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	<title>6 sem mca &#8211; MUJ ASSIGNMENT </title>
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		<title>DCA6208 OPERATING SYSTEM  SEPTEMBER 2025</title>
		<link>https://muj.assignmentsupport.in/product/dca6208-operating-system-september-2025/</link>
		
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		<pubDate>Thu, 04 Dec 2025 15:37:23 +0000</pubDate>
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										<content:encoded><![CDATA[<body><table width="100%">
<tbody>
<tr>
<td width="38%"><strong>SESSION</strong></td>
<td width="61%"><strong>JULY/SEPTEMBER 2025</strong></td>
</tr>
<tr>
<td width="38%"><strong>PROGRAM</strong></td>
<td width="61%"><strong>MASTER OF COMPUTER APPLICATIONS (MCA)</strong></td>
</tr>
<tr>
<td width="38%"><strong>SEMESTER</strong></td>
<td width="61%"><strong>6</strong></td>
</tr>
<tr>
<td width="38%"><strong>COURSE CODE &amp; NAME</strong></td>
<td width="61%"><strong>DCA6208 OPERATING SYSTEM</strong></td>
</tr>
<tr>
<td width="38%"></td>
<td width="61%"></td>
</tr>
<tr>
<td width="38%"></td>
<td width="61%"></td>
</tr>
</tbody>
</table>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong>Set-I</strong></p>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong>Q1.a. Consider a system with five processes P1, P2, P3, P4, P5 and the 4 resource types A, B and C ,D. Following is the snapshot at the time t. </strong></p>
<p><strong> </strong></p>
<table width="100%">
<tbody>
<tr>
<td width="13%"><strong>Process</strong></td>
<td width="31%"><strong>Allocation (A B C D)</strong></td>
<td width="23%"><strong>Max (A B C D)</strong></td>
<td width="30%"><strong>Available (A B C D)</strong></td>
</tr>
<tr>
<td width="13%"><strong>P1</strong></td>
<td width="31%"><strong>0 0 1 2</strong></td>
<td width="23%"><strong>0 0 1 2</strong></td>
<td width="30%"><strong>1 5 2 0</strong></td>
</tr>
<tr>
<td width="13%"><strong>P2</strong></td>
<td width="31%"><strong>1 0 0 0</strong></td>
<td width="23%"><strong>1 7 5 0</strong></td>
<td width="30%"></td>
</tr>
<tr>
<td width="13%"><strong>P3</strong></td>
<td width="31%"><strong>1 3 5 4</strong></td>
<td width="23%"><strong>2 3 5 6</strong></td>
<td width="30%"></td>
</tr>
<tr>
<td width="13%"><strong>P4</strong></td>
<td width="31%"><strong>0 6 3 2</strong></td>
<td width="23%"><strong>0 6 5 2</strong></td>
<td width="30%"></td>
</tr>
<tr>
<td width="13%"><strong>P5</strong></td>
<td width="31%"><strong>0 0 1 4</strong></td>
<td width="23%"><strong>0 6 5 6</strong></td>
<td width="30%"></td>
</tr>
</tbody>
</table>
<p><strong> </strong></p>
<p><strong>Answer the following questions using the banker’s algorithm: </strong></p>
<p><strong>Calculate the Need Matrix?  </strong></p>
<p><strong>Demonstrate that system is safe or not. </strong></p>
<p><strong>Apply algorithm to identify which process will finished last.</strong></p>
<ol>
<li><strong> Explain the Reader Writer problem using suitable pseudocode.</strong></li>
</ol>
<p><strong> </strong></p>
<p><strong>Ans 1.</strong></p>
<p><strong>(a) Banker’s Algorithm</strong></p>
<p><strong>Step 1: Data and Need Matrix</strong></p>
<p><strong>Given:</strong></p>
<table width="100%">
<thead>
<tr>
<td width="13%">Process</td>
<td width="31%">Allocation (A B C D)</td>
<td width="23%">Max (A B C D)</td>
<td width="30%">Available (A B C D)</td>
</tr>
</thead>
<tbody>
<tr>
<td width="13%">P1</td>
<td width="31%">0 0 1 2</td>
<td width="23%">0 0 1 2</td>
<td width="30%">1 5 2 0</td>
</tr>
<tr>
<td width="13%">P2</td>
<td width="31%">1 0 0 0</td>
<td width="23%">1 7 5 0</td>
<td width="30%"> </td>
</tr>
<tr>
<td width="13%">P3</td>
<td width="31%">1 3 5 4</td>
<td width="23%">2 3 5 6</td>
<td width="30%"> </td>
