1. ASSIGNMENT
1. The process of designing conceptual and quantitative models.
2. Find two consecutive integers such that 10 times the smaller number is 5 times the bigger
number.
Solution:
Let the two integers be A, B. Since they are consecutive, A=B-1
Now the given condition is :
10A=5B
Replacing A with B, 10(B-1)=5B
i.e., 10B-10=5B
5B=10
B=2
Hence, A=B-1=2-1=1.
A&B=1 & 2
3. When the solutions of quadratic equation are termed as: (a) rational and equal, (b) real and
distinct and (c) imaginary and distinct.
4. Ash Lubes sells X units of Supreme Lubes each day at the rate of Rs 50 per unit of 100 gm. The
cost of manufacturingandsellingtheseunitsisRs35 perunit plus a fixed daily overhead cost of
Rs 10,000. Determine the profitfunction.How would youinterpretthe situation if the company
manufactures and sells 400 units of the lubes a day?
Solution:
SP=Rs. 50 per Unit(100Gm)
Direct Cost= Rs. 35 per unit
Daily over head=Rs. 10,000
Assuming the company manufactures X units a day:
The total cost to the company is (X*35)+10000=35X+10000
For breaking even: SP = CP
i.e., X*50=35X+10000
15X=10000
The Profit function of the company would be: F(P)= 15X-10000.
Solving this, X=666.6
Hence, the company should manufacture minimum 667 units to get profit
If the company manufactures 400 units a day:
The profit function becomes, F(P)=15(400)-10000
=6000-10000
=-4000

2. The company makes a loss of Rs. 4000 if it manufactures 400 units.
In numerical, Total cost to the company is :
(35*400)+10000= 14000+10000=24000 for 400 units
Cost per unit = 60
Company would lose Rs. 10 per unit.
5. An investorwantstoinvestRs.15,000 in twotypesof bonds.He earns 12% in first type and 15%
in the second. Find his investment in each of his total earning is Rs1950.
Solution:
a. (X*12)/100+(Y*15)/100=1950
b. X+Y=15000
Solving these 2 equations:
X=15000-Y
Replacing X in Eq A.
[(15000-Y)*12+15Y]/100=1950
180000+3Y=195000
3Y=15000
Y=5,000/-
X=10,000/-
6. IshaanPetrochemicalshasintroducedinthe market its latest lube. The marketing manager has
worked out that the demand function of this product, which can be expressed as: Q = 30 – 4P,
Where, Q is the quantity and P is the per kilogram price.
(a) Write the total revenue as a function of price.
(b) Draw the graph of this function.
Solution: Using the demand function given above, Q=30-4P, the relation between Price,
Quantity and Revenue can be tabulated as follows:
Price Quantity Revenue
0 30 0
1 26 26
2 22 44
3 18 54
4 14 56
5 10 50
6 6 36
7 2 14
Revenue as a function of Price: R=30P-4P^2

3. Graph:
7. A switch manufacturer finds that his total monthly production costs are Rs 10,600 when
production is 16,000 units per month, Rs17,800 when it is 26,000 units and Rs 27,000 when the
productionis 36,000 per month. He can sell 16,000 units per month at Rs104each, but has to
reduce the price to Rs94 each in order to sell26,000 pieces. He can sell 36,000 pieces only atRs
80.
Question
Assuming that both cost curve and price curve are quadratic, find
(a) the monthly total cost,
(b) the price,
(c) the monthly revenue, and
(d) the monthly gross profit as functions of the quantity sold.
Find also
0
10
20
30
40
50
60
0 1 2 3 4 5 6 7
Revenue
Revenue

