TAKE HOME EXAMINATION 14 (PROBABILITY) books and three economics books and three

PROBLEM 1: economics books and arrange them on a shelf. PROBLEM 10:

Consider that a cafeteria is serving the following a. 120,820 From 4 chemists and 3 physicists, find the number
vegetables for lunch one day: carrots, broccoli, b. 200,850 of committees that can be formed consisting of 2
spinach, baked beans, corn, and green beans. c. 100,800 chemists and 1 physicist.
Suppose you will to order a vegetable plate with 4 d. 200, 700 a. 18
different vegetables. How many ways can this plate PROBLEM 6: b. 21
be prepared? In how many ways can 5 people line up to pay their c. 41
a. 15 bills if two particular persons refuse to follow each d. 31
b. 16 other? PROBLEM 11:
c. 14 a. 120 Two lottery tickets are drawn from 20 for first and
d. 17 b. 72 second prizes. Find the number of sample points in
PROBLEM 2: c. 24 the space S.
How many different groups of six passengers can fit d. 48 a. 350
into a four – passenger vehicle? PROBLEM 7: b. 422
a. 35 How many line segments can be formed by 15 c. 380
b. 25 distinct point? d. 450
c. 45 a. 156 PROBLEM 12:
d. 15 b. 78 There are 6 copies each of 3 different books. In how
PROBLEM 3: c. 210 many ways can they be arranged on a shelf?
How many ways can 10 canisters be arranged 6 d. 105 a. 13154132
canisters at a time? PROBLEM 8: b.10150136
a. 151,200 In how many ways can a hostess select six luncheon c. 17153136
b. 100,102 guests from 10 men if she is to avoid having a d.11653034
c. 111,100 particular two of them together at the luncheon? PROBLEM 13:
d. 210,233 a. 210 From the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, find the
PROBLEM 4: b. 140 number of seven digit combinations.
How many ways can 8 adults be arranged in 5 c. 84 a. 140
chairs? d. 151200 b.130
a. 6720 PROBLEM 9: c. 120
b. 5760 In how many ways can seven scientists be assigned d.110
c. 1680 to one triple and two double hotel rooms?
d. 3680 a. 220
PROBLEM 5: b. 200
From six history books and eight economics books, c. 190
how many ways can a person select two history d. 210
PROBLEM 14: c. 20,000 PROBLEM 23:
A student has ten posters to pin up on the walls of d. 90,000 How many even three-digit numbers can be formed
her room, but there is space for only seven. In how PROBLEM 19: from the digits 1, 2, 5, 6, and 9 if each digit can be
many ways can she choose the posters to be pinned A football conference consists of eight teams. If used only once?
up? each team plays every other team, how many a. 19
a. 48 conference games are played? b. 27
b. 28 a. 48 c. 30
c. 18 b. 28 d. 24
d. 38 c. 18 PROBLEM 24:
PROBLEM 15: d. 38 Forty electrical engineers, twenty chemical
How many different groups of seven passengers can PROBLEM 20: engineers, thirty mechanical engineers, and ten civil
fit into a five – passenger vehicle? In how many ways can a bowling player score in one engineers attend a banquet. A television station
a. 11 throw of a bowling ball? randomly selects four engineers to interview. What
b. 21 a. 1023 is the probability that someone from all four
c. 41 b. 30, 240 disciplines will be interviewed?
d. 31 c. 1024 a. 1.05 x 10ˉ³
PROBLEM 16: d. 720 b. 1.25 x 10ˉ³
Find the number of permutation than can be PROBLEM 21: c. 2.55 x 10ˉ³
formed from the word MISSISSIPPI. In how many ways can 10 trees be planted in a d. 3.85 x 10ˉ³
a. 51,235 circular lot? PROBLEM 25:
b.34,650 a. 362, 880 A bag contains 4 red balls, 3 green balls and 5 blue
c. 55,230 b. 3628800 balls. The probability of not getting a red ball in the
d.45,360 c. 40, 320 first draw is
PROBLEM 17: d. 326, 880 a. 2/3
Find the number of permutation than can be PROBLEM 22: b. 2/5
formed from the word COMMITTEE. How many lunches consisting of a soup, sandwich, c. 1/3
a. 35,360 dessert, and a drink are possible if we can select d. 1/2
b. 25,330 from 4 soups, 3 kinds of sandwiches, 5 desserts, PROBLEM 26:
c. 55,230 and 4 drinks? The probability that a student passes Mathematics
d. 45,360 a. 180 is 2/3, and the probability that he passes English is
PROBLEM 18: b. 240 4/9. If the probability of passing at least one course
How many 5-digit number divisible by 5 can be c. 350 is 4/5, what is the probability that he will pass both
formed from the digit 0, 1, 3, . . .,9 if repetition is d. 310 courses?
allowed? a. 5/6
a. 18,000 b. 32/135
b. 100,000 c. 14/45
d. 10/27 PROBLEM 31: PROBLEM 35:
How many color combinations are possible in How many ways can a local chapter of the
SITUATION 1: arranging 10 books on a shelf, 2 with red covers, 3 Philippine Institute of Civil Engineers schedule 3
In a certain college, 25% of the students failed with green cover and 5 with blue covers? speakers for 3 different meetings if they are all
mathematics, 15% of the students failed chemistry, a. 3020 available on any of 5 possible dates?
and 10% of the students failed both mathematics b. 2000 a. 80
and chemistry. A student is selected at random. c. 2220 b. 70
d. 2520 c. 50
27. If he failed chemistry, what is the SITUATION 2: d. 60
probability that he failed mathematics? A large jar contains more marbles than you are PROBLEM 36:
a. 1/9 willing to count. Instead, you draw some coins at The probability that Paula passes mathematics is
b. 4/9 random, replacing each coin before the next draw. 2/3, and the probability that she passes English is
c. 2/3 You record the picks in the following table: 4/9. If the probability of passing both courses is ¼,
d. 1/3 what is the probability that Paula will pass at least
28. If he failed mathematics, what is the one of these courses?
probability that he failed chemistry? a. 11/28
32. What is the probability that on your next draw
a. 3/7 b. 31/36
you will obtain a blue marble?
b. 2/5 c. 28/33
a. 0.4006
c. 3/14 d. 25/39
b. 0.3436
d. 5/14 PROBLEM 37:
c. 0.1116
29. What is the probability that he failed One bag contains 4 white balls and 3 black balls and
d. 0.2636
mathematics or chemistry? a second bag contains 3 white balls and 5 black
33. What is the probability that on your next draw
a. 5/9 balls. One ball is drawn from the first bag and
you will obtain a yellow marble?
b. 2/7 placed unseen in the second bag. What is the
a. 0.1811
c. 2/3 probability that a ball now drawn from the second
b. 0.1218
d. 3/10 bag is black?
c. 0.2818
PROBLEM 30: a. 40/56
d. 0.2018
Find the probability that 2 heads and 1 tail will b. 30/72
PROBLEM 34:
show in tossing 3 coins. c. 38/63
With 50 examination questions each of which has 4
a. 2/3 d. 31/36
given answers, how many possible answer patterns
b. 3/8 SITUATION 3:
are there?
c. 1/8 The probability that a regularly scheduled flight
a. 1.27 x 10³⁰
d. 1/3 departs on time is 0.83, the probability that it
b. 2.23 x 10³⁰
arrives on time is 0.92, and the probability that it
c. 1.12 x 10³⁰
departs and arrives on time is 0.78. Find the
d. 23.12 x 10³⁰
probability that a plane:
 42. If he has brown eyes, what is the probability b. 10/25
38. Arrives on time given that it departed on time. that he does not have brown hair? c. 1/2
a. 0.63 a. 3/8 d. 3/5
b. 0.85 b. 2/5 PROBLEM 47:
c. 0.79 c. 1/2 Find the probability of drawing 4 white balls from a
d. 0.94 d. 3/4 container with 10 white, 4 black and 3 red balls.
