4% NOMINAL INTEREST
RATE, COMPOUNDED MONTHLY, SO THAT EACH EMPLOYEE
WILL RECEIVE P1,000 BONUS IF THE PAYMENT ARE DONE
EVERY BEGINNING OF THE MONTH?
0.04 12
(1 +
) −1
0.04
12
1000(15) = x [
] (1 +
)
0.04
12
12
9. WHAT IS THE ACCUMULATED AMOUNT OF A 8 YEAR
ANNUITY PAYING P1,790 AT THE END OF EACH YEAR, WITH
INTEREST RATE OF 5% COMPOUNDED ANNUALLY IF THE
PAYMENT ARE DONE EVERY BEGINNING OF THE MONTH?
1790 [
(1+0.05)8 −1
] (1 + 0.05)
0.05
10. IF MONEY IS WORTH 9% ANNUALLY, WHAT ANNUAL
PAYMENT IS REQUIRED TO RAISE P17,151 AFTER 1 YEAR IF
THE PAYMENT ARE DONE EVERY BEGINNING OF THE
MONTH?
17151 = x [
(1+0.09)1 −1
0.09
] (1 + 0.09)
11. HOW MUCH DO EIGHT P64,055 QUARTERLY PAYMENTS
AMOUNT AT PRESENT, IF THE INTEREST RATE IS 2
COMPOUNDED QUARTERLY?
64055 [
0.02 −8
)
4
0.02
4
1−(1+
]
12. AT 5% INTEREST RATE, HOW MUCH SHOULD YOU INVEST
TODAY TO BE ABLE TO WITHDARW P3,199 ANNUALLY FOR 7
YEARS?
3199 [
1−(1+0.05)−7
0.05
]
13. THE PRESIDENT OF A GROWING FIRM WISHES TO GIVE
EACH OF 15 EPLOYEES A HOLIDAY BONUS. HOW MUCH IS
NEEDED TO INVEST MONTHLY FOR A YEAR AT 4% NOMINAL
INTEREST RATE, COMPOUNDED MONTHLY, SO THAT EACH
1000(15) = x [
0.04 12
) −1
12
0.04
12
(1+
]
14. MONEY BORROWED TODAY IS TO BE PAID IN 7 EQUAL
PAYMENTS AT THE END OF EACH QUARTER. IF THE INTEREST
IS 1% COMPOUNDED QUARTERLY, HOW MUCH WAS INIATLLY
BORROWED IF QUARTERLY PAYMENT OF P23,758?
23758 [
0.01 −7
)
4
0.01
4
1−(1+
]
15. IF MONEY IS WORTH 9.73% ANNUALLY, WHAT ANNUAL
PAYMENT IS REQUIRED TO RAISE P15,434 AFTER 4 YEAR?
15434 = x [
(1+0.0973)4 −1
0.0973
]
M6S1: DEFERRED ANNUITY
FORMULA:
i yn
π¦π
(1 + ) − 1
π
n
F = A[
] (1 + )
i
π
n
i −yn
−yn
1 − (1 + )
i
n
P = A[
] (1 + )
i
n
n
QUESTIONS:
1. A CONDOMINIUM UNIT CAN BE BOUGHT AT A
DOWNPAYMENT OF β±150,000 AND A MONTHLY PAYMENT
OF β±23,347 FOR 10 YEARS STARTING AT THE END OF 5TH YEAR
FROM THE DATE OF PURCHASE. IF MONEY IS WORTH 3%
COMPOUNDED MONTHLY, WHAT IS THE CASH PRICE OF THE
CONDOMINIUM UNIT?
0.03 −12(10)
−12(5)
1 − (1 +
)
0.03
12
P = 23347 [
] (1 +
)
+ 150000
0.03
12
12
2. ENGR. ODON IS PAYING A LOAN β±5,000 PESOS 2%
QUARTERLY FOR 3 YEARS. IF THE PAYMENT IS TO START
AFTER 6 YEARS. WHAT IS THE VALUE TODAY?
P = 5000 [
1−(1+0.02)−4(3)
0.02
] (1 + 0.02)−4(6)
3. YOU ARE TO PAY A BILL IN MERALCO EVERY MONTH OF 5
YEARS. THE ACCUMULATED AMOUNT OF THE PAYMENT YOU
PAID IS β±70,223 AT A RATE OF 4% PER YEAR. THE FIRST
PAYMENT IS DONE AFTER 4 YEARS. HOW MUCH ARE YOU
PAYING EVERY PAYMENT?
70223 = A [
0.04 12(5)
)
−1
12
0.04
12
(1+
]
4. SEVERAL PAYMENTS ARE TO BE MADE BY A PERSON WITH A
LOT OF DEBT.
1ST PAYMENT: β±2,000 COMPOUNDED QUARTERLY AT 6%
AFTER 2 YEARS
2ND PAYMENT: β±4,000 PAYABLE COMPOUNDED SEMI
ANUALLY AT 1% FOR 3 YEARS EVERY END OF THE MONTH.
