TN 12th Question Paper 2024 Business Mathematics and Statistics

TN PUBLIC EXAM

QUESTION
PAPER
2024

No. of Printed Pages : 16
7467
!7467BusiMatheandStati! £vÄ Gs
Register Number

PART - III
ÁoPU Pou® ©ØÖ® ¦Òΰ¯À
BUSINESS MATHEMATICS AND STATISTICS
( uªÌ ©ØÖ® B[Q» ÁÈ / Tamil & English Version)

Põ» AÍÄ : 3.00 ©o ÷|µ® ] [ ö©õzu ©v¨ö£sPÒ : 90
Time Allowed : 3.00 Hours ] [Maximum Marks : 90

AÔÄøµPÒ : (1) AøÚzx ÂÚõUPЮ \›¯õP £vÁõQ EÒÍuõ GߣuøÚ \›£õºzxU
öPõÒÍÄ®. Aa_¨£vÂÀ SøÓ°¸¨¤ß AøÓU PsPõo¨£õÍ›h®
EhÚi¯õPz öu›ÂUPÄ®.
(2) }»® AÀ»x P¸¨¦ ø©°øÚ ©mk÷© GÊxÁuØS®,
AiU÷PõikÁuØS® £¯ß£kzu ÷Ásk®. £h[PÒ ÁøµÁuØS
ö£ß]À £¯ß£kzuÄ®.
Instructions : (1) Check the question paper for fairness of printing. If there is any lack of fairness,
inform the Hall Supervisor immediately.

(2) Use Blue or Black ink to write and underline and pencil to draw diagrams.

£Sv & I / PART - I

SÔ¨¦ : (i) AøÚzx ÂÚõUPÐUS® Âøh¯ÎUPÄ®. 20x1=20

(ii) öPõkUP¨£mkÒÍ |õßS ©õØÖ ÂøhPÎÀ ªPÄ® Hئøh¯
Âøhø¯z ÷uº¢öukzxU SÔ±mkhß Âøh°øÚ²® ÷\ºzx
GÊuÄ®.
Note : (i) Answer all the questions.

(ii) Choose the most appropriate answer from the given four alternatives and write
the option code and the corresponding answer.

[ v¸¨¦P / Turn over

7467 2

1. y=?x? GßÝ® ÁøÍÁøµ 0 &¼¸¢x 2 Áøµ HØ£kzx® Aµ[Pzvß
£µ¨¦ :

(A) 2 \. A»SPÒ (B) 1 \. A»S (C) 4 \. A»SPÒ (D) 3 \. A»SPÒ

Area bounded by y=?x? between the limits 0 and 2 is :

(a) 2 sq. units (b) 1 sq. unit (c) 4 sq. units (d) 3 sq. units

2. y=ex GÝ® ÁøÍÁøµ 0 ¼¸¢x 1 GÝ® GÀø»PÐUSÒ x&Aa_hß HØ£kzx®
Aµ[Pzvß £µ¨¦ :

 1
(A)  1 − e  \. A»SPÒ (B) (e+1) \. A»SPÒ

 1
(C) (e−1) \. A»SPÒ (D)  1 +  \. A»SPÒ
 e

Area bounded by y=ex between the limits 0 to 1 is :

 1
(a)  1 −  sq. units (b) (e+1) sq. units
 e

 1
(c) (e−1) sq. units (d)  1 +  sq. units
 e

3. ö£õ®ø©PÒ u¯õ›US® {ÖÁÚ® \µõ\›¯õP 1% SøÓ£õkÒÍ u¯õ›¨¦PøÍ
AÎUQßÓx. TöÓkzu¼À 100 ö£õ®ø©US 3 ö£õ®ø©PÒ SøÓ£õkÒÍøÁ&
PÍõP C¸¨£uØPõÚ {PÌuPÂß ©v¨£õÚx :

(A) 0.0613 (B) 0.3913 (C) 0.00613 (D) 0.613

In turning out certain toys in a manufacturing company, the average number of defective is
1%. The probability that in the sample of 100 toys there will be 3 defectives is :

(a) 0.0613 (b) 0.3913 (c) 0.00613 (d) 0.613

3 7467

4. vmh C¯À{ø» AmhÁønø¯ £¯ß£kzxøP°À z=2.18 &US Á»¨¦Ó®
©ØÖ® z=−1.75&US Chx ¦Ó® Aø©²® ©v¨¦PÐUPõÚ {PÌuPÄPÎß TkuÀ
__________ .

