AIOU 247 Solved Assignment Autumn 2025

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ALLAMA IQBAL OPEN UNIVERSITY

(Department of Mathematics)

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WARNING

1. Plagiarism or hiring of ghost writer(s) for solving the assignment(s) will debar the student from award of degree/certificate if found at any stage.

2. Submitting assignment(s) borrowed or stolen from other(s) as one's own will be penalized as defined in the "Aiou Plagiarism Policy".

Assignment Submission Schedule
6 Credit Hours	Due Date	3 Credit Hours	Due Date
Assignment 1	15-12-2025	Assignment 1	08-01-2026
Assignment 2	08-01-2026
Assignment 3	30-01-2026	Assignment 2	20-02-2026
Assignment 4	20-02-2026

Course: Mathematics-I (247)	Semester: Autumn,2025
Level: Matric / SSC

Please read the following instructions for writing your assignments. (SSC, HSSC & BA Programmes)
1. All questions are compulsory and carry equal marks but within a question the marks are distributed according to its requirements.
2. Read the question carefully and then answer it according to the requirements of the questions.
3. Late submission of assignments will not be accepted.
4. Your own analysis and synthesis will be appreciated.
5. Avoid irrelevant discussion/information and reproducing from books, study guide of allied material.

Total Marks: 100	Pass Marks: 40

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ASSIGNMENT No. 1

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Q1(a). Two angles in a triangle are 50° and 80°. Find the ratio of the third angle to the sum of the first two. ▶

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1. The sum of the angles in a triangle is always:

$$ 180^\circ $$

2. The first two angles are:

$$ 50^\circ \text{ and } 80^\circ $$

3. Add them:

$$ 50^\circ + 80^\circ = 130^\circ $$

4. Find the third angle:

$$ 180^\circ - 130^\circ = 50^\circ $$

5. Ratio = third angle : sum of first two

$$ 50 : 130 $$

6. Simplify:

$$ 5 : 13 $$

7. As a fraction:

$$ \frac{50}{130} = \frac{5}{13} $$

Final Answer:
The ratio is 5 : 13 or the fraction is $$ \frac{5}{13} $$.

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Q1(b). If the stay of 12 men for 7 days in a hostel costs Rs. 21200, find the cost for the stay of 10 men for 8 days. ▶

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1. Find the total man-days in the given case:

$$ 12 \times 7 = 84 \ \text{man-days} $$

2. Cost for 84 man-days is:

$$ 21200 \ \text{rupees} $$

3. Cost for 1 man-day:

$$ \frac{21200}{84} = 252.38 \ \text{rupees (approx)} $$

4. Now, find the required man-days:

$$ 10 \times 8 = 80 \ \text{man-days} $$

5. Cost for 80 man-days:

$$ 80 \times 252.38 = 20190.4 \ \text{rupees} $$

Final Answer:
The cost for 10 men for 8 days is Rs. 20,190.40.

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Q2(a). Calculate Ushr on a wheat crop amounting to Rs. 15,00,000 produced by artificial resources. ▶

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1. Rule of Ushr:

If irrigated naturally → Ushr = 10%
If irrigated artificially → Ushr = 5%

2. Given data:

Crop value = Rs. 15,00,000
Irrigation = Artificial resources
Ushr rate = 5%

3. Calculation:

$$ \text{Ushr} = \frac{5}{100} \times 15,00,000 $$

$$ \text{Ushr} = 75,000 $$

Final Answer:
The Ushr on the wheat crop is Rs. 75,000.

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Q2(b). Calculate Zakat on gold amounting to Rs. 11,75,000. ▶

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1. Rule of Zakat on Gold:

Zakat is payable at the rate of 2.5% of the total value of gold (if it reaches or exceeds nisab).

2. Given data:

Value of gold = Rs. 11,75,000
Zakat rate = 2.5%

3. Calculation:

$$ \text{Zakat} = \frac{2.5}{100} \times 11,75,000 $$

$$ \text{Zakat} = 29,375 $$

Final Answer:
The Zakat on gold worth Rs. 11,75,000 is Rs. 29,375.

