Figure it Out — The Roman Number System — A Story of Numbers

1. 14 marksGanita Prakash Cl-8 Part 1, Figure it Out, page 59

   Represent the following numbers in the Roman system: (i) 1222 (ii) 2999 (iii) 302 (iv) 715

   Hint. Break the number into landmark numbers M(1000), D(500), C(100), L(50), X(10), V(5), I(1), taking as many of the largest as possible first.

   The method is to peel off landmark numbers from the largest downwards, taking as many of each as possible, and to use the subtractive shortcut (IV, IX, XL, XC, CD, CM) where four of a symbol would otherwise be needed.

   (i) 1222 1222 = 1000 + 100 + 100 + 10 + 10 + 1 + 1 = M + CC + XX + II = MCCXXII

   (ii) 2999 2999 = 1000 + 1000 + 900 + 90 + 9 Here 900 is written as CM (100 less than 1000), 90 as XC (10 less than 100) and 9 as IX (1 less than 10), since each of these would otherwise need four repeated symbols. = MM + CM + XC + IX = MMCMXCIX

   (iii) 302 302 = 100 + 100 + 100 + 1 + 1 = CCC + II = CCCII (There is no 50 or 10 in this number, so no L or X appears at all.)

   (iv) 715 715 = 500 + 100 + 100 + 10 + 5 = D + CC + X + V = DCCXV

   Notice how 2999 needs nine symbols while the Hindu numeral needs four — this bulk is exactly the weakness of a system without place value.

   ✦ (i) MCCXXII (ii) MMCMXCIX (iii) CCCII (iv) DCCXV

2. 23 marksGanita Prakash Cl-8 Part 1, worked example, page 59

   Add CCXXXII + CCCCXIII without converting them to Hindu numerals.

   Hint. Count how many of each symbol you have altogether, then regroup — remember five Cs make a D.

   Step 1 — Pool the symbols of each kind. CCXXXII has 2 Cs, 3 Xs and 2 Is. CCCCXIII has 4 Cs, 1 X and 3 Is. Together: 6 Cs, 4 Xs, 5 Is.

   Step 2 — Regroup, starting from the largest. It looks at first as though C is the largest landmark present, but 5 Cs (500) make a D. So of the 6 Cs, five become one D and one C is left: 6 C → D + C.

   Step 3 — Regroup the smaller symbols. 5 Is make a V, so the 5 Is become V. The 4 Xs are written XL, since 40 is conventionally shown as 10 less than 50.

   Step 4 — Assemble from largest to smallest. D + C + XL + V = DCXLV

   Check in Hindu numerals: 232 + 413 = 645, and DCXLV = 500 + 100 + 40 + 5 = 645 ✓

   The whole difficulty here is that the regrouping size keeps changing — five Is make a V but two Vs make an X — whereas in a base system every regrouping is by the same number.

   ✦ CCXXXII + CCCCXIII = DCXLV (that is, 232 + 413 = 645)

3. 33 marksGanita Prakash Cl-8 Part 1, 'Do it yourself', page 60

   Add LXXXVII + LXXVIII without converting them to Hindu numerals.

   Hint. Pool the Ls, Xs, Vs and Is separately, then regroup: two Vs make an X, five Xs make an L, two Ls make a C.

   Step 1 — Count how many of each symbol the two numerals contribute. LXXXVII has 1 L, 3 Xs, 1 V, 2 Is. LXXVIII has 1 L, 2 Xs, 1 V, 3 Is. Together: 2 Ls, 5 Xs, 2 Vs, 5 Is.

   Step 2 — Regroup from the smallest upwards. 5 Is make a V, so the 5 Is become one V. That V joins the 2 Vs already there, giving 3 Vs. 2 of those Vs make an X, leaving 1 V over. So we now have 6 Xs (the original 5 plus this new one) and 1 V.

   Step 3 — Continue regrouping. 5 of the 6 Xs make an L, leaving 1 X. That new L joins the 2 Ls already there, giving 3 Ls. 2 of those Ls make a C, leaving 1 L.

