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B.Com. LL.B. (Five Years) Sem 3 Business Statistics 2025 Jun Question Paper with Solutions

KSLU Solved Question Papers

Business Statistics

Previous Year Question Paper with Solution

B.Com. LL.B. (Five Years) · Sem 3

2025 Jun Examination

munotes.in

Mumbai

First published on munotes.in on 21 October 2026.

Published by munotes.in, Mumbai.

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The questions below are the paper as Karnataka State Law University set it at the 2025 Jun examination, in the order it was set.

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The Paper as Set

The questions in this volume are the questions asked at the 2025 Jun examination, reproduced as Karnataka State Law University set them, in the order it set them. Long reading passages are shortened to their first and last lines and our summary; the original paper prints them in full. Only the answers are ours. See the original question paper.

Duration 3 hours · Total marks 80 · 20 questions answered

How to use this volume

Solve the paper first, under exam conditions and against the clock. Then read the answers here and mark your own. Reading a solution before attempting the question feels productive and teaches very little, because recognising an answer is not the same as being able to write one.

Read the original question paper

Q1

Unit I

(a)Draw less than and more than Ogives for the following data and determine the median graphically.[10]

Answer

The figures given in the question

Class Interval	Frequency
10-20	2
20-30	5
30-40	7
40-50	12
50-60	9
60-70	4

For full marks, cover: the less than and more than cumulative frequencies; the two sets of plotting points; how the ogives are drawn on one graph; the median read at the point where they cross; and the median by formula as a check on the reading.

An ogive is a cumulative frequency curve. The less than ogive shows how many items lie below each upper class limit. The more than ogive shows how many lie above each lower class limit. Drawn on one graph, the two curves cross at the median.

Step 1: the cumulative frequencies

Class	f	less than cf	more than cf
10-20	2	2	39
20-30	5	7	37
30-40	7	14	32
40-50	12	26	25
50-60	9	35	13
60-70	4	39	4
Total	39

The less than column is built by adding from the top, and the more than column by adding from the bottom. Each reaches N = 39 at its far end.

Step 2: the plotting points

Less than ogive: upper limit	Items below it	More than ogive: lower limit	Items above it
10	0	10	39
20	2	20	37
30	7	30	32
40	14	40	25
50	26	50	13
60	35	60	4
70	39	70	0

Step 3: drawing the ogives

On the X axis mark the class limits from 10 to 70, with 1 cm for 10 units.
On the Y axis mark the cumulative frequencies from 0 to 40, with 1 cm for 5 items.
Plot the less than points and join them by a smooth curve rising from (10, 0) to (70, 39).
Plot the more than points on the same axes and join them by a smooth curve falling from (10, 39) to (70, 0).
From the point where the curves cross, drop a perpendicular to the X axis. Its foot is the median.

Step 4: reading the median

The curves cross between 40 and 50. In that class the less than curve rises from 14 to 26, and the more than curve falls from 25 to 13. Each changes by 12 over the width of 10. Measured t units to the right of 40, their heights are 14 + 1.2t and 25 - 1.2t. These are equal when 2.4t = 11, that is when t = 4.58.

So the curves cross at about X = 44.6, at a height of 19.5, which is N/2. The median read from the graph is about 44.6.

Step 5: the median by formula, as a check

N/2 = 39/2 = 19.5. The less than cf first passes 19.5 in the class 40-50, with 14 items before it.

Median = L + ((N/2 - cf)/f) × h = 40 + ((19.5 - 14)/12) × 10 = 40 + 4.58 = 44.58

This agrees with the graph.

Conclusion. The less than and more than ogives cross at a height of 19.5 (half of N = 39), above the value 44.58 on the X axis. So the median determined graphically is 44.58, which the formula 40 + (5.5/12) × 10 confirms.

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