Industrial Engineering MCQ Quiz - Objective Question with Answer for Industrial Engineering - Download Free PDF

Last updated on Jun 13, 2025

Latest Industrial Engineering MCQ Objective Questions

Industrial Engineering Question 1:

PERT is

  1. Event oriented technique
  2. Work oriented technique
  3. Activity oriented technique
  4. Time oriented technique

Answer (Detailed Solution Below)

Option 1 : Event oriented technique

Industrial Engineering Question 1 Detailed Solution

Explanation:

PERT (Program Evaluation and Review Technique)

  • PERT is a project management tool used to schedule, organize, and coordinate tasks within a project. It is an event-oriented technique designed to analyze and represent the tasks involved in completing a project. By focusing on events or milestones, PERT helps in planning and managing complex projects with numerous interdependent activities.
  • PERT charts are created by identifying all the activities required to complete a project and the sequence in which they need to be executed. Each activity is represented as an arrow, and the events or milestones are represented as nodes. The technique estimates the shortest, longest, and most likely time required to complete each task, allowing for better planning and control over project timelines.

Applications: PERT is widely used in industries such as construction, software development, engineering, and research and development, where complex projects require detailed planning and coordination.

Industrial Engineering Question 2:

Break-even point shows that

  1. Sales revenue = Total cost
  2. Variable cost = Fixed cost
  3. Sales revenue < Total cost
  4. Sales revenue > Total cost

Answer (Detailed Solution Below)

Option 1 : Sales revenue = Total cost

Industrial Engineering Question 2 Detailed Solution

Explanation:

Break-Even Point

  • The break-even point (BEP) is a critical financial concept in business and economics. It refers to the level of sales revenue at which a company's total revenue equals its total costs, resulting in neither profit nor loss. At this point, the business has covered all its expenses, including fixed and variable costs, and begins to generate profit with any additional sales revenue.

Formula:

The break-even point can be calculated using the following formula:

BEP (in units) = Fixed Costs ÷ (Selling Price per Unit − Variable Cost per Unit)

BEP (in sales revenue) = Fixed Costs ÷ Contribution Margin Ratio

At the break-even point:

  • Total Revenue = Total Costs: This includes both fixed costs (costs that do not vary with the level of production, such as rent and salaries) and variable costs (costs that vary directly with production, such as materials and labor).
  • No Profit, No Loss: The company is not earning a profit, but it is also not incurring a loss. The break-even point is the threshold at which profitability begins.

Importance of Break-Even Analysis:

  • Helps businesses determine the minimum sales volume required to avoid losses.
  • Provides insights into the relationship between cost, price, and sales volume.
  • Assists in setting sales targets and pricing strategies.
  • Useful for evaluating the financial viability of new products or projects.

Industrial Engineering Question 3:

A dummy activity

  1. does not consume time
  2. is represented by dotted lines
  3. is artificially introduced
  4. all of the above

Answer (Detailed Solution Below)

Option 4 : all of the above

Industrial Engineering Question 3 Detailed Solution

Explanation:

Dummy Activity

  • A dummy activity is an artificial activity introduced in project management and network analysis to maintain the logical sequence and dependencies of activities. It does not represent any actual task or consume any resources or time. Dummy activities are typically represented by dotted lines in network diagrams.

Purpose and Importance:

  • Maintaining logical relationships: Dummy activities are introduced to maintain the logical dependencies between tasks where no actual activity exists.
  • Clarifying network paths: They help clarify the sequencing and flow of activities in a network diagram, ensuring accurate representation of the project plan.
  • Preventing ambiguity: Dummy activities eliminate ambiguity in situations where two or more activities depend on the completion of a common predecessor but are not directly related to each other.

Representation:

  • Dummy activities are represented by dotted lines in network diagrams.
  • The dotted lines show that the activity is artificial and does not represent any real work or resource consumption.

Characteristics:

  • Does not consume time: Dummy activities do not have any duration or consume any time in the project schedule.
  • Does not consume resources: Since dummy activities are artificial, they do not require resources for completion.
  • Introduced artificially: Dummy activities are created artificially to establish logical relationships and dependencies in the project network diagram.