   Step 4 — Assemble what remains: 1 C, 1 L, 1 X, 1 V. = CLXV

   Check in Hindu numerals: 87 + 78 = 165, and CLXV = 100 + 50 + 10 + 5 = 165 ✓

   Note how the carrying rule alternated between 5 and 2 at every step, which is precisely why Roman arithmetic is so error-prone.

   ✦ LXXXVII + LXXVIII = CLXV (that is, 87 + 78 = 165)

4. 44 marksGanita Prakash Cl-8 Part 1, Math Talk, page 60

   How will you multiply two numbers given in Roman numerals, without converting them to Hindu numerals? Find the product of the following pairs of landmark numbers: V × L, L × D, V × D, VII × IX.

   Hint. Treat the product as repeated addition of one landmark number, then regroup the result into the largest landmark numbers available.

   Without place value, the only way to multiply is repeated addition followed by regrouping into landmark numbers.

   V × L — five Ls, that is L + L + L + L + L. Two Ls make a C, so four of them make CC, and one L is left over. = CCL (checking: 5 × 50 = 250 ✓)

   L × D — fifty Ds, that is D added fifty times. Two Ds make an M, so fifty Ds regroup into twenty-five Ms. = MMM…M, with M written 25 times (checking: 50 × 500 = 25000 ✓) This answer is itself the point of the exercise: the Roman system has no symbol above M, so a perfectly ordinary product has to be written as a string of twenty-five identical symbols.

   V × D — five Ds. Two Ds make an M, so four of them make MM, and one D is left over. = MMD (checking: 5 × 500 = 2500 ✓)

   VII × IX — seven added nine times. VII + VII = XIV; building up to nine sevens gives sixty-three, which regroups as 50 + 10 + 3. = LXIII (checking: 7 × 9 = 63 ✓)

   The contrast with a base system is stark: there, multiplying two landmark numbers always gives another landmark number straight away, because the landmarks are powers of one fixed number.

   ✦ V × L = CCL; L × D = M written 25 times; V × D = MMD; VII × IX = LXIII

5. 53 marksGanita Prakash Cl-8 Part 1, 'Try This', page 60

   Multiply CCXXXI and MDCCCLII. What does this tell you about the Roman system?

   Hint. First read off what each numeral is worth, then think about how many symbols the answer would need.

   Step 1 — Read the two numerals. CCXXXI = 100 + 100 + 10 + 10 + 10 + 1 = 231. MDCCCLII = 1000 + 500 + 100 + 100 + 100 + 50 + 1 + 1 = 1852.

   Step 2 — Find the product. 231 × 1852 = 231 × 1800 + 231 × 52 = 415800 + 12012 = 427812

   Step 3 — Now try to write it in Roman numerals. The largest Roman symbol is M = 1000, so 427812 would need M repeated 427 times, followed by DCCCXII. That is well over four hundred symbols for a single number.

   Step 4 — What the exercise is really showing. Even doing the multiplication inside the Roman system is close to impossible, because there is no way to break the work into digit-by-digit steps — every partial product has to be built by repeated addition and then regrouped by hand, and the regrouping size keeps switching between 5 and 2.

   This is why people using Roman numerals relied on an abacus for calculation, and why the arrival of the Hindu system, where the same product takes a few lines of ordinary long multiplication, was such a transformation.

   ✦ CCXXXI × MDCCCLII = 231 × 1852 = 427812, which in Roman numerals would need M repeated 427 times followed by DCCCXII — over four hundred symbols for one number.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp103.pdf). This chapter is a history of number *systems* — tally marks, Roman, Egyptian, base-5, Mesopotamian, Mayan, Chinese rod and Hindu numerals — not a chapter on rational numbers. Questions appear as numbered 'Figure it Out' blocks plus 'Math Talk'/'Try This' prompts in the running text; every numeric answer here is checked against the book's own printed answer key at the end of the chapter. Four sub-parts whose questions exist only as printed Egyptian/base-5 glyph images (the addition drills on pages 65 and the products on page 68) are deliberately omitted rather than guessed at, since the operands cannot be recovered from the text.. Questions are referenced from the NCERT textbook for identification.