</tr>
<tr>
<td width="13%">P4</td>
<td width="31%">0 6 3 2</td>
<td width="23%">0 6 5 2</td>
<td width="30%"> </td>
</tr>
<tr>
<td width="13%">P5</td>
<td width="31%">0 0 1 4</td>
<td width="23%">0 6 5 6</td>
<td width="30%"> </td>
</tr>
</tbody>
</table>
<p> </p>
<p>Initial Available = (1, 5, 2, 0).</p>
<p>Formula:</p>
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<p> </p>
<p> </p>
<p><strong>Q2. a. Why Multilevel paging is required.                          </strong></p>
<p><strong>Consider </strong></p>
<p><strong>Analyze the 2-level paging scheme in operating system, where given a 32-bit logical address space and a page size of 4 KB:</strong></p>
<ol>
<li><strong> Demonstrate how the logical address is divided into appropriate fields used in the 2-level paging. (4)</strong></li>
<li><strong> Illustrate the step-by-step translation process from a logical address to a physical address with a suitable example. (4)</strong></li>
<li><strong> Justify how this scheme improves memory management compared to single-level paging. (2)</strong></li>
<li><strong> Consider a paging system with the page table stored in memory. </strong></li>
<li><strong> If a memory reference takes 100 milliseconds, how long does a paged memory reference take?</strong></li>
<li><strong> If we add Translation Lookaside Buffer (TLB) s, and 75 percent of all page-table references are found in the TLBs, what is the effective memory reference time? If that finding a page-table entry in the TLBs takes zero time if the entry is there.</strong></li>
</ol>
<p><strong> </strong></p>
<p><strong>Ans 2.</strong></p>
<p><strong>(a) Multilevel Paging – 2-Level Scheme on 32-bit Address, 4 KB Page</strong></p>
<p><strong>Why Multilevel Paging?</strong></p>
<p>In a large virtual address space with single-level paging, the page table can be huge and must be kept in (or at least partly in) memory. Many entries may never be used. Multilevel paging breaks the page table into smaller pieces, loading only the required parts, which saves memory and supports sparse address spaces.</p>
<p><strong>(a) Dividing the 32-bit Logical Address </strong></p>
<p>Page size = 4 KB =  ⇒ offset = 12 bits.</p>
<p>Remaining bits for page number = 32 − 12 = 20 bits.</p>
<p> </p>
<p> </p>
<p> </p>
<p><strong>Q3. a. Explain in detail the process creation in UNIX/Linux using the fork() system call.<br>
Also describe the working of the exec() and wait() system calls.</strong></p>
<p><strong>Support your explanation with a neat and labelled diagram showing the relationship among these three system calls.</strong></p>
<p><strong> </strong></p>
<ol>
<li><strong> We have the following processes with their arrival times and burst times:</strong></li>
</ol>
<p><strong> </strong></p>
<table width="100%">
<tbody>
<tr>
<td width="30%"><strong>Process</strong></td>
<td width="33%"><strong>Arrival Time</strong></td>
<td width="35%"><strong>Burst Time</strong></td>
</tr>
<tr>
<td width="30%"><strong>P1</strong></td>
<td width="33%"><strong>0 ms</strong></td>
<td width="35%"><strong>5 ms</strong></td>
</tr>
<tr>
<td width="30%"><strong>P2</strong></td>
<td width="33%"><strong>1 ms</strong></td>
<td width="35%"><strong>3 ms</strong></td>
</tr>
<tr>
<td width="30%"><strong>P3</strong></td>
<td width="33%"><strong>2 ms</strong></td>
<td width="35%"><strong>8 ms</strong></td>
</tr>
<tr>
<td width="30%"><strong>P4</strong></td>
<td width="33%"><strong>3 ms</strong></td>
<td width="35%"><strong>6 ms</strong></td>
</tr>
</tbody>
</table>
<p><strong> </strong></p>
<p><strong>Calculate the Average waiting time and average turnaround time using Time Quantum 3 ms and Round Robin Scheduling Algorithm.      7+3\</strong></p>
<p><strong>Ans 3.</strong></p>
