PART 3- SOLVING Flashcards
A farmer figured that he needed 8 pieces 2” x 4” x 12’ and 10 pieces 2” x 3” x 12’ lumber to make tables and chairs for their barangay hall. He availed them from a newly-opened local hardware shop, named J.E.R. Hardware Supplies, which charges P25 for every board foot of lumber. However, to gain patronage, the shop gave 15% discount for the first 50 board feet of purchased lumber. How many board feet of lumber are needed for the farmer’s project?
a. 110 b. 100 c. 120 d. 124
8 pcs= 2” x 4” x 12’ 10 pcs = 2” x 3” x 12’ Sol'n: 2” x 4” x 12’/12 = 8x8pcs = 64 pcs 2” x 3” x 12’/12 = 6x10pcs = 60 pcs Total: 60+64 = 124 pcs
A wooden rectangular frame (1.5 m x 3 m) is to be fabricated using 2” x 2” coco lumber. Compute for the total bdft required for this frame.
a. 4 b. 6 c. 8 d. 10
P= 2L+2W (ft)
2x2x ft / 12 = 9.8
Determine the quantity of CHB needed for a 16m x 5m (ground line included) wall.
a. 1000 b. 1200 c. 1400 d. 1600
13 or 12.5 CHB / m2
12.5 CHB x 80 = 1000 (a)
Fifty 40 Watt incandescent lamps are used for lighting livestock at night at an average of 10 hours
per day for 30 days. How much electric bill will be charged in a month if energy cost is 35 cents/W-H
for the 1st 100 kw-Hr, 50 cents for the second and 75 cents for the remaining Kw-Hr.
a. P380.00 b. P385.00 c. P390.00 d. P395.00
b. P385.00
A coconut farmer wanted to buy copra dryers and oil mill for his coconut produced from a 300
hectares area of coconut trees. If the average yield is 10,000 nuts per hectare. How many 1000 nuts
capacity per day copra dryer and 5000 nuts capacity per day coconut oil mill will be required. Assume
operations to be 300 days per annum.
a. 10 and 2 b. 2 and 10 c. 5 and 10 d. 10 and 5
a. 10 and 2
Engr. Macana wants to dry 500 kg fresh fish Sardinella fimbriata using Biomass-fueled flat bed
dryer with a controlled temperature of 50-55 °C. He found out in the laboratory that the fish has an
initial moisture content of 74.28% wet base and he wants to reduce it to 25% wet base.
What is the final weight of the dried fish?
a. 121.47 kg b. 131.47 kg c.161.47 kg d. 171.47 kg
d. 171.47 kg
What must be the original weight of Palay grain at 26% moisture content dry base to provide 50 kg
of material at 14% moisture content wet base
a. 54 kg b.58.108 kg c. 54.18 kg d. 92.8 kg
b.58.108 kg
What is the moisture content wet base of rice straw at 40% moisture content dry base?
a. 20.57% b. 24.57 % c. 26.57% d. 28.57%
d. 28.57%
Md / Md + 1 times 100
The equivalent resistance of three resistors A,B and C connected in parallel is 1.714 ohms. If A is
twice of B and C is half as much as B, find the equivalent resistance when the three of them are
connected in series.
a.11 ohms b.21 ohms c. 31 ohms d.41 ohms
b.21 ohms
A= 2x = 12 B= x = 6 C= x/2 = 3
1/1.714 = 1/2x + 1/X + 1/x/2 X = 5.999 or 6
S = 12 + 6 + 3 = 21 ohms
20,30, and 50 (ohms)-resistors are connected in series to 125-V source .What is the voltage drop across the 50 ohms?
a. 25V
b. 37.5V
c. 62.5V
d. 50.0V
e. 87.5V
SOLN:
For series circuit:
Req = R1 + R2 + R3 = 20Ω + 30Ω + 50Ω = 100Ω
Ieq = Veq/Req = 125V / 100Ω = 1.25A
For the 50-Ω resistor:
V = IR = (1.25A)(50Ω)
V = 62.5V-ANSWER
A battery can deliver 10 joules of energy to move 5 coulombs of charge. What is the potential
difference between the terminals of the battery?
a. 2 V b. 50 V c. 0.5 V d. 15 V
a. 2 V
Suppose you have a micro turbine with blade diameter of 1.50 m (about 5 ft) and efficiency of 19.5%. Calculate the electricity it can generate for your home at air speed of 6 m/sec (about 13.4 mph). Assume air density to be 1.34 kg/m3.
a. 45 W
b. 50 W
c. 55 W
d. 60 W
e. 65 W
SOLN:
Rotor swept area: A= π(D/2)2 = 3.14×(1.5/2)2 = 1.767 m2
Available power in the wind: Pwind= ρ×A×v3/2 = 1.34×1.767×63/2 = 255.74 watt
Then the power that can be extracted at that speed is: Pturbine=0.195×255.74= 49.87 watts ANSWER
The current in an electric lamp is 5 amperes. What quantity of electricity flows towards the
filament in 6 minutes?
