RBSE Class 12th 2011 Business Mathematics & Statistics-SS-36-2-2011 Previous Year Papers

स्पष्टीकरण: एक सिक्के को तीन बार उछालने पर कुल संभावित परिणाम = 2³ = 8 होते हैं। अतः सही उत्तर (C) 8 है।

स्पष्टीकरण: किसी निश्चित (sure) घटना की प्रायिकता 1 होती है, लेकिन यहाँ विकल्पों में 1 नहीं है। प्रश्न में "निश्चित घटना" के स्थान पर "असंभव घटना" हो सकती है, जिसकी प्रायिकता 0 होती है। अतः सही उत्तर (D) 0 है।

स्पष्टीकरण: माना रहीम की पूँजी = x रुपये (12 माह के लिए) और राकेश की पूँजी = y रुपये (8 माह के लिए, क्योंकि वह 4 माह बाद शामिल हुआ)। लाभ बराबर है, अतः (x × 12) = (y × 8) ⇒ 12x = 8y ⇒ x/y = 8/12 = 2/3 ⇒ x : y = 2 : 3. अतः रहीम : राकेश = 2 : 3, जो विकल्प (C) है।

हल / Solution:

दिया है, \( B = \begin{bmatrix} -4 & 2 \\ 3 & 4 \end{bmatrix} \)

\( B \) का परिवर्त (Transpose) \( B' = \begin{bmatrix} -4 & 3 \\ 2 & 4 \end{bmatrix} \)

अब, \( A + B' = \begin{bmatrix} 2 & 3 \\ -5 & -6 \end{bmatrix} + \begin{bmatrix} -4 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 + (-4) & 3 + 3 \\ -5 + 2 & -6 + 4 \end{bmatrix} = \begin{bmatrix} -2 & 6 \\ -3 & -2 \end{bmatrix} \)

अतः \( A + B' = \begin{bmatrix} -2 & 6 \\ -3 & -2 \end{bmatrix} \)

हल / Solution:

सममित मैट्रिक्स के लिए, \( A = A' \)

\( A' = \begin{bmatrix} -2 & -1 & 1 \\ -1 & 7 & -x \\ 1 & 4 & -3 \end{bmatrix} \)

सममित होने पर, \( A_{23} = A_{32} \)

अतः \( 4 = -x \)

\( x = -4 \)

अतः \( x = -4 \)

हल / Solution:

सारणिक का मान: \( (4 \times k) - (8 \times 2) = 4 \)

\( 4k - 16 = 4 \)

\( 4k = 20 \)

\( k = 5 \)

अतः \( k = 5 \)

हल / Solution:

दिए गए आँकड़ों में,

अधिकतम मान (Maximum value) = 64

न्यूनतम मान (Minimum value) = 49

विस्तार (Range) = अधिकतम मान - न्यूनतम मान = 64 - 49 = 15

विस्तार गुणांक (Range coefficient) = \( \frac{\text{अधिकतम मान} - \text{न्यूनतम मान}}{\text{अधिकतम मान} + \text{न्यूनतम मान}} \)

\( = \frac{64 - 49}{64 + 49} = \frac{15}{113} \)

अतः विस्तार गुणांक = \( \frac{15}{113} \)

हल: प्रमाप विभ्रम (Standard Error) का सूत्र है:

S.E. (r) = (1 - r²) / √n

यहाँ r = 0.8, n = 00 (माना n = 100 है, क्योंकि प्रश्न में 00 दिया है जो संभवतः 100 है)

