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CAT 2025 Lesson : Number Systems - Roman Numerals

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So far, we have worked with the Base 1010 number system. In this lesson, we will learn about other number systems.

1. Roman Numerals

This numeric system originated in ancient Rome over
20002000 years ago. Instead of using separate symbols (like in our decimal system) to represent numbers, they utilised the following letters of the Latin alphabet to represent these values.

Symbol Value
I 11
V 55
X 1010
L 5050
C 100100
D 500500
M 10001000

Numbers were expressed by writing these symbols together. The following algorithm/rules are applied to compute the value of the number.

1) The value of a symbol is subtracted, if it is written to the left of a higher-valued symbol.
2) If not, the value of the symbol is added.

For example,
VIII
=5+1+1+1=8= 5 + 1 + 1 + 1 = 8
IX
=1+10=9= -1 + 10 = 9
XLIV
=10+501+5=44= -10 + 50 -1 + 5 = 44
CDLVII
=100+500+50+5+1+1=457= -100 + 500 + 50 + 5 + 1 + 1 = 457

Note that
4949 is written as XLIX and not IL. This is because of the following additional rules.

1) I can be written before V or X only
2) X can be written before L or C only.
3) C can be written before D or M only.

In Roman Numerals,
00 or negative numbers cannot be written and decimals cannot be represented. Large numbers are difficult to write. For instance, the distance between sun and the earth is written as 150,000,000150,000,000 km in the decimal system. To write this in Roman Numerals, the symbol M has to be written 150,000150,000 times.

Upon the advent of the decimal system, Roman Numerals for mathematical operations ceased. Today we use Roman Numerals to indicate smaller numbers, such as the hours in a clock, the class in school, movie sequel numbers, book volumes, etc.

Example 1

Which of the following equals the sum of the roman numbers CDLVI and CCXXXIX?
(11) CDXCVI           (22) DCCXV           (33) DCXCV           (44) DCCXXI          

Solution

CDLVI
=100+500+50+5+1=456= -100 + 500 + 50 + 5 + 1 = 456
CCXXXIX
=100+100+10+10+101+10=239= 100 + 100 + 10 + 10 + 10 - 1 + 10 = 239

456+239=695456 + 239 = \textbf{695}

We do not consider options
11 and 44 as their last digits are 66 and 11 respectively.

Option 2: DCCXV
=500+100+100+10+5=715= 500 + 100 + 100 + 10 + 5 = 715

Option 3: DCXCV
=500+10010+100+5=695= 500 + 100 - 10 + 100 + 5 = \textbf{695}

Answer: (3) DCXCV


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