43. What is the probability that he has neither a. 4/30
39. Departed on time given that it has arrived on brown hair nor brown eyes. b. 2/10
time. a. 3/8 c. 3/4
a. 0.63 b. 2/5 d. 3/34
b. 0.85 c. 1/2 PROBLEM 48:
c. 0.79 d. 3/4 How many different ways can 3 red, 4 yellow, and 2
d. 0.94 PROBLEM 44: blue bulbs be arranged in a string of Christmas
PROBLEM 40: Three fair coins are tossed simultaneously. What is lights with 9 sockets?
The probability that Juan will be able to solve a the probability of getting 3 tails or 3 heads? a. 1180
problem is 1/2 and the probability that Pedro will a. 1/2 b. 1220
be able to solve it is 2/5. What is the probability b. 1/4 c. 1260
that the problem will be solved if both worked c. 3/8 d. 1602
together? d. 1/8 PROBLEM 49:
a. 0.70 PROBLEM 45: Find the probability of getting a prime number
b. 0.25 In the board examination, 75% of examinees passed thrice by tossing a die 5 times.
c. 0.5 in Civil engineering, 85% passed in Electrical a. 0.02455
d. 0.75 engineering, and 90% passed both. If an examinee b. 0.03125
SITUATION 4: is selected at random, what is the probability that c. 0.04245
In a certain town, 40% of the people have brown passed in Civil Engineering or Electrical d. 0.05642
hair, 25% have brown eyes, and 15% have both Engineering? PROBLEM 50:
brown hair and brown eyes. A person is selected at a. 0.2 Three lovely ladies, Jo Ann, Cheryl and Karen rode in
random from the town. b. 0.7 a minibus that has six vacant seats on each side. In
41. If he has brown hair, what is the probability that c. 0.5 how many ways can they be seated if Jo Ann insists
he also has brown eyes? d. 0.8 on sitting at the right side?
a. 3/8 PROBLEM 46: a. 640 ways
b. 2/5 A bag contains 10 white balls and 15 black balls. b. 650 ways
c. 1/2 Two balls are drawn in succession. What is the c. 660 ways
d. ¾ probability that one of them is black and the other d. 670 ways
is white?
a. 9/14
TAKE HOME EXAMINATION 13 PROBLEM 4: c. P14 622.56
(ENGINEERING ECONOMY 2) An engineer bought an equipment for P500 000. He d. P18 055.56
spent an additional amount of P30 000 for SITUATION 2:
PROBLEM 1: installation and other expenses. The estimated The cost of producing a commodity consist of
A new Civil Engineer produces a certain useful value of the equipment is 10 years. The P102.00 per unit for labor and material cost and
construction material at a labor cost of P16.20 per salvage value is x% of the first cost. Using the P54.00 per unit for other variable cost. The fixed
piece; materials cost P38.50 per piece and variable straight line method of depreciation, the book value cost per month amounts to P850, 000.00. The
cost of P7.40 per piece. The fixed charge on the at the end of 5 years will be P291,500. What is the commodity is sold at P740.00 each.
business is P100, 000 per month. If he sells the value of x? 8. What is the break-even quantity per month?
finished product at P95.00 per piece, how many a. 13 a. 1465
pieces must be manufactured each month in order b. 12 b. 1546
to break even? c. 15 c. 1645
a. 3040 d. 10 d. 1456
b. 3050 PROBLEM 5: 9. How many units must be produced each month
c. 3500 A machine costing P60 000 is estimated to have a in order that the net profit equals the cost?
d. 3060 book value of P 4,000 when retired at the end of 8 a. 3874
PROBLEM 2: years. Depreciation cost is computed using a b. 3547
A machine is purchase for P10,000.00. Its constant percentage of the declining book value. c. 3972
estimated life is 10 years, after which it will be sold What is the annual rate of depreciation in %? d. 2912
for P2,000.00. Find the book value at the end of the a. 32.25% 10. What is the net profit for a production of 4000
third year using sum-of-years digit (SOYD) method. b. 28.72% units per month, in pesos?
a. P6 072.72 c. 42.50% a. 1 486 000
b. P7 332.42 d. 18.50% b. 1 254 000
c. P8 493.32 SITUATION 1: c. 1 563 000
d. P3 452.23 The cost of the machine is P140 000 and has a d. 1 632 000
PROBLEM 3: useful life of 8 years. If the salvage value is P10 000; PROBLEM 11:
Calculate the difference of the future amount of 6. Determine the depreciation during the 4th year An electronic balance costs P90 000 and has an
annuity due and ordinary annuity. Given that the using double-declining balance method. estimated salvage value of P8 000 at the end of its
periodic payment is P 12 500 with interest rate of a. P14 765.63 10 years life time. What would be the book value
12 % compounded annually for 20 years. b. P16 250.00 after three years using the straight line method in
a. P86 462.93 c. P14 622.56 solving for depreciation?
b. P108 078.66 d. P18 055.56 a. P65 400
c. P193 231.53 7. Determine the depreciation during the 4th year b. P70 300
d. P273 248.18 using sum of the year’s digit method. c. P63 320
a. P14 765.63 d. P72 230
b. P16 250.00
PROBLEM 12: PROBLEM 15: PROBLEM 18:
A broadcasting corporation purchased an A bond issue of P200 000 in 10 year bonds, in P1 A one-bagger concrete mixer can be purchased with
equipment for P53 000 and paid P1 500 for freight 000 units, paying 16% nominal interest in semi- a down payment of P8 000 and equal installments
and delivery charges to the job site. The equipment annual payments, must be retired by the use of a of P600 each paid at the end of every month for the
has a normal life of 10 years with a trade-in value of sinking fund that earns 12% compounded semi- next 12 months. If money is worth 12%
P5 000 against the purchase of a new equipment at annually. What is the total semi-annual expense? compounded monthly, determine the equivalent
the end of the life. Determine the annual a. P27 421 cash price of the mixer.
depreciation cost by the sinking fund method. b. P21 437 a. P14 753.05
Assume interest at 6.5% compounded annually. c. P24 127 b. P13 475.25
a. P22 590 d. P23 143 c. P17 055.75
b. P25 920 PROBLEM 16: d. P15 755.05
c. P32 250 A man wants to make 14% nominal interest PROBLEM 19:
d. P29 520 compounded semi-annually on a bond investment. A steam boiler is purchased on the basis of
PROBLEM 13: How much should the man be willing to pay now for guaranteed performance. However, initial tests
A plant bought a calciner for P220 000 and used it a 12%, P10 000 bond that will mature in 10 years indicate that the operating cost will be P400 more
for 10 years, the life span of the equipment. What is and pays interest semi-annually? per year than guaranteed. If the expected life is 25
the book value of the calciner after 5 years of use? a. P5 980 years and money is worth 10%, what deduction
Assume a scrap value of P22 000 for declining b. P9 805 from the purchase price would compensate the
balance method. c. P5 089 buyer for the additional operating cost.