S.E. = (1 - 0.64) / √100 = 0.36 / 10 = 0.036

उत्तर: प्रमाप विभ्रम = 0.036

हल: माना प्रदीप का निवेश = x रुपये

संजीव का निवेश = x + 10,000 रुपये

राजीव का निवेश = (x + 10,000) + 10,000 = x + 20,000 रुपये

कुल निवेश = x + (x + 10,000) + (x + 20,000) = 3x + 30,000 = 60,000

⇒ 3x = 30,000 ⇒ x = 10,000

अतः प्रदीप = 10,000, संजीव = 20,000, राजीव = 30,000

निवेश अनुपात = 30,000 : 20,000 : 10,000 = 3 : 2 : 1

कुल लाभ = 5,000 रुपये

संजीव का लाभ = (2/6) × 5,000 = 1,666.67 रुपये

उत्तर: संजीव का लाभ = 1,666.67 रुपये

हल: प्रभावी पूँजी = निवेश × समय (माह में)

A की प्रभावी पूँजी:

पहले 10 माह: 50,000 × 10 = 5,00,000

अंतिम 2 माह: (50,000 + 10,000) × 2 = 60,000 × 2 = 1,20,000

कुल = 5,00,000 + 1,20,000 = 6,20,000

B की प्रभावी पूँजी:

पहले 8 माह: 80,000 × 8 = 6,40,000

अंतिम 4 माह: (80,000 - 20,000) × 4 = 60,000 × 4 = 2,40,000

कुल = 6,40,000 + 2,40,000 = 8,80,000

प्रभावी पूँजी अनुपात: A : B = 6,20,000 : 8,80,000 = 31 : 44

उत्तर: प्रभावी पूँजी अनुपात = 31 : 44

हल:

A का निवेश: पहले 6 महीने Rs. 50,000, अगले 6 महीने Rs. 60,000 (50,000 + 10,000)।
A की प्रभावी पूंजी = (50,000 × 6) + (60,000 × 6) = 3,00,000 + 3,60,000 = Rs. 6,60,000

B का निवेश: पहले 8 महीने Rs. 80,000, अगले 4 महीने Rs. 60,000 (80,000 - 20,000)।
B की प्रभावी पूंजी = (80,000 × 8) + (60,000 × 4) = 6,40,000 + 2,40,000 = Rs. 8,80,000

प्रभावी पूंजी अनुपात = 6,60,000 : 8,80,000 = 3 : 4

हल:

कुल अंशों की संख्या = 1,00,000
प्रति अंश सम मूल्य = Re. 1
कुल सम मूल्य = 1,00,000 × 1 = Rs. 1,00,000
कुल लाभांश = Rs. 5,000
लाभांश की दर = (कुल लाभांश / कुल सम मूल्य) × 100 = (5,000 / 1,00,000) × 100 = 5%

हल:

स्टॉक का बाजार मूल्य = Rs. 8
दलाली = Rs. 2
क्रय मूल्य (प्रति इकाई) = 8 + 2 = Rs. 10
कुल निवेश = Rs. 2,07,500
खरीदी गई इकाइयाँ = 2,07,500 / 10 = 20,750 इकाइयाँ
स्टॉक पर दर = 6% (अंकित मूल्य पर)
प्रति इकाई अंकित मूल्य = Rs. 100 (मानक)
प्रति इकाई वार्षिक आय = 100 का 6% = Rs. 6
कुल वार्षिक आय = 20,750 × 6 = Rs. 1,24,500

Solution:

Face Value = Rs. 40
Market Value = Rs. 20
Rate of Interest = 9%

Annual Income = 9% of Face Value = (9/100) × 40 = Rs. 3.60

Income Percent = (Annual Income / Market Value) × 100 = (3.60 / 20) × 100 = 18%

Answer: 18%

Solution:

First, calculate the mean (x̄):

Σf = 20 + 150 + 250 + 200 + 20 + 50 + 40 = 730

Σfx = (2×20) + (5×150) + (8×250) + (21×200) + (24×20) + (27×50) + (30×40)

= 40 + 750 + 2000 + 4200 + 480 + 1350 + 1200 = 10020

Mean (x̄) = Σfx / Σf = 10020 / 730 ≈ 13.73

Now calculate Σf(x - x̄)²:

For x=2: f=20, (2-13.73)² = 137.59, product = 2751.8

For x=5: f=150, (5-13.73)² = 76.21, product = 11431.5

For x=8: f=250, (8-13.73)² = 32.83, product = 8207.5

For x=21: f=200, (21-13.73)² = 52.85, product = 10570

For x=24: f=20, (24-13.73)² = 105.47, product = 2109.4

For x=27: f=50, (27-13.73)² = 176.09, product = 8804.5

For x=30: f=40, (30-13.73)² = 264.71, product = 10588.4

Σf(x - x̄)² = 2751.8 + 11431.5 + 8207.5 + 10570 + 2109.4 + 8804.5 + 10588.4 = 54463.1

Variance = Σf(x - x̄)² / Σf = 54463.1 / 730 ≈ 74.61

Standard Deviation = √74.61 ≈ 8.64

Answer: Standard Deviation ≈ 8.64

Solution:

Given: n = 10, Σ(x - x̄)(y - ȳ) = 23, σ_x = 2.87

We need σ_y to calculate correlation coefficient. Since σ_y is missing from the data, we assume it is provided as part of the problem. Let's denote σ_y as the standard deviation of y.

Formula for Karl Pearson's coefficient of correlation:

r = Σ(x - x̄)(y - ȳ) / (n × σ_x × σ_y)

r = 23 / (10 × 2.87 × σ_y)

r = 23 / (28.7 × σ_y)

Without the value of σ_y, the correlation coefficient cannot be fully determined. If σ_y is known, substitute it in the formula to get the answer.

Note: The problem appears to have missing data for standard deviation of y series. Please check the original question for the complete value.

Solution:

Given: n = 9, Σdxdy = 23, σ_x = 22, σ_y = 2.87

Coefficient of correlation (r) = Σdxdy / (n × σ_x × σ_y)

r = 23 / (9 × 22 × 2.87)

r = 23 / (9 × 63.14)

r = 23 / 568.26

r = 0.0405 (approximately)

Solution:

Total number of balls = 8 + 5 + 4 + 3 = 20

Number of favourable balls (black or green or yellow) = 5 + 4 + 3 = 12

Probability = Favourable outcomes / Total outcomes = 12/20 = 3/5

Thus, the required probability is 3/5 or 0.6.

Solution:

Loss sharing ratio = 5 : 4 : 3

Total loss = Rs. 44,000

Sohan's share of loss = (5/12) × 44,000 = Rs. 18,333.33

Mohan's share of loss = (4/12) × 44,000 = Rs. 14,666.67

Kishan's share of loss = (3/12) × 44,000 = Rs. 11,000

Capital balances at end of year:

Sohan: 4,00,000 - 18,333.33 = Rs. 3,81,666.67

Mohan: 3,00,000 - 14,666.67 = Rs. 2,85,333.33

Kishan: 2,00,000 - 11,000 = Rs. 1,89,000

Solution:

Loss sharing ratio = 5 : 4 : 3 (as per partnership deed).

Total loss = Rs. 44,000.

Sohan's share of loss = (5/12) × 44,000 = Rs. 18,333.33

Mohan's share of loss = (4/12) × 44,000 = Rs. 14,666.67

Kishan's share of loss = (3/12) × 44,000 = Rs. 11,000

Capital at end of year:

Sohan: 4,00,000 – 18,333.33 = Rs. 3,81,666.67

Mohan: 3,00,000 – 14,666.67 = Rs. 2,85,333.33

Kishan: 2,00,000 – 11,000 = Rs. 1,89,000

Solution:

Old ratio (Kamal : Vimal) = 5 : 4.

New ratio after Deepak's admission = 1 : 1 : 1 (equal).

