ALLAMA IQBAL OPEN UNIVERSITY
| Level: I.Com | Semester: Spring 2021 | ||||||||||||
| Course Code: 1349 - Introduction to Business Mathematics | Maximum Marks: 100 | ||||||||||||
| Time Allowed: 03 Hours | Pass Marks: 40 |
Note: Attempt any Five Questions, Question No. 1 is compulsory.
Q1.(a) During a sale, a store offers a discount of 25% off any purchase. What is the regular price of a dress that a customer purchased for $73.50?
Step 1. Let P be the regular price. That is 100% of the price before the discount.
Step 2. After a 25% discount the customer pays 75% of the regular price.
\[ 0.75P = 73.50 \]
Step 3. Solve for P.
\[ P = \frac{73.50}{0.75} = \frac{7350}{75} = 98 \]
Step 4. State the answer.
\[ P = \$98 \]
Optional. Amount the customer saved.
\[ \text{Savings} = P \times 0.25 = 98 \times 0.25 = 24.50 \]
\[ \text{Customer saved } \$24.50 \]
Q1.(b) Two numbers have the ratio 2:3. The larger is 30 more than 1/2 of the smaller. Find the numbers.
Step 1. Assume the numbers.
Let the smaller number be \(2x\) and the larger number be \(3x\).
Step 2. Translate the condition.
\[ 3x = \tfrac{1}{2}(2x) + 30 \]
Step 3. Simplify.
\[ 3x = x + 30 \]
Step 4. Solve for \(x\).
\[ 3x - x = 30 \quad \Rightarrow \quad 2x = 30 \quad \Rightarrow \quad x = 15 \]
Step 5. Find the numbers.
\[ \text{Smaller} = 2x = 2(15) = 30 \] \[ \text{Larger} = 3x = 3(15) = 45 \]
Final Answer:
\[ \text{The two numbers are } 30 \text{ and } 45 \]
Q1.(c) Jeremy said that if the means and extremes of a proportion are interchanged, the resulting ratios form a proportion. Do you agree with Jeremy? Explain why or why not.
A proportion means that two ratios are equal. Suppose we have:
\[ \frac{a}{b} = \frac{c}{d} \]
Here, \(a\) and \(d\) are the extremes, and \(b\) and \(c\) are the means.
If we interchange the means, we get:
\[ \frac{c}{b} = \frac{a}{d} \]
If we interchange the extremes, we get:
\[ \frac{d}{b} = \frac{c}{a} \]
Both of these are still true proportions because in every case the cross multiplication gives:
\[ ad = bc \]
Conclusion: Yes, Jeremy is correct. Interchanging either the means or the extremes in a proportion still results in a true proportion.
Q2.(a) If $1,000 is invested at 8% compounded.
(i) annually
(ii) semiannually
(iii) quarterly
(iv) monthly
What is the amount after 5 years? Write answers to the nearest cent.
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where \(P = 1000\), \(r = 0.08\), \(t = 5\) years, and \(n\) is the number of compounding periods per year.
(i) Compounded Annually (n = 1)
\[ A = 1000 \left(1 + \frac{0.08}{1}\right)^{1 \times 5} \]
\[ A = 1000 (1.08)^5 = 1000 \times 1.4693 \approx 1469.33 \]
(ii) Compounded Semiannually (n = 2)
\[ A = 1000 \left(1 + \frac{0.08}{2}\right)^{2 \times 5} \]
\[ A = 1000 (1.04)^{10} = 1000 \times 1.4802 \approx 1480.24 \]
(iii) Compounded Quarterly (n = 4)
\[ A = 1000 \left(1 + \frac{0.08}{4}\right)^{4 \times 5} \]
\[ A = 1000 (1.02)^{20} = 1000 \times 1.4859 \approx 1485.95 \]
(iv) Compounded Monthly (n = 12)
\[ A = 1000 \left(1 + \frac{0.08}{12}\right)^{12 \times 5} \]
\[ A = 1000 (1.0067)^{60} = 1000 \times 1.4899 \approx 1489.85 \]
Final Answers:
- (i) Annually: \$1469.33
- (ii) Semiannually: \$1480.24
- (iii) Quarterly: \$1485.95
- (iv) Monthly: \$1489.85
Q2.(b) Which is the better investment and why: 9% compounded quarterly or 9.25% compounded annually? Explain your answer.
We compare the two options by finding the Effective Annual Rate (EAR).
Option 1: 9% compounded quarterly
\[ EAR = \left(1 + \frac{0.09}{4}\right)^4 - 1 \]
\[ EAR = (1 + 0.0225)^4 - 1 \]
\[ EAR = (1.0225)^4 - 1 \approx 0.09308 \]
\[ EAR \approx 9.31\% \]
Option 2: 9.25% compounded annually
\[ EAR = 9.25\% = 0.0925 \]
The better investment is 9% compounded quarterly because it gives a slightly higher effective annual yield than 9.25% compounded annually.
