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FINAL ESTIMATES LEVEL 2 EXAM
2025 | ACTUAL EXAM WITH
VERIFIED ANSWERS
Determine the adjusted plan quantity tonnage and the maximum tonnage that
can be paid for miscellaneous Asphalt based on the current Specifications, which
allows a maximum pay of 105% of the adjusted plan quantity tons. ( pay to the
appropriate accuracy)
Plan Quantity Area (SY) =700 SY ( no revisions)
Plan Quantity Tons: 70.0 Tons based on 2" thick per specs
Adjusted PQ Tonnage Placed =75.2 Tons from report
Gmm fromSpecs = 2.540 and Target Spread Rate =220 Lbs/SY
Gmm from Mix 1: 2.522 and tonnage- weighted Gmm=2.55 Lbs/SY - correct-
answer-69.5 Tons and max pay 73.0 Tons
Find the total thickness on a composite base which includes 4 inches of Limerock
Base and 4 inches of type B (12.5) Asphalt Base given the following information:
Limerock Base Thickness from Core Out Report =4.19"
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Asphalt (12.5) Base: Gmm =2.365 from the design mix
Average Spread Rate from the roadway report= 429 Lbs/SY - correct-answer-Gmm
x 43.3 =102.9
.4 = 4.19"
4.19" + 4.19" = Total 8.38 inches
On an open graded Friction Course (FC-5) Pavement, a straightedge deficiency was
left in place at no pay. Calculate the equivalent tonnage at no pay. ( Pay to the
appropriate accuracy)
Lane Width =12 ft
Location of Deficiency= STA 1+56 to STA 1+67 ( 167-156=11) - correct-answer-Lx
Wx .0044
L=50'+11'+50' = 111'
111' X 12' X 0.0044 = 5.9 TNS
Lot 3 has a Tonnage of 2529.0 tons with a Composite Pay Factor of 1.04. What
would the calculated Composite Pay Factor adjustment amount be if the unit price
is $100.00 per ton? - correct-answer-(CPF-1) = Unit price
(1.04-1) $100= 4
$4/ton x 2529 tn= $10,116
FINAL ESTIMATES LEVEL 2 EXAM
2025 | ACTUAL EXAM WITH
VERIFIED ANSWERS
Determine the adjusted plan quantity tonnage and the maximum tonnage that
can be paid for miscellaneous Asphalt based on the current Specifications, which
allows a maximum pay of 105% of the adjusted plan quantity tons. ( pay to the
appropriate accuracy)
Plan Quantity Area (SY) =700 SY ( no revisions)
Plan Quantity Tons: 70.0 Tons based on 2" thick per specs
Adjusted PQ Tonnage Placed =75.2 Tons from report
Gmm fromSpecs = 2.540 and Target Spread Rate =220 Lbs/SY
Gmm from Mix 1: 2.522 and tonnage- weighted Gmm=2.55 Lbs/SY - correct-
answer-69.5 Tons and max pay 73.0 Tons
Find the total thickness on a composite base which includes 4 inches of Limerock
Base and 4 inches of type B (12.5) Asphalt Base given the following information:
Limerock Base Thickness from Core Out Report =4.19"
, 2|Page
Asphalt (12.5) Base: Gmm =2.365 from the design mix
Average Spread Rate from the roadway report= 429 Lbs/SY - correct-answer-Gmm
x 43.3 =102.9
.4 = 4.19"
4.19" + 4.19" = Total 8.38 inches
On an open graded Friction Course (FC-5) Pavement, a straightedge deficiency was
left in place at no pay. Calculate the equivalent tonnage at no pay. ( Pay to the
appropriate accuracy)
Lane Width =12 ft
Location of Deficiency= STA 1+56 to STA 1+67 ( 167-156=11) - correct-answer-Lx
Wx .0044
L=50'+11'+50' = 111'
111' X 12' X 0.0044 = 5.9 TNS
Lot 3 has a Tonnage of 2529.0 tons with a Composite Pay Factor of 1.04. What
would the calculated Composite Pay Factor adjustment amount be if the unit price
is $100.00 per ton? - correct-answer-(CPF-1) = Unit price
(1.04-1) $100= 4
$4/ton x 2529 tn= $10,116