Industrial Engineering Question 4:

In the CPM technique, the critical path slack is:

  1. zero
  2. negative
  3. either negative or zero
  4. positive

Answer (Detailed Solution Below)

Option 1 : zero

Industrial Engineering Question 4 Detailed Solution

Explanation:

Critical Path Method (CPM):

  • The Critical Path Method (CPM) is a project management technique used to determine the sequence of activities that directly affect the project completion time. It identifies the longest path of dependent activities in the project schedule, known as the critical path. The duration of this path determines the shortest possible project completion time.

Critical Path and Slack:

  • The critical path in a project is defined as the sequence of tasks where any delay in one task would directly result in a delay in the overall project completion. The slack (or float) is the amount of time an activity can be delayed without affecting the project’s completion date. For activities on the critical path, the slack is zero because there is no flexibility to delay these activities without impacting the entire project.

Why Slack on the Critical Path is Zero:

  • The critical path represents the longest duration path in the project network. If any activity on this path is delayed, the entire project completion time will also be delayed.
  • Slack is calculated as the difference between the latest allowable finish time (LF) and the earliest finish time (EF) of an activity:
    Slack = LF - EF
  • For activities on the critical path, the LF equals EF because they determine the project’s end date. Therefore, their slack is zero.

Industrial Engineering Question 5:

Identify the critical path from the given diagram.

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  1. 1-2-3-7
  2. 1-2-4-5-6-7
  3. 1-2-4-5-6
  4. 1-2-4-7

Answer (Detailed Solution Below)

Option 2 : 1-2-4-5-6-7

Industrial Engineering Question 5 Detailed Solution

Explanation:

Critical Path Method (CPM):

Definition: The Critical Path Method (CPM) is a project management technique used to analyze and schedule tasks within a project. It identifies the longest sequence of dependent tasks (known as the critical path) that determines the shortest possible project duration. Any delay in the tasks on the critical path will directly impact the overall project completion time.

How to Identify the Critical Path:

  • List all the tasks required to complete the project.
  • Define dependencies between tasks (i.e., which tasks must be completed before others can start).
  • Determine the duration of each task.
  • Calculate the earliest start (ES) and finish (EF) times for each task by performing a forward pass through the network diagram.
  • Calculate the latest start (LS) and finish (LF) times for each task by performing a backward pass through the network diagram.
  • Identify the tasks with zero slack (i.e., tasks where ES = LS and EF = LF). These tasks form the critical path.

To identify the critical path, we analyze the given network diagram and follow these steps:

  • Step 1: List all paths in the network diagram and calculate the total duration for each path.
  • Step 2: Identify the path with the longest duration. This is the critical path because it governs the minimum time required to complete the project.

Based on the given data:

  • Path 1: 1-2-3-7
    • Total duration = Sum of durations of tasks along this path.
  • Path 2: 1-2-4-5-6-7
    • Total duration = Sum of durations of tasks along this path.
  • Path 3: 1-2-4-5-6
    • Total duration = Sum of durations of tasks along this path.
  • Path 4: 1-2-4-7
    • Total duration = Sum of durations of tasks along this path.

From the calculations, Path 2 (1-2-4-5-6-7) has the longest duration. Therefore, it is the critical path.

Top Industrial Engineering MCQ Objective Questions

Which of the followings, leads to industrial hazards and causes accidents?

  1. Noise and vibrations
  2. Poor lighting and Poor ventilation
  3. Heat and Humidity
  4. All of these

Answer (Detailed Solution Below)

Option 4 : All of these

Industrial Engineering Question 6 Detailed Solution

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Explanation:

  • Accidents which occur from environmental causes refer to those workplace accidents that happen because of the working environment. The environmental factors can be both natural and man-made such as workplace design. Common environmental causes of accidents include:
    • Poor lighting: Low visibility is a common cause of slips, trips, and falls.
    • Ambient temperature: If a workplace is too hot, overheating can occur. If the workplace is too cold, frostbite or hypothermia can occur.
    • Air pollution: Breathing issues can develop if a workplace has poor ventilation or air pollution.
    • Sound pollution: The sound in a workplace can cause injury to a worker’s hearing.​

A PERT network has 9 activities on its critical path. The standard deviation of each activity on the critical path is 3. The standard deviation of the critical path is

  1. 3
  2. 9
  3. 81
  4. 27

Answer (Detailed Solution Below)

