Industrial Engineering MCQ Quiz in मल्याळम - Objective Question with Answer for Industrial Engineering - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 8, 2025
Latest Industrial Engineering MCQ Objective Questions
Top Industrial Engineering MCQ Objective Questions
Industrial Engineering Question 1:
Break even analysis consist of
Answer (Detailed Solution Below)
Industrial Engineering Question 1 Detailed Solution
Breakeven analysis is used to find the minimum level of production required. It evaluates both fixed and variable costs.
A breakeven analysis is used to determine how much sales volume your business needs to start making a profit, based on your fixed costs, variable costs, and selling price.
Break-even point usually means the business volume that balances total costs with total gains. At break-even volume, in other words, total cash inflows equal total cash outflows. At Break-Even, in other words, net cash flow equals zero.
Break even analysis consists of:
- Fixed cost
- Variable cost
- Sales revenue
Industrial Engineering Question 2:
The headquarters of the Eastern Railway Zone is located at _______.
Answer (Detailed Solution Below)
Industrial Engineering Question 2 Detailed Solution
The correct answer is Kolkata.
Key Points
- Indian Railways is divided into 18 zones and 73 divisions.
- A Divisional Railway Manager (DRM) heads the division and he/she reports to General Manager (GM).
- A Railway Division is the smallest administrative unit of Railways.
- North Zone is the largest zone.
Given below is the list of all railway zones and their headquarters:
Railway Zone |
Headquarters |
Central Railway |
Mumbai |
Northern Railway |
Delhi |
North Eastern Railway |
Gorakhpur |
Northeast Frontier Railway |
Guwahati |
Eastern Railway |
Kolkata |
South Eastern Railway |
Kolkata |
South Central Railway |
Secunderabad |
Southern Railway |
Chennai |
Western Railway |
Mumbai |
South Western Railway |
Hubballi |
North Western Railway |
Jaipur |
West Central Railway |
Jabalpur |
North Central Railway |
Allahabad |
South East Central Railway |
Bilaspur |
East Coast Railway |
Bhubaneswar |
East Central Railway |
Hajipur |
Metro Railway |
Kolkata |
South Coast Railway |
Visakhapatnam |
Industrial Engineering Question 3:
The amount of time by which an activity can be delayed without affecting project completion time is
Answer (Detailed Solution Below)
Industrial Engineering Question 3 Detailed Solution
Explanation
Slack or Event Float
- Slack corresponds to the event in PERT.
- Float corresponds to activity in CPM.
Slack
- It is defined as the amount of time by which an event can be delayed without delaying the project schedule.
- Slack of an event = Latest Start Time – Earliest Start Time OR Latest Finish Time – Earliest Finish Time
There are three types of floats.
Total Float (TF) |
|
Free Float (FF) |
· Part of the Total Float, which can be used without affecting the float of succeeding activity. · Extra time by which an activity can be delayed so that the succeeding activity can be started on earliest start time.
|
Independent Float (IF) |
|
Industrial Engineering Question 4:
Pre-planning stage in production planning and control includes which of the following activities?
Answer (Detailed Solution Below)
Industrial Engineering Question 4 Detailed Solution
Explanation:
Production Planning & Control consist of three different stages.
- Planning
- Action
- Monitoring.
Planning Stage: Planning stages include activities such as planning the resources, facilities, etc. They are further divided into two stages.
- Pre-planning Stage: This stage deals with the activities such as product planning, forecasting of the demand on the basis of the past trend, inputs planning, plant and facility planning related to location and layout.
- Planning Stage: After the pre-planning, the quantity, level of quantity, process capacity, production planning like routing, scheduling materials, tools planning, etc. are carried out in the planning stage.
Action Stage: It is the real implementation of the plan. It begins with dispatching functions, which deals with the progress of work or job.
Monitoring: In this stage, the planned activities are controlled and monitored by using various techniques such as inventory control, tool control, cost control, quality control, etc.
Industrial Engineering Question 5:
The original cost of equipment is rupees 1,00,000. Its salvage value at the end of its useful life of five years is 40,000. Its book value at the end of two years of its useful life as per straight line method of evaluation of depreciation will be
Answer (Detailed Solution Below)
Industrial Engineering Question 5 Detailed Solution
Concept
By Straight line method
Depreciation is given by
\({D_m} = \frac{{{C_i} - {C_s}}}{n}\)
Where Dm is depreciation
Ci is Initial cost of an asset
Cs is Salvage or scrap value (Estimated at the end of utility period)
Book value (Bm) at the end of 'm ' years life is given by
\({B_m} = {C_i} - m × {D_m}\)
Calculation
Given,
Ci = 1,00,000 /-, Cs = 40,000 /-
n = 5 years
Depreciation is
\({D_m} = \frac{{1,00,000 - 40000}}{5} = 12,000\)
Book value at the end of 2 years of lifetime is given by
\({B_m} = {C_i} - m × {D_m}\)
= 1,00,000 - 2 × 12,000 = 76,000 /-
Industrial Engineering Question 6:
Fixed quantity systems are also termed as ______ systems of inventory control.