(A) 0.0146 (B) 0.4854 (C) 0.0547 (D) 0.4599
Using the Standard normal table the sum of the probabilities to the right of z=2.18 and to
the left of z=−1.75 is :
(a) 0.0146 (b) 0.4854 (c) 0.0547 (d) 0.4599

5. f (x)=x2+2x+2 ©ØÖ® h=1 GÛÀ ∆ f (x)&ß ©v¨¦ :

(A) x−3 (B) x+3 (C) 2x+3 (D) 2x−3
If f (x)=x2+2x+2 and the interval of differencing is unity, then ∆ f (x) is :
(a) x−3 (b) x+3 (c) 2x+3 (d) 2x−3

6. (1+∆) (1−L)=

(A) −1 (B) 0 (C) ∆ (D) 1
(1+∆) (1−L)=

(a) −1 (b) 0 (c) ∆ (d) 1

7. JxURmk PnUQÀ wº©õÚ ©õÔ xij &ß ©v¨¦ :
(A) 0 (B) 1 AÀ»x 0 (C) −1 (D) 1
In an assignment problem the value of decision variable xij is :
(a) 0 (b) 1 or 0 (c) −1 (d) 1

8. ^µØÓ wºÂÀ JxURmk AøÓPÎß GsoUøP BÚx :
(A) m+n−1&US \©©ØÓx (B) m+n−1&US \©®
(C) m+n+1&US \©©ØÓx (D) m+n+1&US \©®
In a non-degenerate solution number of allocations is :
(a) Not equal to m+n−1 (b) Equal to m+n−1
(c) Not equal to m+n+1 (d) Equal to m+n+1

[ v¸¨¦P / Turn over

7467 4

9. ρ (A)=r GÛÀ ¤ßÁ¸ÁÚÁØÖÒ Gx \› ?

(A) A BÚx SøÓ¢u£m\® J¸ r Á›ø\ §a]¯©ØÓ ]ØÓoU÷PõøÁ¯õÁx
ö£ØÔ¸US®
(B) A BÚx SøÓ¢u£m\® (r+1) Á›ø\²øh¯ ]ØÓoU÷PõøÁ°ß ©v¨¦
§a]¯©õP C¸US®£i¯õP ö£ØÔ¸US®
(C) r Á›ø\²øh¯ AøÚzx ]ØÓoU÷PõøÁPÎß ©v¨¦® §a]¯[PÍõP
C¸UPõx
(D) AøÚzx (r+1) Á›ø\ ©ØÖ® AøuÂh AvP©õÚ Á›ø\ öPõsh
§a]¯©ØÓ ]ØÓoU÷PõøÁPÒ C¸US®
If ρ (A)=r then which of the following is correct ?
(a) A has atleast one minor of order r which does not vanish.
(b) A has atleast (r+1) order minor which vanishes.

(c) All the minors of order r which does not vanish.
(d) All (r+1) and higher order minors should not vanish.

a1 b a b a b1 b c1 c a1
10. + 1 = c1 , 2 + 2 = c 2 , ∆ 1 = 1 , ∆2 = 1 , ∆3 = 1 GÛÀ
x y x y a 2 b2 b2 c2 c2 a2

(x, y)&ß ©v¨¦ :

 ∆3 ∆2   −∆ 1 ∆1 
(B)  ∆ ,−
(A)  ∆ , ∆ 
 1 1  2 ∆ 3 

 ∆1 ∆1   ∆2 ∆3 
(C)  ∆ , ∆  (D) ∆ , ∆ 
 2 3  1 1

a1 b a b a b1 b c1 c a1
If + 1 = c1 , 2 + 2 = c 2 , ∆ 1 = 1 , ∆2 = 1 , ∆3 = 1 then the
x y x y a 2 b2 b2 c2 c2 a2
value of (x, y) is :