a. 1800 C b. 1700 C c. 1600 C d. 1500 C
a. 1800 C
Q = IT = 5 x 6 x 60 sec = 1800 C
( A/sec)
A concrete open fence shall be constructed. Two sides are 25.00 m long while the other side (facing the opening) is 20.00 m long. Estimate the number of hollow blocks needed if the height of the fence is 2.50 m. Allow 5% for breakage. a. 2,188
b. 1,956
c. 2,354
d. 2,487
e. 2,297
SOLN:
A = (perimeter)(height) = (25 + 25 + 20 m)(2.5 m) = 175 m2
Number of hollow blocks = (175 m2)(12.5 pcs/m2)(1.05) = 2,297 ANSWER
20-Ω, 30-Ω, and 50-Ω resistors are connected in series to a 125-V source. What is the voltage drop across the 20-Ω resistor? a. 25.0V
b. 37.5V
c. 62.5V
d. 50.0V
e. 87.5V
SOLN:
For series circuit:
Req = R1 + R2 + R3 = 20Ω + 30Ω + 50Ω = 100Ω
Ieq = Veq/Req = 125V / 100Ω = 1.25A
For the 20-Ω resistor:
V = IR = (1.25A)(20Ω)
V = 25.0V -ANSWER
The power consumed in a household over a 24-hr period is as follows: 8:00AM to 2:00PM – 1.5kW; 2:00PM to 6:00PM – 0.5kW; 6:00PM to 11:00PM – 2.6 kW; and 11:00PM to 8:00AM – 1.0kW. What is the energy consumption in megajoules? a. 181.8MJ
b. 8.118MJ
c. 811.1MJ
d. 118.8MJ
e. 18.18MJ
Schedule Time (hr) Power consumption (kW) Energy consumption (kWhr) 8:00AM to 2:00PM 6 1.5kW 9.0 2:00PM to 6:00PM 4 0.5kW 2.0 6:00PM to 11:00PM 5 2.6 kW 13.0 11:00PM to 8:00AM 9 1.0kW 9.0 TOTAL 33.0
In megajoules:
E = (33,000 Whr) (3,600J/1Whr) (1 MJ/1,000,000J)
E = 118.8 MJ -ANSWER
A biogas plant has the following specifications and utilization:
Number of swine (36-55 kg): 20
Manure production per head: 5.22 kg/day
Gas production per 1 kg of manure: 0.058 m3
Retention time: 25 days
For one day utilization,
Light: 4 light bulbs at 6 hrs operation
Stove: 1 5-cm stove at 3 hrs operation
Refrigerator: 1 unit at 24 hrs operation
If the biogas requirement for light is 0.071 m3/hr utilization, 0.226 m3/hr for the stove, and 0.053 m3/hr for the refrigerator, estimate the gasholder volume of the plant. Allow 30% allowance for biogas fluctuation.
a. 2.874 m3/day c. 4.832 m3/day e. 5.543 m3/day
b. 1.238 m3/day d. 3.122 m3/day
SOLN:
Manure production = (20 heads)(5.22 kg/day per head) = 104.4 kg/day Biogas production rate = (104.4 kg/day)(0.058 m3/day) = 6.0552 m3/day Daily utilization:
Light = (4 lights)(6 hrs/light)(0.071 m3/hr) = 1.704 m3/day
Stove = (1 stove)(3 hrs)(0.226 m3/hr) = 0.678 m3/day
Ref = (24 hrs/light)(0.053 m3/hr) = 1.272 m3/day
Total = 3.654 m3/day
Gas to be stored = (6.0552 – 3.654 m3/day)(1.30 for biogas fluctuation)
= 3.122 m3/day -ANSWER
20-Ω, 100-Ω, and 50-Ω resistors are connected in parallel to a 125-V source. What is the current through the 100-Ω resistor?
a. 10.00A b. 2.50A c. 6.25A d. 7.50A e. 1.25A
SOLN:
For the 100-Ω resistor:
I = V/R = 125V/100Ω = 1.25A -ANSWER
20-Ω, 100-Ω, and 50-Ω resistors are connected in parallel to a 125-V source. What is the power dissipated through the 50-Ω resistor?
a. 312.50W c. 64.625W e. 78.125W
b. 46.875W d. 52.430W
SOLN:
For the 50-Ω resistor:
P = IV = V2/R = (125V)2/50Ω = 312.5W
After 24 hours of bone-drying, a 110-gram sample of fresh paddy is reduced to 83.6 grams. What is the moisture content in dry basis? a. 24.00%
b. 76.00%
c. 68.42%
d. 88.75%
e. 31.58%
SOLN: Given: Fresh weight = 110 grams Weight of dry matter = 83.6 grams Weight of moisture = 110–83.6 = 26.4 grams MCdb = 31.58%
