CBSE Class 8 Mathematics A Story of Numbers Short Answer Questions with Answers

Short Answer Questions

Question 1. Answer the following about the vocabulary of this chapter.
(i) What is meant by a number system?
(ii) What are numerals? Give two examples from the Hindu number system.
Answer:
(i) A number system is a standard sequence of objects, names or written symbols in a fixed order, used to count a collection by matching its objects one by one with the sequence.
(ii) Numerals are the written symbols of a number system. Examples: 7 and 193.
In simple words: A number system is a fixed sequence used for counting; numerals are its written symbols.

Teacher's Note:
1. Any two Hindu numerals are acceptable as examples.
2. Do not confuse a number with the numeral that writes it.

Question 2. A Stone Age family keeps one stick for every cow in its herd. Explain what one-to-one mapping means here, and describe how the collection of sticks helps the family check in the evening whether every cow has returned from grazing.
Answer:
One-to-one mapping means each cow is paired with exactly one stick, and no two cows share a stick.
So the number of sticks equals the number of cows.
In the evening, the family pairs each returning cow with one stick.
If every stick gets a cow, the herd is complete; if some sticks are left over, that many cows are missing.
In simple words: One stick per cow; leftover sticks in the evening show how many cows are missing.

Teacher's Note:
1. Matching works even without number names.
2. This is the basic idea behind all counting.

Question 3. Two early ways of counting were (i) keeping a stick for each object and (ii) using the sounds of the letters ‘a’ to ‘z’ in order. State one clear drawback of each of these two ways of representing numbers.
Answer:
(i) Sticks: one stick is needed for every object, so a large collection needs a huge heap of sticks to carry and match. It works but is very inconvenient.
(ii) Letters ‘a’ to ‘z’: the sequence stops at 26, so a collection of more than 26 objects cannot be counted.
In simple words: Sticks become clumsy for big numbers, and letters run out after 26.

Teacher's Note:
1. A good system must be both unending and convenient.
2. Each method fails one of these two tests.

Question 4. What are tally marks? Write one fact each about the Lebombo bone and the Ishango bone as described in the chapter.
Answer:
Tally marks are notches cut on a surface such as a bone or a cave wall, one mark for each object counted.
Lebombo bone: found in South Africa, it has 29 notches and is about 44,000 years old.
Ishango bone: found in the Democratic Republic of Congo, it is 20,000 to 35,000 years old and its notches are arranged in columns.
In simple words: Tally marks are one-per-object notches; the Lebombo bone has 29 and the Ishango bone has columns of them.

Teacher's Note:
1. Any one correct fact about each bone is enough.
2. Tally marks are the stick method cut into a surface.

Question 5. The Gumulgal of Australia counted in 2s.
(i) Write their number names for 5 and 6 and show how each name is built from ‘urapon’ and ‘ukasar’.
(ii) What did they call any number greater than 6?
Answer:
urapon = 1 and ukasar = 2.
(i) 5 = 2 + 2 + 1, so 5 is ukasar-ukasar-urapon. 6 = 2 + 2 + 2, so 6 is ukasar-ukasar-ukasar.
(ii) Any number greater than 6 was called ras.
In simple words: They built names by adding 2s and a 1; anything above 6 was 'ras'.

Teacher's Note:
1. This is counting in groups of a single size.
2. Such a system cannot go far without new ideas.

Question 6. Write 1263 in Roman numerals. Show the grouping of 1263 as a sum of Roman landmark numbers before writing the numeral.
Answer:
Take as many of each landmark number as possible:
1263 = 1000 + 100 + 100 + 50 + 10 + 1 + 1 + 1
1000 → M, two 100s → CC, 50 → L, 10 → X, three 1s → III.
So 1263 = MCCLXIII. Check: 1000 + 200 + 50 + 10 + 3 = 1263.
In simple words: Break 1263 into 1000, 200, 50, 10 and 3, then write M CC L X III.

Teacher's Note:
1. Work from the largest landmark number down.
2. Always add the parts back to check.

Question 7. Answer both parts.
(i) State the two conditions that make a number system a base-n number system.
(ii) Write the first five landmark numbers of a base-6 system.
Answer:
(i) (a) The first landmark number is 1, and (b) every next landmark number is the current one multiplied by the same fixed number n.
(ii) Powers of 6: 1, 6, 36, 216, 1296.
In simple words: Start at 1 and keep multiplying by the base; for base 6 this gives 1, 6, 36, 216, 1296.

Teacher's Note:
1. Do not forget that the list starts at 1.
2. 6 × 216 = 1296.