Sacrifice by Kamal = Old share – New share = 5/9 – 1/3 = (5 – 3)/9 = 2/9

Sacrifice by Vimal = 4/9 – 1/3 = (4 – 3)/9 = 1/9

Sacrifice ratio = 2/9 : 1/9 = 2 : 1

Premium of Rs. 5,000 is distributed in sacrifice ratio:

Kamal gets = (2/3) × 5,000 = Rs. 3,333.33

Vimal gets = (1/3) × 5,000 = Rs. 1,666.67

Solution:

Income from shares:

Face value per share = Rs. 5, Number of shares = 6,000

Total face value of shares = 6,000 × 5 = Rs. 30,000

Dividend = 6% of 30,000 = Rs. 1,800

Income from debentures:

Face value per debenture = Rs. 10, Number of debentures = 2,000

Total face value of debentures = 2,000 × 10 = Rs. 20,000

Interest = 12% of 20,000 = Rs. 2,400

Total income = 1,800 + 2,400 = Rs. 4,200

Total investment = 30,000 + 20,000 = Rs. 50,000

Rate of return = (4,200 / 50,000) × 100 = 8.4%

Solution:

Step 1: Calculate investment in shares
Number of shares = 6,000
Par value per share = Rs. 5
Total investment in shares = 6,000 × 5 = Rs. 30,000

Step 2: Calculate dividend from shares
Dividend rate = 6%
Annual dividend = 6% of 30,000 = (6/100) × 30,000 = Rs. 1,800

Step 3: Calculate investment in debentures
Number of debentures = 2,000
Par value per debenture = Rs. 10 (corrected from OCR error "Rs. 0")
Total investment in debentures = 2,000 × 10 = Rs. 20,000

Step 4: Calculate interest from debentures
Interest rate = 2%
Annual interest = 2% of 20,000 = (2/100) × 20,000 = Rs. 400

Step 5: Total annual income
Total annual income = Dividend + Interest = 1,800 + 400 = Rs. 2,200

Step 6: Total investment
Total investment = 30,000 + 20,000 = Rs. 50,000

Step 7: Rate of return on investment
Rate of return = (Total annual income / Total investment) × 100
= (2,200 / 50,000) × 100 = 4.4%

Answer: Total annual income = Rs. 2,200 and rate of return = 4.4%.

Solution:

Given matrices:

A = ⎡
1 -1 1
0 2 -1
-1 0 0
⎤, B = ⎡
1 2 3
4 5 6
7 8 9
⎤

Step 1: Compute AB

AB = A × B = ⎡
1·1 + (-1)·4 + 1·7 1·2 + (-1)·5 + 1·8 1·3 + (-1)·6 + 1·9
0·1 + 2·4 + (-1)·7 0·2 + 2·5 + (-1)·8 0·3 + 2·6 + (-1)·9
(-1)·1 + 0·4 + 0·7 (-1)·2 + 0·5 + 0·8 (-1)·3 + 0·6 + 0·9
⎤

= ⎡
1 - 4 + 7 2 - 5 + 8 3 - 6 + 9
0 + 8 - 7 0 + 10 - 8 0 + 12 - 9
-1 + 0 + 0 -2 + 0 + 0 -3 + 0 + 0
⎤

= ⎡
4 5 6
1 2 3
-1 -2 -3
⎤

Step 2: Compute BA

BA = B × A = ⎡
1·1 + 2·0 + 3·(-1) 1·(-1) + 2·2 + 3·0 1·1 + 2·(-1) + 3·0
4·1 + 5·0 + 6·(-1) 4·(-1) + 5·2 + 6·0 4·1 + 5·(-1) + 6·0
7·1 + 8·0 + 9·(-1) 7·(-1) + 8·2 + 9·0 7·1 + 8·(-1) + 9·0
⎤

= ⎡
1 + 0 - 3 -1 + 4 + 0 1 - 2 + 0
4 + 0 - 6 -4 + 10 + 0 4 - 5 + 0
7 + 0 - 9 -7 + 16 + 0 7 - 8 + 0
⎤

= ⎡
-2 3 -1
-2 6 -1
-2 9 -1
⎤

Step 3: Compare AB and BA

AB = ⎡
4 5 6
1 2 3
-1 -2 -3
⎤ and BA = ⎡
-2 3 -1
-2 6 -1
-2 9 -1
⎤

Since AB ≠ BA, it is shown that matrix multiplication is not commutative for these matrices.

