Gate 2003

GA/QB/PI−III

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121

2003 PI: Production and Industrial Engineering Duration: Three hours Maximum Marks: 150 Read the following instructions carefully. 1.

This question paper contains 90 objective questions. Q.1–30 carry one mark each and Q.31–90 carry two marks each.

2.

Answer all the questions.

3.

Questions must be answered on special machine gradable Objective Response Sheet (ORS) by darkening the appropriate bubble (marked A,B,C,D) using HB pencil against the question number on the left hand side of the ORS. Each question has only one correct answer. In case you wish to change an answer, erase the old answer completely using a good soft eraser.

4.

There will be NEGATIVE marking. For each wrong answer 0.25 marks from Q.1–30 and 0.5 marks from Q.31–90 will be deducted. More than one answer marked against a question will be deemed as an incorrect response and will be negatively marked.

5.

Write your registration number, name and name of the Centre at the specified locations on the right half of the ORS.

6.

Using HB pencil, darken the appropriate bubble under each digit of your registration number.

7.

Using HB pencil, darken the appropriate bubble under the letters corresponding to your paper code.

8.

No charts or tables are provided in the examination hall.

9.

Use the blank pages given at the end of the question paper for rough work.

10.

Choose the closest numerical answer among the choices given.

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This question paper contains 24 pages. Please report if there is any discrepancy.

Q. 1–30 carry one mark each.

1.

The rank of the matrix (A) 0

2.

 1 1 1  2 2 2    3 3 3

(B) 1

is (C) 2

(D) 3

The function f of a real variable x is defined as f(x) = max {x, –x}. At x = 0, f is (A) continuous and differentiable

(B) neither continuous nor differentiable

(C) not continuous but is differentiable (D) continuous but is not differentiable

3.

For function f defined on R, if (A) (B)

4.

f ( x 0 +α)
f ( x 0 +α) >f ( x 0 )

f ′( x 0 )

< 0, then

for all α > 0 for all α > 0

(C) there exists δ > 0 such that

f ( x 0 +α)
for 0 > α > δ

(D) there exists δ > 0 such that

f ( x 0 +α) >f ( x 0 )

for 0 < α < δ

The number of initial conditions required to uniquely determine a particular solution of the second order differential equation (A) 2

5.

(B) 1

(C) 0

y ′′ =m

, where m is a constant, is

(D) dependent on the value of m

Stiffness of a metal is characterized by its (A) modulus of elasticity

(B) yield strength

(C) ultimate tensile strength

(D) elongation

GA/QB/PI−III

6.

7.

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During the torsion test, the state of stress in the specimen is defined by (A)

σ , σ2 =0 1 =σ 3 =τ

(B)

σ 1 =σ2 =σ 3 =τ

(C)

σ , σ2 =0 1 = −σ 3 =τ

(D)

σ , σ2 =σ3 =0 1 =τ

The heat treatment process used for hardening of steel is (A) tempering

8.

123

(B) quenching

(C) normalizing

(D) annealing

By doubling the absolute temperature of a body, the amount of heat radiated from it will change by a factor of (A) 2

9.

10.

(B) 4

(C) 8

(D) 16

Inversions of a mechanism can be obtained by (A) adding more links

(B) changing its link lengths

(C) fixing its different links

(D) changing link positions/orientations

A hole of diameter

+ 0.03

25.00 + 0.01

mm is to be inspected by using GO/NO GO gages.

The size of the GO plug gage should be (A) 25.00 mm 11.

(D) 25.03 mm

(B) Collapsibility

(C) Strength

(D) Fluidity

For good weldability, the carbon equivalent (%) of steel should be in the range of (A) 0.2–0.4

13.

(C) 25.02 mm

Which ONE of the following is NOT a property of a sand mold? (A) Permeability

12.

(B) 25.01 mm

(B) 0.5–0.6

(C) 0.7–0.8

(D) 0.9–1.0

Which of the following processes can be used for welding of aluminium alloys? P Submerged arc welding Q Gas metal arc welding R Electroslag welding

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S Gas tungsten arc welding (A) P, Q 14.

15.

16.

17.

(B) Q, S

(C) Q, R

(D) R, S

Formation of built–up edge during machining can be avoided by using (A) tool with low positive rake angle

(B) high feed rate

(C) high cutting speed

(D) large depth of cut

A good cutting fluid should have (A) low thermal conductivity

(B) high specific heat

(C) high viscosity

(D) high density

The function of an interpolator in a CNC machine controller is to provide (A) spindle speed control

(B) position feedback and control

(C) feed drive coordination and control

(D) on/off relay control

While allocating n facilities to n locations, in CRAFT, the local optimality of a solution is checked by interchanging the locations of two facilities. The number of alternatives to be checked is approximately (A) n

18.

(B) n/2

(C) 2n

(D)

n2 / 2

A press costs Rs. 50 lakhs and is to be used for manufacturing a single item. The material and processing cost is Rs. 300 per unit and the anticipated selling price is Rs. 500 per unit. The minimum yearly demand of the product for a payback period of one year for the press is (A) 10,000 units (B) 15,000 units (C) 20,000 units

19.

(D) 25,000 units

The min–max distance criterion in facility location is relevant to the location of a (A) new machining center in a factory

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(B) vehicle sales center in a large city (C) petroleum refinery in coastal region (D) fire extinguisher in a warehouse for chemicals 20.

Which of the following are achieved by a Relational Database System? P Independence of physical database and logical database Q Retrieval of records based on field value and not address in memory R Flexibility for a file to have records of different lengths within it (A) P, R

21.

(B) P, Q

(C) Q, R

(D) P, Q, R

In a continuous pipe welding process, the number of defects in a given length of weld can be expected to have the following type of distribution (A) Normal

22.