Option 2 : 9

Industrial Engineering Question 7 Detailed Solution

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Concept:

In CPM:

The standard deviation of critical path:

σcp \(\sqrt {Sum\;of\;variance\;along\;critical\;path} \)

σcp\(\sqrt {σ _1^2 + σ _2^2 + \ldots + σ _8^2 + σ _9^2} \)

Where, σ1, σ2, ...., σ8, σ9 are the standard deviation of each activity on the critical path   

Calculation:

Given:

σ1, σ2, ...., σ8, σ9 = 3

σcp = \(\sqrt {σ _1^2 + σ _2^2 + \ldots + σ _8^2 + σ _9^2} \)

σcp\(\sqrt {3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2} \)

σcp\(\sqrt {9 \times 9} \) = 9

∴ the standard deviation of the critical path is 9.

If the cost of 157 litre of oil is Rs.  29763.65, then what is the cost per litre (rounded off to two decimal places)?

  1. Rs. 170.08
  2. Rs. 182.06
  3. Rs. 178.31
  4. Rs. 189.58

Answer (Detailed Solution Below)

Option 4 : Rs. 189.58

Industrial Engineering Question 8 Detailed Solution

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Given:

The cost of 157 litre of oil is Rs.  29763.65

Calculation:

Cost price of 157 liters of oil = Rs. 29763.65

Cost price of 1 liter of oil = 29763.65/157

⇒ 189.577 ≈ 189.58

∴ The cost per liter is 189.58 (rounded off to two decimal places).

A building fetches a rent of Rs. 10,000 annually. After repairing, it will last for 2 years. If the rate of interest on capital is 5% and the co-efficient of annual sinking fund is 0.05, then estimate the capitalized value of the building after 2 years.

  1. Rs. 4,50,000
  2. Rs. 8,50,000
  3. Rs. 1,50,000
  4. Rs. 1,00,000

Answer (Detailed Solution Below)

Option 4 : Rs. 1,00,000

Industrial Engineering Question 9 Detailed Solution

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Concept:

Capitalized value:

It is the amount of money whose annual interest at the highest prevailing rate of interest will be equal to the net income from the property.

To determine the capitalized value of a property, it is required to know the net income from the property and the highest prevailing rate of interest.

Capitalized value (V) of the property is given by:

V = Net annual income × Year purchase

Year’s purchase is defined as the capital sum required to be invested in order to receive an annuity of Rs. 1.00 at a certain rate of interest.

\({\rm{Yea}}{{\rm{r}}^{\rm{'}}}{\rm{s\;purchase}} = \frac{1}{{i \;+ \;s}}\)
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(Put i and S in decimals)

Where,

i = rate of interest

S = Sinking fund coefficient

Calculation:

Given: Net annual income = 10000, i = 5% = 0.05, s =0.05

\({\rm{Yea}}{{\rm{r}}^{\rm{'}}}{\rm{s\;purchase}} = \frac{1}{{0.05 \;+ \;0.05}}\)

⇒ Year's purchase = 10

V = 10000 × 10 = 100000

Consider the following statements in respect of PERT and CPM:

1. PERT is event-oriented while CPM is activity-oriented.

2. PERT is probabilistic while CPM is deterministic.

3. Levelling and smoothing are the techniques related to resource scheduling in CPM.

Which of the statements given above are correct?

  1. 1, 2 and 3
  2. 1 and 2 only
  3. 2 and 3 only
  4. 1 and 3 only

Answer (Detailed Solution Below)

Option 1 : 1, 2 and 3

Industrial Engineering Question 10 Detailed Solution

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Explanation:

A project may be defined as a combination of interrelated activities which must be executed in a certain order before the entire task can be completed.

The aim of planning is to develop a sequence of activities of the project so that the project completion time and cost are properly balanced.

To meet the objective of systematic planning, the management has evolved several techniques applying network strategy.

PERT (Programme Evaluation and Review Technique) and CPM (Critical Path Method) are network techniques which have been widely used for planning, scheduling and controlling the large and complex projects.

  • PERT (Project Evaluation and Review Technique) approach takes account of the uncertainties. In this approach, 3-time values are associated which each activity. So it is probabilistic.