3RD PAYMENT: β±3,000 COMPOUNDED ANNUALLY AT 5% FOR
4 YEARS PAYABLE EVERY BEGINNING OF THE MONTH.
HOW MUCH MONEY TODAY MUST BE SET ASIDE TO COVER
ALL THE PERSONS DEBT? NOTE: PAYMENTS ARE MADE
SEQUENTIALLY.
0.06 −4(2)
1 − (1 +
)
4
P1 = 2000 [
] = 14971.8502
0.06
4
0.01 −2(3)
1 − (1 +
)
0.01 −2(2)
2
P2 = 4000 [
] [(1 +
)
] = 23119.6648
0.01
2
2
−4
1 − (1 + 0.05)
P3 = 3000 [
] (1 + 0.05)(1 + 0.05)−(3+2) = 8751.7868
0.05
5. FIND THE PRESENT VALUE OF A CONTRACT THAT REQUIRES
YEARLY PAYMENTS OF β±11,928 FOR 20 YEARS. MONEY IS
WORTH: 1.45% FOR THE 1ST 8 YEARS AND 9.14% FOR THE
LAST 12 YEARS.
−8
1 − (1 + 0.0145)
P1 = 11928 [
] = 89486.92222
0.0145
−12
1 − (1 + 0.0914)
P2 = 11928 [
0.0914
] [(1 + 0.0914)−8 ] = 42130.4511
ANSWER: P1 + P2 = 131617.3733
6. ENGR. ODON IS PAYING A LOAN β±5,000 4% MONTHLY FOR 3
YEARS. IF THE PAYMENT IS TO START AFTER 6 YEARS. WHAT
IS THE AFTER 14 YEARS OF THE LAST PAYMENT?
−12(3)
1 − (1 + 0.04)
P = 5000 [
0.04
] [(1 + 0.04)−12(6) ][(1 + 0.04)−12(3+6+14) ]
7. THE PRESENT VALUE OF AN ANNUITY IS PAYABLE ANNUALLY
FOR 3 YEARS, WITH THE 1ST PAYMENT AT THE END OF 10
YEARS IS β±36,281. IF MONEY IS WORTH 3.30%. WHAT IS THE
VALUE OF ANNUITY?
1 − (1 + 0.033)
36281 = A [
0.033
−(3)
] (1 + 0.033)−(10)
8. FIND THE PRESENT WORTH OF A SERIES OF EQUAL
QUARTERLY PAYMENTS OF β±750.00 FOR 18 YEARS WITH THE
FIRST PAYMENT STARTING THREE MONTHS FROM NOW.
THE APPLIED INTERST ARE:
2% COMPOUNDED QUARTERLY FOR THE 1ST 5 YEARS
2% COMPOUNDED QUARTERLY FOR THE NEXT 5 YEARS
3% COMPOUNDED QUARTERLY FOR THE REMAINING 8 YEARS
0.02 −4(5)
1 − (1 +
)
4
P1 = 750 [
] = 14240.5644
0.02
4
0.02 −4(5)
1 − (1 +
)
0.02 −4(5)
4
P2 = 750 [
] [(1 +
)
] = 12888.6065
0.02
4
4
0.03 −4(8)
1 − (1 +
)
0.03 −4(5+5)
4
P3 = 750 [
] [(1 +
)
] = 15772.4328
0.03
4
4
9. ENGR. VELASCO DECIDED TO SUBSCRIBE IN A GLOBE PROMO
WHICH MUST BE PAID β±3,004 PESOS PER MONTH FOR 5
YEARS. HOWEVER THE PAYMENT WILL START 3 YEARS AFTER
FROM TODAY. IF THE PAYMENT IF 4% COMPOUNDED
MONTHLY. HOW MUCH IS THE VALUE OF THE MONEY
TODAY.
0.04
P = 3004 [
1−(1+ 12 )
0.04
12
−12(5)
] [(1 +
0.04 −12(3)
12
)
]
10. FIND THE PRESENT WORTH OF A SERIES OF EQUAL
QUARTERLY PAYMENTS OF β±750.00 FOR 16 YEARS WITH THE
FIRST PAYMENT STARTING THREE MONTHS FROM NOW.
THE APPLIED INTERST ARE:
2% COMPOUNDED QUARTERLY FOR THE 1ST 5 YEARS
3% COMPOUNDED QUARTERLY FOR THE NEXT 4 YEARS
5% COMPOUNDED QUARTERLY FOR THE REMAINING 7
YEARS.