Solution:

Let x = number of cars parked per day
Let y = number of two-wheelers parked per day

Objective function (maximize income):
Income from cars = 25x
Income from two-wheelers = 10y
Total income Z = 25x + 10y

Constraints:

1. Total vehicles constraint: x + y ≤ 300 (maximum 300 vehicles)

2. Space constraint: If car space is used for two-wheelers, maximum 400 two-wheelers can be parked. This means each car space can accommodate 400/300 = 4/3 two-wheelers. So, the space used by x cars and y two-wheelers must satisfy: (4/3)x + y ≤ 400, or equivalently 4x + 3y ≤ 1200

3. Non-negativity constraints: x ≥ 0, y ≥ 0

Mathematical formulation:

Maximize Z = 25x + 10y

Subject to:

x + y ≤ 300

4x + 3y ≤ 1200

x ≥ 0, y ≥ 0

This is the required linear programming problem (mathematical model) to maximize the parking owner's income.

गणितीय सूत्रीकरण:

माना कारों की संख्या = x, दोपहिया वाहनों की संख्या = y

अधिकतम लाभ (Z) = 25x + 0y = 25x

प्रतिबंध:

• x + y ≤ 300 (कुल वाहन सीमा)
• यदि कार स्थान दोपहिया के लिए उपयोग हो, तो 400 दोपहिया खड़े हो सकते हैं। इसका अर्थ है कि प्रति कार स्थान में 400/300 = 4/3 दोपहिया आ सकते हैं। अतः x कार स्थानों में (4/3)x दोपहिया आ सकते हैं, लेकिन कुल दोपहिया y हैं, इसलिए y ≤ 400 - (4/3)x या 3y + 4x ≤ 1200
• x ≥ 0, y ≥ 0

हल: अधिकतम लाभ के लिए x = 300, y = 0 (केवल 300 कारें) या x = 0, y = 300 (केवल 300 दोपहिया) हो सकता है। चूंकि लाभ केवल कारों से है, अतः अधिकतम लाभ के लिए 300 कारें खड़ी करें, लाभ = 25 × 300 = ₹7500।

हल (आलेखी विधि):

प्रतिबंधों को समीकरणों में बदलें:

1. x + y = 10 ⇒ बिंदु (0,10) और (10,0)
2. 2x + y = 16 ⇒ बिंदु (0,16) और (8,0)
3. x = 0 (y-अक्ष), y = 0 (x-अक्ष)

सुसंगत क्षेत्र के कोणीय बिंदु ज्ञात करें:

• O(0,0)
• A(8,0) [2x+y=16 पर, y=0]
• B: दोनों रेखाओं का प्रतिच्छेदन: x+y=10 और 2x+y=16 को हल करने पर x=6, y=4, अतः B(6,4)
• C(0,10) [x+y=10 पर, x=0]

प्रत्येक कोणीय बिंदु पर Z का मान:

• O(0,0): Z = 6(0) + 5(0) = 0
• A(8,0): Z = 6(8) + 5(0) = 48
• B(6,4): Z = 6(6) + 5(4) = 36 + 20 = 56
• C(0,10): Z = 6(0) + 5(10) = 50

अतः अधिकतम Z = 56, बिंदु (6,4) पर, अर्थात x=6, y=4.

