1. The gold mine is expected to be exhausted within 20 years. With annual costs of $750,000 and $120/ton, it earns $450/ton and has annual output of 25,000 tons. The property valuation is $53,745,535. 2. The timber land was bought for $8,000,000 and sold for $200,000 after 14 years of $1,400,000 average annual profits. The investment rate is 13.2%. 3. For a company with 4 machines, the total annual straight-line depreciation charge is $49,600 by group depreciation and $47,666.67 by composite depreciation.

1. 1. A gold mine in Mount Province has an annual output of 25,000 tons of ore. It is expected that
the ore body will be exhausted within a period of 20 years. The management cost annually is
P750,000 , and the operating cost of the mine and the smelter plants is P120 per ton. The
processed ore produce an income of P450 per ton. If the annual dividend rate is 12% payable
annually,andthe sinkingfundrate is9% annually,determine the valuation of the propertynow.
GIVEN:
n = 20 r = 12% i = 9%
REQ’D:
P = ?
SOLUTION:
Annual gross income = (25000)(P450) = P11, 250, 000
Annual management and operating costs = P750, 000 + 25000(120) = P3, 750, 000
R = P11, 250, 000 - P3, 750, 000 = P7, 500 000
𝑃 =
𝑅
𝑟 + ( 𝐴 𝐹⁄ ,9%, 𝑛)
=
7500000
0.12 + 0.0195 4648
= P53, 745, 535. 31 ans.
2. Timber land in Palawan was purchased for P8,000,000 and earned an average annual profit of
P1,400,000 for 14 years,at the endof whichtime the landwas sold for P200,000. Assuming that
a sinkingfundearning7%wasestablishedto provide for depletion, determine the investment
rate.
GIVEN:
n = 14 years i = 7% R = P1, 400, 000
REQ’D:
r = ?
SOLUTION:
𝑃 = 8,000,000 − 200,000(1.07 )−14
= P7, 922, 436.55
𝑃 =
𝑅
𝑟 + (𝐴 𝐹, 𝑖%, 𝑛)⁄
𝑟 =
𝑅
𝑃
− (𝐴 𝐹⁄ , 𝑖%, 𝑛)
𝑟 =
1400000
7922436.
−
0.07
(0.0771)14 − 1
= 𝟎. 𝟏𝟑𝟐
r = 13.2%
3. The San FernandoManufacturingCompany owns four different production machine with data
tabulated below :
Machine Number Number Owned First Cost Salvage Value Expected Life
1 8 P40, 000 P12, 000 12
2 6 32, 000 8, 000 10
3 4 18, 000 4, 000 10
4 4 24, 000 6, 000 8

