Construction Materials and Management

Question 1
An equipment has been purchased at an initial cost of Rs. 160000 and has an estimated salvage value of Rs. 10000. The equipment has an estimated life of 5 years. The difference between the book values (in Rs, in integer) obtained at the end of 4th year using straight line method and sum of years digit method of depreciation is _____
A
10,500
B
32,000
C
20,000
D
15,000
GATE CE 2021 SET-2      
Question 1 Explanation: 
Initial Cost (P) = 1,60,000
Salvage value (SV) = 10,000
Useful life (n) = 5 Year
Straight Line Method:

Annual depreciation (D),
D=\frac{P-SV}{n}=\frac{1,60,000-10,000}{5}=30,000
Book value after 4 year (BV_4)=P-4D=1,60,000-4(30,000)=40,000
SOY Method (Sum of year Digit)
\begin{aligned} SOY&=\frac{n(n+1)}{2}=\frac{5 \times 6}{2}=15 \\ d_m&=\frac{n-(m-1)}{SOY} \\ D_m&=(P-SV) \times d_m \\ BV_m&=BV_{m-1}-D_m \\ d_1&= \frac{5-(1-1)}{15}=\frac{5}{15}\\ D_1&= (1,60,000-10,000) \times \frac{5}{15} \\ &=50,000\\ d_2&=\frac{4}{15} ; \;\;D_2=40,000\\ d_3&=\frac{3}{15} ; \;\;D_3=30,000\\ d_4&=\frac{2}{15} ; \;\;D_4=20,000\\ d_5&=\frac{2}{15} ; \;\;D_2=10,000\\ BV_0&=1,60,000 \\ BV_1&=1,60,000 -50,000=1,10,000 \\ BV_2&=1,10,000 -40,000=70,000 \\ BV_3&=70,000 -30,000=40,000 \\ BV_4&=40,000 -20,000=20,000 \\ BV_5&=20,000 -10,000=10,000 \\ \end{aligned}

Difference of book value obtained in two methods = 40,000-20,000=20,000
Question 2
The activity details for a small project are given in the Table.
\begin{array}{|l|c|c|} \hline \text { Activity } & \text { Duration (days) } & \text { Depends on } \\ \hline \text { A } & 6 & \text { - } \\ \hline \text { B } & 10 & \text { A } \\ \hline \text { C } & 14 & \text { A } \\ \hline \text { D } & 8 & \text { B } \\ \hline \text { E } & 12 & \text { C } \\ \hline \text { F } & 8 & \text { C } \\ \hline \text { G } & 16 & \text { D, E } \\ \hline \text { H } & 8 & \text { F, G } \\ \hline \text { K } & 2 & \text { B } \\ \hline \text { L } & 5 & \text { G, K } \\ \hline \end{array}
The total time (in days,in integer) for project completion is ________
A
42
B
68
C
56
D
36
GATE CE 2021 SET-2      
Question 2 Explanation: 


Project duration = 56 days.
Question 3
In case of bids in Two-Envelop System, the correct option is
A
Technical bid is opened first
B
Financial bid is opened first
C
Both (Technical and Financial) bids are opened simultaneously
D
Either of the two (Technical and Financial) bids can be opened first
GATE CE 2021 SET-2      
Question 3 Explanation: 
Option 1 technical bid is opened first

Opening of Tender
First technical bid is opened and after ensuring that all the technical aspects of a contractor are in order than only financial bid is opened
1. Envelope 1 ( Technical bid )

1. Cover letter
2. Registration Details
3. Pre-qualification documents
4. Earnest money deposit
5. Assumptions & Deviations in making of tender
6. Drawings

2. Envelope 2 (Financial Bid)

1. Forms of tender
Question 4
Seasoning of timber for use in construction is done essentially to
A
increase strength and durability
B
smoothen timber surfaces
C
remove knots from timber logs
D
cut timber in right season and geometry
GATE CE 2021 SET-2      
Question 4 Explanation: 
Option 1 Increase strength and durability.
The process of drying of timber is known as seasoning.
Natural tree has more the 50% weight of water of its dry weight.
If we directly use this timber the because of irregular drying internal stresses will develop between fibres of timber and it will develop lots of defects (warps, shakes etc).
Question 5
A small project has 12 activities - N, P, Q, R, S, T, U, V, W, X, Y, and Z. The relationship among these activities and the duration of these activities are given in the Table.
\begin{array}{|c|c|c|} \hline \text { Activity } & \begin{array}{c} \text { Duration } \\ \text { (in weeks) } \end{array} & \begin{array}{c} \text { Depends } \\ \text { upon } \end{array} \\ \hline N & 2 & - \\ \hline \mathrm{P} & 5 & \mathrm{~N} \\ \hline \mathrm{Q} & 3 & \mathrm{~N} \\ \hline \mathrm{R} & 4 & \mathrm{P} \\ \hline \mathrm{S} & 5 & \mathrm{Q} \\ \hline \mathrm{T} & 8 & \mathrm{R} \\ \hline \mathrm{U} & 7 & \mathrm{R}, \mathrm{S} \\ \hline \mathrm{V} & 2 & \mathrm{U} \\ \hline \mathrm{W} & 3 & \mathrm{U} \\ \hline \mathrm{X} & 5 & \mathrm{~T}, \mathrm{~V} \\ \hline \mathrm{Y} & 1 & \mathrm{~W} \\ \hline \mathrm{Z} & 3 & \mathrm{X}, \mathrm{Y} \\ \hline \end{array}
The total float of the activity 'V' (in weeks, in integer) is _________
A
0
B
1
C
2
D
3
GATE CE 2021 SET-1      
Question 5 Explanation: 