Answer (Detailed Solution Below)
Industrial Engineering Question 6 Detailed Solution
Explanation:
In inventory management system there are two ways to review the inventory, they are
Fixed order or quantity system:
- In this system the reorder level of the inventory is fixed as soon as the inventory reaches the reorder level a prescribed quantity is ordered in this system the size of order is fixed while the time of order is variable. It is also called reorder level system or two-bin system or Q-system.
Periodic review system/periodic inventory system:
- In this system the period of time after which inventory is reviewed is fixed, after that particular period new order is placed at that point. In this system the time of order is fixed but the size of order is variable. It is also called fixed period system or P-system.
Industrial Engineering Question 7:
Consider the Linear Programming problem:
Maximize: 7X1 + 6X2 + 4X3
subject to:
X1 + X2 + X3 ≤ 5;
2X1 + X2 + 3X3 ≤ 10,
X1, X2, X3 ≥ 0 (Solve by algebraic method).
The number of basic solutions is:
Answer (Detailed Solution Below)
Industrial Engineering Question 7 Detailed Solution
Concept:
For a system of equation, the number of possible basic solution is calculated by - \({n_C}_m\)
n = number of variables.
m = number of equations.
Inequalities must be converted into equalities.
Calculation:
Given:
X1 + X2 + X3 ≤ 5
X1 + X2 + X3 + S1 + 0S2 = 5 (1)
2X1 + X2 + 3X3 ≤ 10
2X1 + X2 + 3X3 + 0S1 + S2 = 10 (2)
n = number of variables = 5
m = number of equations = 2
∴ number of basic solution = \({n_C}_m ⇒ {5_C}_2\)
∴ \(\frac{5!}{2!\;\times\;(5-2)!}\Rightarrow10\)
Industrial Engineering Question 8:
For ship vessel industry the following layout is best suited
Answer (Detailed Solution Below)
Industrial Engineering Question 8 Detailed Solution
Explanation:
Layout planning in manufacturing and service organisations involves the physical arrangement of various resources available in the system to improve the performance of the operating system, thereby providing better customer services.
Layouts can be classified into the following four categories:
- Process layout
- Product layout
- Group layout (Combination layout)
- Fixed-position layout
Fixed-position layout:
In a fixed position layout, the service is performed around a customer that remains stationary while the work is being done.
For example, surgery is performed on a patient, where a patient remains stationary in operation theatre. Different doctors or specialist perform different activities in a sequence on a stationary patient.
Example: Manufacturers of aeroplanes, ships, locomotives, large turbines, heavy machinery, pressure vessels and others which involve heavy materials and sub-assemblies.
This type of layout is suitable:
- When one or a few pieces of an identical product are to be manufactured
- When the assembly consists of a large number of heavy parts, the cost of transportation of which is very high
Product Layout: In product layout, the workstations are located according to the processing sequence for the service.
Process Layout: In process layout, similar activities are grouped together according to the process or function they perform.
Industrial Engineering Question 9:
Which of the following is referred to as MRP II?
Answer (Detailed Solution Below)
Industrial Engineering Question 9 Detailed Solution
Explanation:-
MRP II
- Manufacturing Resources Planning (MRP II) is defined as a method for the effective planning of all resources of a manufacturing company.
- MRP II Serves as an extension of MRP
Additional Information
Materials Requirement Planning, It is a planning technic that converts the master production schedule of end products into a detailed schedule for raw materials and parts used in those end products.
Maximum retail price is the highest price labeled on the product which can be charged by the seller of that product.
ERP - Enterprise Resource Planning
Industrial Engineering Question 10:
Which of the following is NOT the part of 6M’s of Six Sigma?
Answer (Detailed Solution Below)
Industrial Engineering Question 10 Detailed Solution
Explanation:
Six-sigma:
- Six-sigma is an business statistical strategy.
- It is used to identify defects, and removing them from the process of products to improve quality.
- Bill Smith, "Father of six-sigma" introduces this quality improvement methodology to Motorola.
6M's of six-sigma:
Process variation in six-sigma is defined in terms of 6M's i.e. 6 elements contribute to variation in a process. They are:
- Method
- Man
- Machines
- Materials
- Measurement
- Mother nature