2. One-half of the machines of the kindwillbe replacedafter8yearsand the restwill be sold after
12 years. Compute the total annual straight-line depreciation charges by (a) the group
depreciation method, and (b) the composite depreciation method.
SOLUTION:
a.) Group Depreciation Method
𝐴𝑉𝐸𝑅𝐴𝐺𝐸 𝐷𝐸𝑃𝑅𝐸𝐶𝐼𝐴𝑇𝐼𝑂𝑁 =
12 + 10 + 10 + 8
4
= 𝟏𝟎 𝒚𝒆𝒂𝒓𝒔
𝐴𝑁𝑁𝑈𝐴𝐿 𝐺𝑅𝑂𝑈𝑃 𝐷𝐸𝑃𝑅𝐸𝐶𝐼𝐴𝑇𝐼𝑂𝑁 =
𝑇𝑂𝑇𝐴𝐿 𝐷𝐸𝑃𝑅𝐸𝐶𝐼𝐴𝑇𝐼𝑂𝑁 𝐴𝑀𝑂𝑈𝑁𝑇
𝐴𝑉𝐸𝑅𝐴𝐺𝐸 𝐿𝐼𝐹𝐸
=
8(40000) − 12000) + 6(32000 − 8000) + 4(18000 − 4000) + 4(24000 − 6000)
10
=
496000
10
= 𝐏𝟒𝟗, 𝟔𝟎𝟎 𝒑𝒆𝒓 𝒚𝒆𝒂𝒓
𝐴𝑁𝑁𝑈𝐴𝐿 𝐷𝐸𝑃𝑅𝐸𝐶𝐼𝐴𝑇𝐼𝑂𝑁 𝑀𝐴𝐶𝐻𝐼𝑁𝐸⁄ =
49000
8 + 6 + 4 + 4
= 𝐏𝟐𝟐𝟓𝟒. 𝟓𝟓 𝑴𝑨𝑪𝑯𝑰𝑵𝑬⁄
DEPRECIATION FOREACH YEAR IS SHOWN IN THE TABLE BELOW.
YEAR NUMBER IN SERVICE ANNUALDEPRECIATION ACCOMULATED
1 2 3 4 DEPRECIATION
1 – 8 8 6 4 4 P49, 600 P396, 400
9 – 12 4 3 2 2 P24, 000.05 P198, 400.40
P595, 200.05
DEPRECIATION FORYEARS 9 – 12 = P2254.55(4 + 3 + 2 + 2) = P24, 800.05
b.) Composite DepreciationMethod:
The annual depreciationamountsforeachmachine are:
𝐷1 =
8(40000 − 12000)
12
= 𝐏𝟏𝟖, 𝟔𝟔𝟔. 𝟔𝟕 𝒚𝒆𝒂𝒓⁄
𝐷2 =
6(32000 − 8,000)
10
= 𝐏𝟏𝟒, 𝟒𝟎𝟎 𝒚𝒆𝒂𝒓⁄
𝐷3 =
4(18000 − 4000)
10
= 𝐏𝟓, 𝟔𝟎𝟎 𝒚𝒆𝒂𝒓⁄
𝐷4 =
4(24,000 − 6000)
8
= 𝐏𝟗, 𝟎𝟎𝟎 𝐲𝐞𝐚𝐫⁄
Composite Depreciationamounts:
Years 1 – 8:
P18, 600.67 + P14, 400 + P5, 600 + P9, 000 = P47, 666.67
AccumulatedDepreciationafter8years= 8(47, 666.67) = P381, 333.36
Years 9 – 10:
18, 666.67 + P14, 400 + P5, 600 = P38, 666.67
AccumulatedDepreciationafter10years= P381, 333.36 + 2(38, 666.67) = P458, 666.70

3. Years 11 – 12: (Machine 1 only)
AccumulatedDepreciationafter12 years= P450, 666.70 + 2(P18, 666.67) = P496, 000.00
The total accumulateddepreciationbyeithermethodafter12years= P496, 000.00
4. A firm plans to market a new minicomputer that will sell for $10,000. This financing scheme is
being considered by the company: Payments of $872 each year for 20 years would be made by
the purchaser after an initial down payment is made. If their interest is charge is 6%
compounded monthly, what down deposit should the company request?
GIVEN:
Cost = $10, 000
A =$872
n = 20 years
i = 6% compounded monthly
REQ’D:
DOWN DEPOSIT = ?
SOLUTION:
𝐸𝑅 = [1 + (
0.06/12
12
)12 − 1]
𝐶𝑂𝑆𝑇 = 𝐷𝑂𝑊𝑁 𝐷𝐸𝑃𝑂𝑆𝐼𝑇 + 𝐴( 𝑃 𝐴⁄ , 𝑖, 𝑛)
1000 = 𝐷𝑂𝑊𝑁 𝐷𝐸𝑃𝑂𝑆𝐼𝑇 + 872 [
(1 + 0.06167781)20 − 1
0.061617781(1 + 0.06167781)20
]
𝐷𝑂𝑊𝑁 𝐷𝐸𝑃𝑂𝑆𝐼𝑇 = 10000 − 9866.95
𝑫𝑶𝑾𝑵 𝑫𝑬𝑷𝑶𝑺𝑰𝑻 = $𝟏𝟑𝟑.𝟎𝟓
5. A man deposit $2,000 an a savings account when his son was born. The nominal interest rate
was 8% per year, compounded continuously. On the son’s birthday, the accumulated sum is
withdrawn from the account. How much would this accumulated amount be?
GIVEN:
P = $2000
Nominal rate of interest = 8% per year compound continuously
N = 18
REQ’D:
F = ?
SOLUTION:
𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 𝑒0.08 – 1 = 0.0832 8708
𝐹 = 𝑃(1 + 𝑖) 𝑛
𝐹 = $2000(1 + 0.08333)18
F = $8, 441