a. P62 450 d. P8 950 a. P3 630.82
b. P70 130 PROBLEM 17: b. P3 038.63
c. P69 570 Mr. Smith bought a bond having a face value of P1 c. P2 382.08
d. P58 230 000 for P970. The bond rate was 14% nominal and d. P6 330.28
PROBLEM 14: interest payments were made to him semi-annually PROBLEM 20:
A structure costs P12 000 new. It is estimated to for a total of 7 years. At the end of the seventh year, How much money would you have to deposit for
have a life of 5 years with a salvage value at the end he sold the bond to a friend at a price that resulted five consecutive years starting one year from now if
of life P1 000. Determine the book value at the end a yield of 16% nominal on his investment. What was you want to be able to withdraw P50 000 ten years
of 5 years of life using Sum-of-the-years digits the selling price? from now? Assume the interest is 14% compounded
method. a. P1 139.50 annually.
a. P2 500 b. P1 531.13 a. P3 829.50
b. P2 000 c. P1 153.90 b. P3 638.30
c. P1 500 d. P1 113.35 c. P3 928.60
d. P1 000 d. P4 023.28
PROBLEM 21: PROBLEM 24: PROBLEM 27:
A company sets aside P200 000 at the end of each It costs P50 000 at the end of each year to maintain Annual maintenance costs for a machine are P1 500
year for plant expansion. If the fund drawn 8% a section of Kennon road in Baguio City. If money is this year and are estimated to increase 10% each
compounded annually, how long will it take before worth 10%, how much would it pay to spend year every year. What is the present worth of the
P2 500 000 can be saved? immediately to reduce the annual cost to P10 000? maintenance costs for six years if i = 8%.
a. 7 a. P260 500 a. P9 779.72
b. 12 b. P350 000 b. P8 728.79
c. 9 c. P500 500 c. P8 297.77
d. 10 d. P400 000 d. P7 927.28
PROBLEM 22: PROBLEM 25: SITUATION 3:
A certain annuity pays P80 at the end of every 3 To maintain a bridge, P5 000 will be required at the A contractor imported a bulldozer for his job, paying
months. If the present value of the annuity is P1 end of 3 years and annually thereafter. If money is P250 000 to the manufacturer. Freight and
200 and accumulated amount is P2 000, determine worth 8%, determine the capitalized cost of all insurance charges amounted to P18 000; broker’s
the nominal rate. future maintenance. fees and arrastre services, P8 500; taxes, permits,
a. 11.33% compounded yearly a. P55 339.93 and other expenses, P25 000. If the contractor
b. 12.17% compounded monthly b. P59 333.55 estimates the life of the bulldozer to be 10 years
c. 9.% compounded quarterly c. P53 583.68 with a salvage value of P20 000, determine the
d. 10.67% compounded quarterly d. P56 385.40 book value at the end of 6 years using;
PROBLEM 23: PROBLEM 26: 28. straight-line formula.
A lathe for a machine shop costs P60 000 if paid in The capitalized cost of a piece of equipment was a. P132 600
cash. On the installment plan, a purchaser should found to be P142 000. The rate of interest used in b. P123 800
pay P20 000 down payment and 10 quarterly the computations was 12%, with a salvage value of c. P120 500
installments, the first due at the end of the first year P10 000 at the end of a service life of 8 years. d. P130 000
after purchase. If money is worth 15% compounded Assuming that the cost of perpetual replacement 29. Sinking fund formula at 8%.
quarterly, determine the quarterly installment. remains constant, determine the original cost of the a. P149 581.94
a. P5 439.18 equipment. b. P194 694.49
b. P4 319.20 a. P86 788.24 c. P158 949.69
c. P3 518.35 b. P78 688.78 d. P164 185.14
d. P5 615.40 c. P82 842.68
d. P88 687.42
PROBLEM 30: c. P271.92 PROBLEM 36:
An engineering firm purchased 12 years ago a heavy d. P212.79 Corrosive liquids are transported through pipes in a
planer for P50 000. As the life of the planer was 20 PROBLEM 33: factory. Ordinary pipes will have an installed cost of
years, a straight-line depreciation reserve has been A man invests P10 000 now for the college P30 000 and their useful life is 3 years. Stainless
provided on that basis. Now the firm wishes to education of his 2-year old son. If the fund earns steel pipes are highly resistant to the corrosive
replace the old planer with newly-designed planer 14% effective, how much will the son get each year effect of the liquids and are being considered as an
possessing several advantages. It can sell the old starting from his 18th to the 22nd birthday? alternative. These pipes are estimated to have an
planer with newly-designed planer possessing a. P21 916.21 installed cost of P55 000. Scrap value is zero in each
several advantages. It can sell the old planer for P10 b. P29 116.90 case. If money is worth 8% and assuming
000. The new machine will cost P70 000. How much c. P20 791.64 replacement cost to be the same as the original
new capital will be required to make the purchase? d. P22 619.79 prices, what should be the useful life of the
a. P60 000 PROBLEM 34: stainless steel pipes to have equal capitalized cost
b. P30 000 A farmer bought a tractor costing P25 000 payable as the ordinary pipes?
c. P50 000 in 10 semi-annual payments, each installment a. 7.4 yrs
d. P40 000 payable at the beginning of each period. If the rate b. 5.5 yrs
SITUATION 4: of interest is 26% compounded semi-annually, c. 6.2 yrs
The cost of constructing and developing an determine the amount of each installment. d. 4.8 yrs
irrigation system averages P1 600 per hectare. If the a. P6 214.35 PROBLEM 37:
money borrowed to build it is to be repaid by end b. P4 077.20 A fund is to be donated by a wealthy man to
users in 15 years, c. P2 774.77 provide annual scholarships to deserving students.
31. What is the contribution to a sinking fund the d. P3 275.15 The fund will grant P5 000 for each of the first 5
end-users would have to pay in annual irrigation PROBLEM 35: years, P8 000 for each of the next 5 years, and P10
fees per hectare if the rate of interest on the fund is A manufacturing plant installed a new boiler at a 000 each year thereafter. The scholarship will start
7% compounded annually? total cost of P150 000 and is estimated to have a one year after the fund is established. If the fund
a. P175.67 useful life of 10 years. It is estimated to have a scrap earns 8% interest, what is the amount of the
b. P156.75 value at the end of its useful life of P5 000. If donation?
c. P167.17 interest is 12% compounded annually. Determine its a. P96 710.17
d. P177.55 capitalized cost. b. P79 697.70
32. If the interest on the loan is 12% per annum, a. P187 582.21 c. P99 601.71
and the operation and maintenance expenses, b. P255 128.78 d. P90 690.66
estimated to be P37 per hectare per year, were also c. P158 949.69
assumed by the end-users instead of by the d. P218 855.87
government, what would be the total annual
irrigation charge per hectare.
a. P219.27
b. P277.21
PROBLEM 38: PROBLEM 41: PROBLEM 44:
Juan borrowed P2 400 at 1% per month payable in The JIZ Construction Co. agreed to pay P100 000 for If P1 000 becomes P1 811.36 after 5 years when
24 equal end-of-the-month payments. How much of a new bulldozer with an expected life of 20 years. invested at an unknown rate of interest
the loan remains unpaid immediately after he has After delivery, it was found that the dozer had only compounded bimonthly, determine the
paid the 12th payment? 85% of the guaranteed horsepower. If a more corresponding effective rate.