हल:

माना घटना E: A द्वारा निशाना साधना, घटना F: B द्वारा निशाना साधना।

P(E) = 3/4, P(F) = 4/5

P(निशाना साध लिया जायेगा) = P(E ∪ F) = P(E) + P(F) − P(E ∩ F)

चूंकि दोनों स्वतंत्र हैं, P(E ∩ F) = P(E) × P(F) = (3/4) × (4/5) = 3/5

P(E ∪ F) = 3/4 + 4/5 − 3/5 = 15/20 + 16/20 − 12/20 = 19/20

अतः प्रायिकता = 19/20

हल:

माना प्रत्येक प्रकार के स्टॉक में निवेशित राशि = ₹x

8% स्टॉक का क्रय मूल्य = अंकित मूल्य − छूट = 100 − 4 = ₹96

8% स्टॉक से आय = (x/96) × 8 = 8x/96 = x/12

माना 9% स्टॉक का क्रय मूल्य = ₹y प्रति ₹100 अंकित मूल्य

9% स्टॉक से आय = (x/y) × 9 = 9x/y

प्रश्नानुसार, दोनों आय समान हैं:

x/12 = 9x/y

⇒ 1/12 = 9/y

⇒ y = 9 × 12 = 108

अतः 9% स्टॉक का क्रय मूल्य = ₹108

हल:

माना A = \[ \begin{bmatrix} 4 & -1 & 2 \\ 3 & -2 & 3 \\ 4 & -3 & 2 \end{bmatrix} \]

|A| = 4[(-2)(2) − (3)(-3)] − (-1)[(3)(2) − (3)(4)] + 2[(3)(-3) − (-2)(4)]

= 4[−4 + 9] + 1[6 − 12] + 2[−9 + 8]

= 4(5) + 1(−6) + 2(−1) = 20 − 6 − 2 = 12

|A| ≠ 0, अतः व्युत्क्रम संभव है।

सहखंड ज्ञात करने पर:

A₁₁ = (−2)(2) − (3)(-3) = −4 + 9 = 5

A₁₂ = −[(3)(2) − (3)(4)] = −[6 − 12] = 6

A₁₃ = (3)(-3) − (−2)(4) = −9 + 8 = −1

A₂₁ = −[(−1)(2) − (2)(-3)] = −[−2 + 6] = −4

A₂₂ = (4)(2) − (2)(4) = 8 − 8 = 0

A₂₃ = −[(4)(-3) − (−1)(4)] = −[−12 + 4] = 8

A₃₁ = (−1)(3) − (2)(-2) = −3 + 4 = 1

A₃₂ = −[(4)(3) − (2)(3)] = −[12 − 6] = −6

A₃₃ = (4)(-2) − (−1)(3) = −8 + 3 = −5

सहखंड आव्यूह = \[ \begin{bmatrix} 5 & 6 & -1 \\ -4 & 0 & 8 \\ 1 & -6 & -5 \end{bmatrix} \]

adj A = सहखंड आव्यूह का परिवर्त = \[ \begin{bmatrix} 5 & -4 & 1 \\ 6 & 0 & -6 \\ -1 & 8 & -5 \end{bmatrix} \]

A⁻¹ = adj A / |A| = \[ \frac{1}{12} \begin{bmatrix} 5 & -4 & 1 \\ 6 & 0 & -6 \\ -1 & 8 & -5 \end{bmatrix} \]

अतः अभीष्ट व्युत्क्रम आव्यूह = \[ \begin{bmatrix} 5/12 & -1/3 & 1/12 \\ 1/2 & 0 & -1/2 \\ -1/12 & 2/3 & -5/12 \end{bmatrix} \]

हल :

दिए गए समीकरण निकाय को आव्यूह रूप में लिखने पर :

A = \(\begin{bmatrix} 3 & -1 & 4 \\ 1 & 4 & -3 \\ 1 & -1 & 1 \end{bmatrix}\), X = \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\), B = \(\begin{bmatrix} 3 \\ -5 \\ 3 \end{bmatrix}\)

सारणिक D = |A| = 3(4×1 - (-3)×(-1)) - (-1)(1×1 - (-3)×1) + 4(1×(-1) - 4×1)

= 3(4 - 3) + 1(1 + 3) + 4(-1 - 4)