 ∆3 ∆2   −∆ 1 ∆1 
(a) ∆ , ∆  (b)  ∆ ,− ∆ 
 1 1  2 3

 ∆1 ∆ 1   ∆2 ∆3 
(c) ∆ , ∆  (d) ∆ , ∆ 
 2 3  1 1

5 7467

11. ""\©Áõ´¨¦U TÓõÚx •Êø©z öuõSv°À EÒÍ JÆöÁõ¸ EÖ¨¦® ©õv›°À
Ch® ö£ÖÁuØPõÚ \©Áõ´¨ø£¨ ö£ØÔ¸US® EÖ¨¦PÍõÀ BÚx'' GÚU
TÔ¯Áº :
(A) íõº£º (B) L¤åº
(C) PõºÀ ¤¯õºéß (D) hõUhº ÷¯mì
“A random sample is a sample selected in such a way that every item in the population has
an equal chance of being included” is said by :
(a) Harper (b) Fisher
(c) Karl Pearson (d) Dr. Yates

12. ©v¨¥mk AÍøÁ¯õÚx £s£ÍøÁ°À SÔzu AøÚzx ©v¨¥kPøͲ®
EÒÍhUQ¯ uµÄPøͨ ö£ØÔ¸¢uõÀ Ax __________ Áõ´¢ux BS®.
(A) ¤øǯØÓ ußø© (B) vÓß ußø©
(C) {ø»z ußø© (D) {øÓÄz ußø©
An estimator is said to be __________ if it contains all the information in the data about the
parameter it estimates.
(a) unbiased (b) efficient
(c) consistent (d) sufficient

sin 2x
13. ∫ 2cosx dx &ß ©v¨¦a \õº¦ :
(A) cosx+c (B) sinx+c (C) −sinx+c (D) −cosx+c
sin 2x
The value of ∫ dx is :
2cosx
(a) cosx+c (b) sinx+c (c) −sinx+c (d) −cosx+c

14. Γ(3/2)&ß ©v¨¦ :

3 π
(A) 2 π (B) 2 (C) (D) π
2
The value of Γ(3/2) is :

3 π
(a) 2 π (b) (c) (d) π
2 2

[ v¸¨¦P / Turn over

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15. E(X)=5 ©ØÖ® E(Y)=−2 GÛÀ E(X−Y)&ß ©v¨£õÚx :
(A) 3 (B) 7 (C) 5 (D) −2
Given E(X)=5 and E(Y)=−2 then E(X−Y) is :
(a) 3 (b) 7 (c) 5 (d) −2

16. J¸ uÛzu £µÁÀ \õº¤À AøÚzx {PÌuPÄPÎß Tmkz öuõøP¯õÚx
G¨ö£õÊx® :
(A) JßÖ (B) «a]Ö©® (C) §ä¯® (D) «¨ö£¸©®
In a discrete probability distribution the sum of all the probabilities is always equal to :
(a) one (b) minimum (c) zero (d) maximum

d2 y dy
17.
2
= + 5 &GßÓ ÁøPUöPÊ \©ß£õmiß Á›ø\ ©ØÖ® £i •øÓ÷¯ :
dx dx

(A) 2 ©ØÖ® 1 (B) 2 ©ØÖ® 3 (C) 2 ©ØÖ® 2 (D) 3 ©ØÖ® 2

d2 y dy
The order and degree of the differential equation = + 5 are respectively :
2 dx
dx

(a) 2 and 1 (b) 2 and 3 (c) 2 and 2 (d) 3 and 2

dy
18. = cos x &GßÓ ÁøPUöPÊa \©ß£õmiß ö£õxz wºÄ :
dx
(A) y=sinx−2
(B) y=cosx+c, c ©õÓzuUP ©õÔ¼
(C) y=sinx+1
(D) y=sinx+c, c ©õÓzuUP ©õÔ¼
dy
The general solution of the differential equation = cos x is :
dx
(a) y=sinx−2
(b) y=cosx+c, c is an arbitrary constant
(c) y=sinx+1
(d) y=sinx+c, c is an arbitrary constant

7 7467

19. J¸ Põ»®\õº öuõh›À __________ TÖPÒ EÒÍÚ.
(A) |õßS (B) Cµsk (C) ‰ßÖ (D) I¢x
A time series consists of __________ components.
(a) four (b) two (c) three (d) five

20. Gs ÁiÂÀ AÍÂhUTi¯ AÍÄPÒ SÔUP¨£kÁx :
(A) x & Áøµ£h® (B) p & Áøµ£h®
(C) np & Áøµ£h® (D) c & Áøµ£h®
The quantities that can be numerically measured can be plotted on a :

(a) x - chart (b) p - chart
(c) np - chart (d) c - chart

£S-v & II / PART - II
SÔ¨¦ : GøÁ÷¯Ý® HÊ ÂÚõUPÐUS Âøh¯ÎUPÄ®. ÂÚõ Gs 30 &US
Pmhõ¯©õP Âøh¯ÎUPÄ®. 7x2=14

Note : Answer any seven questions. Question No. 30 is compulsory.