  • CPM (Critical Path Method) involves the critical path which is the largest path in the network from starting to ending event and defines the minimum time required to complete the project. So it is deterministic.

Difference between PERT and CPM (Critical Path Method)

PERT

CPM

1. Probabilistic approach

1. Deterministic approach

2. Three-time estimate

2. One - time estimate

3. Event oriented network model

3. Activity-oriented network model

4. The slack concept is used

4. Float concept is used

5. Project crashing is not possible

5. Project crashing is possible

6. Deals with probabilistic time estimates

6. Deals with deterministic time estimates

Which one of the following is NOT a technique of inventory control?

  1. ABC analysis
  2. FSN analysis
  3. GOLF analysis
  4. FTMN analysis

Answer (Detailed Solution Below)

Option 4 :

FTMN analysis

Industrial Engineering Question 11 Detailed Solution

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Explanation:

Various techniques of inventory control are explained in the table below:

ABC analysis(Always Better Control)

Inventory items are classified based on their annual usage value in monetary terms.

Class A - item: 10 % of the item accounts 75% costs.

Class B - item: 20% of the item accounts 15% costs.

Class C - item: 70% of the item accounts 10% costs.

VED Analysis (Vital, Essential, Desirable)

Inventory items are classified on the basis of their criticality i.e. according to the cost of incurring a stock out

V-Vital: Without which the production process would come to standstill

E-Essential: Their non-availability will adversely affect the efficiency of the production system. It should be given second priority.

D-Desirable: Without which the process is unaffected but is good if they are available for better efficiency.

GOLF Analysis GOLF analysis is carried out mainly on the basis of material.  

      GOLF stands for,

      G → government

      O → Ordinary

      L → Local

      F → foreign

SDE Analysis (Scarce, Difficult, Easily Available)

This type of analysis is useful in the study of those items which are scarce in availability

S-Scarce: Imported items which are generally in short supply

D-Difficult: These are available in market but not always traceable or immediately supplied

E-Easily: Easily available in the market

HML Analysis (High, Medium, Low Cost)

 

This type of analysis is similar to ABC analysis, except that cost per item is taken.

H-Highest: Items whose unit cost is very high, or maximum are given top priority

M-Medium: Items whose unit cost is of medium value

L-Low: Items whose unit cost is low

 

FSND Analysis (Fast, Slow, Non-moving, Dead items)

Inventory items are classified in the descending order of their usage (Consumption rate/ movement value).

F-Fast moving items: That are consumed in short span of time

N-Normal moving items: That are consumed over a period of one year

S-Slow moving items: These items are not frequently issued and consumed over a period of two years or more.

D-Dead items: Consumption of such items are almost nil. It can also be taken as obsolete items

Pessimistic time and optimistic time of completion of an activity are given as 10 days and 4 days respectively, the variance of the activity will be 

  1. 1
  2. 6
  3. 12
  4. 18

Answer (Detailed Solution Below)

Option 1 : 1

Industrial Engineering Question 12 Detailed Solution

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Concept:

Project Evaluation and Review Technique (PERT) is probabilistic in nature and is based upon three-time estimates to complete an activity.

Optimistic Time (to): It is the minimum time that will be taken to complete an activity if everything goes according to the plan.

Pessimistic Time (tp): It is the maximum time that will be taken to complete an activity when everything goes against the plan.

Most likely time (tm): It is the time required to complete a project when an activity is executed under normal conditions.

Average or most expected time is given by \({t_E} = \left( {\frac{{{t_p}\; + \;4{t_m}\; + {t_o}}}{6}} \right)\)

The variance gives the measure of uncertainty of activity completion. The variance of the activity is given by 

Variance, \(V = {\left( {\frac{{{t_p} - {t_0}}}{6}} \right)^2}\)

Standard duration, \(\sigma = \sqrt {variance} \)

Calculation:

Given:

tp = 10 days, to = 4 days

\({\rm{V}} = {\left( {\frac{{{{\rm{t}}_{\rm{p}}} - {{\rm{t}}_{\rm{o}}}}}{6}} \right)^2} = {\left( {\frac{{10 - 4}}{6}} \right)^2} = 1\)

The variance of the activity is 1.

Which of the following keeps a record of receipts, issues and running balance of certain items of stock, especially of fitting items?