\begin{aligned} F_{T_{V}} &=T_{L_{j}}-T_{E_{i}}-t_{i_{j}} \\ &=20-18-2 \\ &=0 \end{aligned}
Question 6
Contractor X is developing his bidding strategy against Contractor Y. The ratio of Y's bid price to X's cost for the 30 previous bids in which Contractor X has competed against Contractor Y is given in the Table

\begin{array}{|l|l|} \hline \begin{matrix} \text{Ratio of Y's bid}\\ \text{price to X's cost } \end{matrix} & \text{Number of bids}\\ \hline 1.02 & 6 \\ \hline 1.04 &12\\ \hline 1.06 & 3\\ \hline 1.10 & 6\\ \hline 1.12 & 3\\ \hline \end{array}

Based on the bidding behaviour of the Contractor Y, the probability of winning against Contractor Y at a mark up of 8% for the next project is
A
0%
B
more than 0% but less than 50%
C
more than 50% but less than 100%
D
100%
GATE CE 2021 SET-1      
Question 6 Explanation: 
SD(\sigma )=\frac{6 \times (1.02-1.058)^2+ 12 \times (1.04-1.058)^2 +3 \times (1.06-1.058)^2 + 6 \times (1.10-1.058)^2 +3 \times (1.12-1.058)^2}{30}=0.034
Markup is 8%
Z=\frac{T-mean}{SD}=\frac{1.08-1.058}{0.034}=0.647
By general understanding of normal distribution table.
Corresponding to Z=0.647 and 8% mark up we get probability of 0.2668=26.68%
Which is more than 0% bu less tahn 50%.
Question 7
The direct and indirect costs estimated by a contractor for bidding a project is Rs.160000 and Rs.20000 respectively. If the mark up applied is 10% of the bid price, the quoted price (in Rs.) of the contractor is
A
200000
B
198000
C
196000
D
182000
GATE CE 2021 SET-1      
Question 7 Explanation: 
Direct Costs = 160000
Indirect costs = 20000
Mark up applied is 10% of the Bid price
Bid price = Direct cost + Indirect cost + Markup
Bid price = Direct cost + Indirect cost + \frac{10}{100} \times Bid price
\text{Bid price}\left (1-\frac{10}{100} \right )=DC+IC
\text{Bid price}=\frac{160000+20000}{0.9}=2,00,000
Question 8
Gypsum is typically added in cement to
A
prevent quick setting
B
enhance hardening
C
increase workability
D
decrease heat of hydration
GATE CE 2021 SET-1      
Question 8 Explanation: 
The Gypsum is added to cement at the end of grinding clinker it is added to prevent quick setting.
Question 9
The network of a small construction project awarded to a contractor is shown in the following figure. The normal duration, crash duration, normal cost, and crash cost of all the activities are shown in the table. The indirect cost incurred by the contractor is INR 5000 per day.

If the project is targeted for completion in 16 days, the total cost(in INR) to be incurred by the contractor would be _______
A
149500
B
147896
C
123654
D
987654
GATE CE 2019 SET-1      
Question 9 Explanation: 


Indirect cost = Rs. 5000
Crashing possibility = 18 - 17 = 1 day
To reduce the project duration by 1 day, the following options available.

Best option : Crashing 'R' by 1 day.
After crashing 'R' by 1 day, the new network is as for follows.

The reduce the project duration by 1 day

After crashing Q & R by 1 day, the new network is as follows.

Total project cost = Total normal cost + crashing cost of 'R' by 1 day + crashing cost of Q & R by 1 day + Indirect cost/day ? Project duration
= 15000 + 6000 + 8000 + 7000 + 6000 + 4000 + 20000 + 750 ? 1 + 2750 x 1 + 5000 x 16 = 1,49,500 /-
Question 10
The setting time of cement is determined using
A
Le Chatelier apparatus
B
Briquette testing apparatus
C
Vicat apparatus
D
Casagrande's apparatus
GATE CE 2018 SET-2      
Question 10 Explanation: 
Vicat apparatus is used to determine the normal consistancy, IST, FST of cement.
There are 10 questions to complete.

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