21. 3x−2y=6, 6x−4y=10 GßÓ \©ß£õkPÒ J¸[Pø©Ä AØÓx GÚU PõmkP.
Show that the equations 3x−2y=6, 6x−4y=10 are inconsistent.

22. ©v¨¤kP : ∫sin2x dx
Evaluate : ∫sin2x dx

23. ÷uøÁa \õº¦ p=122−5x−2x 2 ©ØÖ® x=20 GÝ®÷£õx ~Pº÷Áõº E£›ø¯U
PõsP.
Calculate consumer’s surplus if the demand function p=122−5x−2x2 and x=20.

24. wºUP : log  dy  = ax + by
 dx 

Solve : log 
dy 
 = ax + by
 dx 
[ v¸¨¦P / Turn over

7467 8

25. ©v¨¤kP : ∆2ex

Evaluate : ∆2ex

26. D¸Ö¨¦¨ £µÁ¼ß \µõ\› 5 ©ØÖ® vmh»UP©õÚx 2 GÛÀ, £µÁø»
wº©õÛUPÄ®.

The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution.

27. J¸ TÔß AÍÄ 50 Eøh¯ J¸ ©õv›°ß vmh»UP® 6.3. AuØS›¯ •Êø©z
öuõøP°ß vmh»UP® 6 GÛÀ ©õv›°ß vmh¨ ¤øÇ PõsP.

The standard deviation of a sample of size 50 is 6.3. Determine the standard error whose
population standard deviation is 6 ?

28. }ÒPõ»¨ ÷£õUS & Áøµ¯ÖUPÄ®.

Define Secular trend.

29. ÷£õUSÁµzx PnUSPÒ GßÓõÀ GßÚ ?

What is transportation problem ?

30. E(X)=0 GÛÀ V(X)=E(X2) GÚ {¹¤UPÄ®.

Prove that if E(X)=0, then V(X)=E(X2)

9 7467

£S-v & III / PART - III
SÔ¨¦ : GøÁ÷¯Ý® HÊ ÂÚõUPÐUS Âøh¯ÎUPÄ®. ÂÚõ Gs 40 &US
Pmhõ¯©õP Âøh¯ÎUPÄ®. 7x3=21

Note : Answer any seven questions. Question No. 40 is compulsory.

31. \¢øu°À EÒÍ A ©ØÖ® B BQ¯ uµ Aøh¯õÍ® öPõsh ö£õ¸ÒPÐUPõÚ
A B
A  0.9 0.1 
©õÖuÀ {PÌuPÄ Ao B  0.3 0.7  GÛÀ \©{ø»°À uµ Aøh¯õÍ® öPõsh

JÆöÁõ¸ ö£õ¸ÐUPõÚ \¢øu £[RkPøÍU PõsP.
Consider the matrix of transition probabilities of a product available in the market in two
A B
A  0.9 0.1 
brands A and B B  0.3 0.7  . Determine the market share of each brand in equilibrium

position.

1
32. ©v¨¤kP : ∫ dx
x+ 1 + x− 1

1
Evaluate : ∫ dx
x+ 1 + x− 1

dy
33. wºUP : + y tanx = cos 3 x
dx

dy
Solve : + y tanx = cos 3 x
dx

34. U0=1, U1=11, U2=21, U3=28 ©ØÖ® U4=29 GÛÀ ∆4U0&I PõsP.

Given U0=1, U1=11, U2=21, U3=28 and U4=29, find ∆4U0.

[ v¸¨¦P / Turn over

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35. öPõkUP¨£mkÒÍ AoUPõÚ EP¢u ³Pzøu (i) «a]ÖÂß «¨ö£¸ ©ØÖ®
(ii) «¨ö£¸Âß «a]Ö BQ¯ÁØøÓ £¯ß£kzvU PõsP.