(B) Exponential

(C) Geometric

(D) Poisson

CUSUM chart for statistical process control is more effective than

X

chart when

there is a

23.

(A) rapid drift in process mean

(B) slow drift in process mean

(C) high variability in process output

(D) low variability in process output

Which among the following are used for discrete event simulation? P SLAM Q GPSS R SMALLTALK S SIMAN (A) R, S

24.

(B) P, Q, R

(C) P, Q, S

(D) Q, S

In the programming language C, which among the following statements are correct?

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P Array dimensions must be determined at compile time Q Array dimensions may be determined at run time R An array name is a pointer variable whose value cannot be changed (A) P, R 25.

(C) Q

(D) P

Which one among the following is a Therbligs? (A) Get

26.

(B) Q, R (B) Step

(C) Put

(D) Position

Which among the following are correct rules in Business Process Reengineering? P Organize around outcomes, not tasks Q Capture information through at least two sources R Those who use the output of a process should perform the process (A) P, Q

(B) Q, R

(C) P, R

(D) P

Q. 27 – 30 are “Matching” exercises. Choose the correct one out of the following alternatives A, B, C, D. 27.

28.

Gear type

Drive configuration

P Helical gear

1.

Planetary

Q Bevel gear

2.

Enplane, perpendicular shafts

R Worm and wheel

3.

Out of plane, not intersecting shafts

4.

Inplane, parallel shafts

(A)

P– 4, Q– 2, R– 1

(B) P– 4, Q – 2, R–3

(C)

P – 1, Q – 3, R– 2

(D) P – 3, Q– 4, R– 2

Instrument

Principle of inspection

P Dial. Indicator

1.

Non contact

Q Pneumatic gage

2.

Limit of size

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R GO/NO GO gage

3.

127

Comparator

(A) P–2, Q–3, R–1

(B) P – 3, Q– 1, R– 2

(C) P – 1, Q– 2, R– 3

(D) P – 2, Q– 1, R– 3

29.

Product

Manufacturing process

P Food cans

1.

Forging

Q Connecting rods

2.

Rolling

R Metal foil

3.

Deep drawing

4.

Extrusion

(A) P–3, Q–1, R–2

(B) P – 1, Q– 2, R– 4

(C) P – 4, Q– 3, R– 3

(D) P – 2, Q– 4, R– 1

30

Process

Mechanism of material removal

P EDM

1.

Erosion

Q ECM

2.

Thermal evaporation

R AJM

3.

Anodic dissolution

4.

Etching

(A) P–2, Q–3, R–1

(B) P – 2, Q– 4, R– 1

(C) P – 1, Q– 3, R– 4

(D) P – 4, Q– 3, R– 1 Q. 31–90 carry two marks each.

31.

In a plane stress system

σx

= 700 MPa,

σy

= 300 MPa,

MPa. What is the magnitude of the principal stresses? (A)

σ1

= 860 MPa,

= 140 MPa,

(B)

σ1

= 1721 MPa,

σ2

= 279 MPa,

σ3

=0

(C)

σ1

= 1360 MPa,

σ2

= 640 MPa,

σ3

=0

σ2

=0

σ3

σz

= 0 and

τ xy

= 300

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32.

33.

σ1

= 760 MPa,

= 40 MPa,

σ3

=0

Microstructure of annealed hypereutectoid plain carbon steel consists of (A) pearlite with grain boundary cementite

(B) full pearlite

(C) ferrite and pearlite

(D) tempered martensite

Proper gating design in metal casting P

influences the freezing range of the melt

Q

compensates the loss of fluidity of the melt

R

facilitates top feeding of the melt

S

avoids misruns

(A) P, R 34.

σ2

GA/QB/PI−III

(B) Q, R

(C) R, S

(D) P, S

A 100 mm thick steel bar is to be horizontally cast with two correctly spaced top risers of adequate feeding capacity. Assuming end effect without chill, what should be the theoretical length of the bar? (A) 96 mm

35.

(C) 192 mm

(D) 156 mm

In sheet metal working, the spring back increases when P

ratio of bend radius to sheet thickness is small

Q

Young’s modulus of the sheet is low

R

yield strength of the sheet is high

S

tension is applied during bending

(A) P, Q, R 36.

(B) 132 mm

(B) Q, R

(C) P, Q, S

(D) P, S

A 900 mm long steel plate is welded by manual metal arc welding process using welding current of 150 A, arc voltage of 20 V and welding speed of 300 mm/min. If the process efficiency is 0.8 and surface resistance is 36 micro–ohm, the heat input will be (A) 600 J/mm

(B) 480 J/mm

(C) 146 kJ

(D) 116 kJ

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The cold cracking susceptibility of the heat affected zone in an arc weld is influenced by P

entrapped hydrogen

Q

residual stresses

R

martensitic transformation

S

slag inclusion

(A) P, S 38.

(B) P, Q, S

(C) P, Q, R

(D) R, S

The following operations are involved in producing powder metallurgy parts. Arrange them in sequence P

Forging

Q

Sintering

R

Compaction

S

Blending

(A) S, R, P, Q 39.

(B) S, P, R, Q

(C) S, Q, R, P

(D) S, R, Q, P

Strength of a polymer can be enhanced by (A) increasing its crystallinity (B) decreasing the cross–linking of the polymer chains (C) increasing the branching of the polymer chains (D) decreasing the length of the polymer chains

40.