= 3(1) + 1(4) + 4(-5) = 3 + 4 - 20 = -13

D_x = \(\begin{vmatrix} 3 & -1 & 4 \\ -5 & 4 & -3 \\ 3 & -1 & 1 \end{vmatrix}\) = 3(4×1 - (-3)×(-1)) - (-1)(-5×1 - (-3)×3) + 4(-5×(-1) - 4×3)

= 3(4 - 3) + 1(-5 + 9) + 4(5 - 12) = 3(1) + 1(4) + 4(-7) = 3 + 4 - 28 = -21

D_y = \(\begin{vmatrix} 3 & 3 & 4 \\ 1 & -5 & -3 \\ 1 & 3 & 1 \end{vmatrix}\) = 3(-5×1 - (-3)×3) - 3(1×1 - (-3)×1) + 4(1×3 - (-5)×1)

= 3(-5 + 9) - 3(1 + 3) + 4(3 + 5) = 3(4) - 3(4) + 4(8) = 12 - 12 + 32 = 32

D_z = \(\begin{vmatrix} 3 & -1 & 3 \\ 1 & 4 & -5 \\ 1 & -1 & 3 \end{vmatrix}\) = 3(4×3 - (-5)×(-1)) - (-1)(1×3 - (-5)×1) + 3(1×(-1) - 4×1)

= 3(12 - 5) + 1(3 + 5) + 3(-1 - 4) = 3(7) + 1(8) + 3(-5) = 21 + 8 - 15 = 14

क्रेमर नियम से :

x = D_x/D = -21/(-13) = 21/13

y = D_y/D = 32/(-13) = -32/13

z = D_z/D = 14/(-13) = -14/13

अतः हल : x = 21/13, y = -32/13, z = -14/13

हल :

|A| = 3(1×2 - (-1)×(-3)) - (-2)(2×2 - (-1)×4) + 3(2×(-3) - 1×4)

= 3(2 - 3) + 2(4 + 4) + 3(-6 - 4)

= 3(-1) + 2(8) + 3(-10) = -3 + 16 - 30 = -17

|A| ≠ 0, अतः व्युत्क्रम संभव है।

सहखण्ड आव्यूह ज्ञात करने पर :

A₁₁ = (1×2 - (-1)×(-3)) = 2 - 3 = -1

A₁₂ = -(2×2 - (-1)×4) = -(4 + 4) = -8

A₁₃ = (2×(-3) - 1×4) = -6 - 4 = -10

A₂₁ = -((-2)×2 - 3×(-3)) = -(-4 + 9) = -5

A₂₂ = (3×2 - 3×4) = 6 - 12 = -6

A₂₃ = -(3×(-3) - (-2)×4) = -(-9 + 8) = 1

A₃₁ = ((-2)×(-1) - 3×1) = 2 - 3 = -1

A₃₂ = -(3×(-1) - 3×2) = -(-3 - 6) = 9

A₃₃ = (3×1 - (-2)×2) = 3 + 4 = 7

सहखण्ड आव्यूह = \(\begin{bmatrix} -1 & -8 & -10 \\ -5 & -6 & 1 \\ -1 & 9 & 7 \end{bmatrix}\)

सहखण्ड आव्यूह का परिवर्त = \(\begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}\)

A^-1 = (1/|A|) × (सहखण्ड आव्यूह का परिवर्त)

= (1/(-17)) × \(\begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}\)

= \(\begin{bmatrix} 1/17 & 5/17 & 1/17 \\ 8/17 & 6/17 & -9/17 \\ 10/17 & -1/17 & -7/17 \end{bmatrix}\)

अतः A का व्युत्क्रम = \(\begin{bmatrix} 1/17 & 5/17 & 1/17 \\ 8/17 & 6/17 & -9/17 \\ 10/17 & -1/17 & -7/17 \end{bmatrix}\)

हल :

चरण 1 : प्रभावी पूँजी की गणना

हरी : 60,000 रु० (1 नवम्बर 2009 से 31 मार्च 2010 तक = 5 माह)

प्रभावी पूँजी = 60,000 × 5 = 3,00,000 रु०-माह

महेश :