a. P1 219.27 efficient machine would increase the earnings of a. 13.25%
b. P1 727.21 the company by an estimated P12 000 annually, b. 18.72%
c. P1 271.60 what would the company be willing to pay for the c. 12.62%
d. P1 212.79 more efficient machine? Interest is 12% d. 15.48%
PROBLEM 39: compounded annually. PROBLEM 45:
A back pay certificate is offered for sale on which a. P189 633.32 How many years are required for P1 000 to increase
the yearly payments are P996, partly principal and b. 178 688.78 to P2 000 if invested at 9% per year compounded
partly interest. There are 8 annual payments still c. P182 842.68 monthly?
due, the first of these due one year from now. How d. P168127.42 a. 7.7 yrs
much should one pay for this note in order to get PROBLEM 42: b. 5.5 yrs
5% interest, compounded annually, on the Determine the ordinary simple interest on P10 000 c. 6.2 yrs
investment. for 9 months and 10 days if the rate of interest is d. 4.8 yrs
a. P4 637.46 12%. PROBLEM 46:
b. P6 437.36 a. P743.12 A motor that cost P100 000 has an estimated life of
c. P4 433.74 b. P833.79 I0 years and can be sold at P20 000 at the end of 10
d. P7 346.64 c. P933.33 years. Compute the book value at the end of 7
PROBLEM 40: d. P903.75 years using sinking fund method if i = 10%.
A man bought a Ford Fairlane, for P21 000 on PROBLEM 43: a. P64 862.59
installment basis at the rate of 12% per annum on A man borrows P6 400 from a loan association. In b. P47 122.78
the unpaid balance. If he paid a down payment of P repaying thus debt he has to pay P400 at the end of c. P55 354.34
6000 in cash and proposes to pay the balance in 20 every 3 months on the principal and a simple d. P51 567.94
monthly payments, what should these monthly interest of 16% on the principal outstanding at that SITUATION 5:
payments be? time. Determine the total amount he has paid after An annual payment of P14 000 is made with
a. P643.12 paying all his debt. interest rate of 10% compounded quarterly for 16
b. P831.23 a. P8 576 years.
c. P774.77 b. P7 578 47. Determine the present worth considering
d. P903.75 c. P6 758 annuity due.
d. P5 577 a. P441 808.97
b. P455 808.97
c. P545 808.97
d. P554 808.97
48. Compute the difference of the future amounts
of the annuity due and ordinary annuity.
a. P35 919.62
b. P53 991.62
c. P46 199.62
d. P29 491.62
49. If the first payment was made at the end of the
10th year, determine the present worth considering
ordinary annuity.
a. P2 159 665.00
b. P199 328.44
c. P171 447.89
d. P182 809.96
PROBLEM 50:
Anne buys a television set from a merchant who
asks for Php1 250 if paid at the end of 60 days.
However, Anne wishes to pay immediately and the
merchant offers to compute the cash price today if
money is worth 8% simple interest?
a. P 1335.23
b. P 1233.56
c. P 1632.33
d. P 1523.23
 d. 0.0019 m 8. What is the correction for temperature?
TAKE HOME EXAMINATION 23 4. Which of the following nearly gives the a. -0.00522
(SURVEYING 1) value of the error due to sag? b. -0.00904
SITUATION 1 a. - 0.0625 c. 0.00132
A 50 m steel tape was standardized and supported b. - 0.1114 d. 0.00125
throughout its whole length and found to be c. + 0.27 9. What is the pull correction?
0.00329 m longer at an observed temperature of d. + 1.08 a. -0.00522
41.6°C and a pull of 12 kilos. This tape was used to 5. Find the corrected distance between the b. -0.00904
measure a line that was found to be 683.21 m at two points. c. 0.00132
average temperature of 49.8°C at the same pull. a. 3116.84 m d. 0.00125
Coefficient of linear expansion is 0.0000116 m per b. 3134.97 m 10. What is the correction due to sag?
m. degree centigrade. c. 3094.42 m a. 0.00522
1. Determine the standard temperature. d. 3121.19 m b. -0.00904
a. 50.76°C SITUATION 3: c. 0.00132
b. 35.93°C A 30 m metallic tape 6 mm too long was used to d. 0.00125
c. 44.63°C measure a rectangular lot. The dimensions recorded 11. What is the true horizontal distance?
d. 68.25°C were 70.5 m long by 37.10 wide. a. 472.229
2. What is the correct length of the line? 6. What was the error in area due to the error in the b. 472.921
a. 683.32 m length of the tape? c. 472.561
b. 663.28 m a. 1.025 d. 472.765
c. 638.23 m b. 1.029 SITUATION 5:
d. 633.88 m c. 1.046 The angles in a triangle were as follows:
SITUATION 2: d. 1.008
A 100 m tape weighing 3 kg was standardized and 7. What are the actual dimensions of the lot?
found to be 0.04 m short at temperature 15°C and a a. 70.514 m x 37.107m
30 N pull when supported throughout. It was b. 70.526 m x 37.109 m
measure a distance which was recorded as 3125.68 c. 70.519 m x 37.112 m 12. Compute the most probable value of angle A.
m when the temperature was 20°C and the pull 45N d. 70.522 m x 37.116 m a. 41°85’00”
supported only at quarter points. E = 200GPa. SITUATION 4: b. 46°9’23”
Density of material is 7600 kg/m³, coefficient of A line recorded as 472.90 m is measured with a 0.65 c. 40°46’9”
thermal expansion = 11.6 x 10ˉ⁶ m/m °C. kg tape 30.005 m long with cross-sectional area 3 d. 39°41.2’
3. Which of the following nearly gives the value of mm² at 20°C under a 50 N pull. At the time of 13. Compute the most probable value of angle
the error due to pull? measurement the temperature dropped to 5°C and C.
a. 0.0045 m the applied pull was 75 N. The line was 3% grade, E a. 62°25’23”
b. 0.0065 m = 200 GPa, α= 0.0000116 m/°C. b. 76°48’27”
c. 0.0035 m c. 64°46’29”
d. 50°40’25” c. 472.56 m b. 25.71°C
PROBLEM 14: d. 472.211m c. 28.27°C
A line was determined to be 2395.25 m, when SITUATION 6: d. 15.61°C
measured with a 30 m steel tape supported The distance recorded for a baseline was 430.60 21. Compute the total error due to change in
throughout its length under a pull of 4 kg at a mean using a 100 m steel tape standardized at 15°C and temperature.
temperature of 35°C. Tape used is standard length with a standard pull of 10 kg. At the time the *
at 20°C under a pull of 5 kg. Cross-sectional area of temperature was measured to be 20°C and the pull 1 point
tape is 0.03 sq. cm. Coefficient of thermal exerted was 16 kg. Unit weight of steel is 7.86 g/cc a. 0.0145 m
expansion is 0.0000116/°C. Modulus of elasticity of and weight of steel is 2.67 kg, E = 200000 MPa, α = b. 0.0012 m
tape is 2 x 10⁶ kg/cm².Determine the corrected 0.70 x 10ˉ⁶ m/°C. c. 0.0124 m
length of the line. 17. What is the correction for temperature? d. 0.0213 m
a. 2395.1514 m a. 0.0082 22. Compute the corrected length of the line.
