Business Statistics and Mathematics Solved Past Papers, B.com Part 1 Punjab University 2009-2018 PDF

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This is the Solution of Past Exams of Business Statistics and Mathematics(B.com part-I) relating to Punjab University,Pakistan. It contains solution of all past questions from 2009 to 2018....

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers

20 092018 200909-2

SOLVED PAPER 2009 QUESTION NO. 1: SOLUTION Weekly Wages 30-39 40-49 50-59 60-69 70-79 80-89 90-99 Total

(a)

No.of Works f 6 10 11 12 fm 32 fm 18 f2 8 ∑f = 97

x

fx

fx2

C.B.

C.F

34.5 44.5 54.5 64.5 74.5 84.5 94.5

207 445 599.5 774 2384 1521 756 6696.5

7141.5 19802.5 32672.75 49923 177608 128524.5 71442 487114.25

29.5-39.5 39.5-49.5 49.5-59.5 59.5-69.5 69.5-79.5 79.5-89.5 89.5-99.5

6 16 27 39 71 89 97

=+

Mode



( )( )

=69.5 + =69.5 +

ℎ

   () ()

 10

   

= 75.38 (b)



 

=  +  ( − )

Median

=69.5 +







(48.5 − 39)

=





= 48.5

 = 69.5, h = 10, C = 39

= 72.47 (c)

$

C.V where x

= x 100% =

∑'( ∑'

S

%

=

))).* 

=+ =+

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= 68.93

∑' ∑'

− (

 .* 

∑  ) ∑

− (

))).* 

)

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers

20 092018 200909-2

= ,5021.796 − (68.93) = 16.45 $

= % x 100%

C.V

).*

= ). x 100% = 23.86% QUESTION NO. 2 (a)

Consider the year 1950 as base year for the price relatives of commodities A, B, C and D. Price Relatives Link Relatives Chain G.M. Year Indices A B C D A B C D 1950 1951 1952 1953 1954 1955

(b)

(i)

79.93 85.93 145.68 100.65 77.62

100 79.93 68.68 100.06 100.71 78.12

Since one card is drawn from 52 playing cards: 52 0 = 52 1

n(S) =

/

A

Black Card

=

n(A) =

(ii)

119 77 55 70.1 107.6 94.5 161.1 86.7 192.3 86.2 138.7 96.0 57.3 107.1 91.7

100 100 100 100 81 77 119 55 81 62 54 128 52 76.5 104 87 111 100 167.7 93 75 154 96 89.4 60 43 165 88 64.5

/

 26 26 0 / 0 = 1 0 )

P(A) =

(1)

B

Black Card

=

n(B) = Web Websit sit site: e: ww www. w. w.pak pak paksi si sights ghts ghts.co .co .com m

($)

/

=

*

=





16 36 0 / 0 = 16 1 0

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030 0302-5 2-5 2-514 14 14884 884 8843 3

Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers (2)

P(B) =

Sr.No. 1 2 3 4 5 8

=

($)

)

*

=

20 092018 200909-2





X

QUESTION NO. 3 = given population = 2, 4, 6, 10

N

= population size = 4

n

= Sample size

=2

5 3 All possible samples (w.o.r) = / 0 = / 0 = 6 4 6 Means (x44 ) Samples 3 2, 4 4 2, 6 6 2, 10 5 4, 6 7 4, 10 8 6, 10

Probability distribution of samples means: x44 3 4 5 6 7 8 Total

Tally Sheet 1 1 1 1 1 1

f 1 1 1 1 1 1 ∑f = 6



= P(x44 )∑ 1/6 1/6 1/6 1/6 1/6 1/6

x44 p(x 44 ) 3/6 4/6 5/6 6/6 7/6 8/6 33/6

2 44xp(x44 ) 9/6 16/6 25/6 36/6 49/6 64/6 199/6

Mean and variance of sampling distribution of means: µ x44 2

 x44

= ∑ x44 P(x 44 ) 2



=) = 5.5 2

)4 = ∑ x44P(x44 ) – x4(µ

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers =

 )

20 092018 200909-2

– (5.5) 2

= 2.92

Mean and Variance of population: x x2 4 2 16 4 36 6 100 10 2 ∑x = 22 ∑x = 156 µ

=

2

= =

 ∑( = = 5.5 7   ∑ – µ2 8 *) – (5.5) 2 

= 8.76 Verification: (i)

(ii)

µ x44

=

µ

5.5

=

5.5

2

=

 9

 x44



. 9





2.92 =

. *

2.92 =

2.92





QUESTION NO. 4 (a)

H 0: There is no association between general ability and mathematical ability.

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers

20 092018 200909-2

H 1: There is some association between general ability and mathematical ability. Level of significance:

α

= 0.05

Test statistics:

  = ∑;?) @ >?