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Q3(a). Distribute Rs. 19600 as profit among three friends, so that their shares are in the ratio 2:4:7. ▶

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1. The given ratio is:

$$ 2 : 4 : 7 $$

2. Total parts:

$$ 2 + 4 + 7 = 13 $$

3. Value of one part:

$$ \frac{19600}{13} \approx 1507.69 $$

4. Now calculate each friend’s share:

Friend A (2 parts):

$$ \frac{2}{13} \times 19600 = 3015.38 $$

Friend B (4 parts):

$$ \frac{4}{13} \times 19600 = 6030.77 $$

Friend C (7 parts):

$$ \frac{7}{13} \times 19600 = 10553.85 $$

Final Answer:
Friend A’s share ≈ Rs. 3015.38
Friend B’s share ≈ Rs. 6030.77
Friend C’s share ≈ Rs. 10553.85

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Q3(b). The marked price of a clock is Rs. 800. The shopkeeper offers a discount of 6% and still gains 12%. Find the price at which the shopkeeper purchased it. ▶

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1. Marked Price (MP): 800

2. Discount = 6% of 800:

$$ \frac{6}{100} \times 800 = 48 $$

3. Selling Price (SP):

$$ 800 - 48 = 752 $$

4. Formula:

$$ SP = CP \times \left(1 + \frac{\text{Gain%}}{100}\right) $$

Substitute values:

$$ 752 = CP \times \left(1 + \frac{12}{100}\right) $$

$$ 752 = CP \times 1.12 $$

5. Find CP:

$$ CP = \frac{752}{1.12} $$

$$ CP = 671.43 $$

Final Answer:
The shopkeeper purchased the clock at Rs. 671.43.

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Q4(a). Find the compound profit on Rs.2500 for 2 1/2 years at 4% per annum. ▶

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1. Formula for Compound Amount:

$$ A = P \times \left(1 + \frac{r}{100}\right)^n $$

Where:

• Principal (P) = 2500
• Rate (r) = 4%
• Time (n) = 2 \tfrac{1}{2} years = 2 years + \tfrac{1}{2} year

2. After 2 years:

$$ A_2 = 2500 \times \left(1 + \frac{4}{100}\right)^2 $$

$$ A_2 = 2500 \times (1.04)^2 $$

$$ A_2 = 2500 \times 1.0816 = 2704 $$

3. For next half year (rate = 2%):

$$ A = 2704 \times \left(1 + \frac{2}{100}\right) $$

$$ A = 2704 \times 1.02 $$

$$ A = 2758.08 $$

4. Compound Profit (CI):

$$ CI = A - P $$

$$ CI = 2758.08 - 2500 $$

$$ CI = 258.08 $$

Final Answer:
The compound profit is Rs. 258.08.

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Q4(b). At what annual rate of profit would a sum of Rs. 840 will increase to Rs. 1050 in 1 year 8 months? ▶

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1. Given Data:

• Principal (P) = 840
• Amount (A) = 1050
• Time (T) = 1 year 8 months = \( 1 \tfrac{8}{12} = \tfrac{5}{3} \, \text{years} \)

2. Formula for Simple Interest:

$$ A = P \left(1 + \frac{r \times T}{100}\right) $$

3. Substitution:

$$ 1050 = 840 \left(1 + \frac{r \times \tfrac{5}{3}}{100}\right) $$

$$ \frac{1050}{840} = 1 + \frac{5r}{300} $$

$$ 1.25 = 1 + \frac{5r}{300} $$

4. Solve for r:

$$ 1.25 - 1 = \frac{5r}{300} $$

$$ 0.25 = \frac{5r}{300} $$

$$ 0.25 \times 300 = 5r $$

$$ 75 = 5r $$

$$ r = 15 $$

Final Answer:
The annual rate of profit is 15%.

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Q5(a). If the gross pay of a person is Rs.70,000, then calculate his net take-home salary, after deductions of Rs.350 as income tax, Rs.900 as a benevolent fund, Rs.1600 s G.P fund and Rs.300 as group insurance. ▶

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1. Gross Pay: 70,000

2. Deductions:

$$ \text{Total Deductions} = 350 + 900 + 1600 + 300 $$

$$ \text{Total Deductions} = 3150 $$

3. Net Salary:

$$ \text{Net Salary} = \text{Gross Pay} - \text{Deductions} $$

$$ \text{Net Salary} = 70000 - 3150 $$

$$ \text{Net Salary} = 66850 $$

Final Answer:
The net take-home salary is Rs. 66,850.

