Frank Learning Maths Class 5 Solutions Chapter 2

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.

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Updated: June 6, 2023 — 9:50 am