0.02 −4(5)
1 − (1 + 4 )
P1 = 750 [
] = 14240.5644
0.02
4
0.03 −4(4)
1 − (1 +
)
0.03 −4(5)
4
P2 = 750 [
] [(1 +
)
] = 9704.0892
0.03
4
4
0.05 −4(7)
1 − (1 +
)
0.05 −4(5+4)
4
P3 = 750 [
] [(1 +
)
] = 11270.7938
0.05
4
4
M7S1: PERPETUITY ANNUITY &amp; CONTINUOUS COMPOUND
INTEREST
FORMULA:
PERPETUITY ANNUITY:
P=
A
i
CONTINUOUS COMPOUND INTEREST:
1 − e−ry
P = A[ r
]
e −1
ery − 1
F = A[ r
]
e −1
QUESTIONS:
1. A BUILDING TO BE PAID INDEFINITELY AT THE RATE OF 5%.
THE BUILDING’S PRICE IS P1M IF PAID IN CASH. WHAT IS THE
AMOUNT OF THE BUILDING TODAY FOR 12YEARS AT THE
RATE OF 4.83% COMPOUNDED CONTINUOUSLY?
A = 1M(0.05) [
1−e−0.0483(12)
e0.0483 −1
]
2. A TRUCK IS PAYABLE P44,223 FOR 7 YEARS AT THE RATE OF
4% COMPOUNDED CONTINUOUSLY. CALCULATE THE FUTURE
WORTH?
F = 44223 [
e0.04(7) −1
e0.04 −1
]
3. GIVEN THAT THE PRESENT WORTH OF CAR IS P32,969 WITH
AN INTEREST OF 2%. HOW MUCH IS THE ANNUITY OF IT IS
PAID FOREVER?
π΄ = 32969(0.02)
4. AN AMOUNT OF P1,888 IS TO BE PAID INDEFINITELY AT THE
RATE OF 1.22%. HOW MUCH IS THE PRESENT WORTH?
π=
1888
0.0122
5. AN ANNUITY OF P17,089 IS TO BE PAID IN 5 YEARS AT THE
RATE 5% COMPOUNDED CONTINUOUSLY. WHAT IS THE
PRESENT WOTH?
P = 17089 [
1−e−0.05(5)
e0.05 −1
]
6. FIFTHY THOUSAND PESOS IS DEPOSITED ANNUALLY AT AN
INREST RATE OF 7% PER ANNUM COMPOUNDED
CONTINUOUSLY, HOW MANY YEARS WILL IT ACCUMMULATE
TO P455,324?
455324 = 50000 [
e0.07(x) −1
]
e0.07 −1
7. IF MONEY IS WORTH 0.05% DETERMINE THE PRESENT VALUE
OF PERPETUITY OF P4,154 PAYABLE ANNUALLY, WITH THE
FIRST PAYMENT DUE AT THE END OF FIVE YEARS?
π=
4154
0.0005
(1 + 0.0005)5
8. IF AN ELECTRIC APPLIANCE PRESENT WORTH IS P11,658 AND
IS BEING PAID P2,734 INDEFINITELY, CALCULATE THE
INTEREST RATE.
π=
2734
11658
9. A MAN DEPOSITS P58,454 EACH YEAR INTO HAS SAVINGS
ACCOUNT THAT PAYS 2% NOMINAL INTEREST COMPOUNDED
CONTINUOUSLY. HOW MUSH WILL BE THE WORTH OF THE
ACCOUNT AT THE END OF 5 YEARS?
F = 58454 [
e0.02(5) −1
e0.02 −1
]
10. A PRODUCT IS TO BE PAID PERPETUALLY FOR P10,551 AT
THE RATE OF 1%. HOW MUCH IS THE PRESENT WORTH OF
THE PRODUCT?
π=
10551
0.01
1. THE MAINTENANCE COST FOR A SEWING MACHINE AFTER
ONE YEAR IS EXPECTED TO BE P500. THE COST WILL INCREASE
BY P50 EACH YEAR FOR THE SUBSEQUENT 9 YEARS. THE
INTEREST IS 11% COMPOUNDED ANNUALLY. WHAT IS THE
PRESENT WORTH OF THE MAINTENANCE FOR THE MACHINE
OVER THE FULL 10 YEAR.
10
π = ∑ ((500 + 50(π₯ − 1))(1 + 0.11)−π₯ )
π₯=1
2. A MAN PAYS HIS DEBT IN THE FOLLOWING MANNER: P1,000
AFTER 1 YEAR, P900 AFTER 2 YEARS, P800 AFTER 3 YEARS
AND SO ON UP TO 6TH YEAR. FIND THE ACCUMULATE
AMOUNT OF THESE PAYMENTS BUT THE RATE OF 10%
COMPOUNDED ANNUALLY.
G = 1000 - 900 = 100
*DECREASING*
6
π = ∑ ((1000 + 100(π₯ − 1))(1 + 0.11)6−π₯ )
π₯=1
3. AN ANNUAL MAINTENANCE COST A GEENRATOR P1,000
AFTER 1 YEAR AND IT IS ESTIMATED TO INCREASE BY 10%
EACH YEAR FOR THE SUBSEQUENT 7 YEARS. FIND THE
PRESENT WORTH OF THE MAINTENANCE COST IF THE RATE
OF INTEREST IS 15% COMPOUNDED ANNUALLY.