b. 2395.5426 m b. 0.00097 a. 662.674 m
c. 2395.6269 m c. 0.0090 b. 662.841 m
d. 2395.2468 m d. 0.00035 c. 662.741 m
PROBLEM 15: 18. What is the pull correction? d. 662.561 m
The transit is inclined at 0°05’00” with respect to a. 0.0082 PROBLEM 23:
the horizontal axis. If the first sight is at vertical b. 0.00097 A stadia intercept of 1.02 m was recorded for a
angle of 50° and the second is -25°, what is the c. 0.00866 horizontal sight stadia surveying. If the stadia
error in the horizontal angle? d. 0.00035 interval factor and stadia constant are 98.36 and
a. 0°8’17.42” 19. What is the true length of the baseline? 0.30 respectively, compute for the horizontal
b. 0°7’4.6” a. 430.61 distance.
c. 0°6’23.12” b. 430.64 a. 100.63 m
d. 0°5’30.55” c. 430.62 b. 106.03 m
PROBLEM 16: d. 430.65 c. 600.13 m
A line recorded as 472.90 m long. It was measured SITUATION 7: d. 601.30 m
with 6.37N tape which is 30.005 m long at 20°C A 50 – m tape was standardized and supported SITUATION 8:
under a 50N pull supported at both ends. During throughout its whole length and found to be A 100 m tape weighing 4 kg was standardized and
measurement the temperature is 5°C and the tape 0.00205 m long at an observed temperature of found to be 0.04 m short at temperature 15°C and a
is suspended under a 75N pull. The line was 31.8°C and a pull of 10 kilos. This tape was used to 30 N pull when supported throughout. It was
measured on 3% grade. E = 200GPa, coefficient of measure a line which was found to be 662.702 m at measure a distance which was recorded as 3125.68
linear expansion is 0.0000116 m/m °C and cross- an average temperature of 24.6°C using the same m when the temperature was 20°C and the pull 45N
sectional area of tape is 3 mm². Compute the true pull. Coefficient of thermal expansion is supported only at quarter points. E = 200GPa.
horizontal distance of the line. 0.0000116m/m°C. Density of material is 7600 kg/m³, coefficient of
a. 473.51 m 20. Compute the standard temperature. thermal expansion = 11.6 x 10ˉ⁶ m/m°C
b. 472.703 m a. 18.51°C
24. Which of the following nearly gives the value of 28. Determine the error due to temperature. c. 118.46 m
the error due to pull? a. 0.05812 m d. 142.65 m
a. 0.0045 m b. 0.04572 m PROBLEM 34:
b. 0.0035 m c. 0.04958 m The interior angles A, B and C of a triangular
c. 0.0065 m d. + 0.05684 m traverse were measured with the same precision.
d. 0.001425 m 29. Find the error due to sag. The result was as follows:
25. Which of the following nearly gives the a. -0.579 m
value of the error due to sag? b. 0.489 m
a. -0.0495 c. -0.639 m
b. +0.27 d. -2.555 m
c. -0.198 30. What is the corrected distance?
d. +1.08 a. 12134.884m a. 57°24° 10.85’
26. Find the corrected distance between the b. 12420.995 m b. 57°27° 41.54’
two points. c. 13004.265m c. 57°34° 20.45’
a. 3118.87 m d. 12386.953 m d. 57°14° 56.75’
b. 3094.42 m SITUATION 10: PROBLEM 35:
c. 3134.97 m Two points A and B are 2000 m each away from The error of scale on the map is 0.02 mm. The map
d. 3125.91 m point C, from which the measured vertical angle to has a scale of 1:250,000 meters. Find the error of
PROBLEM 27: A is +1°30’ and to B is -2°30’.Considering the effect measurement on the ground.
A distance measured with a 50 m steel tape is of the earth curvature and refraction. a. 2 m
recorded as 696.41 m. What is the correct length of 31. Compute the difference in elevation b. 4 m
the line if the tape is known to be 0.015 m short? between A and C. c. 5 m
a. 696.6189 a. 52.64 m d. 6m
b. 696.2027 b. 52.23 m PROBLEM 36:
c. 696.2019 c. 52.12 m Two hills 100 km apart has elevations of 80 m and
d. 696.2011 d. 52.84 230 m respectively. What would be the minimum
SITUATION 9: 32. Compute the difference in elevation height of tower that could be constructed at B so
A 700 m tape weighing 7 kg was standardized and between A and B. that it would be visible from A considering the
found to be 0.05 m short at temperature of 30°C a. 135.46 m effect of curvature and refraction correction.
and a 50 N pull when supported all throughout . It b. 139.69 m a. 564 m
was used to measure a distance which was c. 126.51 m b. 834 m
recorded to be 12431.84 m when the temperature d. 124.81 m c. 456 m
was 37°C and the pull was 58 N supported only at 33. Compute the at point C if elevation B is 55.6 d. 526 m
quarter points . E = 200 GPa. Density of the material m
is 7500 kg/m³, coefficient of thermal expansion is a. 156.14 m
11.6x10ˉ⁶. b. 138.45 m
PROBLEM 37: SITUATION 11: coefficient of thermal expansion equal to
Lines of levels are run from B11 to B12 over four Two hills A and C have an elevation of 680 m and 0.0000116m/°C.
different routes. BM12 is above BM11 and elevation 620 m respectively. In between A and C is another a. 0.01
of BM11 is 220 m above sea level. Compute the hill B of elevation 645 m and located 12 km from A b. 0.02
probable elevation of BM12 above sea level. and 15 km from C. Consider the effects of earth’s c. 0.03
curvature and atmospheric refraction. d. 0.04
40. Compute the elevation of the line of sight at B SITUATION 12:
so that A and C are intervisible. The observed interior angles of a triangle and their
a. 641.1 m corresponding number of measurements were as
b. 604.81 m follows.
a. 374.97 m c. 481.20 m
b. 376.51 m d. 3412.1 m
c. 376.81 m 41. Find the equal height of towers constructed
d. 65.03 m at A and C so that the three hills are intervisible.
PROBLEM 38: a. 5.817 m
A 50 m tape that was 0.05 m too short was used to b. 4.817 m 45. Compute the most probable value of angle C.
measure a square lot that was 0.05 m too short. If c. 3.835 m a. 66°27’16”
the area of the lot is known to be 5000 m², what is d. 2.817 m b. 66°10’11”
the error in the computed area? 42. Find the height of tower constructed at A so c. 67°27’11”
a. 11.005 m² that B and C are still intervisible. d. 67°06’45”
b. 10.015 m² a. 5.87 m 46. Compute the most probable value of angle A.
c. 9.004 m² b. 6.9024 m
d. 12.003 m² c. 4.85 m a. 41°27’16”
PROBLEM 39: d. 8.71 m b. 40°10’11”
A 50 m tape is 0.04 m too short used to measured a PROBLEM 43: c. 40°43’38”
line AB. The measured distance was known to be The error of scale on the map is 0.02 mm. The map d. 40°06’45”
130.45 m. What is the adjusted distance of the line has a scale of 1:200,000 meters. Find the error of 47. Compute the most probable value of angle
AB. measurement on the ground. B.
a. 130.712 m a. 2 m a. 72°27’16”
b. 130.346 m b. 4 m b. 72°10’11”
c. 130.215 m c. 5 m c. 72°43’38”
d. 130.534 m d. 6m d. 72°49’5”
PROBLEM 44:
A 50 m length of tape is of standard length at
temperature of 50.76 °C. What would be the error
in the length of tape at a temp. of 68°C. Use
PROBLEM 48:
Point A is in between points B and C. The distances
of B and C from point A are 1000m and 2000m
respectively. Measured from point A, the angle of
elevation of point B is 18°30’ while that of C is 8°15’.