Q3. Determine the solution of the system of equations by (i) Matrix Method (ii) Cramer's rule
10x1 + 4x2 = 46
-5x1 + 6x2 = 9
Let x1 = x and x2=y
So, equation is:
10x + 4y = 46
-5x + 6y = 9
\[ 10x + 4y = 46 \]
\[ -5x + 6y = 9 \]
(i) Matrix Method
Write the system as \(A \mathbf{x} = \mathbf{b}\):
\[ A = \begin{pmatrix}10 & 4 \\ -5 & 6\end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix}x \\ y\end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix}46 \\ 9\end{pmatrix} \]
Determinant:
\[ \det(A) = 10 \cdot 6 - (-5)\cdot 4 = 60 + 20 = 80 \]
Inverse:
\[ A^{-1} = \frac{1}{\det(A)} \begin{pmatrix}6 & -4 \\ 5 & 10\end{pmatrix} = \frac{1}{80}\begin{pmatrix}6 & -4 \\ 5 & 10\end{pmatrix} \]
Now:
\[ \mathbf{x} = A^{-1}\mathbf{b} = \frac{1}{80}\begin{pmatrix}6 & -4 \\ 5 & 10\end{pmatrix} \begin{pmatrix}46 \\ 9\end{pmatrix} \]
\[ = \frac{1}{80}\begin{pmatrix}6\cdot46 - 4\cdot9 \\ 5\cdot46 + 10\cdot9\end{pmatrix} = \frac{1}{80}\begin{pmatrix}240 \\ 320\end{pmatrix} = \begin{pmatrix}3 \\ 4\end{pmatrix} \]
\[ x = 3, \quad y = 4 \]
(ii) Cramer's Rule
\[ D = \det(A) = 80 \]
Replace the first column with \(\mathbf{b}\):
\[ D_x = \det\begin{pmatrix}46 & 4 \\ 9 & 6\end{pmatrix} = 46\cdot6 - 9\cdot4 = 276 - 36 = 240 \]
Replace the second column with \(\mathbf{b}\):
\[ D_y = \det\begin{pmatrix}10 & 46 \\ -5 & 9\end{pmatrix} = 10\cdot9 - (-5)\cdot46 = 90 + 230 = 320 \]
So:
\[ x = \frac{D_x}{D} = \frac{240}{80} = 3, \quad y = \frac{D_y}{D} = \frac{320}{80} = 4 \]
Final Answer:
\[ x = 3, \quad y = 4 \]
Q4.(a) Simplify the following in binary system.
{(10001101)2 × (235)10} - (27)10
Step 1: Convert the binary number to decimal
(10001101)2 = 1×27 + 0×26 + 0×25 + 0×24 + 1×23 + 1×22 + 0×21 + 1×20
= 128 + 8 + 4 + 1 = 141
Step 2: Multiply by 235
141 × 235 = 141 × (200 + 30 + 5)
= 141×200 + 141×30 + 141×5
= 28200 + 4230 + 705 = 33135
Step 3: Subtract 27
33135 - 27 = 33108
Step 4: Convert 33108 to binary
Divide by 2 repeatedly and record remainders:
| Division | Quotient | Remainder |
|---|---|---|
| 33108 ÷ 2 | 16554 | 0 |
| 16554 ÷ 2 | 8277 | 0 |
| 8277 ÷ 2 | 4138 | 1 |
| 4138 ÷ 2 | 2069 | 0 |
| 2069 ÷ 2 | 1034 | 1 |
| 1034 ÷ 2 | 517 | 0 |
| 517 ÷ 2 | 258 | 1 |
| 258 ÷ 2 | 129 | 0 |
| 129 ÷ 2 | 64 | 1 |
| 64 ÷ 2 | 32 | 0 |
| 32 ÷ 2 | 16 | 0 |
| 16 ÷ 2 | 8 | 0 |
| 8 ÷ 2 | 4 | 0 |
| 4 ÷ 2 | 2 | 0 |
| 2 ÷ 2 | 1 | 0 |
| 1 ÷ 2 | 0 | 1 |
Reading the remainders from bottom to top:
3310810 = (1000000101010100)2
Final Answer:
{ (10001101)2 × (235)10 } - (27)10 = (1000000101010100)2
Q4.(b) Solve the equation:
\( \sqrt{2x^2 + 7x + 3} + \sqrt{x^2 + 5x + 6} = \sqrt{x + 3} \)
Solving a Square Root Equation
We solve the equation:
\[ \sqrt{2x^{2} + 7x + 3} + \sqrt{x^{2} + 5x + 6} = \sqrt{x + 3} \]
Step 1. Domain
Inside square roots must be non-negative:
\(x + 3 \ge 0\) and \(x^{2} + 5x + 6 \ge 0\).