`Ì{ø»PÎß
³P® {ø»¨£õkPÒ
E1 E2
S1 40 60
S2 10 −20
S3 −40 150

Given the following pay-off matrix (in rupees) for three strategies and two states of nature.
States of Nature
Strategy
E1 E2
S1 40 60
S2 10 −20
S3 −40 150

Select a strategy using each of the following rule :
(i) Maximin (ii) Minimax

36. J¸ ©õnÁß £mh® ö£ÖÁuØPõÚ {PÌuÄ 0.4 BS®. CÆÁõÓõP C¸¨¤ß
5 (I¢x) ©õnÁºPÐÒ
(i) J¸Áº ©mk® £mhuõ›¯õP
(ii) SøÓ¢u£m\® J¸Áº £mhuõ›¯õP
C¸¨£uØPõÚ {PÌuPøÁ PõsP.
A probability that a student get a degree is 0.4. Determine the probability that out of
5 students
(i) one will be graduate
(ii) atleast one will be graduate

37. £¸zv ¡¼ß Á¼ø© (AÖ® ußø©) AÔ¯ 100 AÍÃkPÒ öPõsh J¸
öuõSv°øÚz öu›Ä ö\´x AÁØÔß \µõ\› ©ØÖ® vmh»UP® •øÓ÷¯
7.4 Qµõ® ©ØÖ® 1.2 Qµõ® GÛÀ £¸zv ¡¼ß \µõ\› Á¼ø©°ß 95% |®¤UøP
CøhöÁÎø¯U PõsP.
A sample of 100 measurements at breaking strength of cotton thread gave a mean 7.4 g and
a standard deviation of 1.2 g. Find 95% confidence limits for the mean breaking strength of
cotton thread.

11 7467

38. J¸ C¯¢vµ®, SÇõø¯ 0.532 ö\.«. \µõ\›¯õÚ Âmhzxhß, vmh»UP®
0.002 ö\.«. A;® xøÍ°kQÓx. Pmk¨£õk \µõ\›UPõÚ Áµ®¦PøÍ 5
TÖPÐUS PnUQkP.

A machine drills hole in a pipe with a mean diameter of 0.532 cm and a standard deviation
of 0.002 cm. Calculate the control limits for mean of 5 samples .

39. J¸ öuõhºa]¯õÚ \©Áõ´¨¦ ©õÔ X BÚx ¤ßÁ¸® £µÁÀ \õºø£¨
ö£ØÖÒÍx

 0 , x
1


F( x ) = k( x − 1)4 , 1 < x
3 GÛÀ


 1 , x>3

(i) k ©ØÖ® (ii) {PÌuPÄ Ahºzva \õºø£U PõsP.

A continuous random variable X has the following distribution function.

 0 , if x
1


F( x ) = k( x − 1)4 , if 1 < x
3


 1 , if x>3

Find (i) k and (ii) the probability density function.

40. öuõøP°hø»¨ £¯ß£kzv y−1=x GßÓ ÷Põk x−Aa_ x=−3 ©ØÖ® x=3
GßÝ® GÀø»USÒ HØ£kzx® Aµ[Pzvß £µ¨ø£U PõsP.

Using integration find the area of the region bounded by the line y−1=x, the x−axis and
the ordinates x=−3 and x=3.

[ v¸¨¦P / Turn over

7467 12

£S-v & IV / PART - IV
SÔ¨¦ : AøÚzx ÂÚõUPÐUS® Âøh¯ÎUPÄ®. 7x5=35
Note : Answer all questions.
41. (A) ö©õzuz öuõøP ` 8,500 BÚx Ámi Á¸©õÚ® u¸® ‰ßÖ Âu©õÚ
PnUSPÎÀ •u½k ö\´¯¨£mhx. JÆöÁõ¸ •u½mkUPõÚ Ámi Ãu®
2%, 3%©ØÖ® 6% BPÄ® J¸ Á¸hzvØPõÚ ö©õzu Ámi ` 380 BPÄ®
EÒÍx. ÷©¾® 6% •u½mkz öuõøP¯õÚx ©ØÓ Cµsk •u½kPÎß
TmkzöuõøPUS \©® GÛÀ JÆöÁõ¸ ¤›Â¾® ö\´u •u½mkz öuõøP
GÆÁÍÄ ? (Q÷µ©º Âvø¯ £¯ß£kzxP)
AÀ-»-x
(B) J¸ ö£õ¸Îß Bsk EØ£zv R÷Ç öPõkUP¨£mkÒÍx. «a]Ö ÁºUP
•øÓ°À ÷|ºU÷Põmk ÷£õUQøÚ¨ ö£õ¸zxP.
BskPÒ 1995 1996 1997 1998 1999 2000 2001
EØ£zv
155 162 171 182 158 180 178
(hßPÎÀ)
(a) A total of ` 8,500 was invested in three interest earning accounts. The interest rates
were 2%, 3% and 6%. If the total simple interest for one year was ` 380 and the
amount invested at 6% was equal to the sum of the amounts in the other two accounts,
then how much was invested in each account ? (Use Cramer’s rule).
OR
(b) The annual production of a commodity is given as follows. Fit a straight line trend by
the method of least squares.
Year 1995 1996 1997 1998 1999 2000 2001
Production
155 162 171 182 158 180 178
(in tones)