= (r – 1) (c – 1) =(3 – 1) (3 – 1) =2x2=4

 = ,.*= 9.49 ,C

Critical Value:

2

Critical region: Decision rule:

x > 9.49 2 Reject H0, if x > 9.49, Otherwise accept H0.

Critical region

Acceptance region

0 Observed Frequency (Oij ) General Ability

9.49

Mathematical Ability Good Fair Poor

Good 44 265 41

Fair 22 257 91

Poor 4 178 98

Total

350

370

280

1000

o ij 44 265 41 22 257 91 4

eij 24.5 245 80.5 25.9 259 85.1 19.6

oij – eij 19.5 20 -39.5 -3.9 -2 5.9 -15.6

(oij – eij) 2 380.25 400 1560.25 15.21 4 34.81 243.36

(oij – eij)2 /eij 15.5204 1.6327 19.3820 0.5873 0.0154 0.4090 12.4163

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Total 70 700 230

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers 178 98 Total

196 64.4

-18 33.6

324 1128.96

20 092018 200909-2

1.6531 17.5304 69.1466

2 Conclusion: Since x = 69.1466 >,.*= 9.49

So, we reject H 0. Regression coefficient of y on x: byx

=

∑(D ∑( . ∑D

=

×.   .)×.

=

.

∑ (∑() 

×.)  ( .)) 

 .)

= 2.064

Regression coefficient of x on y: bxy

=

∑(D ∑( . ∑D ∑F (∑D)

=

×.   .)×.

=

).

×  (.)

.)

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= 0.474

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers

20 092018 200909-2

QUESTION NO. 5 a) Solve the Equation for x √5 − 4 − √3 + 1 = 1 Taking square on both sides: (√5 − 4 − √3 + 1) = (1)2

(5x + 4) + (3x + 1) – 2√5 − 4√3 + 1 = 1

5x + 4 + 3x + 1 – 2,(5 − 4)(3 + 1) = 1 8x + 5 – 2√15  + 5 + 12 + 4 = 1 – 2√15  + 17 + 4

= 1 – 8x – 5

– 2√15  + 17 + 4

= –4 – 8x

−2√15  + 17 + 4

= –2(2 + 4x)

√15  + 17 + 4

= 2 + 4x

Again, taking square on both sides: 15  + 17 + 4 = (2 + 4x) 2

15  + 17 + 4 = 4 + 16x2 + 16x 2

–x + x

=0

–x(x – 1)

=0

Solution set is {0, 1}

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030 0302-5 2-5 2-514 14 14884 884 8843 3

Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers (b)

  =  

 

–



=

20 092018 200909-2







x+1=3–x x+x=3–1 2x = 2 x=1 Solution set is {1} QUESTION NO. 6 (a)

2x + 6y + 4z

=

320

…..(i)

6x + 6y + 4z

=

480

…..(ii)

3x + 2y + 4z

=

192

…..(iii)

Subtract equation (i) from (ii), we get: 6x + 6y + 4z

=

480

_ 2x + 6y + 4z

=

_ 320

=

160

4x x

=

160/4

=

40

Subtract equation (iii) from (ii), we get: 6x + 6y + 4z

=

480

_ 3x + 2y + 4z

=

192

=

288

3x + 4y

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…..(iv)

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers

20 092018 200909-2

Put x = 40 in eq. (iv): 3(40) + 4y

=

288

4y

=

288 – 120

y

=

42

=

320

4z

=

320 – 120

z

=

–3

Put x = 40 and y = 42 in eq. (i): 2(40) + 6(42) + 4z

Solution set is {(40, 42, – 3)} (b)

We have: a10 = 20 and a20 = 40, Find a7 of the A.P. Since

an

=

a + (n – 1)d

a 10

=

a + (10 – 1)d

20

=

a + 9d

a 20

=

a + (20 – 1)d

40

=

a + 19d

…..(i)

…..(ii)

Subtract equation (i) from (ii), we get: 40

=

a + 19d

_ 20

=

_ a + 9d

20

=

10d

d

=

2

Put d = 2 in eq. (i)

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers 20

=

a + 9 (2)

a

=

20 – 18

a

=

2

20 092018 200909-2

=2

Now, 7 th term of the A.P. is a7

=

a + (7 – 1)d

=

2 + 6 x 2 = 14 QUESTION NO. 7

(a)

P

=

Principal amount =

?

i

=

Internal rate

=

5% p.a.

n

=

No. of periods

=

3 years

=

0.05

Now, difference between compound interest and simple interest = Rs. 61 P[(1 + i)n – 1] – P x i x n =Rs. 61 P[(1 + 0.05)3 – 1] – P x 0.05 x 3 =61 3

P[(1.05) – 1] – 0.15P =61 0.157625P– 0.15P =61 0.007625P =61 )

P

= . )*

P

= Rs. 8000

Thus, principal amount is Rs. 8000. (b)

R

= Rs. 5000 (Payable at the end of the each quarter. It is ordinary annuity)

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers n

= 5 years

i

=

20 092018 200909-2

= 5 x 4 = 20 quarters

8% p.a. =

. 