  1. Stock items
  2. Bin card
  3. Quantity account
  4. Value account

Answer (Detailed Solution Below)

Option 2 : Bin card

Industrial Engineering Question 13 Detailed Solution

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Concept

Stock items:

  • Stock items are defined as material resources that are held in storerooms and issued to activities that require the materials to be completed.
  • The stock item record determines whether or not the type of stock can be purchased, repaired, tracked, and so on.

Bin cards:

  • Bin means a rack, container, or room where goods are kept. Bin cards are printed cards used for accounting for the stock of material, in stores.
  • A bin card is a quantitative record of receipts, issues, and closing balance of each item of stores. For every item of material, separate bin cards are kept.

ABC inventory control focuses on those

  1. Items not readily available 
  2. Items which consume less money
  3. Items which have more demand 
  4. Items which consume more money

Answer (Detailed Solution Below)

Option 4 : Items which consume more money

Industrial Engineering Question 14 Detailed Solution

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Concept:

The inventory comprises of a large number of items. All items are not of equal importance. The firm, therefore, should pay more attention and care to those items whose usage value is high and less attention to those whose usage value is low.

There are different types of selective inventory control:

ABC analysis(Always Better Control)

Inventory items are classified based on their annual usage value in monetary terms.

Class A - item: 10 % of the item accounts 75% costs.

Class B - item: 20% of the item accounts 15% costs.

Class C - item: 70% of the item accounts 10% costs.

VED Analysis (Vital, Essential, Desirable)

Inventory items are classified on the basis of their criticality i.e. according to the cost of incurring a stock out

V-Vital: Without which the production process would come to a standstill

E-Essential: Their non-availability will adversely affect the efficiency of the production system. It should be given second priority.

D-Desirable: Without which the process is unaffected but is good if they are available for better efficiency.

SDE Analysis (Scarce, Difficult, Easily Available)

This type of analysis is useful in the study of those items which are scarce in availability

S-Scarce: Imported items which are generally in short supply

D-Difficult: These are available in the market but not always traceable or immediately supplied

E-Easily: Easily available in the market

HML Analysis (High, Medium, Low Cost)

 

This type of analysis is similar to ABC analysis, except that cost per item is taken.

H-Highest: Items whose unit cost is very high, or maximum are given top priority

M-Medium: Items whose unit cost is of medium value

L-Low: Items whose unit cost is low

 

FSND Analysis (Fast, Slow, Non-moving, Dead items)

Inventory items are classified in the descending order of their usage (Consumption rate/ movement value).

F-Fast moving items: That is consumed in a short span of time

N-Normal moving items: That is consumed over a period of one year

S-Slow moving items: These items are not frequently issued and consumed over a period of two years or more.

D-Dead items: Consumption of such items are almost nil. It can also be taken as obsolete items

The demand rate for a particular item is 12000 units/year. The ordering cost is Rs.100 per order and the holding cost is Rs.0.80 per item per month. If no shortages are allowed and the replacement is instantaneous, then the economic order quantity is

  1. 1500 units
  2. 2000 units
  3. 500 units
  4. 1000 units

Answer (Detailed Solution Below)

Option 3 : 500 units

Industrial Engineering Question 15 Detailed Solution

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Concept:

F1 Krupalu 26.10.20 Pallavi D1

This model is used when the replacement is instantaneous and no shortage is allowed. The Economic Order Quantity for this model is given by Wilson Formula.

\({Q^*} = \sqrt {\frac{{2D{C_o}}}{{{C_h}}}} \)

where Q* = Economic Order Quantity (Units), D = Demand rate (Units/month or Units/year), Co = Ordering cost/order (Rs.), Ch = Handling cost (Rs./unit/year)

[Note: Time unit of Demand & Handling Cost must be same i.e. units/year or units/month]

Calculation:

Given:

D = 12000 units/year, Co = Rs. 100, Ch = Rs. 0.80/unit/month ⇒ Rs. 0.80 × 12/unit/year

\(\;{Q^*} = \sqrt {\frac{{2D{C_o}}}{{{C_h}}}} \)

\( \Rightarrow {Q^*} = \sqrt {\frac{{2 \times 12000 \times 100}}{{0.80 \times 12}}} \)

⇒ Q* = 500 units.

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