<p><strong>(a) Process Creation with fork(), exec() and wait() in UNIX/Linux</strong></p>
<p>In UNIX/Linux, new processes are created using the fork() system call. When a running process (parent) calls fork(), the kernel creates a new child process that is an almost exact copy of the parent. Both processes continue execution from the instruction following the fork() call. The return value distinguishes them: fork() returns 0 to the child and child PID to the parent. The child gets its own PID, separate address space, and copies of the parent’s code, data, and open file descriptors.</p>
<p>Often, the child does not want to keep running the same program; it wants to run a new one. For that, it uses an exec() family function (execl, execv, etc.). exec() replaces the child’s current memory image with a new program loaded from disk. The PID remains the same, but the code,</p>
<p> </p>
<p> </p>
<p><strong>SET-II</strong></p>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong>Q4. a. Explain the Disck architecture in detail</strong></p>
<ol start="67">
<li><strong> Consider a disk queue with requests for I/O to blocks on cylinders 98, 183, 41, 122, 14, 124, 65, 67. The C-SCAN scheduling algorithm is used. The head is initially at cylinder number 53 moving towards larger cylinder numbers on its servicing pass. The cylinders are numbered from 0 to 199. Calculate the total head movement (in number of cylinders) incurred while servicing these requests using following algorithms;</strong></li>
<li><strong> C-SCAN</strong></li>
<li><strong> SSTF 5+5 </strong></li>
</ol>
<p><strong>Ans 4.</strong></p>
<p><strong>(a) Explain the Disk Architecture </strong></p>
<p>A traditional magnetic disk consists of one or more platters mounted on a common spindle, rotating at a constant speed. Each platter surface is coated with magnetic material and logically divided into tracks, which are concentric circles. Tracks on all surfaces with the same radius form a cylinder. Every track is further divided into small arcs called sectors, which are the minimum addressable units on the disk and usually store 512 bytes or 4 KB of data. To read or write, the disk uses read–write heads, one per surface, mounted on an actuator arm assembly. All heads move together radially; the movement to position the head over the required cylinder is</p>
<p> </p>
<p> </p>
<p><strong>Q5. a. Discuss how file allocation methods affect file system performance and disk utilization.</strong></p>
<ol>
<li><strong> Differentiate between sequential, direct, and indexed file access methods. 5+5 </strong></li>
</ol>
<p><strong>Ans 5.</strong></p>
<ol>
<li><strong> How File Allocation Methods Affect Performance and Disk Utilization</strong></li>
</ol>
<p>The method a file system uses to allocate blocks on disk directly influences access speed, storage efficiency, fragmentation, and overall system responsiveness. Every file allocation technique—contiguous, linked, or indexed—carries a different strategy for mapping logical file blocks to physical disk blocks, and this choice determines both performance and disk utilization patterns. In contiguous allocation, the file occupies a sequence of consecutive blocks. This produces excellent read performance, particularly for large files, because the disk head moves linearly without repeated seeks. However, it tends to suffer from external fragmentation. As files grow or shrink, finding continuous free space becomes difficult, resulting in inefficient disk utilization</p>
<p> </p>
<p> </p>
<p><strong>Q6. a. Explain the sequence of actions taken by the operating system from issuing an I/O request to its completion.</strong></p>
<p><strong>Illustrate the steps with a neat diagram.</strong></p>
<ol>