1000
0.15−0.10
[1 − (
1+0.10 7+1
)
1+0.15
]
M9S1: STRAIGHT LINE METHOD
FORMULA:
DEPRECIATION @ YEAR n:
d=
C0 − CL
L
TOTAL DEPRECIATION @ YEAR n:
Dn = dn
BOOK VALUE:
Cn = CO − dn
QUESTIONS:
1. A HYDRAULIC PRESS COATING P125,000 HAS ESTIMATED
LIFE OF 6 YEARS WITH A BOOK VALUE OF P20,000 AT THE
END OF THE PERIOD. COMPUTE THE DEPRECIATION PER
YEAR USING STRAIGHT LINE METHOD.
d=
125k−20k
6
2. A HEAVY DUTY CONSTRUCTION EQUIPMENT COSTING
P500,000 HAS AN ESTIMATED LIFE OF 25 YEARS WITH
BOOK VALUE OF P100,000 AT THE END OF THE PERIOD.
COMPUTE ITS BOOK VALUE AFTET 20 YEARS USING
STRAIGHT LINE METHOD.
500k − 100k
d=
= 16π
25
C20 = 500k − (16k)(20)
3. A BULLDOZER HAS AN INITIAL COST OF P2,000,000. IT
SALVAGE VALUE AFTER 10 YEARS IS P255,000. AS A
PERCENTAGE OF THR INITIAL COST, WHAT IS THE STRAIGHT
LINE DEPRECIATION RATE OF THE EQUIPMENT?
2π − 250π
= 175000
10
175000
π₯100
2π
4. A BROADCASTING EQUIPMENT COSTING P200,000 HAS AN
ESTIMATED LIFE OF 15 YEARS WITH A BOOK VALUE OF
P25,000 AT THE END OF THE PERIOD. COMPUTE ITS BOOK
VALUE AFTER 11 YEARS USING STRAIGHT LINE METHOD.
d=
200k − 25k
= 11666.6667
15
C11 = 200k − (11666.6667)(11)
5. A DATE SERVER IS PURCHASED P450,000. THE SALVAGE
VALUE IN 10 YEARS IS P25,000. WHAT IS THE TOTAL
DERECIATION IN THE FIRST 3 YEARS USING STRAIGHT LINE
METHOD?
450π − 25k
d=
= 42500
10
D3 = (42500)(3)
6. A DATE SERVER IS PURCHASED P450,000. THE SALVAGE
VALUE IN 18 YEARS IS P25,000. WHAT IS THE TOTAL
DERECIATION IN THE FIRST 15 YEARS USING STRAIGHT LINE
METHOD?
450π − 25k
d=
= 23611.1111
18
D15 = (23611.1111)(15)
7. A BULLDOZER HAS AN INITIAL COST OF P2,000,000. IT
SALVAGE VALUE AFTER 10 YEARS IS P175,000. AS A
PERCENTAGE OF THR INITIAL COST, WHAT IS THE STRAIGHT
LINE DEPRECIATION RATE OF THE EQUIPMENT?
2π − 175π
= 182500
10
199982.5
π₯100
2π
8. A BROADCASTING EQUIPMENT COSTING P200,000 HAS AN
ESTIMATED LIFE OF 15 YEARS WITH A BOOK VALUE OF
P25,000 AT THE END OF THE PERIOD. COMPUTE ITS BOOK
VALUE AFTER 7 YEARS USING STRAIGHT LINE METHOD.
200k − 25k
d=
= 11666.6667
15
C7 = 200k − (11666.6667)(7)
M9S2: SINKING FUND METHOD
FORMULA:
CONSTANT DEPRECIATION:
(CO − CL )(i)
d=
(1 + i)L − 1
DEPRECIATION @ YEAR n:
dn = d(1 + i)n−1
TOTAL DEPRECIATION @ YEAR n:
(1 + i)n − 1
(CO − CL ) [
]
(1 + i)L − 1
BOOK VALUE:
Cn = CO − Dn
1. A TV IS COST P90,000. ITS ESTIMATED SCRAP VALUE IS
P10,000 AFTER 9 YEARS. SOLVE FOR THE CONSTANT
DEPRECIATION PER YEAR(d) WITH 13% INTEREST RATE. USE
SINKING FUND METHOD.
(90k − 10k)(0.13)
d=
(1 + 0.13)9 − 1
2. AN EQUIPMENT COST P60,000 TODAY AND ITS SALVAGE
VALUE IS P20,000 AFTER 6 YEARS. CALCULATE ITS TOTAL
DEPRECITAION AT YEAR 2 WITH 4% INTEREST RATE. USE
SINKING METHOD.
(1 + 0.04)2 − 1
(60k − 20k) [
]
(1 + 0.04)6 − 1
3. A SALA SET COSTS P80,000. ITS ESTIMATED SCRAP VALUE IS
P5,000 AFTER 3 YEARS. SOLVE FOR THE CONSTANT
DEPRECIATION PER YEAR(d) WITH 6% INTEREST RATE. USE
SINKING FUND METHOD.