4. 6. Maintenance costs for a new bridge with an expected 50-year life are estimated to be $1,000
each yearfor the first5 years,followed by a $10,000 expenditure in the 15th
year and a $10,000
expenditure inyear30. If I = 10% per year, what is the equivalent uniform annual cost over the
entire 50-year period?
GIVEN:
i = 10%
n = 50 years
A1 = $ 1,000 maintenance costeachyearfor the first5 years
C1 = 1,000 [
(1+0.10)5−1
0.10
] = $ 6,105.1 moneyworthafter5 years
C2 = $ 10,000 expenditure duringthe 15th
year
C3 = $ 10,000 expenditure duringthe 30th
year
Required:
P = ? & A = ?
SOLUTION:
𝐹 = 6,105.1 (1 + 0.10)45 + 10,000 (1 + 0.10)35 + 10,000 (1 + 0.10)20
𝐹 = $ 793,303.0599
Since,
𝐹 = 𝑃 (1 + 𝑖) 𝑛
𝑃 =
793,303.0599
(1+0.10)50
𝑷 = $ 𝟔, 𝟕𝟓𝟕. 𝟕𝟗
Therefore;
𝐴 =
𝐹𝑖
(1+𝑖) 𝑛−1
=
793,303.0599 (0.10)
(1+0.10)50−1
= $ 𝟔𝟖𝟏. 𝟓𝟗
7. Uncle Wilbur’s trout ranch is now for sale for $40,000. Annual property taxes, maintenance,
supplies, and so on, are estimated to continue to be $3,000 per year. revenues from the ranch
are expectedtobe $10,000 nextyearand then decline by $500 per year thereafter through the
tenthyear.if you boughtthe ranch, youwouldplanto keepitonly5 yearsand at the time sell it
for the value of the land, which is $15,000. If your desired annual rate of return is 12% should
you become a trout rancher?
SOLUTION:
TOTAL EXPENSES = 15, 000+ 40, 000 = $55, 000 for 5 years
An expenses every year = 11, 000
TOTAL REVENGE = 45, 000 + 15, 000 = $60, 000 for 5 years
An expenses every year = 12, 000
FPROFIT EVERY YEAR = 12, 000 – 11, 000 = 1, 000
𝑅𝐴𝑇𝐸 𝑂𝐹 𝑅𝐸𝑇𝑈𝑅𝑁 =
1000
12000
x 100 = 𝟗. 𝟎𝟗% < 12% (𝒅𝒐 𝒏𝒐𝒕 𝒃𝒖𝒚 𝒕𝒉𝒆 𝒓𝒂𝒏𝒄𝒉)
8. The heat lossthroughthe exteriorwallsof acertainpoultryprocessingplantisestimated tocost
the owner $3,00 next year. a salesman from super fier Insulation, Inc. has told you, the plant
engineer,thathe canreduce the heatlossby 80% withthe insallationof $15,000 worth of Super
fiber now. If the cost of heat loss rises by $200 per year (gradient) after next year and the

5. ownderplanstokeepthe presentbuildingfor 10 more years, what would yoy recommended if
the money is 12% per year?
Illustration:
1 2 3 4 5 6 7 8 9 10
WithoutSuperFiberInsulation
1 2 3 4 5 6 7 8 9 10
WithSuperFiberInsulation
SOLUTION:
Wheni = 12%
WithoutSuperFiberInsulation
𝐹 = 3000 [
(1+0.12)10−1
0.12
] +
200
0.12
[
(1+0.12)10−1
0.12
− 10] = $ 𝟔𝟓,𝟐𝟐𝟕. 𝟒𝟑𝟎𝟑𝟑
𝑃 =
65,227.43033
(1 + 0.12)10 = $ 𝟐𝟏, 𝟎𝟎𝟏.𝟒𝟗
WithSuperFiberInsulation
𝐹 = 600[
(1+0.12)10−1
0.12
] +
200
0.12
[
(1+0.12)10−1
0.12
− 10] + 15000(1 + 0.12)10 = $ 𝟔𝟗,𝟔𝟗𝟖. 𝟏𝟖𝟗𝟐𝟗
𝑃 =
69,698.18929
(1+0.12)10
= $ 𝟐𝟐,𝟒𝟒𝟎. 𝟗𝟓
Therefore:
The ownermust not accept the offerofthe Salesman.
9. Solve forthe value of F below so that the left-hand cash flow diagram is equivalent to the one
on the right. Let I = 8% per year.
P = ?
3000
F = ?
3200
3600
3400
3800
4000
4200
4400
4600
P = ?
F = ?
600
800
1200
1000
1400
1600
1800
2000
2200
2400
15000
4800