42. (A) ©v¨¤kP : 4x 2 + 2x + 6
∫ ( x + 1)2 ( x − 3 ) dx
AÀ-»-x
(B) 500 ©õnÁºPÎß E¯µ©õÚx C¯À{ø»¨ £µÁ¼À \µõ\›¯õP 68
A[S»•® Auß vmh»UP® 3 A[S»©õP PnUQh¨£kQÓx.
(i) 72 A[S»zvØS® AvP©õP
(ii) 64 A[S»zvØS® SøÓÁõP
(iii) 65 ©ØÖ® 71 A[S»zvØS® Cøh¨£mh ©õnÁºPÎß
GsoUøP°øÚU PnUQkP.
Z 1 1.33
£µ¨£ÍÄ 0.3413 0.4082

13 7467

(a) Evaluate : 4x 2 + 2x + 6
∫ ( x + 1)2 ( x − 3 ) dx
OR
(b) If the heights of 500 students are normally distributed with mean 68 inches and standard
deviation 3 inches, how many students have height
(i) greater than 72 inches
(ii) less than 64 inches
(iii) between 65 and 71 inches
Z 1 1.33
Area 0.3413 0.4082

43. (A) öPõkUP¨£mh CÖv{ø»a ö\»Ä ©ØÖ® CÖv{ø» Á¸Áõ´ \õº¦PÒ
x
•øÓ÷¯ C
(x ) = 50 + 50 ©ØÖ® R’(x)=60. ©õÓõaö\»Ä ` 200 GÛÀ «¨ö£¸
C»õ£zøuU PõsP.
AÀ-»-x
(B) ÷ÁõP¼ß ÷uõµõ¯ •øÓø¯¨ £¯ß£kzv ¤ßÁ¸® ÷£õUSÁµzxU
PnUQß Bµ®£ Ai¨£øh \õzv¯©õÚz wºøÁU PõsP.
D1 D2 D3 D4 Aئ
O1 2 3 11 7 6
O2 1 0 6 1 1
O3 5 8 15 9 10
÷uøÁ 7 5 3 2

x
(a) The marginal cost C9(x) and marginal revenue R’(x) are given by C
(x ) = 50 + and
50
R’(x)=60. The fixed cost is ` 200. Determine the maximum profit.
OR
(b) Using Vogel’s approximation method, obtain the initial feasible solution of the following
transportation problem.
D1 D2 D3 D4 Supply
O1 2 3 11 7 6
O2 1 0 6 1 1
O3 5 8 15 9 10
Demand 7 5 3 2

[ v¸¨¦P / Turn over

7467 14

44. (A) wºUP : (y2−2xy)dx=(x2−2xy)dy
AÀ-»-x
(B) ¤ßÁ¸® ÂÁµ[PÐUS L¤åº Âø»U SÔ±mk GsønU Pmhø©UPÄ®.
÷©¾® Ax Põ»©õØÖa ÷\õuøÚ, Põµo ©õØÖa ÷\õuøÚ BQ¯ÁØøÓ¨
§ºzva ö\´²® GÚ {¹¤UPÄ®.
Ai¨£øh Bsk |h¨¦ Bsk
ö£õ¸ÒPÒ
Âø» AÍÄ Âø» AÍÄ
A›] 40 5 48 4