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Q5(b). The price of a motorcycle is Rs.65,000. If 18% sales tax is charged, then calculate the amount of sales tax on 30 such motorcycles. ▶

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1. Price of one motorcycle: 65,000

2. Sales Tax on one motorcycle:

$$ \text{Sales Tax} = \frac{18}{100} \times 65,000 $$

$$ \text{Sales Tax} = 11,700 $$

3. Sales Tax on 30 motorcycles:

$$ \text{Total Sales Tax} = 11,700 \times 30 $$

$$ \text{Total Sales Tax} = 351,000 $$

Final Answer:
The amount of sales tax on 30 motorcycles is Rs. 351,000.

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ASSIGNMENT No. 2

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Q1(a). If A.M between a and 10 is 40, then find the value of a. ▶

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1. Formula of Arithmetic Mean (A.M):

$$ A.M = \frac{a + b}{2} $$

2. Given values: A.M = 40, b = 10

$$ 40 = \frac{a + 10}{2} $$

3. Multiply both sides by 2:

$$ 80 = a + 10 $$

4. Subtract 10 from both sides:

$$ a = 70 $$

The value of a is 70.

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Q1(b). The positive G.M between two numbers is 6 and the A.M between them is 10. Find the numbers. ▶

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1. Formula of Arithmetic Mean (A.M):

$$ A.M = \frac{x + y}{2} $$

Given A.M = 10:

$$ \frac{x + y}{2} = 10 $$

$$ x + y = 20 $$

2. Formula of Geometric Mean (G.M):

$$ G.M = \sqrt{xy} $$

Given G.M = 6:

$$ \sqrt{xy} = 6 $$

$$ xy = 36 $$

3. Form the quadratic equation using sum and product of roots:

$$ t^2 - (x+y)t + xy = 0 $$

$$ t^2 - 20t + 36 = 0 $$

4. Solve the quadratic:

$$ t = \frac{20 \pm \sqrt{20^2 - 4(36)}}{2} $$

$$ t = \frac{20 \pm \sqrt{400 - 144}}{2} $$

$$ t = \frac{20 \pm \sqrt{256}}{2} $$

$$ t = \frac{20 \pm 16}{2} $$

5. Values of t:

$$ t = \frac{20 + 16}{2} = 18 $$

$$ t = \frac{20 - 16}{2} = 2 $$

The two numbers are 18 and 2.

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Q2(a). If logx = 0.1597, find x. ▶

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1. Given:

$$ \log x = 0.1597 $$

2. Rewrite in exponential form:

$$ x = 10^{0.1597} $$

3. Calculate the value:

$$ x \approx 1.445 $$

The value of x is approximately 1.445.

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Q2(b). Prove that log(2 x 4 x 6) = log2 + log4 + log6. ▶

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1. Start with the left-hand side:

$$ \log(2 \times 4 \times 6) $$

2. Group the terms:

$$ \log((2 \times 4) \times 6) $$

3. Apply product rule of logarithms:

$$ \log(2 \times 4) + \log(6) $$

4. Apply the rule again:

$$ \log(2) + \log(4) + \log(6) $$

$$ \log(2 \times 4 \times 6) = \log 2 + \log 4 + \log 6 $$

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Q3(a). If A = {1,2,3,4,5,6,7,8,9}, B = {2,3,4,5} and C = {3,4,6}. Verify that (A∩B)∩C = An(B∩C). ▶

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Given:

$$ A = \{1,2,3,4,5,6,7,8,9\}, \; B = \{2,3,4,5\}, \; C = \{3,4,6\} $$

We need to verify:

$$ (A \cap B) \cap C = A \cap (B \cap C) $$

1. Find intersection of A and B:

$$ A \cap B = \{2,3,4,5\} $$

2. Now, (A ∩ B) ∩ C:

$$ (A \cap B) \cap C = \{2,3,4,5\} \cap \{3,4,6\} = \{3,4\} $$

3. Find intersection of B and C:

$$ B \cap C = \{2,3,4,5\} \cap \{3,4,6\} = \{3,4\} $$

4. Now, A ∩ (B ∩ C):

$$ A \cap (B \cap C) = A \cap \{3,4\} = \{3,4\} $$

$$ (A \cap B) \cap C = A \cap (B \cap C) = \{3,4\} $$

Hence, verified.

