Question 1 |
The activities of a project are given in the following table along with their durations and dependency.
\begin{array} {|c|c|c|}\hline \text{Activities} &\text{Duration (days)}&\text{Depends on}\\ A&10&-\\ \hline B&12&-\\ \hline C&5&A\\ \hline D&14&B\\ \hline E&10&B,C\\ \hline \end{array}
The total float of the activity E (in days) is ________. (in integer)
\begin{array} {|c|c|c|}\hline \text{Activities} &\text{Duration (days)}&\text{Depends on}\\ A&10&-\\ \hline B&12&-\\ \hline C&5&A\\ \hline D&14&B\\ \hline E&10&B,C\\ \hline \end{array}
The total float of the activity E (in days) is ________. (in integer)
0 | |
1 | |
2 | |
3 |
Question 1 Explanation:
For activity E,
Total float (F_T)=T_L^5-T_L^4-t^E=26-15-10=1 day
Question 2 |
The project activities are given in the following table along with the duration and dependency.
\begin{array}{|c|c|c|} \hline \text{Activities}& \text{Duration (days)} & \text{Depends on}\\ \hline P & 10 & -\\ \hline Q & 12 &- \\ \hline R & 5 &P \\ \hline S & 10 &Q \\ \hline T & 10 & P,q \\\hline \end{array}
Which one of the following combinations is correct?
\begin{array}{|c|c|c|} \hline \text{Activities}& \text{Duration (days)} & \text{Depends on}\\ \hline P & 10 & -\\ \hline Q & 12 &- \\ \hline R & 5 &P \\ \hline S & 10 &Q \\ \hline T & 10 & P,q \\\hline \end{array}
Which one of the following combinations is correct?
Total duration of the project = 22 days, Critical path is Q \rightarrow S | |
Total duration of the project = 20 days, Critical path is Q \rightarrow T | |
Total duration of the project = 22 days, Critical path is P \rightarrow T | |
Total duration of the project = 20 days, Critical path is P \rightarrow R |
Question 2 Explanation:
Critical path are Q-S and Q-T=22 days
Question 3 |
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 _____
10,500 | |
32,000 | |
20,000 | |
15,000 |
Question 3 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
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 4 |
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 ________
\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 ________
42 | |
68 | |
56 | |
36 |
Question 4 Explanation:
Project duration = 56 days.
Question 5 |
In case of bids in Two-Envelop System, the correct option is
Technical bid is opened first | |
Financial bid is opened first | |
Both (Technical and Financial) bids are opened simultaneously | |
Either of the two (Technical and Financial) bids can be opened first |
Question 5 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
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 6 |
Seasoning of timber for use in construction is done essentially to
increase strength and durability | |
smoothen timber surfaces | |
remove knots from timber logs | |
cut timber in right season and geometry |
Question 6 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).
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 7 |
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 _________
\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 _________
0 | |
1 | |
2 | |
3 |
Question 7 Explanation:
\begin{aligned} F_{T_{V}} &=T_{L_{j}}-T_{E_{i}}-t_{i_{j}} \\ &=20-18-2 \\ &=0 \end{aligned}
Question 8 |
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
\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
0% | |
more than 0% but less than 50% | |
more than 50% but less than 100% | |
100% |
Question 8 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%.
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 9 |
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
200000 | |
198000 | |
196000 | |
182000 |
Question 9 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
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 10 |
Gypsum is typically added in cement to
prevent quick setting | |
enhance hardening | |
increase workability | |
decrease heat of hydration |
Question 10 Explanation:
The Gypsum is added to cement at the end of grinding clinker it is added to prevent quick setting.
There are 10 questions to complete.