CBSE Class 8 Mathematics A Story of Numbers Practice Paper Worksheet Set 01 with Answers

This is Class 8 Maths Ganita Prakash Chapter 3 A Story of Numbers Practice Paper 11, a 40-mark test of 20 questions in Sections A to E, with step-by-step answers for every question. It checks how well you understand tally sticks and one-to-one mapping, additive and subtractive Roman numerals, binary to decimal conversion, landmark numbers in base 5 and base 8, Mesopotamian base-60 grouping, and how zero and place value make the Hindu numeral system so powerful.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. An ancient Egyptian uses a system where a single symbol represents a value. He has 8 symbols for 100, and 12 symbols for 1. What is the total number he has? Show your working.
Answer:
The Egyptian system is additive, so we add the values of all the symbols.
8 symbols for 100 = 8 × 100 = 800.
12 symbols for 1 = 12 × 1 = 12.
Total = 800 + 12 = 812.
In simple words: Eight hundreds and twelve ones make 812.

Teacher's Note:
1. In the Egyptian system, value = (number of symbols) × (value of the symbol).
2. An Egyptian scribe would normally regroup ten 1-symbols into one 10-symbol, giving 8 hundreds, 1 ten and 2 ones.

Question 12. What is the Roman numeral for the year 1995? Show your working.
Answer:
Break 1995 into thousands, hundreds, tens and ones: 1995 = 1000 + 900 + 90 + 5.
1000 = M
900 = 1000 − 100 = CM
90 = 100 − 10 = XC
5 = V
So 1995 = MCMXCV.
In simple words: Write each place separately and join them: M + CM + XC + V.

Teacher's Note:
1. Always split the number by place value first.
2. Use the subtractive pairs for 4s and 9s (CM, XC, IX, etc.).

Question 13. In a fictional number system with a base of 8, what would be the first three landmark numbers?
Answer:
In a base-n system the landmark numbers are the powers of n: 1, n, n², n³, …
For base 8: 8⁰ = 1, 8¹ = 8, 8² = 8 × 8 = 64.
So the first three landmark numbers are 1, 8 and 64.
In simple words: Start at 1 and keep multiplying by 8: 1, 8, 64.

Teacher's Note:
1. Each landmark number is the base times the previous one.
2. The next landmark in base 8 would be 512.

Question 14. Convert the binary number 1101 to its decimal equivalent.
Answer:
Place values in base 2 (from the right): 1, 2, 4, 8.
1101 = 1 × 8 + 1 × 4 + 0 × 2 + 1 × 1
= 8 + 4 + 0 + 1 = 13.
In simple words: Add the place values where the digit is 1: 8 + 4 + 1 = 13.

Teacher's Note:
1. Binary digits are only 0 and 1.
2. Check: 13 = 8 + 4 + 1, which gives back 1101.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Explain how the Hindu number system's use of zero allows for the unambiguous writing of all numbers. Use the numbers 3, 30, and 300 as examples.
Answer:
In the Hindu system the value of a digit depends on its place, and zero is used as a placeholder for an empty place.
3: the digit 3 is in the ones place, so it means three.
30: the 0 shows that there are no ones, which pushes 3 into the tens place, so it means 3 × 10 = thirty.
300: two zeros fill the ones and tens places, so 3 is in the hundreds place and means 3 × 100 = three hundred.
Without zero, all three would be written as just “3” and could not be told apart. With zero, every number has exactly one clear way of being written.
In simple words: Zero holds empty places, so 3, 30 and 300 look different and cannot be confused.

Teacher's Note:
1. Zero is both a placeholder and a number in its own right.
2. Older systems left a gap for an empty place, which was easy to misread.

Question 16. The Roman numeral system is not as efficient for complex calculations as the Hindu number system. Explain why this is so, using the subtraction of LXXX from C as an example.
Answer:
Roman numerals have no place value and no zero, so symbols cannot be lined up in columns and subtracted digit by digit.
To find C − LXXX, the symbols do not match column by column, so we usually convert: C = 100 and LXXX = 50 + 10 + 10 + 10 = 80.
100 − 80 = 20, which is then converted back: 20 = XX.
So C − LXXX = XX.
In the Hindu system we simply write 100 − 80 = 20 using a standard column method; in Roman numerals we must regroup or convert, which is slow. This is why the Roman system is less efficient for calculation.
In simple words: Roman numerals have no columns or zero, so even a simple subtraction needs converting back and forth.

Teacher's Note:
1. Place value lets us use the same simple steps for any numbers.
2. Romans actually did their calculations on counting boards (abacus), not with numerals.

Question 17. A Mesopotamian scribe uses two vertical wedges to represent 2, and four horizontal wedges to represent 40. How would the scribe represent the number 164? Explain the process of grouping.
Answer:
The Mesopotamian system has base 60: a vertical wedge stands for 1 and a sideways (horizontal) wedge for 10 within each place, and the next place counts sixties.
Group 164 into sixties: 164 = 2 × 60 + 44.
Then split the remainder: 44 = 4 × 10 + 4 × 1.
So the scribe writes two vertical wedges in the sixties place, then, in the ones place, four horizontal wedges (40) and four vertical wedges (4).

