Roman Numerals

The Roman numerals is the numeral system of ancient Rome. It uses combinations of letters from the Latin alphabet to signify values. The numbers 1 to 10 can be expressed in Roman numerals as follows:

I, II, III, IV, V, VI, VII, VIII, IX, and X.

This followed by XI for 11, XII for 12, ... tilll XX for 20. Some more roman numerals are :

1.0Rules to Form Roman Numerals

We can form different roman numerals using the symbols and the following rules. 

Rule 1

If a symbol is repeated one after the other, its value is added as many times as it occurs. For example,

III = 1 + 1 + 1 = 3

XX = 10 + 10 = 20

Rule 2

The symbols I, X, C and M can be repeated up to a maximum of three times. 

For example,

I = 1, II = 2, III = 3

X = 10, XX = 20, XXX = 30

C = 100, CC = 200, CCC = 300

M = 1000, MM = 2000, MMM = 3000

Rule 3

The symbols V, L and D (i.e., 5, 50, and 500, respectively) can never be repeated in a roman numeral.

Rule 4

If a symbol with a smaller value is written on the right of a symbol with a greater value, then its value is added to the value of the greater symbol. 

For example,

XII = 10 + 2 = 12, LX = 50 + 10 = 60, 

DCCCX = 500 + 300 + 10 = 810

Rule 5

If a symbol with a smaller value is written on the left of symbol with a larger value, then its value is subtracted from the value of the greater symbol. 

For example, 

IV = 5 – 1 = 4, IX = 10 – 1 = 9, CD = 500 – 100 = 400

VI = 5 + 1 = 6, XI = 10 + 1 = 11, DC = 500 + 100 = 600

2.0Important Points

• V, L and D are never subtracted. 
• I can be subtracted from V and X once only. 
• X can be subtracted from L and C once only.
• C can be subtracted from D and M once only. Thus, I or V is never written to the left of L or C. 
• L is never written to the left of C. 
• There is another symbol ”–“ which is called bar. If bar is place over a numeral, it is multiplied by 1000. Thus, and stand for 5000 and 12000 respectively. 
• With the help of the symbols I, V, X, L, C we can write numbers upto 399. 

3.0Fun Facts

Zero is the only Number that can’t be represented in Roman Numerals.

Romans Numerals were invented as a means of trading.

4.0Use of Brackets

Raju brought 6 pencils from the market, each at Rs. 2 His brother Ramu also bought 8 pencils of the same type. Raju and Ramu both calculated the total cost but in their own ways. Raju found that they both spent Rs. 28 and he used the following method:

(6 × 2) + (8 × 2)

= (12 + 16)

= 28

Here number of operations are two times multiplication and one time addition

But Ramu found an easier way. He did 6 + 8 = 14 and then (2 × 14) = 28. The use of brackets makes this sum easy. It can be done as follows :

Rs. 2 × (6 + 8)

= Rs. (2 × 14)

= Rs. 28

Here first solve the operation inside the bracket and then multiply it by the number outside.

Now number of operations are one addition and one multiplication.

So, second method takes less time.

5.0BODMAS Explanation

Note 

Start Divide/Multiply from left side to right side since they perform equally.

Start Add/Subtract from left side to right side since they perform equally.

A vinculum is a horizontal line which is placed over a mathematical expression to indicate that the expression is to be considered as a group. 

Fun Fact

18 is the only number that is twice the sum of its digits

6.0Solved Examples

1. Simplify: 78 – [5 + 3 of (25 – 2 × 10)]

Solution

78 – [5 + 3 of (25 – 2 × 10)]

= 78 – [5 + 3 of (25 – 20)] (Simplifying ‘multiplication’ 2 × 10 = 20)

= 78 – [5 + 3 of 5] (Simplifying ‘subtraction’ 25 – 20 = 5)

= 78 – [5 + 3 × 5] (Simplifying ‘of’)

= 78 – [5 + 15] (Simplifying ‘multiplication’ 3 × 5 = 15)

= 78 – 20 (Simplifying ‘addition’ 5 + 15 = 20)

= 58 (Simplifying ‘subtraction’ 78 – 20 = 58)

2. Write the following in Hindu-Arabic numerals:

(i) XLV

(ii) LXIII

(iii) LXXVI

(iv) XCII

(v) XXXVIII

Solution

(i) XLV = XL + V = (50 – 10) + 5 = 40 + 5 = 45

(ii) LXIII = L + X + III = 50 + 10 + 3 = 63

(iii) LXXVI = L + XX + VI= 50 + (2 × 10) + 6 = 76

(iv) XCII = XC + II = (100 – 10) + 2 = 90 + 2 = 92

(v) XXXVIII = XXX + VIII = (3 × 10) + 8 = 30 + 8 = 38

3. Simplify: 17 – [19 – {(11 – 7 – 4)}]

Solution

= 17 – [19 – {(11 – 7 – 4)}]

= 17 – [19 – {(11 – 3)}]

= 17 – [19 – 8]

= 17 – 11 = 6

4. Set the following numbers 7, 9, 14, 18, 54 which will make the given number sentence true?

……×......+......÷……–……= 86

Solution

Here, 7 × 14 + 54 ÷ 9 – 18

= 7 × 14 + 6 – 18

= 98 + 6 – 18

= 104 – 18

= 86

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