Find the difference in elevation between B and C
considering effects of curvature and refraction.
a. 44.4 m
b. 32.21 m
c. 46.43 m
d. 42.14 m

SITUATION 13:
The following data were obtained by stadia
measurement: vertical angle = +18°23’, and
observed stadia intercept = 2.20 m. The stadia
interval factor of the instrument used is 95.5 and C
= 0.30 m. If the height of the instrument is 1.62 m,
and the rod is taken at 1.95 m.
49. Determine the vertical distance from the
center of the instrument to point on the rod
bisected by the horizontal cross hair.
a. 66.26 m
b. 62.88 m
c. 63.63 m
d. 68.35 m
50. Determine the inclined or slope distance
from the instrument center to the portion on the
rod bisected by the horizontal cross hair.
a. 189.48 m
b. 201.43 m
c. 199.38 m
d. 172.78 m
TAKE HOME EXAMINATION 24 (SURVEYING 2) and weight of steel is 3 kg, E = 200000 MPa, α = PROBLEM 9:
11.6 x 10ˉ⁶ m/°C. The data of a closed traverse are as follows:
PROBLEM 1: 4. What is the correction for temperature? Find the corrected latitude of line 1-2 using
The DMD of the line BC is equal to 426.20. The a. 0.0082 transit rule.
departure of the line AB is 117.56 m. What is then b. 0.00097
the departure of line BC? c. 0.00866
a. 156.87 m d. 0.0029
b. 191.08 m 5. What is the pull correction?
c. 141.24 m a. 0.0082
d. 123.45 m b. 0.00193 a. 9.312
PROBLEM 2: c. 0.0090 b. 9.215
The total error in latitude is -15.97 while the sum of d. 0.00035 c. 9.012
all positive and negative latitudes 1870.97. What is 6. What is the true length of the baseline? d. 9.1985
the correction to be applied to a line which has a. 430.6119 SITUATION 2:
latitude of 640.13 using transit rule? b. 430.6403 Given the data shown:
a. 5.46 c. 510.479
b. 6.42 d. 510.6505
c. 4.82 PROBLEM 7:
d. 4.35 A lot ABCDEFA is a closed traverse forming a regular
PROBLEM 3: hexagon with sides equal to 100m. What is the 10. What is distance AB?
A series of perpendicular offsets were taken from bearing of FA if the bearing of AB is N 25° E? a. 28.51
the transit line to a curved boundary line. These a. N 35° W b. 19.25
offsets were taken 10 m apart and were recorded as b. N 25° W c. 24.21
3.2 m, 4.8 m, 6.2 m and 7.1 m. Determine the area c. S 35° W d. 18.02
included between the transit line and the curved d. S 40° W 11. What is distance BC?
boundary using trapezoidal rule. PROBLEM 8: a. 13.21
a. 187.60 m² In an old survey made when the declination was b. 15.92
b. 161.50 m² 2°10’ W, the magnetic bearing of a given line was N c. 12.03
c. 318.45 m² 35°45’ E. The declination of the same place is d. 17.52
d. 436.45 m² 3°30’E. What is the present magnetic bearing that 12. Find the DMD for side AB.
SITUATION 1: would be used in retracing the line? a. -7.85
The distance recorded for a baseline was 510.43 a. N 33°05’ E b. -6.51
using a 50 m steel tape standardized at 15°C and b. N 35°05’ E c. -5.99
with a standard pull of 10 kg. At the time the c. N 32°15’ E d. -8.55
temperature was measured to be 20°C and the pull d. N 30°05’ E
exerted was 16 kg. Unit weight of steel is 7.86 g/cc
13. Find the DMD for side BC. 18. Compute the total error due to change in d. 3.45
a. -1.85 temperature. SITUATION 5:
b. -2.02 a. 0.0116 m Three independent lines of levels are run from BM₁
c. -0.96 b. 0.0012 m to BM₂. Route A is 6 km long, route B is 4 km long
d. -1.75 c. 0.0124 m and route C is 8 km long .By route A, BM₂ is 82.27 m
14. Find the DMD for side CD. d. 0.0282 m above BM₁, by route B, BM₂ is 82.40 m above BM₁
a. 85.65 19. Compute the corrected length of the line. and by route C, BM₂ is 82.10 m above BM₁.
b. 72.05 a. 1662.674 m 23. Compute the probable difference in elevation
c. 82.40 b. 1662.841 m between BM₁ and BM₂*
d. 80.66 c. 1662.98 m 1 point
15. Find the DMD for side DA. d. 1662.561 m a. 82.29 m
a. 75.65 SITUATION 4: b. 82.95 m
b. 74.03 For a closed traverse, the data are as follows; c. 82.56 m
c. 73.56 d. 82.63 m
d. 72.76 24. Compute the probable elevation of BM₂ if the
16. Find the area using DMD Method. elevation of BM₁ is 30.5 m*
a. 1215.98 1 point
b. 1625.31 a. 113.13 m
c. 1449.66 b. 112.79 m
d. 1315.04 c. 113.45 m
SITUATION 3: d. 113.06 m
20. Find the bearing of course DA.
A 100 – m tape was standardized and supported SITUATION 6:
a. S 11°24 W
throughout its whole length and found to be 0.02 m The data for a closed traverse are as follows:
b. S 10° W
long at an observed temperature of 31.8°C and a
c. N 8°48’34” W
pull of 10 kilos. This tape was used to measure a
d. S 16°15’20” W
line which was found to be 1662.783 m at an
21. Find the distance of course DA.
average temperature of 24.6°C using the same pull.
a. 110.8 m
Coefficient of thermal expansion is
b. 124.3 m
0.0000116m/m°C.
c. 95.21 m
17. Compute the standard temperature 25. Compute the distance of course CD.
d. 172.3 m
a. 14.56°C
22. Find the area of the closed traverse in acres. a. 295.94 m
b. 25.71°C
Clue: 1 acre = 0.405 hectare b. 107.98 m
c. 28.27°C
a. 2.39 c. 297.50 m
d. 15.61°C
b. 5.61 d. 179.26 m
c. 1.41
26. Determine the area of the traverse in acre. a. 7°30’
a. 3.19 c. 8°15’
b. 1.41 b. 5°20’
c. 9.13 d. 6°45’
d. 6.25 SITUATION 9:
SITUATION 7: A magnetic bearing of N 54°30’E was observed
Two points A and B are 3000 m each away from a. 3548.39 m² along a line XY in July 1982. The declination for the
middle point C, from which the measured vertical b. 4077.18 m² area surveyed is found by interpolation from
angle to A is +2°30’ and to B is -1°30’.Considering c. 3977.56 m² isogonic chart dated 1975 to be 17°30’E with an
the effect of the earth curvature and refraction. d. 1746.42 m² annual change of 0°1’ westward.