In simple words: 164 is two sixties and 44 more, and 44 is four tens and four ones.

Teacher's Note:
1. First make the biggest groups (sixties), then tens, then ones.
2. Within one place, Mesopotamian numerals are additive; across places they are positional.

Section D: Long Answer Question (5 Marks)

Question 18. Design a number system with a base of 5. Create a set of symbols for the numbers 1, 5, and 25. Then, use these symbols to represent the number 142. Explain how this process demonstrates the efficiency of a base-n system.
Answer:
Base: 5, so the landmark numbers are 1, 5, 25 (and then 125, …).
Symbols (our own choice): ● = 1, ▲ = 5, ■ = 25.
Grouping 142: the largest available landmark is 25; 142 ÷ 25 gives 5 groups, since 5 × 25 = 125, and 142 − 125 = 17.
17 ÷ 5 gives 3 groups, since 3 × 5 = 15, and 17 − 15 = 2.
So 142 = 5 × 25 + 3 × 5 + 2 × 1. Check: 125 + 15 + 2 = 142.
Representation: ■■■■■ ▲▲▲ ●● (5 squares, 3 triangles, 2 dots).
(If we also made a symbol for the next landmark 125 = 5³, the five ■ would become one symbol, and 142 would be written as 1032 in base 5.)
Efficiency: a tally needs 142 separate marks, but grouping by powers of 5 needs only 10 symbols here. Grouping into landmark numbers makes numbers shorter, quicker to write and easier to read and compare.
In simple words: Group 142 into twenty-fives, fives and ones: 5 big symbols, 3 middle symbols and 2 small symbols instead of 142 marks.

Teacher's Note:
1. Any symbols are fine as long as each landmark number has its own symbol.
2. Always check the grouping by adding the values back.
3. In a full base-5 system, no symbol is ever needed more than 4 times in a row.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study-1: The Abacus Master
Rohan's grandfather, a retired accountant, shows him an old abacus. He tells Rohan, "This was my calculator before you had a phone." He explains that the abacus is a form of a place value system. On the board, the first line from the right represents ones (10⁰), the next line represents tens (10¹), and so on. A single bead on a line has a value of one times the line's value, while a special bead, a "five-bead," has a value of 5 times the line's value. He challenges Rohan to represent numbers on the abacus. Rohan realizes how this simple tool made complex calculations possible for people who did not have calculators.
(i) On this abacus, what is the value of a bead on the fifth line from the right? (1)
(ii) How would you represent the number 39 on this abacus, using the beads described? (1)
(iii) If there is one five-bead on the thousands line, what is its value? (1)
(iv) How would you represent the number 17 on this abacus? (1)
Answer:
(i) The lines from the right are 10⁰, 10¹, 10², 10³, 10⁴. The fifth line is 10⁴, so one bead there is worth 10,000.
(ii) 39 = 3 tens + 9 ones, and 9 = 5 + 4. So put 3 single beads on the tens line, and 1 five-bead + 4 single beads on the ones line.
(iii) Value = 5 × 1000 = 5000.
(iv) 17 = 1 ten + 7 ones, and 7 = 5 + 2. So put 1 single bead on the tens line, and 1 five-bead + 2 single beads on the ones line.

In simple words: Each line is ten times the line to its right, and a five-bead saves using five single beads.

Teacher's Note:
1. The five-bead means no line ever needs more than 4 single beads.
2. Bead value = (1 or 5) × (value of its line).

Question 20. Case Study-2: The Hindu Numerals
A historian, Dr. Priya, is researching the spread of the Hindu numeral system. She finds that the system, with its ten digits, including zero, was so powerful that it enabled a new way of doing mathematics. Unlike the Roman system, it was perfect for performing multiplication and division. She discovers that the word "algorithm" is named after the Persian mathematician Al-Khwarizmi, who wrote a book explaining the system. This system's efficiency ultimately led to it becoming the global standard for mathematics and science, a testament to its elegance and power.
(i) What is the base of the Hindu number system? (1)
(ii) What is a key arithmetic operation that the Hindu numeral system made easy, but was difficult in the Roman system? (1)
(iii) Convert the number 4,321 to the Roman number system. (1)
(iv) The word "algorithm" is named after which mathematician? (1)
Answer:
(i) The base is 10 (ten digits 0 to 9).
(ii) Multiplication (and division), which are hard with Roman numerals because they have no place value.
(iii) 4321 = 4000 + 300 + 20 + 1 = MMMM + CCC + XX + I = MMMMCCCXXI. (Standard Roman numerals stop at 3999, so 4000 is written by repeating M four times.)
(iv) The Persian mathematician Al-Khwarizmi.
In simple words: Base 10, easy multiplication, 4321 = MMMMCCCXXI, and “algorithm” comes from Al-Khwarizmi.

Teacher's Note:
1. Convert to Roman numerals place by place.
2. Al-Khwarizmi's book helped carry Indian numerals to the Arab world and then Europe.