A metric thread with 2 mm pitch and 60° thread angle is inspected for pitch diameter using the three wire method. The diameter of the best size of wire is (A) 0.577 mm

(B) 1.0 mm

(C) 1.155 mm

(D) 2.0 mm

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GA/QB/PI−III

During the inspection of a M 10 × 2.5 screw thread, the following observations were made Thread Number 1 2 3 + 10 –2 –4

Pitch error

4 +8

5 –3

(microns) The cumulative pitch for the thread is (A) 1.8 microns 42.

(B) 5.4 microns

(C) 9 microns

In an interchangeable assembly, a shaft of

20.00

+ 0.025 + 0.010

20.00

− 0.015 − 0.005

(D) 14 mcirons

mm diameter and a hole of

mm diameter form a mating pair. In the worst assembly condition,

the clearance between them will be (A) 40 microns

43.

(B) 30 microns

(C) 25 microns

(D) 15 microns

A typical 2D transformation operation to convert a point P1 ( x 1 , y1 ) to the point P2 ( x 2 , y 2 ) in a CAD package is represented below

[x 2

y 2 1] = [ x 1

y1

1.75 0 0 1]  0 0.5 0  2.5 1.5 1

The transformation operation(s) envisaged around the origin will be

44.

(A) rotation only

(B) scaling only

(C) rotation and translation

(D) scaling and translation

Tool life equations for two tools under consideration are as follows HSS

:

VT 0.2 = 150

Carbide

:

VT 0.3 = 250

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where V is the cutting speed in m/min and T is the tool life in min. The breakeven cutting speed above which the carbide tool will be beneficial is (A) 54 m/min 45.

(B) 150 m/min

(C) 194 m/min

(D) 250 m/min

The following part program refers to the machining of a circular slot on a CNC milling machine

−−−−−−−−−−−−− Metric Units, Absolute Coordinate System | | N0050

G01

X 50.0

Y 75.0

Z− 1.0

F 100.0;

N0060

G02

X 70.0

Y 55.0

R 20.0

F 80.0;

| | The center of the circular slot is (A) (50, 75) 46.

(B) (70, 55)

(C) (50, 55)

(D) (70, 75)

A small pipe with a cross sectional area of 0.001 m 2 discharges water from an open tank resting on the ground with the level of water 1 m above the center line of the pipe. If the tank contains 0.2 m 3 of water and the flow through the pipe is steady, the velocity of water flowing through the pipe will be (A) 4.43 m/s

47.

(B) 8.86 m/s

(C) 9.81 m/s

(D) 19.62 m/s

A hydraulic press is used to produce circular blanks of 10 mm diameter from a sheet of 2 mm thickness. If the shear strength of the sheet material is 400 N/ mm 2 , the force required for producing a circular blank is (A) 8kN

(B) 25.13 kN

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GA/QB/PI−III

(D) 125.66 kN

A single point cutting tool with a nose radius of 0.4 mm was used to turn a component in a lathe employing a feed rate of 0.3 mm/rev. If the feed−rate is doubled, the ideal surface roughness (peak−to−valley height) produced on the component will increase by a factor of (A) 2

49.

(B) 4

(C) 8

(D) 16

In continuous time, a Base Stock policy always orders against each unit of demand, and ensures that Inventory Level + Amount on Order = S = Base Stock Suppose S = 3 and that demand process is Poisson with rate λ = 0.3 per day. Assume lead time = 4 days. Then in steady of the system, the Expected Inventory Level equals (A) 1.2

50.

(B) 1.65

(C) 1.8

(D) 1.9526

The solution to the differential equation y ′′ + y = 2 cos x on [0, ∞) such that y (0) = y ′ (0) = 1 is

51.

(A) (1 + x) sin x + cos x

(B) (1 − x) sin x + cos x

(C) sin x + cos x

(D) sin x − cos x

The number of defectives X in a lot of size n has Binomial distribution B(n, p). We are given that n = 10. The number p is the defective fraction for the process that produced individual pieces in the lot. A priority we are told that 0.10 with probability 0.3 p=  0.15 with probability 0.7  Upon observing the event that X = 0, the conditional probability that p = 0.1 is

GA/QB/PI−III

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(A) 0.34 52.

(B) 0.43

133

(C) 0.54

(D) 0.66

If demand data for a product over time is expected to have high statistical correlation, then which of the following models can be useful? P

A Least Squares Regression model

Q

A model having linear and seasonal trends

R

A model based on auto-regression

(A) P, Q 53.

(B) Q, R

(C) Q

(D) R

For estimating the fraction of time that a machine is busy (i.e., machine utilization), 100 observations were made at random times for work sampling and at 20 of these times, the machine was found busy. The 95% confidence interval for machine utilization is (A) (0.11, 0.29)

54.

(B) (0.12, 0.28)

(C) (0.13, 0.27)

(D) (0.14, 0.26)

In the following pseudo−code, the input x 0 , a 0 , a 1 , a 2 and a 3 are real numbers

Set b i = 0 Set i = 2 Repeat b i = a i +1 + x o b i +1 i = i−1 Until i < 0 The value of b −1 is

134

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GA/QB/PI−III

(A) a 3 + a 2 x 0 + a 1 x 02 + a 0 x 30

(B) a 0 + a 1 x 0 + a 2 x 02 + a 3 x 30

(C) ( a 0 + a 1 + a 2 + a 3 ) x 3

(D) 3 a 0 a 1 a 2 a 3 x

Consider the n eigenvalues of n by n real valued matrix. Which ONE of the following is necessarily true? (A) The eigenvalues are n distinct real numbers (B) The eigenvalues are n distinct complex numbers (C) If any eigenvalue is the real number p, then − p is also an eigenvalue (D) If any eigenvalue is the complex number p + iq, then p − iq is also an eigenvalue

56.