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Q3(b). If X = {3,5,7} and Y = {3,6}. Find X×Y and Y×X and also the domains and ranges of the two binary relations established on our own for each case. ▶

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Given:

$$ X = \{3,5,7\}, \quad Y = \{3,6\} $$

1. Cartesian Product X × Y:

$$ X \times Y = \{(3,3), (3,6), (5,3), (5,6), (7,3), (7,6)\} $$

2. Cartesian Product Y × X:

$$ Y \times X = \{(3,3), (3,5), (3,7), (6,3), (6,5), (6,7)\} $$

3. Binary Relation from X × Y:

Let us define a relation R from X to Y as:

$$ R = \{(x,y) \in X \times Y \;|\; x < y\} $$

So:

$$ R = \{(3,6), (5,6)\} $$

Domain of R:

$$ \text{Dom}(R) = \{3,5\} $$

Range of R:

$$ \text{Ran}(R) = \{6\} $$

4. Binary Relation from Y × X:

Let us define a relation S from Y to X as:

$$ S = \{(y,x) \in Y \times X \;|\; y > x\} $$

So:

$$ S = \{(6,3), (6,5)\} $$

Domain of S:

$$ \text{Dom}(S) = \{6\} $$

Range of S:

$$ \text{Ran}(S) = \{3,5\} $$

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$$ X \times Y = \{(3,3),(3,6),(5,3),(5,6),(7,3),(7,6)\} $$

$$ Y \times X = \{(3,3),(3,5),(3,7),(6,3),(6,5),(6,7)\} $$

$$ \text{For relation R: Domain = {3,5}, Range = {6}} $$

$$ \text{For relation S: Domain = {6}, Range = {3,5}} $$

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Q4(a). Draw the graph of: $$ \frac{y}{5} = \frac{x}{3} - \frac{7}{4} $$ ▶

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Q4(b). Draw the graph of: $$ \frac{3y}{2} = \frac{2x}{3} - \frac{7}{2} $$ ▶

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Q5(a). Find the harmonic mean of the values 5, 25 and 75. ▶

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1. Formula of Harmonic Mean (H.M):

$$ H.M = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \cdots + \frac{1}{x_n}} $$

2. Given values: \(x_1 = 5, \; x_2 = 25, \; x_3 = 75, \; n = 3\)

$$ H.M = \frac{3}{\frac{1}{5} + \frac{1}{25} + \frac{1}{75}} $$

3. Simplify the denominators:

$$ \frac{1}{5} = \frac{15}{75}, \quad \frac{1}{25} = \frac{3}{75}, \quad \frac{1}{75} = \frac{1}{75} $$

$$ \frac{1}{5} + \frac{1}{25} + \frac{1}{75} = \frac{15+3+1}{75} = \frac{19}{75} $$

4. Substitute back:

$$ H.M = \frac{3}{\tfrac{19}{75}} = \frac{3 \times 75}{19} $$

5. Final value:

$$ H.M = \frac{225}{19} = 11.84 $$

The harmonic mean of 5, 25 and 75 is
$$ \frac{225}{19} \; = 11.84 $$

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Q5(b). Find the median for the set of values: 8,6,5,2,3,1,7,4. ▶

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1. Arrange the data in ascending order:

$$ 1, 2, 3, 4, 5, 6, 7, 8 $$

2. Number of values:

$$ n = 8 \quad (\text{even}) $$

3. Formula for median when n is even:

$$ \text{Median} = \frac{\text{Value at } \tfrac{n}{2} \text{ position} + \text{Value at } (\tfrac{n}{2}+1) \text{ position}}{2} $$

4. Identify the two middle values:

Value at position $$ \tfrac{8}{2} = 4 $$ is $$ 4 $$

Value at position $$ \tfrac{8}{2} + 1 = 5 $$ is $$ 5 $$

5. Calculate the median:

$$ \text{Median} = \frac{4 + 5}{2} = \frac{9}{2} = 4.5 $$

The median of the given data is 4.5.

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