(80k − 5k)(0.06)
d=
(1 + 0.06)3 − 1
4. A MACHINE WAS BOUGHT P100,000 AND THE SCRAP VALUE
IS P15,000 AFTER 30 YEARS. CALCULATE ITS BOOK VALUE AT
YEAR 3 WITH 6% INTEREST RATE. USE SINKING FUND
METHOD.
(1 + 0.06)3 − 1
(100k − 15k) [
] = 3422.8713
(1 + 0.06)30 − 1
C3 = 100k − 3422.8713
5. A MACHINE WAS BOUGHT P100,000 AND THE SCRAP VALUE
IS P15,000 AFTER 30 YEARS. CALCULATE ITS TOTAL
DEPRECIATION AT YEAR 8 WITH 5% INTEREST RATE. USE
SINKING FUND METHOD.
(1 + 0.05)8 − 1
(100k − 15k) [
]
(1 + 0.05)30 − 1
6. A MACHINE WAS BOUGHT P100,000 AND THE SCRAP VALUE
IS P15,000 AFTER 30 YEARS. CALCULATE ITS DEPRECIATION
AT YEAR 7 WITH 6% INTEREST RATE. USE SINKING FUND
METHOD.
(100k − 15k)(0.06)
d=
= 1075.1575
(1 + 0.06)30 − 1
d7 = 1075.1575(1 + 0.06)7−1
7. A BEDROOM COSTS P300,000. ITS ESTIMATED SCRAP VALUE
IS P15,000 AFTER 7 YEARS. SOLVE FOR THE CONSTANT
DEPRECIATION PER YEAR(d) WITH 6% INTEREST RATE. USE
SINKING FUND METHOD.
(300k − 15k)(0.06)
d=
(1 + 0.06)7 − 1
8. AN EQUIPMENT COST P60,000 TODAY AND ITS SALVAGE
VALUE IS P20,000 AFTER 6 YEARS. CALCULATE ITS
DEPRECITAION AT YEAR 4 WITH 6.94% INTEREST RATE. USE
SINKING METHOD.
(60k − 20k)(0.0694)
d=
= 5600.2943
(1 + 0.0694)6 − 1
d7 = 1075.1575(1 + 0.0694)4−1
9. A BUFFING MACHINE IS COSTS P10,000. ITS ESTIMATED
SCRAP VALUE IS P1,000 AFTER 10 YEARS. SOLVE FOR THE
CONSTANT DEPRECIATION(d) WITH 4% INTEREST RATE. USE
SINKING FUND METHOD.
(10k − 1k)(0.04)
d=
(1 + 0.04)10 − 1
10. AN EQUIPMENT COST P60,000 TODAY AND ITS SALVAGE
VALUE IS P20,000 AFTER 6 YEARS. CALCULATE ITS BOOK
VALUE AT YEAR 4 WITH 6.2% INTEREST RATE. USE SINKING
METHOD.
(1 + 0.062)4 − 1
(60k − 20k) [
] = 25034.3711
(1 + 0.062)6 − 1
C4 = 100k − 3422.8713
11. A NETWORK SWITCH COSTING P300,000 HAS ESTIMATED
LIFE 12 YEARS WITH A BOOK VALUE OF P75,000 AT THE END
OF THE PERIOD. COMPUTE THE DEPRECIATION PER YEAR
USING STRAIGHT LINE METHOD.
d=
300k−75k
12
12. A BULLDOZER HAS AN INITIAL COST OF P2,000,000. IT
SALVAGE VALUE AFTER 10 YEARS IS P375,000. AS A
PERCENTAGE OF THR INITIAL COST, WHAT IS THE STRAIGHT
LINE DEPRECIATION RATE OF THE EQUIPMENT?
2π − 375π
= 162500
10
162500
π₯100
2π
13. A HEAVY DUTY CONSTRUCTION EQUIPMENT COSTING
P500,000 HAS AN ESTIMATED LIFE OF 25 YEARS WITH BOOK
VALUE OF P100,000 AT THE END OF THE PERIOD. COMPUTE
ITS BOOK VALUE AFTET 15 YEARS USING STRAIGHT LINE
METHOD.
500k − 100k
d=
= 16000
25
C20 = 500k − (16k)(15)
14. A HYDRAULIC PRESS COATING P125,000 HAS ESTIMATED
LIFE OF 6 YEARS WITH A BOOK VALUE OF P38,000 AT THE
END OF THE PERIOD. COMPUTE THE DEPRECIATION PER YEAR
USING STRAIGHT LINE METHOD.
π=
125π−38π
6
M10S1: SUM OF YEAR’S DIGIT METHOD
FORMULAS:
DEPRECIATION @ YEAR n:
2(πΏ − π + 1)
(πΆπ − πΆπΏ )
ππ =
πΏ(πΏ + 1)
TOTAL DEPRECIATION @ YEAR n:
π(2(πΏ) − π + 1)
π·π = [
] (πΆπ − πΆπΏ )
πΏ(πΏ + 1)
BOOK VALUE:
πΆπ = πΆπ − π·π
QUESTIONS:
1. A COMPUTER RIG COSTS P150,000. IF THE USEFULL LIFE IS 20
YEARS AND THE SCRAP VALUE IS P10,000. CALCULATE THE
DEPRECIATION AT YEARS 14. USE SUM OF YEAR’S DIGIT
METHOD.