<li><strong> Discuss various file consistency models used in Distributed File Systems.</strong></li>
</ol>
<p><strong>Explain the difference between strict, session, and loose consistency. 5+5          </strong></p>
<p><strong>Ans 6.</strong></p>
<ol>
<li><strong> OS Actions from I/O Request to Completion</strong></li>
</ol>
<p>When a process issues an I/O request, the operating system initiates a coordinated sequence of actions involving the device driver, disk controller, interrupt mechanism, and kernel-level data structures. First, the request enters the system-call interface, where parameters such as device ID, operation type, memory buffer address, and block numbers are validated. The OS then queues the request in the device’s I/O queue, applying scheduling algorithms such as FCFS or SCAN to optimize access. The device driver formats the request into device-specific commands and sends them to the disk controller. At this stage, the controller performs seek operations, waits for</p>
<p> </p>
</body>]]></content:encoded>
					
		
		
		<post-id xmlns="com-wordpress:feed-additions:1">4702</post-id>	</item>
		<item>
		<title>DCA2204 PRINCIPLES OF FINANCIAL ACCOUNTING AND MANAGEMENT SEPTEMBER 2025</title>
		<link>https://muj.assignmentsupport.in/product/dca2204-principles-of-financial-accounting-and-management-september-2025-2/</link>
		
		<dc:creator><![CDATA[dEEpak]]></dc:creator>
		<pubDate>Thu, 04 Dec 2025 15:36:05 +0000</pubDate>
				<guid isPermaLink="false">https://muj.assignmentsupport.in/?post_type=product&#038;p=4701</guid>

					<description><![CDATA[<strong><span lang="EN-IN">Match your questions with the sample provided in description</span></strong>

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										<content:encoded><![CDATA[<body><table width="99%">
<tbody>
<tr>
<td width="32%"><strong>SESSION</strong></td>
<td width="67%"><strong>JULY/SEPTEMBER 2025</strong></td>
</tr>
<tr>
<td width="32%"><strong>PROGRAM</strong></td>
<td width="67%"><strong>BACHELOR OF COMPUTER APPLICATIONS (BCA) / MASTER OF COMPUTER APPLICATIONS (MCA)</strong></td>
</tr>
<tr>
<td width="32%"><strong>SEMESTER</strong></td>
<td width="67%"><strong>VI</strong></td>
</tr>
<tr>
<td width="32%"><strong>COURSE CODE &amp; NAME</strong></td>
<td width="67%"><strong>DCA2204 PRINCIPLES OF FINANCIAL ACCOUNTING AND MANAGEMENT</strong></td>
</tr>
<tr>
<td width="32%"></td>
<td width="67%"></td>
</tr>
<tr>
<td width="32%"></td>
<td width="67%"></td>
</tr>
</tbody>
</table>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong>SET-I</strong></p>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong>Q1. a. Explain the dual aspect concept and the matching concept in accounting.</strong></p>
<ol>
<li><strong> Explain the importance of financial statements. 5 + 5 </strong></li>
</ol>
<p><strong>Ans 1.</strong></p>
<ol>
<li><strong> Dual Aspect Concept</strong></li>
</ol>
<p>The dual aspect concept is the foundation of modern double-entry accounting. It states that every business transaction has two aspects—a giving aspect and a receiving aspect—which must be recorded equally. This concept ensures that the accounting equation Assets = Liabilities + Capital always remains in balance. For example, when a business purchases machinery worth ₹50,000 for cash, there is an increase in assets (machinery) and a decrease in another asset (cash). Similarly, if goods are sold on credit, the business receives a debtor and gives away stock. This dual effect maintains accuracy, prevents manipulation, and ensures that financial books reflect the true financial position. It also forms the basis for preparing trial balances and financial</p>
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<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong>Q2. Record journal entries for the following transactions of Priya Traders:</strong></p>