The linear programme [LP] is defined as

[LP]

Min

− 2 x 1 + 3x 2

s.t.

x 2 ≤ 10 x1 , x 2 ≥ 0

[LD] is the dual linear program of [LP]. Then

57.

(A)

[LP] is feasible and unbounded whereas [LD] is infeasible

(B)

[LP] and [LD] are feasible and unbounded

(C)

[LP] and [LD] are feasible and bounded

(D)

[LP] is infeasible whereas [LD] is feasible and bounded

The assignment problem is the optimization problem n

Min

n

∑ ∑ c ij x ij i −1 j =1 n

s.t.

∑ x ij = 1 i =1

all j

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n

∑ x ij = 1

all i

x ij ∈ { 0, 1}

all i, j

j =1

Suppose [LP] is the same problem with the constraints x ij ∈ {0, 1} replaced by x ij ≥ 0. Then (A) [LP] does not have a feasible solution (B)

every solution to [LP] is a valid solution to the assignment problem

(C) every solution to the assignment problem is a degenerate basic feasible solution of [LP] (D) every solution to the assignment problem is a non−degenerate basic feasible solution of [LP] 58.

In M/M/I queue with arrival rate λ and service rate µ > λ, the Expected Busy Period duration is (A)

59.

1 µ−λ

(B)

µ+λ (µ − λ ) 2

(C)

λ µ (µ − λ)

(D)

λ (µ − λ ) 2

In the network shown below components 1, 2, and 3 are similar, with p as reliability of a component. All components work or fail independently

The reliability of connectivity of the network between P and Q is (A) p (1 − p) 2

(B) p 2 (2 − p)

(C) 2p 2 (1 − p)

(D) p 2 (1 − p)

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Which of the following statements are true for Relational Database Systems? P

Non−binary relations can be stored

Q

A given key value determines a unique record

R

Design of the databases precludes redundancy of information

S

Index files can be stored on trees

(A) P, R 61.

GA/QB/PI−III

(B) Q, S

(C) Q, R, S

(D) P, Q, S

In the context of economic lot size, with the usual notation we have TC (Q) =

Kd hQ + Q 2

If TC (300) = TC (600), what is the economic lot size Q*? (A) 348

(B) 379

(C) 424

(d ) 450

Q. 62−69 are "Matching" exercises. Choose the correct one out of the following alternatives a, b, c, d 62.

Material P Q R S

Grey cast iron White cast iron Medium carbon steel Stainless steel

Product 1. 2. 3. 4.

Mill rolls Crank shaft Surgical instrument Machine tool bed

(A) P − 4, Q− 1, R − 2, S− 3

(B) P − 1, Q− 4, R − 2, S− 3

(C) P − 2, Q− 4, R − 3, S− 1

(D) P − 2, Q− 1, R − 4, S− 3

63.

Process P Q R S

Green sand molding Shell molding Investment molding Ceramic molding

Mold making technique 1. 2. 3. 4.

Pouring Dipping Compaction Resin bonding

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(A) P − 2, Q− 1, R − 4, S− 3

(B) P − 1, Q− 3, R − 2, S− 4

(C) P − 3, Q− 4, R − 2, S− 1

(D) P − 4, Q− 2, R − 3, S− 1

64.

Group I P Q R S

Arc welding Friction welding Solid state welding Laser welding

Group II 1. 2. 3. 4.

Diffusion Polarity Focussing Kinetic energy

(A) P − 4, Q− 1, R − 3, S− 2

(B) P − 3, Q− 2, R − 4, S− 1

(C) P − 1, Q− 2, R − 4, S− 3

(D) P − 2, Q− 4, R − 1, S− 3

65.

Group I P Q R S

Grain growth Dilution Dendrite Porosity

Group II 1. 2. 3. 4.

(A) P − 2, Q− 4, R − 1, S− 3 (C) P − 1, Q− 2, R − 3, S− 4 66.

Weld deposit HAZ Gas Solidification (B) P − 4, Q− 2, R − 1, S− 3

(D) P − 2, Q− 1, R − 4, S− 3

Group I

Group

II

P

Automatic lathe

1.

Simultaneous control of several

Q

CNC

2.

machine tools Batch production of a family of

R

FMS

3.

parts Mass production of a single

S

DNC

4.

variety of parts Flexible control of a single machine tool

(A) P − 3, Q− 1, R − 4, S− 2

(B) P − 3, Q− 4, R − 2, S− 1

(C) P − 4, Q− 3, R − 1, S− 2

(D) P − 4, Q− 1, R − 3, S− 2

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GA/QB/PI−III

Group I P Q R S

Group II

Dressing Loading Glazing Truing

1. 2. 3. 4.

Blunting of grinding wheels Shaping of grinding wheels Sharpening of grinding wheels Clogging of grinding wheels

(A) P − 2, Q− 1, R − 4, S− 3

(B) P − 3, Q− 1, R − 4, S− 2

(C) P − 3, Q− 4, R − 1, S− 2

(D) P − 4, Q− 3, R − 1, S− 2

68.

Life cycle phases P Q R S

Start–up Rapid growth Maturation Stabilization (or decline)

Management decisions 1. 2. 3. 4.

Cost control and variety reduction Clearance schemes and discounts Revenue maximization and premiums Market feedback and positioning

(A) P − 4, Q− 3, R − 1, S− 2

(B) P − 4, Q− 1, R − 2, S− 3

(C) P − 1, Q− 3, R − 2, S− 4

(D) P − 1, Q− 2, R − 4, S− 3

69.

Group I P Q R S

Group II

Manufacture of dies Manufacture of families of gears Manufacture of steel sheets Manufacture of cylinder blocks

1. 2. 3. 4.