π14 =
2(20−14+1)
20(20+1)
(150,000 − 10,000)
2. AN EQUIPMENT BOUGHT BY AN ENGINEER COST P250,000
AND THE SALVAGE VALUE IS P25,000 AFTER 6 YEARS.
CALCULATE THE DEPRECIATION AT YEAR 2. USE SUM OF
YEAR’S METHOD.
π2 =
2(6−2+1)
6(6+1)
(250,000 − 25,000) ANSWER: = 53571.4286
3. A SEWING MACHINE BOUGHT BY AN ENGINEER COSTS
P100,000 AND THE SALVAGE VALUE IS P30,000 AFTER 6
YEARS. CALCULATE THE BOOK VALUE AT YEAR 4. USE SUM OF
YEAR’S DIGIT METHOD.
π·4 = [
4(2(6)−4+1)
6(6+1)
] (100,000 − 30,000) = 60,000
πΆπ = 100,000 − 60,000
4. A COMPUTER RIG COSTS P150,000. IF THE USEFULL LIFE IS 20
YEARS AND THE SCRAP VALUE IS P10,000. CALCULATE THE
BOOK VALUE AT YEARS 16. USE SUM OF YEAR’S DIGIT
METHOD.
π·16 = [
16(2(20)−16+1)
20(20+1)
] (150,000 − 10,000) = 133333.3333
πΆπ = 150,000 − 133333.3333
5. A COMPUTER RIG COSTS P150,000. IF THE USEFULL LIFE IS 20
YEARS AND THE SCRAP VALUE IS P10,000. CALCULATE THE
TOTAL DEPRECIATION AT YEARS 17. USE SUM OF YEAR’S
DIGIT METHOD.
π·17 = [
17(2(20)−17+1)
20(20+1)
] (150,000 − 10,000)
6. A MACHINE COST P80,000 AND THE SALVAGE VALUE IS
P20,000 AFTER 10 YEARS. CALCULATE THE BOOK VALUE AT
YEAR 7. USE SUM OF YEAR’S DIGIT METHOD.
π·7 = [
7(2(10)−7+1)
10(10+1)
] (80,000 − 20,000) = 53454.5455
πΆπ = 80,000 − 53454.5455
7. AN EQUIPMENT BOUGHT BY A ENGINEER COSTS P250,000
AND THE SALVAGE VALUE IS P25,000 AFTER 6 YEARS.
CALCULATE THE DEPRECIATION AT YEAR 5. USE SUM OF
YEAR’S METHOD.
π5 =
2(6−5+1)
6(6+1)
(250,000 − 25,000) ANSWER: = 21428.5714
8. AN EQUIPMENT BOUGHT BY A ENGINEER COST P250,000
AND THE SALVAGE VALUE IS P25,000 AFTER 6 YEARS.
CALCULATE THE BOOK VALUE AT YEAR 1. USE SUM OF YEAR’S
METHOD.
π·1 = [
1(2(6)−1+1)
6(6+1)
] (250,000 − 25,000) = 64285.7143
πΆπ = 250,000 − 64285.7143
9. A MACHINE COST P80,000 AND THE SALVAGE VALUE IS
P20,000 AFTER 10 YEARS. CALCULATE THE TOTAL
DEPRECIATION AT YEAR 7. USE SUM OF YEAR’S DIGIT
METHOD.
π·7 = [
7(2(10)−7+1)
10(10+1)
] (80,000 − 20,000)
10. A MACHINE COST P80,000 AND THE SALVAGE VALUE IS
P20,000 AFTER 10 YEARS. CALCULATE THE DEPRECIATION AT
YEAR 7. USE SUM OF YEAR’S DIGIT METHOD.
π7 =
2(10−7+1)
10(10+1)
(80,000 − 20,000)
M10S2: DECLINING BALANCE METHOD
FORMULA:
πΏ
π =1− √
πΆπΏ
πΆπ
DEPRECIATION @ YEAR n:
ππ = πΆπ πΎ (1 − π)π−1
TOTAL DEPRECIATION @ YAER n:
π·π = πΆπ [1 − (1 − π)π ]
BOOK VALUE:
πΆπ = πΆπ (1 − π)π
CONSTANT DECLINING BALANCE (k):
πΏ
π = (1 − √
πΆπΏ
) π₯ 100
πΆπ
QUESTIONS:
1. AN EQUIPMENT COST P60,000 TODAY AND IT SALVAGE
VALUE IS P20,000 AFTER 6 YEARS. CALCULATE ITS BOOK
VALUE AT YEAR 3. USE DECLINING BALANCE METHOD.