÷Põxø© 45 2 42 3

ÁõhøP 90 4 95 6

G›ö£õ¸Ò 85 3 80 2

÷£õUSÁµzx 50 5 65 8

Cuµ ö\»ÄPÒ 65 1 72 3

(a) Solve : (y2−2xy)dx=(x2−2xy)dy
OR
(b) Construct Fisher’s price index number and prove that it satisfies both Time Reversal
Test and Factor Reversal Test for the following data.
Base Year Current Year
Commodities
Price Quantity Price Quantity
Rice 40 5 48 4
Wheat 45 2 42 3
Rent 90 4 95 6
Fuel 85 3 80 2
Transport 50 5 65 8
Miscellaneous 65 1 72 3

π
2
45. (A) ©v¨¤kP : ∫ 1
dx
1 + cotx
0

AÀ-»-x
(B) öPõkUP¨£mkÒÍ ©v¨¦Pμ¸¢x C¸£i £À¾Ö¨¦U ÷PõøÁø¯U
PõsP.
x 0 1 2 3 4 5 6 7
y 1 2 4 7 11 16 22 29

15 7467

π
2
1
(a) Evaluate :
∫ 1 + cotx dx
0

OR
(b) Find a polynomial of degree two, which takes the values.

x 0 1 2 3 4 5 6 7
y 1 2 4 7 11 16 22 29

46. (A) J¸ öuõhºa]¯õÚ \©Áõ´¨¦ ©õÔ X BÚx ¤ßÁ¸® {PÌuPÄa \õºø£¨
ö£ØÖÒÍx GÛÀ,
X= x 0 1 2 3 4 5 6 7
2 2 2
P(x ) 0 k 2k 2k 3k k 2k 7k +k

(i) k -ß ©v¨ø£U PõsP.
(ii) p(x < 6), p(x / 6) ©ØÖ® p(0 < x < 5)&IU PõsP.

1
(iii) P(X አ x ) > &UPõÚ x&Cß SøÓ¢u£m\ ©v¨ø£U Psk¤iUPÄ®.
2
AÀ-»-x
(B) 400 uÛ|£ºPøÍU öPõsh J¸ TÔÀ EÒÍÁºPÎß \µõ\› E¯µ®
67.47 A[S»® GÛÀ, 0.05 ªøPPõs {ø»°À AUTÓõÚx \µõ\› E¯µ®
67.39 A[S»•® vmh»UP® 1.30 A[S»•® öPõsh •Êø©z
öuõSv°¼¸¢x GkUP¨£mhuõPU P¸u»õ©õ ?
(a) A continuous random variable X has the following probability function.

X= x 0 1 2 3 4 5 6 7
2 2 2
P(x ) 0 k 2k 2k 3k k 2k 7k +k

(i) Find k.
(ii) Evaluate p(x < 6), p(x / 6) and p(0 < x < 5).
1
(iii) If P(X አ x ) > , then find the minimum value of x.
2
OR
(b) A sample of 400 individuals is found to have a mean height of 67.47 inches. Can it be
reasonably regarded as a sample from a large population with mean height of
67.39 inches and standard deviation of 1.30 inches at 0.05 level of significance ?

[ v¸¨¦P / Turn over

7467 16

7
− x
47. (A) wºUP : (3D 2
+ D − 14)y = 4 − 13 e 3

AÀ-»-x
(B) J¸ SÔ¨¤mh C¯¢vµ® Áõ°»õP EØ£zv ö\´¯¨£k® v¸S©øÓ°À
EÒÍ SøÓ£õkPÒ 18 \uÂQu® GÛÀ \©Áõ´¨¦ •øÓ°À
÷uº¢öukUP¨£k® |õßS v¸S©øÓ°À
(i) \›¯õP J¸ SøÓ£õkÒÍ v¸S©øÓ
(ii) SøÓ£õk CÀ»õ v¸S©øÓ
(iii) AvP£m\® 2 SøÓ£õkPÒ EÒÍ v¸S©øÓ C¸¨£uØPõÚ {PÌuPÄ
PõsP.
7
− x
(a) Solve : (3D 2
+ D − 14)y = 4 − 13 e 3
OR
(b) If 18% of the bolts produced by a machine are defective, determine the probability that
out of the 4 bolts chosen at random
(i) exactly one will be defective
(ii) none will be defective
(iii) atmost 2 will be defective

-oOo-

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