27. Compute the difference in elevation SITUATION 8: 34. Compute the change of declination of the
between A and C. Three independent lines of levels are run from BM₁ line.
a. 52.64 m to BM₂. Route A is 4 km long, route B is 6 km long a. 0°07’W
b. 131.591 m and route C is 8 km long .By route A, BM₂ is 82.27 m b. 7°30’E
c. 52.12 m above BM₁, by route B, BM₂ is 82.40 m above BM₁ c. 0°37’W
d. 52.84 and by route C, BM₂ is 82.10 m above BM₁. d. 0°23’W
28. Compute the difference in elevation 31. Compute the probable difference in elevation 35. Compute the true bearing of the line.
between A and B. between BM₁ and BM₂. a. 71°53’
a. 135.46 m a. 82.29 m b. 37°00’
b. 209.541 m b. 82.708 m c. 72°07’
c. 126.51 m c. 82.56 m d. 37°07’
d. 124.81 m d. 82.63 m PROBLEM 36:
29. Compute the elevation at point C if 32. Compute the probable elevation of BM₂ if Transit is set up at A observed a stadia intercept of
elevation B is 70 m. the elevation of BM₁ is 40 m 2.2 m at a vertical angle of +18°23’. The stadia
a. 156.14 m a. 113.13 m interval factor of the instrument used is 95.5 and
b. 138.45 m b. 122.271 m stadia constant of 0.3 m. If the height of
c. 118.46 m c. 113.45 m instrument is 1.62 m, and the rod reading is 1.95 m.
d. 147.949 m d. 113.06 m Determine the horizontal distance from the transit
PROBLEM 30: PROBLEM 33: set up at A to the rod @ point B
An engineer set up a transit inside a quadrilateral Point A is between point B and C, the distances of a. 201.44 m
lot and observes the bearings and distances of the point B and C from point A are 1km and 2 km b. 189.203 m
corner A,B,C and D of the lot as follows: Compute respectively. Measured from point A, the angle of c. 198.45 m
the area of the lot. elevation of point B is 18°30’ while that of C is θ.The d. 186.56 m
difference in the elevation of B and C is 44.4 m.
Considering the effects of curvature and refraction.
What is the value of θ?
PROBLEM 37: PROBLEM 41:
Three independent lies of levels are run from BM1 From the given differential leveling notes, find the
to BM2. By route A, BM2 is 82.27 m above BM1, elevation of BM₂.
route B 82.4, and by C 82.10. If the probable error
in the measurement of the difference in elevation
for route A, B and C are respectively 0.40 m, 0.25 m
and 0.5 m., compute the elevation of BM2 a. 374.97 m
considering elevation of BM1 is 30.5 m. b. 376.51 m
a. 182.561 m c. 376.81 m
a. 44.938 m
b. 112.823 m d. 65.03 m
b. 42.938 m
c. 82.323 m PROBLEM 45:
c. 40.938 m
d. 72.451 m In an old survey made when the declination was
d. 41.938 m
SITUATION 10: 2°20’ W, the magnetic bearing of a given line was N
SITUATION 11:
The side BC of a triangular lot ABC is 400 m. Angle 30°45’ E. The declination of the same place is
Given the data of the closed traverse:
ABC is 50°. A 150 m line DE, parallel to BC, divides 3°30’E.What is the present magnetic bearing that
the lot making area BCED to be 50 977.80 m². would be used in retracing the line?
38. Determine the measure of angle A. a. N 33°05’ E
a. 63° b. N 35°05’ E
b. 67° c. N 24°55’ E
c. 69° d. N 30°05’ E
d. 71° PROBLEM 46:
39. Determine area of triangle ABC. 42. Find the distance CD in meters. The following are bearings and distances taken on a
a. 59319.37m² a. 70.4626 closed traverse. A line from vertex A to point F on
b. 53991.67m² b. 77.4626 the side BC divides the traverse and makes area ABF
c. 51 399.67 m² c. 75.4626 to be 280 000 m².Find the length of the line BF.
d. 59913.67m² d. 80.4626
40. Determine the length of the line AE. 43. Find the distance DA in meters.
a. 124.83 m a. 70
b. 142.38 m b. 77
c. 148.23 m c. 75
d. 134.28 m d. 80
a. 538.86 m
PROBLEM 44:
b. 750.12 m
Lines of levels are run from B11 to B12 over four
c. 802.33 m
different routes. BM12 is above BM11 and elevation
d. 878.35 m
of BM11 is 220 m above sea level. Compute the
probable elevation of BM12 above sea level.
PROBLEM 47: c. 99.01 m
A closed traverse has the ff. data: What is the length d. 88.14 m
of course 3-4? PROBLEM 50:
The following are bearings and distances taken on a
closed traverse. A line from vertex A to point F on
the side BC divides the area and makes area ABF to
be 280 000 m². Find the length of the line AF.

a. 35.2 m
b. 38.65 m
c. 39.3 m
a. 878.35 m
d. 37.5 m
b. 750.12 m
PROBLEM 48:
c. 646.63 m
Given the data of the closed traverse. Find the
d. 593.30 m
distance DA in meters.

a. 70
b. 75
c. 77
d. 80
PROBLEM 49:
A triangular lot was formed by three straight lines.
Bearing of line AB is N45°E and is 160 m long; line
AC and BC are 190 m long. Point E is 100 m
from A on side AB connects point D on side AC
which divides the area, making area AED to be 2/5
of the whole area. Find the distance DE.
a. 120.6 m
b. 109.1 m
TAKE HOME EXAMINATION 25 (SURVEYING 3) d. 267.37 m 9. Compute the arc length from PC to A.
PROBLEM 1: 4. Compute the stationing of P.C.C. a. 40.00 m
The perpendicular distance between two parallel a. 30 + 106.97 b. 37.43 m
tangents is equal to 8 m, central angle equal to 8° b. 30 + 110.73 c. 33.74 m
and the radius of curvature of the first curve is 175 c. 30 + 118.46 d. 37.34 m
m. find the radius of second curve of the reversed d. 30 + 235.45 10. Compute the chord length from PC to B.
curve. SITUATION 2: a. 84.59 m
a. 647 m Given a compound curve with I₁ = 18°, I₂ = 23.5° , D₁ b. 89.54 m
b. 800 m = 2° and D₂ = 4°. c. 99.50 m
c. 555 m 5. Determine the length of the common tangent d. 89.45 m
d. 250 m a. 310.73 m PROBLEM 11:
PROBLEM 2: b. 253.69 m A triangular lot was formed by three straight lines.
Determine the most probable value of the c. 150.34 m Bearing of line AB is N45°E and is 160 m long; line
tabulated measurements as shown: d. 90.75 m AC and BC is 190 m long. Point E is 100 m
6. Determine the length of chord from PC to PT of from A on side AB connects point D on side AC
the compound curve. which divides the area, making area AED to be 2/5
a. 295.94 m of the whole area. Find the distance DE.
b. 291.32 m a. 120.6 m
c. 297.50 m b. 109.1 m
d. 179.26 m c. 99.01 m
a. 520.229
7. Determine the stationing of PT if PI is at station d. 88.14 m
b. 520.305
40 + 120. SITUATION 4:
c. 520.208
a. 40 + 326.75 AB and BC are tangents of a simple curve. AB is due
d. 520.276
b. 40 + 236.28 north and BC is N50°E. Degree of simple curve is 4°
SITUATION 1:
c. 40 + 237.78 and PT is at station 20 + 130.