Product layout Process layout Cellular layout Transfer line

(A) P − 2, Q− 3, R − 1, S− 4

(B) P − 4, Q− 3, R − 2, S− 1

(C) P − 1, Q− 4, R − 2, S− 3

(D) P − 3, Q− 2, R − 4, S− 1

Data for Q. 70– 71 is given below. Solve the problems and choose correct answers In a single pass flat rolling operation, a 400 mm wide steel strip having a thickness of 10 mm is reduced to 8 mm by using a roll of 600 mm diameter 70.

The roll–strip contact length is (A) 24.5 mm

(B) 34.6 mm

(C) 17.3 mm

(D) 49 mm

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139

Considering the situation for a maximum draft of the process, what is the coefficient of friction? (A) 0.082

(B) 0.007

(C) 0.164

(D) 0.014

Data for Q. 72–73 is given below. Solve the problems and choose correct answers The following data refers to a slab milling operation

72.

Diameter of cutter

:

50 mm

Number of teeth on cutter

:

12

Cutter spindle speed

:

300 rpm

Depth of cut

:

2 mm

Length of job

:

500 mm

Longitudinal table feed

:

200 mm/min

The feed per tooth during the milling operation is (A) 0.056 mm

73.

(B) 0.167 mm

(C) 8.662 mm

(D) 17.333 mm

If the cutter overtravel is 2 mm, the machining time for the single pass milling operation will be (A) 0.67 min

(B) 1.50 min

(C) 1.67 min

(D) 2.56 min

Data for Q. 74–75 is given below. Solve the problems and choose correct answers The following data refers to an orthogonal machining of mild steel with a single point HSS tool Rake angle of tool

:

+ 10°

Uncut chip thickness

:

0.3 mm

Width of cut

:

2.0 mm

Shear plane angle

:

36°

140

Elite Academy Shear strength of mild steel

:

GA/QB/PI−III 450 N/mm2

Using Merchant’s analysis 74.

The coefficient of friction between the chip and tool will be (A) 0.141

75.

(B) 0.344

(C) 0.532

(D) 0.688

The shear force in cutting will be (A) 270 N

(B) 333.75 N

(C) 450 N

(D) 459.34 N

Data for Q. 76–77 is given below. Solve the problems and choose correct answers A single sampling plan has sample size n = 200 and acceptance number c = 1. Lot size N is 5000. Assume Poisson distribution for number of defectives in the sample. 76.

For producer’s risk of 5% the Acceptable Quality Level (AQL) in terms of defective fraction is (A) 0.1%

77.

(B) 0.125%

(C) 0.15%

(D) 0.175%

When defective fraction is 0.4%, assuming 100% inspection of the rejected lots, the Average Total Inspection (ATI) per lot is (A) 1118 pieces

(B) 1193 pieces (C) 1264 pieces

(D) 1287 pieces

Data for Q. 78–79 is given below. Solve the problems and choose correct answers The lifetime Y (in months) of a typical light bulb made by a certain company has a probability density function f (y) = 0.4 exp {– (04) y}. 78.

The variance of Y is (A) 1.25

(B) 6.25

(C) 12.5

(D) 16.0

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The probability that Y < 10 is (A) 0.3297

(B) 0.4909

(C) 0.9817

(D) 0.9891

Data for Q. 80 – 82 is given below. Solve the problems and choose correct answers Overtime a certain system alternates between working periods and repair periods. A typical working period is a continuous random variable distributed uniformly over the interval (10, 15). A typical repair period is one unit with probability 0.3, two units with probability 0.4 and three units with probability 0.3. All these random variables are independent. 80.

The availability of the system which is the long run fraction of time the system is working is (A) 17/145

81.

(B) 125/145

(C) 20/125

(D) 10/125

If each repair costs Rs. 100, then the long run expected repair cost per unit time is (A) 1700/145

(B) 12500/145

(C) 1000/145

(D)

1000/125 82.

It was observed that on the first repair period, the repair is not completed at the end of one unit of time. Given this, the probability that repair will be completed by the next unit of time, is (A) 4/10

(B) 3/10

(C) 4/7

(D) 3/7

Data for Q. 83– 84 is given below. Solve the problems and choose correct answers A person takes a fishing lake on lease for a period of 2 years by paying Rs. 1 crore now. He considers two choices. Choice I is to fish after one year when the

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receipts will be Rs. 2 crores. Choice II is to fish at the end of 2 years when the receipts will be Rs. 3 crores. 83.

Using the net present value criterion, which of the choices is preferable, if the annual risk–free interest rate is 10% and the interest is compounded yearly (A) Choice I is better

(B) Choice II is better

(C) Both choices are equally attractive

(D) Not enough data to decide

Data for Q. 85 –86 is given below. Solve the problems and choose correct answers For assembly of a product, ten tasks with processing times 6, 5, 2, 1, 5, 5, 4, 3, 3 and 2 minutes respectively are to be performed on an assembly line.

The

production target is to be achieved in a single shift of 7 hours. This is to be done by allocating tasks to a certain number of workstations. 85.

The theoretical minimum number of workstations to produce 35 units per day is (A) 2

86.

(B) 3

(C) 4

(D) 5

For certain task precedences, the following is a feasible allocation of tasks to workstations. Tasks

Tasks

Workstation

1

{1}

Workstation

4

{5}

Workstation

2

{2, 4}

Workstation

5

{7, 10}

Workstation

3

{3, 6}

Workstation

6

{8, 9}

The efficiency of this line in percent (%) is (A) 76.33

(B) 79.21

(C) 83.33

(D) 85.71

Data for Q. 87–88 is given below. Solve the problems and choose correct answers

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Activities K, L, M, N and O constitute a project.

The activity times and

precedence constraints are indicated in the accompanying activity-on-arc network.