6. 10. A corporations floats P200,000 worth of ten-year callable bonds is P1,000 denominations. The
bond rate is 7% compounded annually. Prepare an amortization.
GIVEN:
F = P200, 000 n = 10 i = 7%
SOLUTION:
𝑛𝑜. 𝑜𝑓 𝑏𝑜𝑛𝑑𝑠 =
200000
1000
= 𝐏𝟐𝟎𝟎 𝐛𝐨𝐧𝐝𝐬
A = [
𝑖 ( 1+𝑖 ) 𝑛
( 1+𝑖 ) 𝑛− 1
] 𝐹 = [
0.07( 1+ 0.07 )10
( 1+0.07 )10− 1
]200000 = 𝐏𝟐𝟖, 𝟒𝟕𝟓. 𝟓𝟎
28475.50 – 14000 = P14, 475.50
𝑛𝑜. 𝑜𝑓 𝑏𝑜𝑛𝑑𝑠 =
14475.50
1000
= 𝟏𝟓 𝒃𝒐𝒏𝒅𝒔
YEAR PRINCIPAL
INTEREST AT
7%
NO. OF BONDS
RETIRED
AMOUNT OF
PRINCIPAL
REPAID
YEAR END
PAYMENT
1 P200, 000 P14, 000 14 P14, 000 P28, 000
2 186, 000 13, 020 15 15, 000 28, 020
3 171, 000 11, 970 17 17, 000 28, 970
4 154, 000 10, 780 18 18, 000 28, 780
5 136, 000 9, 520 19 19, 000 28, 520
6 117, 000 8, 190 20 20, 000 28, 190
7 97, 000 6, 790 22 22, 000 28, 790
8 75, 000 5, 250 23 23, 000 28, 250
9 52, 000 3, 640 25 25, 000 28, 640
10 27, 000 1, 890 27 27, 000 28, 890
TOTALS P1, 215, 000 P85, 120 200 P200, 000 P285, 120
11. A man borrowed P150, 000 from a bank for home improvement, to be repaid by month-end
payment for 60 months. The current rate of the interest charge by banks is 19% compounded
monthly. Based on this rate, prepare an amortization schedule.
GIVEN:
P = 150, 000.00
n = 60 months (period)
I = 19% compounded monthly =
19%
12
= 1.5833 3333%
SOLUTON:
A = [
𝑖
1− 𝑖)−𝑛] = 150000[
0.015833333
1 ∓(0.01583333 )−60] = P3891.08
PERIOD
PRINCIPAL AT
THE BEGINNING
OF EACH 6
M0NTHS
INTEREST AT 4%
PER PERIOD
PAYMENT AT
END OF EACH
PERIOD
PERIODIC
PAYMENT TO
PRINCIPAL
1 150000 2374.999995 3891.08 1516.080005

7. 2
148483.92 2350.995395 3891.08 1540.084605
3
146943.8354 2326.610722 3891.08 1564.469278
4
145379.3661 2301.839959 3891.08 1589.240041
5
143790.1261 2276.676991 3891.08 1614.403009
6
142175.7231 2251.11561 3891.08 1639.96439
7
140535.7587 2225.149508 3891.08 1665.930492
8
138869.8282 2198.772275 3891.08 1692.307725
9
137177.5205 2171.977403 3891.08 1719.102597
10
135458.4179 2144.758278 3891.08 1746.321722
11
133712.0961 2117.108184 3891.08 1773.971816
12
131938.1243 2089.020297 3891.08 1802.059703
13
130136.0646 2060.487685 3891.08 1830.592315
14
128305.4723 2031.503307 3891.08 1859.576693
15
126445.8956 2002.06001 3891.08 1889.01999
16
124556.8756 1972.150526 3891.08 1918.929474
17
122637.9461 1941.767477 3891.08 1949.312523
18
120688.6336 1910.903362 3891.08 1980.176638
19
118708.457 1879.550565 3891.08 2011.529435
20
116696.9275 1847.701349 3891.08 2043.378651
21
114653.5489 1815.347854 3891.08 2075.732146
22
112577.8168 1782.482095 3891.08 2108.597905
23
110469.2188 1749.095961 3891.08 2141.984039
24
108327.2348 1715.181214 3891.08 2175.898786
25
106151.336 1680.729483 3891.08 2210.350517
26
103940.9855 1645.732267 3891.08 2245.347733
27
101695.6378 1610.180928 3891.08 2280.899072
28
99414.7387 1574.066693 3891.08 2317.013307
29
97097.72539 1537.380649 3891.08 2353.699351
30
94744.02604 1500.113743 3891.08 2390.966257
31
92353.05979 1462.256777 3891.08 2428.823223
32
89924.23656 1423.800409 3891.08 2467.279591

