CMDLearn

Learn

Class 8 Maths: A Story of Numbers, 8 practice questions with answers

CMDLearn ·

These are 8 practice questions on Chapter 3, A Story of Numbers, of the new NCERT Class 8 Maths textbook, Ganita Prakash, Part 1. Each has a worked answer and the textbook page it comes from. The chapter's own exercises mostly ask you to write numbers in old systems, so these questions go the other way: they ask you to read old numerals, count how many symbols a system needs, read counters on an abacus, and spot mistakes. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and none of the numbers is taken from the chapter, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Ganita Prakash, Grade 8, Part 1, Chapter 3, printed pages 48 to 81):

Roman and Egyptian numerals

Question 1. A stone above an old school gate says it was built in MDCCLXIII. Which year is that?

Answer: 1763. Read the symbols from the largest: M = 1000, D = 500, C = 100 twice, L = 50, X = 10, and I = 1 three times. So MDCCLXIII = 1000 + 500 + 100 + 100 + 50 + 10 + 1 + 1 + 1 = 1763. No smaller symbol stands before a larger one, so there is nothing to subtract. (Roman landmark numbers, page 58; grouping, page 59.)

Question 2. Sita writes 99 as IC and 98 as IIC. Using the chapter's way of writing 4 as IV and 40 as XL, explain what is wrong and write both numbers correctly.

Answer: 99 = XCIX and 98 = XCVIII. In the chapter's examples, a smaller symbol is put in front only of the next one or two landmark numbers: I before V (4) and X (9), and X before L (40). Following the same pattern, 90 is X before C, as in the chapter's answer key, which writes 2999 as MMCMXCIX. Taking I away from C is not part of that pattern. So group 99 as 90 + 9 = XC + IX = XCIX, and 98 as 90 + 8 = XC + VIII = XCVIII. (Writing 4 and 40 as IV and XL, page 58; the table of numerals, page 53; the chapter's answer key writes 2999 as MMCMXCIX, with XC for 90.)

Question 3. How many symbols does the Egyptian numeral for 9973 use? How many does the Hindu numeral use?

Answer: 28 symbols in Egyptian; 4 digits in Hindu. The Egyptian system repeats the symbol for each power of 10: 9 thousand-symbols, 9 hundred-symbols, 7 ten-symbols and 3 one-symbols, which is 9 + 9 + 7 + 3 = 28 symbols. The Hindu system writes one digit for each place: 9, 9, 7, 3, so 4 digits. Place value is what saves the extra symbols. (Egyptian numerals, pages 61 and 62; the Hindu number system, page 78.)

Bases and place value

Question 4. In the chapter's base-5 system, a numeral shows 2 symbols for 125, 3 symbols for 25, none for 5 and 4 symbols for 1. What number is it?

Answer: 329. In base 5 the landmark numbers are 1, 5, 25 and 125. So the number is 2 × 125 + 3 × 25 + 0 × 5 + 4 × 1 = 250 + 75 + 0 + 4 = 329. (Base-5 system and landmark numbers, pages 62 and 63.)

Question 5. A Mesopotamian numeral has 3 in the 3600s place, 37 in the 60s place and 45 in the 1s place. What number is it?

Answer: 13065. The places are worth 3600, 60 and 1. So the number is 3 × 3600 + 37 × 60 + 45 = 10800 + 2220 + 45 = 13065. Each place holds a number from 1 to 59, as in the chapter. (The base-60 place value system, pages 70 to 73.)

Question 6. Ravi reads a Mayan numeral that has, from top to bottom, 3 in the top place, 7 in the middle place and 9 at the bottom, as 3 × 400 + 7 × 20 + 9 = 1349. What mistake has he made, and what is the right number?

Answer: the top place is worth 360, not 400, so the number is 1229. Ravi treated the Mayan system as a true base-20 system, where the third landmark number would be 20 × 20 = 400. But the Mayans' third landmark number is 360, with the 1s at the bottom and the 20s above them. So the number is 3 × 360 + 7 × 20 + 9 = 1080 + 140 + 9 = 1229. (The Mayan number system and its third landmark number 360, page 76.)

Question 7. The later Mesopotamians used a placeholder symbol for a blank place, but mainly in the middle of a numeral, not at its end. Which of these numbers could such a scribe write without any confusion, and which not: 187, 10807, 180? Explain.

Answer: 187 and 10807 are clear; 180 is not. 187 = 3 × 60 + 7 needs no blank place. 10807 = 3 × 3600 + 0 × 60 + 7 has its blank in the middle, so the placeholder goes between the 3 and the 7 and the numeral is clear. But 180 = 3 × 60 + 0 has its blank at the end, where the placeholder was not used, so it looks just like 3. Only a zero that can stand in any place, as in the Hindu system, removes this problem. (Blank spaces and the placeholder, pages 73 and 74; zero in the Hindu system, pages 78 and 79.)

Question 8. On the chapter's line abacus, the thousands line has 1 counter above it and 3 on it, the hundreds line is empty, the tens line has 4 counters on it, and the ones line has 1 counter above it and 2 on it. What number is shown?

Answer: 8047. A counter above a line counts 5, and each counter on a line counts 1. Thousands: 5 + 3 = 8. Hundreds: 0. Tens: 4. Ones: 5 + 2 = 7. So the number is 8 × 1000 + 0 × 100 + 4 × 10 + 7 = 8047. The empty hundreds line does the job a 0 does in 8047. (The abacus, pages 68 and 69.)

Practise more, and check your marks

Where CMDLearn fits

CMDLearn is building an AI tutor for Classes 1 to 10 that explains a step and checks the child's working, but leaves the writing to the child. It is still in development, and no child uses it yet. If you would like to hear when there is something to try, you can join the waitlist.

Sources

All opened and read on 22 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 47.

The number systems are described on the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.