In terms of the variables VX representing the earliest start times of activities starting from node X, consider the linear programme

[LP]

Min

VS − VP

s.t.

VQ − VP ≥ 2 VR − VP ≥ 3 VR − VQ ≥ 3 VS − VQ ≥ 6 VS − VR ≥ 4 VP , VQ , VR , VS ≥ 0

87.

The value of the optimal solution to [LP] is (A) 7

(B) 8

(C) 9

(D) 15

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The number of binding constraints of [LP] at optimality is (A) 1

(B) 2

(C) 3

(D) 4

Data for Q. 89–90 is given below. Solve the problems and choose correct answers In the following discrete time production planning problem, assume Just–In–Time policy : i.e., produce as late as possible, while meeting all demands on time. Suppose stock at the beginning of week 1 equals 100, weekly demands and available production capacities are as follows:

89.

Week

:

1

2

3

4

Demand

:

200

200

200

200

Capacity :

600

100

000

300

What should be the inventory at the end of period 2? (A) 0

90.

(B) 100

(C) 200

(D) 300

What should be the production in period 1? (A) 400

(B) 500

(C) 600

***********

(D) 700

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2003 SOLUTIONS 1.

(B) The determinant of the given matrix is zero and hence the rank of the matrix is 1.

4.

(A)

6.

(C)

5.

(A)

The state of stress and Mohr circle 3 for the same is given below.

7.

(B)

8.

(D) E = σ T4 where

E = amount of heat radiated. σ = Stefan Boltzman constant

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T = absolute temperature of the body. 9.

(C)

10.

(B)

11.

(D)

12.

(A)

13.

(B)

14.

(C)

15.

(B)

16.

(C)

17.

(C)

18.

(D) The press cost = Rs. 50 lakh = Rs. 50 × 10 5 profit earned per unit = Rs 500 – Rs. 300 ∆ P = Rs. 200 ∴ Investment = profit per unit × demand = ∆P × Q ∴ 50 × 10 5 = 200 × Q Q = minimum yearly demand = 25,000 units

19.

(C)

21.

(D)

22.

(B)

20.

(D)

The cumulative sum control chart frequently called cesium chart is designed to identify slight but sustained shift in a universe. 23.

(C)

24.

(A)

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25.

(D)

26.

(C)

27.

(B)

28.

(B)

29.

(A)

30.

(A)

31.

(A) σ y = 300 MPa

σ x = 700 MPa σx + σy σ1 = + 2

 σx − σy      2  

700 + 300 = + 2

τ xy = 300 Mpa

2

+ τ 2xy

 700 − 300    2  

2

+ (300) 2

= 500 + 360 = 860 Mpa.

σx + σy σ2 = − 2

 σx − σy      2  

700 + 300 = − 2

2

+ τ × y2

 700 − 300    2  

2

+ (300) 2

= 500 – 360 = 140 Mpa 32.

(A)

34.

(C)

33.

(C)

The top risering is extensively used for light metals as it enables to take the benefit of metallostatic pressure in the riser. Frequently, the number of risers has

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to be more than one so as to derive its most effective use. In such cases, their spacing should be carefully arranged so as to minimize the shrinkage. The feeding range, which is the distance a riser can feed the metal in casting, thus becomes an important consideration in riser design. It is found that the casting thickness is the main parameter affecting the feeding range. The riser diameter and riser height have only limited effect on it. It is usual practice to maintain a feeding range of about 4.5 the thickness of the plate type castings and 6 T for bar type casting.

Each riser can feed 6 T feed range. ∴ spacing between two riser = 2 × 6 T

= 12

T

where T = thickness of bar End effect without chill = 3.6 ∴ Two end effects = 2 × 3.6

T T = 7.2

T.

∴ Total length of bar that can be feed sufficiently by two correctly spaced risers will be = 12 T + 7.2 T = 19.2 T T = 100 = 19.2 100 = 192 mm.

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35.

(D)

36.

(D)

37.

(C)

38.

(D)

39.

(C)

40.

(C) The diameter of the best size of wire = where

P 1 × 2 cos θ / 2

P = pitch = 2mm θ = thread angle = 60°

∴ the diameter of the best size of wire =

41.

(C)

42.

(A)

2 × 2

1 3/2

= 1.1547 mm

The interchangeable assembly of the shaft and the hole will always give clearance fit. In the worst assembly condition, there will be largest hole of size 20.025 mm and the smallest shaft of size 19.085mm. Hence clearance between them will be 20 – 0.25 – 19.085 = 0.040 mm 44.

or 40 micron.

(A) For HSS

VT 0.2 = 150 1  150  0.2

T =    V 

=

(150) 5 V

5

 150  ∴ T 0.2 =    V  (1)

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VT 0.3 = 250

For Carbide

1

(150) 3.33  250  0.3 T= =   V  V 3.33

(2)

For equal tool life, we will find the cutting speed by equating (1) and (2) (150) 5 V5

=

(150) 5 (250) ∴

3.33

(250) 3.33 V 3.33 = V1.67

V = 53.6 m/min

It is the breakeven cutting speed above which the carbide tool life will be beneficial. 45.

(B)

46.

(A) V =

2gh

h = 1m g = 9.81 m/s2

V=

2 × 9.81 × 1

= 4.43 m/s. 47.

(B) The force required to shear a circular blank = cutting perimeter × sheet thickness × shear strength = π d × t × λs = π × 10 × 2 × 400 (N/mm2)

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= 25132 N = 25.13 KN. 48.

(B) The ideal surface roughness produced on a component =

f2 8R

where f = feed rate R = nose radius on single point cutting tool. Hence the ideal surface roughness produced on the component will increase by a factor of 4, if feed rate is doubted. 49.

