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 _____

10,500 | |

32,000 | |

20,000 | |

15,000 |

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

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 ________

\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 2 Explanation:

Project duration = 56 days.

Question 3 |

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 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

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

increase strength and durability | |

smoothen timber surfaces | |

remove knots from timber logs | |

cut timber in right season and geometry |

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).

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 _________

\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 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

\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 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%.

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

200000 | |

198000 | |

196000 | |

182000 |

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

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

prevent quick setting | |

enhance hardening | |

increase workability | |

decrease heat of hydration |

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 _______

If the project is targeted for completion in 16 days, the total cost(in INR) to be incurred by the contractor would be _______

149500 | |

147896 | |

123654 | |

987654 |

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

Le Chatelier apparatus | |

Briquette testing apparatus | |

Vicat apparatus | |

Casagrande's apparatus |

Question 10 Explanation:

Vicat apparatus is used to determine the normal consistancy, IST, FST of cement.

There are 10 questions to complete.