4. (e) the quantity sold,
(f) the price and
(g) the monthlyrevenue atthe breakevenpointandconfirmthat the monthly total cost is then
equal to the monthly revenue.
Solution:
A:
a(16)^2+b(16)+c=10.61
a(26)^2+b(26)+c=17.82
a(36)^2+b(36)+c=273
Solving the quadratic quations:
(3-2)
620a+10b=9.2
(2-1)
420a+10b=7.2
Solving these, a=0.01, b=0.3, c=3.24.
Hence the equation for cost can be written as: 0.01(X)^2+0.3(X)+3.24
B:
Similarly for price:
a(16)^2+b(16)+c=0.104
a(26)^2+b(26)+c=0.094
a(36)^2+b(36)+c=0.080
Solving these:
620a+10b=-0.014
420a+10b=-0.01
200a=-0.004
A=-0.00002
10b=-0.01-(420*-0.00002)=-0.01+0.0084=-0.0016
b=-0.00016
c=0.11168
The price equation can be written as : -0.00002(X)^2+-0.00016(X)+0.11168

5. C:
Monthly Revenue=Price* Quantity
=(-0.00002(X)^2+-0.00016(X)+0.11168)*(X)
=-0.00002(X)^3+-0.00016(X)^2+0.11168(X)
D: Gross profit as function of Qty sold:
Gross Profit = Revenue-Cost
Revenue=Price*Quantity sold
=0.00002(X)^3+-0.00016(X)^2+0.11168(X)-(0.01(X)^2+0.3(X)+3.24)
=0.0000(X)^3-0.0116(X)^2-0.18832(X)-3.24
E: Breakeven point:
Revenue=Cost
-0.00002(X)^3+-0.00016(X)^2+0.11168(X)= 0.01(X)^2+0.3(X)+3.24
I cant solve the rest of this Q :P
And you people done copy this also :D
8. Nahar Chemical Millsproducesthreevarietiesof base oil,Super fine Grade (A grade),finegrade
(B grade) andcoarse grade (C grade).The total annual salesinlacsof rupeesof these products
for the year1999 and 2000 in the fourcitiesisgivenbelow,findthe total salesof three varieties

6. of base oil fortwo years.
Solution:
Elaborate using matrices:
Sale of A Grade: 160 Lakhs
Sale of B Grade: 172 lakhs
Sale of C grade: 224 Lakhs
9. A 2T oil manufacturerproducesthree productsA,B,C whichhe sellsinthe market.Annual sale
volumes are indicated as follows:
If the unit sale price of A, B and C are Rs 2.25, 1.50 and Rs 1.25 respectively, find the total
revenue ineach market with the help of matrices. (ii) If the unit costs of above three products
are Rs 1.60, Rs1.20 and Rs0.90 respectively, find the gross profit with the help of matrices.
Ans:
Total Revenue in market I:
(2.25 1.5 1.25)*(8000 10000 15000)=(2.25*8000 1.5*10000 1.25*15000)
=(18000 15000 18750)
Revenue of A in Market I = 18000
Revenue of A in Market I = 15000

7. Revenue of A in Market I = 18750
Total Revenue in market II:
(2.25 1.5 1.25)*(10000 2000 20000)=(2.25*10000 1.5*2000 1.25*20000)
=(22500 3000 25000)
Revenue of A in Market II = 22500
Revenue of A in Market II = 3000
Revenue of A in Market II = 25000
Gross Profit:
(Sale price-cost price)*(Revenue)
(SP-CP)=(2.25 1.5 1.25)-(1.6 1.2 0..9)=(2.25-1.6 1.5-1.2 1.25-0.9)
=(0.75 0.3 0.35)
Gross Profit=(0.75 0.3 0.35)*(8000 10000 15000)+
(0.75 0.3 0.35)*(10000 2000 20000)
=(6000 3000 5250)+(7500 600 7000)=(13500 3600 12250)
Gross profit of all products put together=(13500+3600+12250)=Rs. 29350/-
10. Robin Singh & Company Ltd. stocks lubes of Castrol brand and Mak brand. The matrix of
transition probabilities of the lubes is shown below:
Determine the market share of each of the brand in equilibrium position.
Solution:
Assuming the market share of Castrol and Mak at Equillibrium are P1 and P2.
At equilibrium position,
R=RP
[P1 P2] 0.9 0.1 = [P1 P2]
0.3 0.7
Where, P1+P2=1
We get the following equations:
0.9P1+0.3P2=P1
0.1P1+0.7P2=P2
Solving these equations, we get:
3P2=P1