6
20,000
πΎ = 1 − √60,000
= 0.1673
πΆ3 = 60,000(1 − 0.1673)3
2. A BEDROOM COSTS P30,000. ITS ESTIMATED SCRAP VALUE IS
P15,000 AFTER 5 YEARS. SOLVE FOR THE CONSTANT
DECLINING PERCENTAGE(k). USE DECLINING BALANCE
METHOD. 12.9449
5
15,000
πΎ = (1 − √
) π₯ 100 ANSWER: = 12.9449%
30,000
3. A MACHINE WAS BOUGHT P100,000 AND THE SCRAP VALUE
IS P15,000 AFTER 30 YEARS. CALCULATE ITS DEPRECIATION
AT YEAR 9. USE DECLINING BALANCE METHOD.
30
15,000
πΎ =1− √
100,000
= 0.0613
π9 = 100,000(0.0613)(1 − 0.0613)9−1
4. A BUFFING MACHINE IS COSTS P10,000. ITS ESTIMATED
SCRAP VALUE IS P1,000 AFTER 12 YEARS. SOLVE FOR THE
CONSTANT DECLINING PERCENTAGE(k). USE DECLINING
BALANCE METHOD.
12
1,000
πΎ = (1 − √
) π₯ 100
10,000
5. AN EQUIPMENT COST P60,000 TODAY AND ITS SALVAGE
VALUE IS P20,000 AFTER 6 YEARS. CALCULATE ITS
DEPRECIATION AT YEAR 2. USE DECLINING BALANCE
METHOD.
6
20,000
πΎ =1− √
= 0.1673
60,000
π2 = 60,000(0.1673)(1 − 0.1673)2−1
6. A MACHINE WAS BOUGHT P100,000 AND THE SCRAP VALUE
IS P15,000 AFTER 30 YEARS. CALCULATE ITS BOOK VALUE AT
YEAR 15. USE DECLINING BALANCE METHOD.
30
15,000
πΎ =1− √
100,000
=0.0613
πΆπ = 100,000(1 − 0.0613)15
7. A MACHINE WAS BOUGHT P100,000 AND THE SCRAP VALUE
IS P15,000 AFTER 30 YEARS. CALCULATE ITS TOTAL
DEPRECIATION AT YEAR 27. USE DECLINING BALANCE
METHOD. 81877.2507
30
15,000
πΎ =1− √
100,000
=0.0613
π·27 = 100,000[1 − (1 − 0.0613)27 ]
8. A SALA SET IS COSTS P80,000. ITS ESTIMATED SCRAP VALUE IS
P5,000 AFTER 11 YEARS. SOLVE FOR THE CONSTANT
DECLINING PERCENTAGE(k). USE DECLINING BALANCE
METHOD. 22.2797
11
5,000
πΎ = (1 − √
) π₯ 100 ANSWER: = 22.2797%
80,000
9. AN EQUIPMENT COST P60,000 TODAY AND ITS SALVAGE
VALUE IS P20,000 AFTER 6 YEARS. CALCULATE ITS TOTAL
DEPRECIATION AT YEAR 3. USE DECLINING BALANCE
METHOD. 25319.4278
6
20,000
πΎ =1− √
=0.1673
60,000
π·3 = 60,000[1 − (1 − 0.1673)3 ]
10. A TV IS COSTS P90,000. ITS ESTIMATED SCRAP VALUE IS
P10,000 AFTER 2 YEARS. SOLVE FOR THE CONSTANT
DECLINING PERCENTAGE(k). USE DECLINING BALANCE
METHOD.66.6667
2
10,000
πΎ = (1 − √
) π₯ 100
90,000
11. AN EQUIPMENT BOUGHT BY A ENGINEER COST P250,000
AND THE SALVAGE VALUE IS P25,000 AFTER 6 YEARS.
CALCULATE THE BOOK VALUE AT YEAR 3. USE SUM OF YEAR’S
METHOD.
π·3 = [
3(2(6)−3+1)
] (250,000 − 25,000) = 160714.2857
6(6+1)
πΆπ = 250,000 − 160714.2857
12. A MACHINE COST P80,000 AND THE SALVAGE VALUE IS
P20,000 AFTER 10 YEARS. CALCULATE THE BOOK VALUE AT
YEAR 2. USE SUM OF YEAR’S DIGIT METHOD.
π·2 = [
2(2(10)−2+1)
] (80,000 − 20,000) = 20727.2727
10(10+1)
πΆπ = 80,000 − 20727.2727
13. A SEWING MACHINE BOUGHT BY AN ENGINEER COSTS
P100,000 AND THE SALVAGE VALUE IS P30,000 AFTER 6
YEARS. CALCULATE THE BOOK VALUE AT YEAR 2. USE SUM OF
YEAR’S DIGIT METHOD.
π·2 = [
2(2(6)−2+1)
6(6+1)
] (100,000 − 30,000) = 36666.6667
πΆπ = 100,000 − 36666.6667
14. A COMPUTER RIG COSTS P150,000. IF THE USEFULL LIFE
IS 20 YEARS AND THE SCRAP VALUE IS P10,000. CALCULATE
THE BOOK VALUE AT YEAR 10. USE SUM OF YEAR’S DIGIT
METHOD.