8. 33
87456.95697 1384.735149 3891.08 2506.344851
34
84950.61212 1345.051356 3891.08 2546.028644
35
82404.58348 1304.739236 3891.08 2586.340764
36
79818.24271 1263.78884 3891.08 2627.29116
37
77190.95155 1222.190064 3891.08 2668.889936
38
74522.06162 1179.93264 3891.08 2711.14736
39
71810.91426 1137.00614 3891.08 2754.07386
40
69056.8404 1093.399971 3891.08 2797.680029
41
66259.16037 1049.10337 3891.08 2841.97663
42
63417.18374 1004.105407 3891.08 2886.974593
43
60530.20914 958.3949761 3891.08 2932.685024
44
57597.52412 911.9607966 3891.08 2979.119203
45
54618.40492 864.7914094 3891.08 3026.288591
46
51592.11633 816.8751734 3891.08 3074.204827
47
48517.9115 768.2002638 3891.08 3122.879736
48
45395.03176 718.7546681 3891.08 3172.325332
49
42222.70643 668.5261837 3891.08 3222.553816
50
39000.15261 617.5024151 3891.08 3273.577585
51
35726.57503 565.6707701 3891.08 3325.40923
52
32401.1658 513.0184574 3891.08 3378.061543
53
29023.10426 459.5324831 3891.08 3431.547517
54
25591.55674 405.1996475 3891.08 3485.880352
56
18564.60293 293.9395458 3891.08 3597.140454
57
14967.46248 236.984822 3891.08 3654.095178
58
11313.3673 179.1283152 3891.08 3711.951685
59 7601.415613 120.3557469 3891.08 3770.724253
60
3830.69136 60.65261307 3891.08 3830.427387
TOTAL
5271477.736 83465.06397 233464.8 150000
12. (M.E. Board,November1983) On January1, 1978 the purchasingmanagerof a cementcompany
boughta new machine costing P140,000. Depreciation has been computed by the straight-line
method, based on an estimated useful life of 5 years and residual scrap value) 12, 800.

