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**Frank Learning Maths Class 5 Solutions Chapter 2 Roman Numerals**

Welcome to NCTB Solution. Here with this post we are going to help 5th class students for the Solutions of Frank Learning Maths Class 5 Book, Chapter 2 Roman Numerals. Here students can easily find step by step solutions of all the problems for Roman Numerals. Here students will find solutions for Exercise 2.1. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

**Roman Numerals Exercise 2.1 Solution**

**Question no – (1) **

**Solution : **

**(a)** The value of V is = 5

**(b)** The value of X is = 10

**(c)** The value of L is = 50

**(d)** The value of C is = 100

**(e)** The value of D is = 500

**(f)** The value of M is = 1000

**(g)** The value of MM is = 2000

**(h)** The value of CCC is = 300

**Question no – (2)**

**Solution : **

**(a) XLV**

= (50 – 10) + 5

= 45

**(b) XLIX**

= 950 – 10) + 10 – 1)

= 40 + 9

= 45

**(c) LVIII**

= 50 + 8

= 58

**(d) LXXX**

= 50 + 10 + 10 + 10

= 80

**(e) LXXXVIII**

= 80 + 8

= 88

**(f) CCXC**

= 200 + (100 – 10)

= 290

**(g) CCC** = 300

**(h) D** = 500

**(i) CMXII**

= (100 – 100) + 12

= 912

**(j) MMDCXXIX**

= 2000 + 500 + 100 + 20 + 9

= 2629

**(k) MMXLIII**

= 2000 + 50 – 10 + 3

= 2043

**(l) MMMDCCCL**

= 300 + 500 + 300 + 50

= 3850

**Question**** no – (3)**

**Solution : **

**(a) 55 = LV**

= (50 + 5) = L + V

= LV

**(b) 80 = LXXX**

= (50 + 10 + 10 + 10)

= L + X + X + X

= LXXX

**(c) 83 = LXXXIII**

= (50 + 10 + 10 + 10 + 3)

= L + X + X + X + III

= LXXXIII

**(d) 99 = XCIX**

= (90 + 9)

= XC + IX

= XCIX

**(e) 103 = CIII**

= (100 + 3)

= C + III

= CIII

**(f) 305 = CCCV**

= (100 + 100 + 100 + V)

= C + C + C + V

= CCCV

**(g) 600 = DC**

= (500 + 100)

= DC

**(h) 801 = DCCCI**

= (500 + 100 + 100 + 100 + 1)

= D + C + C + C + I

= DCCCI

**(i) 1043 = MXLIII**

= (1000 + 40 + 3)

= M + XL + III

= MXLIII

**(j) 2560 = MMDLX**

= 1000 + 1000 + 500 + 60

= M + M + D + L + X

= MMDLX

**(k) 3211 = MMMCCXI**

= 1000 + 1000 + 1000 + 100 + 100 + 11

= M + M + M + C + C + XI

**(l) 3872 = MMMDCCCLXXII**

= (3000 + 800 + 70 + 2)

= 1000 + 1000 + 1000 + 500 + 100 + 100 + 100 70 + 2

= M + M + M + D + C + C + C + LXX + II

= MMMDCCCLXXII

**Question no – (4) **

**Solution : **

**(a)** XL/40 **<** LIII/53

**(b)** LXXIX/79 **<** XC/90

**(c)** XCC III/193 **<** CXIII/113

**(d)** CXX/120 **>** CXIX/119

**(e)** MC/1900 **>** LMXXX/1080

**(f)** CMXXIL **<** CMCXVII

**Question no – (5) **

**Solution : **

**(a)** M can be repeated thrice – **True.**

= 1000 + 1000 + 1000

= M + M + M

**∴** MMM = 30000

**(b)** C stands for 100 – True

= C = 100

**(c)** IX is bigger than X – **False.**

= IX = 9

= X = 10

**(d)** 1100 can be written as MC – **True.**

= 1000 + 100

= M + C

**∴** MC = 1100

**Question no – (6) **

**Solution :**

**(a)** XXXIII + __XVII__ = L

**(b)** LXXX – __XXI__ = LIX

**(c)** CXXV + __CXXIX__ = CXXIX

**(d)** CXXI – __LXI__ = LX

**(e)** LXIII + __XXVII/27__ = C

**Question**** no – (7) **

**Solution :**

**(a) XXXVIII = 38**

= XXV = 25

**∴** (38 + 25 ) = 63

= 63 = LXIII

Therefore, the sum of XXXVIII and XXV is LXIII.

**(b) LXXXVI = 86**

XI = 11

**∴** (86 + 11) = 97

= 97 = XCVII

Hence, the sum of LXXXVI and XI is XCVII.

**(c) XC = 110**

CX = 90

**∴** (110 – 90) = 20

**∴** 20 = XX

Thus, the difference of XC and CX is XX.

**(d) XXIII = 23**

XXVII = 27

**∴** (27 – 23) = 4

**∴** 4 = IV

So, the difference of XXIII and XXVII is IV

**(e) CC = 200**

VI = 6

**∴** (200 × 6) = 1200

**∴** 1200 = MCC

Therefore, the product of CC and VI is MCC.

**(f) LXI = 61**

XVII = 17

**∴** (61 × 17) = 1037

**∴** 1037 = MXXXVII

Thus, the product of LXI and XVII is MXXXVII.

**Previous Chapter Solution : **

👉 Chapter 1 👈