<ol>
<li><strong> Commenced business with ₹1,20,000.</strong></li>
<li><strong> Bought furniture for ₹10,000 in cash.</strong></li>
<li><strong> Sold goods for ₹30,000 on credit to Anil.</strong></li>
<li><strong> Paid salary ₹5,000.</strong></li>
<li><strong> Received commission ₹2,000 by cheque 2*5 </strong></li>
</ol>
<p><strong>Ans 2.</strong></p>
<p><strong>Journal Entries of Priya Traders </strong></p>
<p><strong>JOURNAL OF PRIYA TRADERS</strong></p>
<table width="100%">
<tbody>
<tr>
<td width="9%"><strong>Date</strong></td>
<td width="52%"><strong>Particulars</strong></td>
<td width="8%"><strong>L.F.</strong></td>
<td width="14%"><strong>Debit (₹)</strong></td>
<td width="15%"><strong>Credit (₹)</strong></td>
</tr>
<tr>
<td width="9%">1</td>
<td width="52%"><strong>Cash A/c Dr</strong></td>
<td width="8%"></td>
<td width="14%">1,20,000</td>
<td width="15%"></td>
</tr>
<tr>
<td width="9%"></td>
<td width="52%"><strong>To Capital A/c</strong></td>
<td width="8%"></td>
<td width="14%"></td>
<td width="15%">1,20,000</td>
</tr>
<tr>
<td width="9%"></td>
<td width="52%">(Being business commenced with cash)</td>
<td width="8%"></td>
<td width="14%"></td>
<td width="15%"></td>
</tr>
<tr>
<td width="9%">—</td>
<td width="52%">—</td>
<td width="8%">—</td>
<td width="14%">—</td>
<td width="15%">—</td>
</tr>
<tr>
<td width="9%">2</td>
<td width="52%"><strong>Furniture A/c Dr</strong></td>
<td width="8%"></td>
<td width="14%">10,000</td>
<td width="15%"></td>
</tr>
</tbody>
</table>
<p> </p>
<p> </p>
<p><strong>Q3a. Explain the steps in financial planning</strong></p>
<ol>
<li><strong> Explain factors affecting the Capital structure. 5+5 </strong></li>
</ol>
<p><strong>Ans 3.</strong></p>
<p><strong>(a) Steps in Financial Planning</strong></p>
<p>Financial planning is the systematic estimation and arrangement of financial resources to ensure smooth business operations. It ensures that adequate funds are available at the right time to support both short-term and long-term business goals.</p>
<ol>
<li><strong> Estimating Financial Requirements</strong></li>
</ol>
<p>The first step is identifying how much money the business needs for fixed assets (machinery,</p>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p> </p>
<p><strong>SET-II</strong></p>
<p><strong> </strong></p>
<p><strong> </strong></p>
<p><strong>Q4. From the following information, calculate:</strong></p>
<ol>
<li><strong> Contribution per unit and Profit/Volume (P/V) Ratio</strong></li>
<li><strong> Break-even point (in units)</strong></li>
</ol>
<p><strong>Selling Price per unit: ₹50</strong></p>
<p><strong>Variable Cost per unit: ₹35</strong></p>
<p><strong>Fixed Cost: ₹45,000   </strong></p>
<p><strong>Ans 4.</strong></p>
<p><strong>Calculate Contribution per Unit, P/V Ratio, and Break-Even Point</strong></p>
<p><strong>Given:</strong></p>
<ul>
<li>Selling Price per unit (SP) = ₹50</li>
<li>Variable Cost per unit (VC) = ₹35</li>
<li>Fixed Cost (FC) = ₹45,000</li>
</ul>
<p>(a) Contribution per Unit &amp; P/V Ratio</p>
<p><strong>Step 1: Contribution per Unit</strong></p>
<p>Formula</p>
<p> </p>
<p> </p>
<p><strong>Q5.a. Explain the meaning and objectives of budgetary control.</strong></p>
<ol>
<li><strong> Discuss its advantages and limitations. 4+6 </strong></li>
</ol>
<p><strong>Ans 5.</strong></p>
<p><strong>(a) Meaning and Objectives of Budgetary Control</strong></p>
<p><strong>Budgetary Control</strong></p>