(C)

52.

(A)

53.

(B)

50.

(A)

P = fraction of time, a machine is busy =

20 = 0.2 100

q = 1 – P = 0.8, that machine is idle. σ P = standard deviation =

Pq = n

0.2 × 0.8 = 0.04 100

For 95% confidence level., the limits for P will be (P – 2 σ P , P + 2 σ P ) = (0.2 – 2 × 0.04, 0.2 + 2 × 0.04)

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= (0.12, 0.28). 54.

(B)

55.

(D)

56.

(B) The LP problem can be understood by drawing solution space.

The value of the objective function decreases with increase in x1 and there is no limit on the value of x1. Hence solution to LP is both feasible and unbounded. When LP has feasible and unbounded, the solution to LD is also feasible and unbounded. 58.

(D)

59.

(B) System reliability of parallel configuration components = 1 – [ (1 − p1 ) (1 − p 2 ) (1 − p 3 )....... (1 − p n ) ]

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where p1 , p 2 , p 3 ,........p n is reliability of each component 1, 2, 3, ……n in parallel configuration. In this problem, reliability of system having parallel components = 1 – (1 – p) (1 – p) = 1 – (1 – p)2 = (2p – p2) system reliability of series configuration = p1 × p 2 × p 3 × ....... × p n where p1 , p 2 , p 3 , .......p n is reliabilities of component 1,2,3…….n in series. In this problem, the two component in series have reliability of p and (2p − p 2 ) = p × ( 2p − p 2 ) = p 2 (2 – p). 60.

(B)

61.

(C) Total cost has two components – (1) order cost and (2) inventory holding cost. TC(Q) =

kd hQ + Q 2

At economic lot size Q*, the order cost is equal to inventory holding cost. ∴

kd hQ * = Q* 2

∴

Q* =

2kd h

where TC(300) = TC(600) h × 300 h × 600 kd kd + = + 300 2 600 2

(1)

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kd kd − = 150 h 300 600 kd = 150 × h 600 ∴

kd = 600 × 150 h

substituting the value of Q* =

600 × 150 × 2

kd in equation (1) h = 424 units

62.

(A)

63.

(C)

64.

(D)

65.

(D)

66.

(B)

67.

(C)

68.

(A)

69.

(A)

70.

(A)

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h0 = initial thickness of the strip = 10 mm h1 = strip thick on rolling = 8mm ∆h = draught = amount of reduction given = h 0 − h 1 = 10 – 8 = 2mm R = roll radius =

600 = 300 mm 2

Roll strip contact length = =

71.

R. ∆ h 300 × 2 = 24.5 mm.

(A) ∆h max = maximum draft   = D 1 −   where

   1 + µ 2  1

D = roll diameter µ = the coefficient of friction between the strip and the roll.   2 = 600 1 −  

∴ 72.

   1 + µ 2  1

µ = 0.082

(A) Feed per tooth during the milling operation (sz) = sz =

73.

(D)

Longitudinal table feed (mm / min) Cutter spindle speed (rpm) × Number of teeth on cutter

f 200 = = 0.056 mm N × Z 300 × 12

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A milling cutter does not make a full cut when it make first contact with the workpiece. Hence the distance traveled between the first contact and cutting to the full depth is known as approach distance (A). This distance is to be added to the length of the cut required. The same distance (A) is required to be added again for the cutter to clear the work at the other end. To these, must be added clearance on both the sides.

A = Approach distance = = where

D   2

2

D  −  − d 2 

2

d (D − d)

D = diameter of the milling cutter d = depth of cut A =

2 × (50 − 2) = 9.8 mm.

c = clearance = cutter over travel = 2 mm (given) l = length of job = 500 mm L = total length of cut = l + 2 A + 2c = 500 + 2 × 9.8 + 2 × 2 = 523.6 mm

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Machining time T =

total length of cut Table feed rate

=

74.

523.6 = 2.618 min 200

(C) Merchant equation 2 φ + λ − γ = 90  where φ = shear angle = 36° (given) γ = rake angle of tool = +10° λ = Friction angle = ∴

F N

2 × 36 + λ – 10° = 90°

∴ λ = 28° µ = the coefficient of friction between the chip and tool = tan λ = tan 28° = 0.532. 75.

(D)

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t 0 = uncut chip thickness = 0.3mm A 0 = length of shear plane = ∴

t0 sin φ

∆θ = length of shear plane =

0.3 sin 36 

= 0.51 mm

width of cut = 2mm ∴ Area of shear plane = 0.51 × 2 = 1.02 mm2. shear force = shear plane area × shear strength of M.S. = 1.02 × 450 = 459.34N 76.

(D) n = sample size N = lot size c = acceptance number Pa = Probability of acceptance of a lot ∝ = producer’s risk. AQL and producer’s risk are interrelated, since probability of acceptance Pa of a lot of specified AQL is nothing but (1 – ∝ ), ∝ being producer’s risk. In this example, producer’s risk is fixed at 5%, then the probability of acceptance of a good lot automatically is 1 – ∝ = 0.95. Further, ∴

Pa = P0 + P1

where P0 = probability of zero defect in a sample size n (n = 200) =

z 0 e −z 0!

(using poisson distribution)

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= e −z where z = n × p where p = percent defective in the lot P1 = probability of one defect in a sample size (n) =

∴

z1 e − z = z e −z 1! Pa = 0.95 = e − z + z e − z 0.95 = (1 + z) e − z 0.175 ; n = 200 100

when p = 0.175% = z = 0.35

with P = 0.175%, the above equation is satisfied and hence defective percentage is 0.175%. 77.