= 0.02 per quarters

The accumulated value is: Sn

(;)H 

=

R

=

5000

=

5000(24.29737) = Rs. 121486.85

;

(.)I .

QUESTION NO. 8 (i)

A matrix is defined as the set of real numbers arranged in the form of rectangular array of numbers enclosed in brackets. Generally, matrices are represented by capital letters such as A, B. c. e etc. For example: A=J

1 2 K , 3 4

B=J

2 46 K etc. 8 0 1

(ii)

A specific number which is multiplied to every next term in a geometric sequence. It is represented by “r”.

(iii)

Compound interest is an interest paid on the initial principal and previously earned interest. C.I

=

n

P[(1 + i) – 1]

where C.I = Compound interest i = Interest rate n = Number of periods (iv)

When the payments are made at beginning of each period and continue for a definite period, it is called annuity due. Sum of annuity due:

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030 0302-5 2-5 2-514 14 14884 884 8843 3

Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers Sn Where

RJ

=

(;)HL  ;

20 092018 200909-2

K–R

R = Regular installment i = Interest rate n = Number of periods

(v)

The totality of observation in particular situation is called population.

(vi) A sample is a subgroup of the population that will represent the characteristics of the population whereas sampling is the procedure of selecting a representative sample from a given population. (vii) Correlation is a measure of the degree to which any two variables vary together. (viii) The square root of the average of all squared deviations taken from A.m. is called standard deviation. S (ix)

+

=

∑()%  

and

S

=

+

∑(% ) ∑

The tendency of the values to concentrate at their centre is called central tendency and any measure indicating the centre of their distribution is called measured central tendency.

(x) If x 1, x2 ,……, xn are n observations with their respective weights w1 , w 2,……, wn . Then weighted mean is defined as: MMM = =

 N  N  …... H NH N N  …... NH

∑P ∑N

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers

200909-2 20 092018

SOLVED PAPER 2010 QUESTION NO. 1 Weekly Wages 117-124 124-131 131-138 138-145 145-152 152-159 159-166 166-173 173-180 180-187 187-194

No.of Works f 13 17 33 47 56 73 81 65 55 40 20

x

fx

fx2

f/x

120.5 127.5 134.5 141.5 148.5 155.5 162.5 169.5 176.5 183.5 190.5

1566.5 2167.5 4438.5 6650.5 8316 11351.5 13162.5 11017.5 9707.5 7340 3810

188763.25 276356.25 596978.25 941045.75 1234926 1765158.25 2138906.25 1867466.25 1713373.75 1346890 725805 2 ∑fx = 12795669

0.1078 0.1333 0.2453 0.3321 0.3771 0.4694 0.4985 0.3835 0.3116 0.2180 0.1050 ∑(f/x) = 3.1816

∑f = 500

A.M. =

∑fx = 79528 ∑

= ∑

x

=

 

A.M. = 159.056 H.M. =

∑

 

∑  

.

=

H.M. = 157.154 S.D

∑ 

=  ∑ −  ∑  ∑

=



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

− 159.056

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers

200909-2018 20 092018

= √25591.338 − 25298.81 S.D

= √292.528 = 117.10

Coefficient of variation: C.V. =

#.$ %

x 100%

.

= . x 100%

C.V. = 10.75% Hence Arithmetic mean

= 159.05

Harmonic mean

= 157.154

Standard Deviation

= 17.10

Coefficient of variation = 10.75% QUESTION NO. 2. x 16 72 73 63 83 80 66 66 74 62 ∑ x = 655

y 40 52 43 49 61 58 44 58 50 45 ∑y = 500

xy 640 3744 3139 3087 5063 4640 2904 3828 3700 2790 ∑xy= 33535

x2 256 5184 5329 3969 6889 6400 4356 4356 5476 3844 2 ∑x = 46059

&∑'(∑∑'

y2 1600 2704 1849 2401 3721 3364 1936 3364 2500 2025 2 ∑ y =25464

)*+∑ (∑ ,*+∑' (∑' ,

Correlation Coefficient =

r =

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Business Statistics & Mathematics Punjab University B.Com Part 1 Solved Past Papers = = = =

200909-2018 20 092018

(

)*-(  ,*--( , (

)*-(-,*--(, 

)*,*--, 

.-

...