So possible values are \(x = -3\) or \(x \ge -2\).
Step 2. Try \(x = -3\)
Substitute \(x = -3\):
\[ \sqrt{2(-3)^{2} + 7(-3) + 3} + \sqrt{(-3)^{2} + 5(-3) + 6} = \sqrt{0} + \sqrt{0} = 0 \]
Right side: \(\sqrt{-3+3} = 0\). So \(x = -3\) is correct.
Step 3. Other values
For \(x \ge -2\), the left side (two square roots added) is always bigger than the right side (one square root). So no other solution exists.
Final Answer
\[ {x = -3} \]
Q5.(a)
Q5.(b)
Q6.(a) Find six Arithmetic Means between 5 and 12.
Step 1: Formula
The general term of an arithmetic progression (AP) is: \[ a_n = a_1 + (n-1)d \]Step 2: Known values
Here, \[ a_1 = 5, \quad a_8 = 12, \quad n = 8 \]Step 3: Put values into the formula
\[ a_8 = a_1 + (8-1)d \] \[ 12 = 5 + 7d \]Step 4: Solve for \(d\)
\[ 12 - 5 = 7d \] \[ 7 = 7d \] \[ d = 1 \]Step 5: Write all terms
\[ a_1 = 5 \] \[ a_2 = a_1 + d = 5 + 1 = 6 \] \[ a_3 = a_1 + 2d = 5 + 2(1) = 7 \] \[ a_4 = a_1 + 3d = 5 + 3(1) = 8 \] \[ a_5 = a_1 + 4d = 5 + 4(1) = 9 \] \[ a_6 = a_1 + 5d = 5 + 5(1) = 10 \] \[ a_7 = a_1 + 6d = 5 + 6(1) = 11 \] \[ a_8 = a_1 + 7d = 5 + 7(1) = 12 \]Final Answer
The six arithmetic means are: \[ 6, \; 7, \; 8, \; 9, \; 10, \; 11 \]Q6.(b) Find the nth term of a Geometric Progression,
\[ \text{If } \frac{a_5}{a_3} = \frac{16}{9} \quad \text{and} \quad a_2 = \frac{6}{7} \]\[ \text{Given: } \quad \frac{a_5}{a_3} = \frac{16}{9}, \quad a_2 = \frac{6}{7} \] \[ \text{In a G.P., } \quad a_n = a \cdot r^{\,n-1} \] \[ a_5 = a \cdot r^4, \quad a_3 = a \cdot r^2 \] \[ \frac{a_5}{a_3} = \frac{a \cdot r^4}{a \cdot r^2} = r^2 \] \[ r^2 = \frac{16}{9} \quad \Rightarrow \quad r = \pm \frac{4}{3} \] \[ a_2 = a \cdot r \] \[ \frac{6}{7} = a \cdot r \quad \Rightarrow \quad a = \frac{6}{7r} \] \[ \text{Case 1: } r = \frac{4}{3} \] \[ a = \frac{6}{7 \cdot \tfrac{4}{3}} = \frac{6}{\tfrac{28}{3}} = \frac{18}{28} = \frac{9}{14} \] \[ \text{So, } a = \frac{9}{14}, \quad r = \frac{4}{3} \] \[ \text{Case 2: } r = -\frac{4}{3} \] \[ a = \frac{6}{7 \cdot \left(-\tfrac{4}{3}\right)} = \frac{6}{-\tfrac{28}{3}} = \frac{18}{-28} = -\frac{9}{14} \] \[ \text{So, } a = -\frac{9}{14}, \quad r = -\frac{4}{3} \] \[ \text{Final Answer: } \quad (a, r) = \left(\tfrac{9}{14}, \tfrac{4}{3}\right) \quad \text{or} \quad (a, r) = \left(-\tfrac{9}{14}, -\tfrac{4}{3}\right) \]
Q7.(a) \(\text{If } U=\{1,2,3,\dots,20\},\ A=\{2,4,6,\dots,20\} \text{ and } B=\{1,3,5,\dots,19\} \text{ then verify } (A \cup B)^{c} = A^{c} \cap B^{c}\)
\(U = \{1,2,3,\dots,20\}\)
\(A = \{2,4,6,\dots,20\}\)
\(B = \{1,3,5,\dots,19\}\)
\(A^{c} = U - A = \{1,3,5,\dots,19\} = B\)
\(B^{c} = U - B = \{2,4,6,\dots,20\} = A\)
\(A \cup B = U\)
\((A \cup B)^{c} = U^{c} = \varnothing\)
\(A^{c} \cap B^{c} = B \cap A = \varnothing\)
\(\therefore (A \cup B)^{c} = A^{c} \cap B^{c}\)
Q7.(b)
Q8.(a)
Q8.(b)