(A) ATI = average total inspection per lot = n. Pa + N (1 – Pa) = n + (N – n) (1 – Pa) P = fraction defective = 0.4% (given) = z = n × p = 200 ×

0.4 100

0.4 = 0.8 100

Since sampling plan must accept a lot if the sample contains 0 or 1 defective as c = 1. Hence we need to add individual

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Probabilities i.e. P0 + P1 to obtain probability of acceptance of the lot Pa = P0 + P1 z 0 e −z P0 = 0! P1 = ∴

= e −z

z1 e − z = z e −z 1!

Pa = e − z + z e − z = (1 + z) e − z = (1 + 0.8) e −0.8 = (1.8) e −0.8 = 0.808

ATI = n + (N – n) (1 – Pa) = 200 + (500 – 200) (1 – 0.808) = 1117.7 = 1118 pieces. 78.

(B) Variance = E ( x 2 ) − [ E( x )] 2 where E(x) = Expected value of x E(x2) = Expected value of x2. ∞

E (x) = ∫ x. f ( x ) dx 0

∞

−0.4 x dx = 0.4 ∫ x e −0.4 x dx = ∫ x . (0.4) e 0

∞

 x e 0.4 x  e −0.4 x = 0.4   + 0.4 ∫ 0.4  − 0.4  0 ∞

= ∫e 0

−0.4 x

dx =

e −0.4 x 0.4

0 ∞

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=

1 [1 − 0] 0.4

E(x) = 2.5 ∞

∞

0

0

E ( x 2 ) = 0.4 ∫ x 2 . e −0.4 dx = 0.4 ∫ x 2 e −0.4 x dx x 2 . e −0.4 x E( x 2 ) = 0.4 0.4

0

∞

x . e −0.4 x dx 0.4 0

+2 ∫ ∞

E( x 2 ) ∞ = ∫ x . e −0.4 x dx 2 0 0 x e −0.4 x 1 + e −0.4 x dx = ∫ 0.4 0.4 ∞

1 = 0.4

e −0.4 x 0.4

0 ∞

  −0.4 x 0 1  e =  ∞  0.4 × 0.4  E( x 2 ) = ∴

2 = 12.5 0.4 × 0.4 variance of X = E ( x 2 ) − [ E ( x ) ] 2 = 12.5 – (2.5)2 = 6.25

79.

(C) P (Y < 10) = 1 – Probability that life is equal to or greater than 10. Probabilities that life of bulb light is equal to or greater than 10 ∞

= ∫ f ( y) . dy 10

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∞

− 0 .4 y .dy = ∫ 0 .4 e 10

e −0.4 y 0 . 4 = − 0.4 =

∞

= e −0.4 y

0

10 ∞

e −4 − e −∞ = 0.0183

∴ The probability that Y < 10 = 1 – 0.0183 = 0.9817 80.

(B) Typical working period is a continuous random variable distributed uniformly over interval of 10 and 15.

Hence mean time between failures = =

10 + 15 2 25 units. 2

Expected mean time to repair = 0.3 × 1 + 0.4 × 2 + 0.3 × 3 = 2 units

The availability of the system =

MTBF MTBF + MTTR

25 / 2 25 125 = = 25 29 145 +2 2 81.

(D)

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The mean time between failures is 12.5 time units. It means there is one failure per 12.5 time units. ∴ Failure rate =

1 2 = 12.5 25

Each repair cost = Rs. 100 ∴ long run expected repair cost per unit time is =

82.

(A)

83.

(B)

2 200 1000 × 100 = = 25 25 125

Choice I: NPV = – 1 +

2 (1 + 0.1)1

= 0.82 cr

Choice II: NPV = − 1 +

3 (1 + 0.1) 2

= 1.48 cr

Choice II is better because NPV is more compared to choice I. 84.

(A) C =

A1 (1 + r )

+

A2 (1 + r ) 2

+ ....... +

An (1 + r ) n

where C = Capital out lay = 1cr A i = net cash proceeds at the end of ith year n = number of years of the expected net cash proceeds. r = internal rate of return (also called discount rate, yield on investment)

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Choice I C = 1 =

A1 (1 + r ) 2 (1 + r )

∴ r = 1 i.e., r = 100% Choice II C =

1 =

A2 (1 + r ) 2 3

∴ (1 + r ) 2 = 3

(1 + r ) 2

1 + r =

3

r = 0.73 or 73% Choice I is better since IRR is 100%, better than choice II. 85.

(B) Working minutes in one shift = 7 × 60 = 420 production rate = 35 units ∴ cycle time =

420 = 12 minutes 35

Total work content of = 36 min. then tasks ∴ Theoretical minimum number of work stations to produce 35 units per day =

total work content 36 = cycle time 12

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= 3. 86.

(D) Line efficiency = =

sum of the time utilised at all stations Time available at all work stations sum of the time utilised at all sations number of stations × cycle time n

∑W = 1 i n×C =

36 6×7

= 0.8571

= 85.71%

C = 7 minutes as it is equal to the time of bottle neck operation or the maximum station time. n = 6 stations. 87.

(C)

Critical path p→ Q→R→S. The value of the optimal solution to [LP] is = 9 which is equal to earliest start time of activities Vs = 9 88.

(C)

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The number of binding constraints at optimality is equal to 3. The binding constrains are VQ – VP ≥ 2 VR – VQ ≥ 3 VS – VR ≥ 4 By substituting the values of these variables, right hand side of equation equal to the left hand size.

89.

(C) Week

:

Demand

1

2

3

4

:

200

200

200

200

Capacity :

600

100

000

300

Production

400

100

000

200

300

200

000

000

Stock

0

100

∴ Inventory at the end of period 2 = 200 90.

(A)

Production in period 1 is 400 units.

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