π·10 = [
10(2(20)−10+1)
20(20+1)
] (150,000 − 10,000) = 103333.3333
πΆπ = 150,000 − 103333.3333
15. AN EQUIPMENT BOUGHT BY A ENGINEER COST P250,000
AND THE SALVAGE VALUE IS P25,000 AFTER 6 YEARS.
CALCULATE THE DEPRECIATION AT YEAR 3. USE SUM OF
YEAR’S METHOD.
π3 =
2(6−3+1)
6(6+1)
(250,000 − 25,000)
M11S1: DOUBLE DECLINING BALANCE METHOD
FORMULAS:
DEPRECIATION @ YEAR n:
2πΆπ
2 π−1
ππ =
(1 − )
πΏ
πΏ
TOTAL DEPRECIATION @ YEAR n:
2 π
π·π = πΆπ [1 − (1 − ) ]
πΏ
BOOK VALUE:
2 π
πΆπ = πΆπ (1 − )
πΏ
SALVAGE VALUE:
2 πΏ
πΆπΏ = πΆπ (1 − )
πΏ
CONSTANT DECLINING PERCENTAGE (k):
2
πΏ
QUESTIONS:
1. A PRODUCT IS BROUGHT FOR AN AMOUNT P50,000
WHICH WILL BE SOLD AFTER 15 YEARS. CALCULATE ITS
TOTAL DEPRECIATION AT YEAR 14. USE THE DOUBLE
DECLINING BALANCE METHOD.
π·14 = 50,000 [1 − (1 −
2 14
) ]
15
2. ENGR. VELASCO BOUGHT A COMPUTER WITH AN AMOUNT
OF P200,000. ASSUMING THAT THE LIFE SPAN OF HIS
COMPUTER IS 10 YEARS. CALCULATE ITS DEPRECIATION AT
YEAR 7. USE THE DOUBLE DECLINING BALNCE METHOD.
π7 =
2(200,000)
10
(1 −
2 7−1
)
10
3. A PRODUCT IS BROUGHT FOR AN AMOUNT P50,000 WHICH
WILL BE SOLD AFTER 15 YEARS. CALCULATE ITS BOOK
VALUE AT YEAR 8. USE THE DOUBLE DECLINING BALANCE
METHOD.
πΆ8 = 50,000 (1 −
2 8
)
15
4. AN EQUIPMENT COST P100,000 AND HAS AN ESTIMATED
LIFE OF 6 YEARS. CALCULATE IST BOOK VALUE AT YEAR 4.
USE THE DECLINING BALANCE METHOD.
2 4
πΆ4 = 100,000 (1 − )
6
5. ENGR. VELASCO BOUGHT A COMPUTER WITH AN AMOUNT
OF P200,000. ASSUMING THAT THE LIFE SPAN OF HIS
COMPUTER IS 10 YEARS. CALCULATE ITS TOTAL
DEPRECIATION AT YEAR 2. USE THE DOUBLE DECLINING
BALNCE METHOD.
π·2 = 200,000 [1 − (1 −
2 2
) ]
10
6. AN EQUIPMENT COST P100,000 AND HAS AN ESTIMATED
LIFE OF 6 YEARS. CALCULATE ITS DEPRECIATION AT YEAR 2.
USE THE DECLINING BALANCE METHOD.
π2 =
2(100,000)
6
2 2−1
(1 − )
6
7. A PRODUCT IS BROUGHT FOR AN AMOUNT P50,000 WHICH
WILL BE SOLD AFTER 15 YEARS. CALCULATE THE CONSTANT
DECLINING PERCENTAGE (k). USE THE DOUBLE DECLINING
BALANCE METHOD.
2
15
= 0.1333 x 100 ANSWER: = 13.33%
8. A PRODUCT IS BROUGHT FOR AN AMOUNT P50,000 WHICH
WILL BE SOLD AFTER 15 YEARS. CALCULATE ITS
DEPRECIATION AT YEAR 13. USE THE DOUBLE DECLINING
BALANCE METHOD.
π13 =
2(50,000)
15
(1 −
2 13−1
)
15
9. AN EQUIPMENT COST P100,000 AND HAS AN ESTIMATED
LIFE OF 6 YEARS. CALCULATE ITS TOTAL DEPRECIATION AT
YEAR 3. USE THE DECLINING BALANCE METHOD.
2 3
π·3 = 100,000 [1 − (1 − ) ] ANSWER: = 70370.3704
6
10. ENGR. VELASCO BOUGHT A COMPUTER WITH AN
AMOUNT OF P200,000. ASSUMING THAT THE LIFE SPAN OF
HIS COMPUTER IS 10 YEARS. CALCULATE BOOK VALUE AT
YEAR 5. USE THE DOUBLE DECLINING BALNCE METHOD.
πΆ5 = 200,000 (1 −
2 5
)
10