9. On January 2, 1981 extraordinary repairs (almost equivalent to a rebuilding of the
machinery) were performedatacost of P30,400. Because of the thoroughgoingnature of these
repairs,the normal life of machinerywasextendedmaterially;the revisedestimate of useful life
was 4 years in 1981.
Determine the annual provision for depreciation for the years 1978 to 1980 and the
adjustedprovisionforthe depreciationof December 31, 1981. Assume payment in cash for the
machine and the repairs.
13. (ECE Board, August 1975) A broadcasting corporation purchased equipment worth P53,000 and
paidP1, 500 forfreightanddeliviery charges to the site. The equipment has a normal life of 10
yearswitha trade-invalue of P5,000 agianst the purchase of new equipment at the end of life.
(a) Determine the annual depreciation cost by the straight-line method.
(b) Determine the annual depreciation cost by the sinking fund method. Assume
interest is 6% compounded annually.
GIVEN:
𝐶 𝑂 = 53000 + 1500 = P54,500
Delivery Change = P1, 800
N = 10 years
𝐶 𝑛 = P5,000
SOLUTION:
a.) Straight – line Formula
𝑑 =
𝐶 𝑜 − 𝐶 𝑛
𝑛
=
54500 − 5000
10
= 𝑷𝟒, 𝟗𝟖𝟎
b.) Sinking Fund Formula
𝑑 = ( 𝐶 𝑜 − 𝐶 𝑛)(𝐴 𝐹, 𝑖%, 𝑛)⁄
𝑑 = (54500 − 5000) [
0.06
(1.06)10 − 1
]
𝒅 = 𝐏𝟑, 𝟕𝟓𝟓. 𝟒𝟔
14. (M.E. Board, June 1990) A machinery supplier is offering a certain machine on a 10% down
payment and the balance payable in equal year-end payments without interest for 2 years.
Under this arrangement the price is pegged at {150,000. However, for cash purchase the
machine would only cost P195, 000. What is the equivalent interest rate that is charged on the
two-year payment plan of interest is compounded quarterly?
GIVEN:
C1 = P 250,000 Price of the machine for2-yearpayment
C0 = P 195,000 Original costof machine whenprice purchased
n = 2 years
d = downpayment= .10(250,000) = 25,000
C1-new = 250,000 – 25,000 = 225,000 Remainingbalance
CA = 225,000/2 = 112,500 Cash to be paideveryyear-end
C0 = 195,000 – 25,000 = 170,000 Remainingamounttobe paidandto be compounded
Therefore:
𝐹 = 112,500 [
(1+𝑖)2−1
𝑖
] Expectedtotal compoundedamountof moneyafter2years
𝐹 = 170,000 (1 + 𝑖)2 Total compoundedmoneyafter2years

10. SOLUTION:
170,000 (1 + 𝑖)2 = 112,500 [
(1+𝑖)2−1
𝑖
]
170,000
112,500
(1 + 𝑖)2 𝑖 = (1 + 𝑖)2 − 1
68
45
(1 + 𝑖)2 𝑖 = (1 + 𝑖)2 − 1
68(1 + 𝑖)2 𝑖 = 45(1 + 𝑖)2 − 45
68(1 + 𝑖)2 𝑖− 45(1 + 𝑖)2 = −45
(1 + 𝑖)2(68𝑖 − 45) + 45 = 0
By trial and error:
𝒊 = 𝟎. 𝟐𝟎𝟗𝟏effectiverate compoundedannually
(1 + .2091)2(68(.2091) − 45) + 45 = 0.0002616 ≅ 0
To get forthe equivalenteffectiverate compoundedquarterly:
0.2091 = (1 +
𝑖
4
)
4
− 1
1.2091 = (1 +
𝑖
4
)
4
√1.20914
= ∜(1 +
𝑖
4
)
4
1.08614 = 1 +
𝑖
4
𝑖 = 0.194456 ≅ 0.1945
THEREFORE:
15. A debtof P10, 000 withinterestatthe rate law of 8% payable semi-annually is to be amortized
by equal paymentsatthe end of each 6 months for 4 years. Find the semi-annual payment and
contract and amortization schedule.
GIVEN:
P = 10, 000 n = 4(2) = 8 quarters 𝑖 =
8%
2
= 4%
SOLUTION:
A = P (𝐴 𝑃⁄ , 𝑖, 𝑛) = 10000 [
0.04
1−(1.04)−8
] = P1,487.28
PERIOD
PRINCIPALATTHE
BEGINNING OF
EACH 6 M0NTHS
INTEREST AT 4%
PER PERIOD
PAYMENT AT END
OF EACH PERIOD
PERIODIC
PAYMENT TO
PRINCIPAL
1 P10, 000.00 P400.00 P1,485.28 P1, 085.28
2 8, 917.72 356.59 P1,485.28 1,228.69
3 8, 786.03 311.44 P1,485.28 1, 173.84
4 6, 612.19 264.49 P1,485.28 1, 220.79
5 5, 391.70 451.66 P1,485.28 1, 269.62
6 4, 121.78 164.87 P1,485.28 1, 320.41
𝒊 = 𝟏𝟗. 𝟒𝟓 % 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅𝒆𝒅 𝒒𝒖𝒂𝒓𝒕𝒆𝒓𝒍𝒚

11. 7 2, 801.37 112.05 P1,485.28 1, 373.23
8 1, 428.14 57. 13 P1,485.28 1, 428.15
TOTALS P47, 055.63 P1882.23 P11, 882.24 P10, 000.00