8. We already know, P1+P2=1
Thus, P1=3/4=0.75 and P2=0.25.
Thus the marketshare of Castrol andMak at equilibriumpositionare 0.75 and 0.25 respectively.
11. Four boys order in a fish-and-chips restaurant. A orders fish, chips and coke. B orders two fish
with chips. C orders fish and coke. D orders chips and coke. The prices are Rs 50 for fish, Rs 18
for chips, and Rs 15 for coke.
(a) Express each boy’s order as a row vector.
(b) Addtogetherthese fourvectorstoobtaina fifthrow vector representingthe total quantities
ordered.
(c) Express the prices as a column vector.
(d) Multiply each of the five row vectors by the price vector, to obtain the amount owed by
each boy and the total amount owed.
(e) Check that the fifth result in (d) is equal to the sum of the other four results.
Ans: Order of A: O(A): (1 1 1)
O(B): (2 1 0)
O(C): (0 1 1)
O(D): (0 1 1)
Total Order:
O(A)+O(B)+O(C)+O(D)=(3 4 3)
Prices:
50
18
15
Amount owed by each boy:
A= (1 1 1)* 50 =(83)
18
15
B=(118)
C=(33)
D=(33)
Total: (267) 50
Total Order*Prices=(3 4 3)* 18 =(150+72+45)=(267)
15
12. Distinguishbetweenobjective probabilityandsubjectiveprobability. Give one example of each
concept.
Objective Probability:
The probability that an event will occur based an analysis in which each measure is
based on a recorded observation, rather than a subjective estimate. Objective

9. probabilities are a more accurate way to determine probabilities than observations based
on subjective measures, such as personal estimates.
For example, one could determine the objective probability that a coin will land "heads"
up by flipping it 100 times and recording each observation. When performing any
statistical analysis, it is important for each observation to be an independent event that
has not been subject to manipulation. The less biased each observation is, the less
biased the end probability will be.
Subjective Probability:
A probability derived from an individual's personal judgment about whether a specific
outcome is likely to occur. Subjective probabilities contain no formal calculations and
only reflect the subject's opinions and past experience
Subjective probabilities differ from person to person. Because the probability is
subjective, it contains a high degree of personal bias. An example of subjective
probability could be asking Indian Cricekt fans, before the world cup season starts, the
chances of India winning the world cup. While there is no absolute mathematical proof
behind the answer to the example, fans might still reply in actual percentage terms, such
as the Indians having a 50% chance of winning the world cup.
For example, let's say Anand buys a lottery ticket to support a local school troop. The
troop sells 1,000 tickets. From an objective probability perspective, John has a 1 in
1,000 chance of winning. But subjectively, Anand thinks his chances of winning are
much higher because "he has a good feeling about it." Nevertheless, his chances are
still 1 in 1,000.
13. State and prove theorem of addition of probabilities for two events when (a) they are not
independent, (b) they are independent.
Solution:
The additiontheoreminthe Probabilityconceptisthe processof determinationof the
probabilitythateitherevent‘A’orevent‘B’occursor both occur. The notationbetweentwo
events‘A’and‘B’the additionisdenotedas '∪' andpronouncedasUnion.
The resultof this additiontheoremgenerallywrittenusingSetnotation.
P (A ∪ B) = P(A) + P(B) – P(A ∩ B),
Where,
P (A) = probabilityof occurrence of event‘A’
P (B) = probabilityof occurrence of event‘B’
P (A ∪ B) = probabilityof occurrence of event‘A’orevent‘B’.
P (A ∩ B) = probabilityof occurrence of event‘A’andevent‘B’.
Additiontheoremprobabilitycanbe definedandprovedasfollows:

10. Let ‘A’and ‘B’are Subsetsof a finite nonemptyset‘S’thenaccordingtothe additionrule
P (A ∪ B) = P (A) + P (B) – P (A).P(B),
On dividingbothsidesbyP(S),we get
P (A ∪ B) / P(S) = P (A) / P(S) + P (B) / P(S) – P (A ∩B) / P(S) Eq:(1).
If the events‘A’and‘B’correspondto the twoevents‘A’and‘B’ of a randomexperimentandif
the set ‘S’correspondstothe Sample Space ‘S’of the experimentthenthe equation(1)
becomes
P (A ∪ B) = P (A) + P (B) – P (A).P(B)
Thisequationisknownasthe additiontheoreminprobability.
Here the eventA ∪ B referstothe meaningthateitherevent‘A’orevent‘B’occursor bothmay
occur simultaneously.
If two eventsA andB are MutuallyExclusive(Independent) EventsthenA ∩ B = 0,
Therefore
P (A ∪ B) = P (A) + P(B) [since P(A ∩B) = 0],
In language of settheoryA ∩ B̅ issame as A / B.
14. A verylarge lotof manufacturedgoodsare to be sampledasa check onits quality. Suppose it is
assumed that 10 percent of the items in the lot are defective and that a sample of 20 items is
drawn from the lot. What are the following probabilities:
1. Probability of exactly zero defectives in the sample?
2. Probability of more than one defective in the sample?
3. Probability of fewer than two defectives in the sample?
Solution:
Assuming the lot size is 100. 10% of the items are defective, i.e., 10 items are defective out of
100. A sample of 20 items is drawn from the lot.
1. Probability of exactly 0 defectives in the sample:
For this event to happen, the sample of 20 has to be from the 90 non defective
pieces. This can be done in 90c3 ways.
The total numberof waysinwhich 20 items can be drawn from a lot of 100 is 100c3.
So the probability of having exactly Zero defectives is:
90c20/100c20=0.095116.
2. Probability of more than one defective in the sample:
For this we have to calculate the probability with 0 defectives and 1 defectives and
subtract it from 1.
The probability of 0 defectives as calculated above is 0.095116
The number of ways of picking 1 defective is: No. of ways of picking 19 pieces from 90
non defective pieces * no. of ways of picking 1 piece from the 10 defective pieces
That is: (90c19)*(10c1)
The probability for this is: (90c19)*(10c1)/100c20=0.026793
Subtracting these from 1:

11. 1-0.095116-0.026793=0.878091.
3. Probability of fewer than 2 defectives:
Probability of having 0 or 1 defectives. The values are calculated above and are to be
added.
P(0)+P(1)= 0.095116+0.026793=0.121909
15. A batch of 5,000 electric lamps has a mean life of 1,000 hours and a standard deviation of 75
hours. Assume a Normal Distribution.
(a) How many lamps will fail before 900 hours?
(b) How many lamps will fail between 950 and 1,000 hours?
(c) What proportion of lamps will fail before 925 hours?
(d) Giventhe same meanlife,whatwouldthe standard deviation have to be to ensure that not
more than 20% of lamps fail before 916 hours?
Solution:
a) Z=(X- µ)/ Σ

12. =(900-1000)/75=-1.33
Area covered by Z=0 and Z=-1.33 is 0.9176(as per Z table)
Number of bulbs which will fail before 900 hrs= 1000*0.09176=91.76=92(approx)
b) Lamps failing between 950 and 1000
a. Z value for bulbs failing before 950:
Z=(950-1000)/75=-50/75=-0.66
Probability Value corresponding to Z=-0.66 = 0.2546
Probabilityvalueforbulbsfailingat1000 is 0.5 as itis the mean life.
Net probability: 0.5-0.2546=0.2454
No. of bulbs failing between 950 and 1000 = 0.2454*1000=245.4=246(Approx)
c) Proportion of lamps failing before 925.
Z value : Z=(925-1000)/75=-1
Probability value for -1: 0.1586
Proportion of lamps: 15.86%
d) Let D be the standard deviation:
For 20% of lamps to fail: Z value corresponding to 0.2 is -0.84.
Substituting the values in the equation:
Z=(X- µ)/ Σ
-0.84=(916-1000)/ Σ
Σ=-84/-0.84=100
The standard deviation is 100