A compound curve has the following
d. 40 + 273.87 12. Determine the external distance.
data:
SITUATION 3: a. 29.62 m
I₁ = 28° D₁ =
The distance between two intermediate points A b. 30.23 m
3°
and B on a simple curve is 60 m. Their deflection c. 14.66 m
I₂ = 31° D₂ =
angles are respectively 4° and 10° measured from d. 10.4 m
4°
tangent through PC. 13. Determine the length of the line from point D
Sta. P.I @ 30 + 120.5
8. Compute the radius of the simple curve. on the simple curve to PI that makes an angle of 54°
3. Determine the length of the common tangent.
a. 264.90 m with the tangent passing thru the PC.
a. 381.97 m
b. 229.47 m a. 29.62 m
b. 174.68 m
c. 286.48 m b. 78.22 m
c. 286.48 m
d. 224.92 m c. 14.66 m
d. 10.4 m 18. Compute the radius of the simple curve. d. 17°3’42”
14. Determine the stationing of point D. a. 264.90 m SITUATION 7:
a. 20 + 249.3 b. 229.47 m Given a compound curve with I₁ = 20°, I₂ = 30° , D₁ =
b. 19 + 999.25 c. 381.97 m 2° and D₂ = 4°.
c. 20 + 074.1 d. 224.92 m 23. Determine the length of the common tangent.
d. 18 + 888.25 19. Compute the arc length from PC to A. a. 310.73 m
SITUATION 5: a. 34.37 m b. 253.69 m
A simple curve having a degree of curve of 4° has an b. 53.33 m c. 177.79 m
angle of intersection of 50°. A line passing through c. 33.74 m d. 90.75 m
the center of the curve makes an angle of 70° with d. 37.34 m 24. Determine the length of chord from PC to PT of
the tangent passing through P.C. intersecting the 20. Compute the arc length from PC to B. the compound curve.
curve at point A. a. 166.67 m a. 295.94 m
15. Determine the external distance of the simple b. 171.54 m b. 339.224 m
curve. c. 133.33 m c. 297.50 m
a. 17.76 m d. 189.45 m d. 179.26 m
b. 26.84 m PROBLEM 21: 25. Determine the stationing of PT if PI is at station
c. 29.62 m A simple curve having a degree of curve of 4° has an 40 + 120.
d. 18.94 m angle of intersection of 50°. A line passing through a. 40 + 326.75
16. Compute the perpendicular offset distance of the center of the curve makes an angle of 65° with b. 40 + 236.28
point A from the tangent drawn from P.C. the tangent passing through P.T. intersecting the c.40 + 237.78
a. 22.51 m curve at point A. If the stationing of the P.T. is 0 + d. 40 + 252.93
b. 26.84 m 820, what is the stationing of point A. Use arc basis SITUATION 8:
c. 29.62 m a. 0 + 720 The following data were obtained by stadia
d. 18.94 m b. 0 + 695 measurement: vertical angle = +20°, and observed
17. If the stationing of the P.C. is 0 + 820, what is the c. 0 + 520 stadia intercept = 2.0 m. The stadia interval factor of
stationing of point A. Use arc basis d. 0 + 780 the instrument used is 95.5 and C = 0.30 m. If the
a. 0 + 101 PROBLEM 22: height of the instrument is 1.62 m, and the rod is
b. 0 + 695 Two parallel roads were connected by two equal taken at 1.95 m.
c. 0 + 710 simple curves and form a reverse curve. If the 26. Determine the vertical distance from the
d. 0 + 934 length of the intermediate tangent located between center of the instrument to point on the rod
SITUATION 6: the two simple curves is 360 m. and the two curves bisected by the horizontal cross hair.
The distance between two intermediate points A have equal radii of 1200 m., find the central angle a. 66.26 m
and B on a simple curve is 80 m. Their deflection common to the two simple curves. b. 61.39 m
angles are respectively 4° and 10° measured from a. 20°34’09” c. 63.63 m
tangent through PC. b. 18°06’24” d. 68.35 m
c. 23°76’52”
27. Determine the inclined or slope distance SITUATION 10: 37. Compute the total length of curve.
from the instrument center to the portion on the A parabolic sag curve has tangent grades of –2% a. 1208.81 m
rod bisected by the horizontal cross hair. and +3%. The stationing of PT is at 10+300 and b. 1887.4 m
a. 189.48 m elevation of 100 m. The total allowable change of c. 1662.1 m
b. 201.43 m grades per 20 m station is 0.002 d. 1631.8 m
c. 199.68 m 32. Compute the total length of the curve. SITUATION 12:
d. 179.50 m a. 500 m Two parallel tangents 12 m apart are connected by
SITUATION 9: b. 600 m a reverse curve having equal radii. The long chord
AB and BC are tangents of a simple curve. AB is due c. 400 m connecting the P.C to P.T is 140 m.
north and BC is N40°E. Degree of simple curve is 6°. d. 700 m 38. Determine the radius of the reverse curve.
28. Determine the external distance. 33. Find the stationing of PC. a. 408.33 m
a. 29.62 m a. 9+900 b. 813.64 m
b. 30.23 m b. 9+600 c. 200.41 m
c. 12.26 m c. 9+700 d. 512.41 m
d. 10.4 m d. 9+800 39. Find the length of the curve.
29. Determine the middle ordinate. 34. Find the elevation of PC. a. 175.75 m
a. 29.62 m a. 97.5 m b. 131.29 m
b. 30.23 m b. 125 m c. 279.30 m
c. 14.66 m c. 102 m d. 140.17 m
d. 11.52 m d. 120 m SITUATION 13:
30. Determine the chord from PC to PT. SITUATION 11: The middle ordinate of a 3° curve is 12 meters.
a. 180.173 The length of common tangent of a compound 40. Find the radius of the curvature.
b. 130.642 horizontal curve is 630 m. D₁ = 30, I₁ = 50°, I₂ = 35°. a. 327.406 m
c. 109.446 35. Compute the radius of the second curve. b. 381.972 m
d. 152.142 a. 1523.m c. 396.122 m
PROBLEM 31: b. 1302.25 m d. 284.519 m
The distance from the midpoint of the curve to the c. 912.12 m 41. Find the central angle.
point of intersection of the tangent of a simple d. 1433.18 m a. 28.8°
curve is 40 m. If radius is 220 m, determine the 36. Compute the length of chord connecting PC b. 26.8°
length of curve. and PT. c. 23.4°
a. 247.31 m a. 1861.45 m d. 31.5°
b. 132.66 m b. 1121.39 m 42. Find the external distance.
c. 123.66 m c. 1764.14 m a. 6.95 m
d. 163.26 m d. 1945.45 m b. 12.39 m
c. 14.90 m
d. 11.09 m
SITUATION 14: c. 110.61°
The external distance of a simple curve is 12 m. If its d. 154.32°
central angle is 40°, 48. Find I₂.
43. Find its radius a. 120°
a. 125.752 b. 141°
b. 218.144 c. 110.61°
c. 145.039 d. 154.32°
d. 186.981 49. Find R₁.
44. Find the middle ordinate. a. 125.752
a. 7.58 b. 205.130
b. 11.28 c. 145.039
c. 13.589 d. 173.205
d. 20.44 50. Find R₂.
45. Find the chord from PC to PT. a. 205.130
a. 184.383 b. 357.526
b. 127.902 c. 139.892
c. 199.212 d. 173.205
d. 106.290
PROBLEM 46:
The distance from the middle of the curve to the
point of intersection of tangent of a circular curve is
19.65 m. If the radius of the curve is 400 m,
Compute the middle ordinate.
a. 18.73 m
b. 36.35 m
c. 169.37 m
d. 73.20 m
SITUATION 15:
The common tangent of a reversed curve is 120
meters long. The PRC is at ¼ distance measured
from the point of intersection of tangent to PC and
PRC. If the curve of the two curves measures 300m
and 400m,
47. Find I₁.
a. 120°
b. 141°