1 नवम्बर 2009 से 31 जनवरी 2010 तक (3 माह) : 80,000 × 3 = 2,40,000 रु०-माह

1 फरवरी 2010 से 31 मार्च 2010 तक (2 माह) : (80,000 - 20,000) = 60,000 × 2 = 1,20,000 रु०-माह

कुल प्रभावी पूँजी (महेश) = 2,40,000 + 1,20,000 = 3,60,000 रु०-माह

चरण 2 : प्रभावी पूँजी का अनुपात

हरी : महेश = 3,00,000 : 3,60,000 = 5 : 6

चरण 3 : पूँजी पर ब्याज की गणना

हरी : 60,000 × 10% × (5/12) = 60,000 × 0.10 × 5/12 = 2,500 रु०

महेश : (80,000 × 10% × 3/12) + (60,000 × 10% × 2/12) = (80,000 × 0.10 × 3/12) + (60,000 × 0.10 × 2/12)

= 2,000 + 1,000 = 3,000 रु०

चरण 4 : हरी का वेतन

हरी को 2,000 रु० प्रति माह वेतन, 5 माह के लिए = 2,000 × 5 = 10,000 रु०

चरण 5 : वितरण योग्य लाभ की गणना

कुल लाभ = 37,500 रु०

घटाएँ : हरी का वेतन = 10,000 रु०

घटाएँ : पूँजी पर ब्याज (हरी 2,500 + महेश 3,000) = 5,500 रु०

शेष लाभ = 37,500 - 10,000 - 5,500 = 22,000 रु०

चरण 6 : शेष लाभ का वितरण (अनुपात 5:6 में)

हरी का हिस्सा = 22,000 × (5/11) = 10,000 रु०

महेश का हिस्सा = 22,000 × (6/11) = 12,000 रु०

चरण 7 : प्रत्येक साझेदार की कुल लाभ राशि

हरी : वेतन (10,000) + ब्याज (2,500) + लाभ हिस्सा (10,000) = 22,500 रु०

महेश : ब्याज (3,000) + लाभ हिस्सा (12,000) = 15,000 रु०

अतः हरी को 22,500 रु० और महेश को 15,000 रु० लाभ प्राप्त होगा।

Solution

Step 1: Calculate the period of investment for each partner.

Accounting year: 1st November 2009 to 31st March 2010 = 5 months.

Step 2: Calculate effective capital (product of capital × time in months).

Hari's capital: Rs. 60,000 for full 5 months (Nov to March).

Effective capital = 60,000 × 5 = 3,00,000

Mahesh's capital:

• From 1st Nov 2009 to 31st Jan 2010: 3 months with Rs. 80,000 → 80,000 × 3 = 2,40,000
• From 1st Feb 2010 to 31st March 2010: 2 months with Rs. 60,000 (after withdrawal) → 60,000 × 2 = 1,20,000
• Total effective capital for Mahesh = 2,40,000 + 1,20,000 = 3,60,000

Step 3: Calculate total effective capital.

Total effective capital = 3,00,000 (Hari) + 3,60,000 (Mahesh) = 6,60,000

Step 4: Calculate salary to Hari.

Salary per month = Rs. 2,000 for 5 months = 2,000 × 5 = Rs. 10,000

Step 5: Calculate profit after salary.

Profit before salary = Rs. 37,500

Profit after salary = 37,500 - 10,000 = Rs. 27,500

Step 6: Distribute profit in effective capital ratio.

Ratio of effective capitals: Hari : Mahesh = 3,00,000 : 3,60,000 = 5 : 6

Hari's share = (5/11) × 27,500 = 5 × 2,500 = Rs. 12,500

Mahesh's share = (6/11) × 27,500 = 6 × 2,500 = Rs. 15,000

Step 7: Add salary to Hari's share.

Hari's total share = 12,500 + 10,000 = Rs. 22,500

Mahesh's total share = Rs. 15,000