<p>Budgetary control is a systematic managerial technique that involves preparing budgets for various departments, comparing actual performance with these budgeted figures, and taking corrective actions wherever necessary. It ensures that every activity operates within planned limits and helps management coordinate future operations. Through continuous monitoring, budgetary control provides a structured approach for planning, executing, and evaluating financial and operational performance.</p>
<p><strong>Objectives of Budgetary Control</strong></p>
<p>The primary objective of budgetary control is planned operations, ensuring that all future activities are clearly defined and financial requirements are estimated in advance. Another major</p>
<p> </p>
<p> </p>
<p><strong>Q6. The summarized final accounts of two companies are as follows:</strong></p>
<p><strong>Balance Sheet </strong></p>
<table width="100%">
<tbody>
<tr>
<td width="20%"><strong>Liabilities </strong></td>
<td width="13%"><strong>X Ltd </strong>
<p><strong>Rs. </strong></p></td>
<td width="13%"><strong>Y Ltd</strong>
<p><strong>Rs. </strong></p></td>
<td width="23%"><strong>Assets </strong></td>
<td width="14%"><strong>X Ltd</strong>
<p><strong>Rs.</strong></p></td>
<td width="14%"><strong>Y Ltd</strong>
<p><strong>Rs. </strong></p></td>
</tr>
<tr>
<td width="20%"><strong>Share Capital </strong>
<p><strong>Reserves</strong></p>
<p><strong>8% debentures </strong></p></td>
<td width="13%"><strong>88,000</strong>
<p><strong>42,900</strong></p>
<p><strong>22,000</strong></p></td>
<td width="13%"><strong>88,000</strong>
<p><strong>35,200</strong></p>
<p><strong>22,000</strong></p></td>
<td width="23%"><strong>Fixed Assets</strong>
<p><strong>Current Assets</strong></p>
<p><strong>Less: Current Liabilities </strong></p></td>
<td width="14%"><strong>1,21,000</strong>
<p><strong>1,25,400</strong></p>
<p><strong>93,500</strong></p></td>
<td width="14%"><strong>96,800</strong>
<p><strong>1,03,400</strong></p>
<p><strong>55,000</strong></p></td>
</tr>
<tr>
<td width="20%"><strong> </strong></td>
<td width="13%"><strong>1,52,900</strong></td>
<td width="13%"><strong>1,45,200</strong></td>
<td width="23%"><strong> </strong></td>
<td width="14%"><strong>1,52,900</strong></td>
<td width="14%"><strong>1,45,200</strong></td>
</tr>
</tbody>
</table>
<p><strong> </strong></p>
<p><strong>Revenue Statement for the year</strong></p>
<p><strong> </strong></p>
<table width="100%">
<tbody>
<tr>
<td width="63%"><strong>Particulars</strong></td>
<td width="16%"><strong>X Ltd (Rs.)</strong></td>
<td width="19%"><strong>Y Ltd. (Rs.)</strong></td>
</tr>
<tr>
<td width="63%"><strong>Sales </strong>
<p><strong>Less: Cost of Sales </strong></p>
<p><strong>Gross Profit </strong></p>
<p><strong>Less: Operating Expenses </strong></p>
<p><strong>Net Profit before Tax </strong></p>
<p><strong>Less: Tax </strong></p>
<p><strong>Profit after Tax </strong></p></td>
<td width="16%"><strong>3,30,000</strong>
<p><strong>2,37,600</strong></p>
<p><strong>92,400</strong></p>
<p><strong>63,800</strong></p>
<p><strong>28,600</strong></p>
<p><strong>12,100</strong></p>
<p><strong>16,500</strong></p></td>
<td width="19%"><strong>2,64,000</strong>
<p><strong>1,98,000</strong></p>
<p><strong>66,000</strong></p>
<p><strong>44,000</strong></p>
<p><strong>22,000</strong></p>
<p><strong>9,240</strong></p>
<p><strong>12,760</strong></p></td>
</tr>
</tbody>
</table>
<p><strong> </strong></p>
<p><strong>You are required to calculate the following ratios for X and Y Ltd.</strong></p>
<p><strong>(1) Current Ratio (2) Capital gearing ratio (3) Gross profit ratio (4) Net profit ratio</strong></p>
<p><strong>Ans 6.</strong></p>
<p>Ratio for <strong>X Ltd</strong> and <strong>Y Ltd</strong>.</p>
<p> </p>
<ol>
<li>Current Ratio</li>
</ol>
<p>Formula</p>
<p>X Ltd</p>
<ul>
<li>Current Assets = ₹1,25,400</li>
<li>Current Liabilities = ₹93,500</li>
</ul>
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