Question 8. Why is the Egyptian number system called a base-10 system? Also state how many symbols in all are needed to write the number 230 in that system.
Answer:
Its first landmark number is 1 and each landmark number is 10 times the one before: 1, 10, 10², 10³, … These are the powers of 10, so it is a base-10 system.
230 = 100 + 100 + 10 + 10 + 10: 2 hundred-symbols and 3 ten-symbols, that is 5 symbols in all.
In simple words: Egyptian landmark numbers go up by ×10, and 230 needs 2 + 3 = 5 symbols.

Teacher's Note:
1. The 0 in 230 needs no symbol in the Egyptian system.
2. Count symbols by adding the digits.

Question 9. Express 342 in the base-5 system by grouping it into the landmark numbers 125, 25, 5 and 1. Write the base-5 numeral and check your answer by adding the groups back.
Answer:
Start with the largest landmark number not more than 342, which is 125.
342 = 2 × 125 + 92
92 = 3 × 25 + 17
17 = 3 × 5 + 2
So 342 = 2 × 125 + 3 × 25 + 3 × 5 + 2 × 1, and the base-5 numeral is 2332.
Check: 250 + 75 + 15 + 2 = 342.
In simple words: 342 has 2 one-twenty-fives, 3 twenty-fives, 3 fives and 2 ones, so it is 2332 in base 5.

Teacher's Note:
1. Every digit in base 5 must be from 0 to 4.
2. The remainder at each step goes to the next smaller landmark number.

Question 10. In a base-4 place value system the numeral 3122 is written. Find the number it stands for in the Hindu (base-10) system, writing down the place value of each digit.
Answer:
Place values from the right: 1, 4, 16, 64.
3 in the 64s place: 3 × 64 = 192
1 in the 16s place: 1 × 16 = 16
2 in the 4s place: 2 × 4 = 8
2 in the 1s place: 2 × 1 = 2
Total = 192 + 16 + 8 + 2 = 218.
In simple words: Multiply each digit by 64, 16, 4 and 1 and add to get 218.

Teacher's Note:
1. Place values in base 4 are powers of 4.
2. Start from the right-hand digit with value 1.

Question 11. Express 3700 in the Mesopotamian base-60 system.
(i) Write 3700 as a sum of multiples of 3600, 60 and 1.
(ii) State how many groups of symbols the Mesopotamian numeral for 3700 will have, reading from the left.
Answer:
(i) 3700 = 1 × 3600 + 100 = 1 × 3600 + 1 × 60 + 40 × 1.
(ii) Three groups: from the left, a group for 1 (3600s), a group for 1 (60s) and a group for 40 (1s). Check: 3600 + 60 + 40 = 3700.
In simple words: 3700 is one 3600, one 60 and forty 1s, so it needs three groups of symbols.

Teacher's Note:
1. Each group can show only 1 to 59.
2. Divide by 3600 first, then by 60.

Question 12. The Mayan landmark numbers were 1, 20 and 360, a dot stood for 1 and a bar for 5. Show how 500 is grouped in their system, and describe the dots and bars written at each of the three positions.
Answer:
500 = 1 × 360 + 140 = 1 × 360 + 7 × 20 + 0 × 1. Check: 360 + 140 + 0 = 500.
Symbols are written one below the other:
Top (360s): one dot, for 1.
Middle (20s): one bar and two dots, since 7 = 5 + 1 + 1.
Bottom (1s): the shell-like placeholder, since there are no 1s.
In simple words: 500 is one 360 and seven 20s: a dot on top, a bar with two dots in the middle and a shell at the bottom.

Teacher's Note:
1. The bottom position counts 1s in Mayan numerals.
2. The shell shows the empty units place.

Question 13. Multiplying two Roman numerals is difficult, but multiplying in the Egyptian system is easy. Give the reason, using the products V × L and “a landmark number times the symbol for 10” to support your answer.
Answer:
In the Roman system the landmark numbers are not powers of one number, so the product of two of them need not be a landmark number: V × L = CCL, that is 250, which needs three symbols.
In the Egyptian system every landmark number is a power of 10, so a landmark number times the symbol for 10 just raises the power by one and gives the next landmark number (e.g. 100 × 10 = 1000).
Since the product of two landmark numbers is again a single landmark number, Egyptian multiplication is quick.
In simple words: Powers of 10 multiply into powers of 10, but Roman symbols do not multiply neatly.

Teacher's Note:
1. V × L = 250 needs three Roman symbols, so it is not a landmark number.
2. Powers of one base always multiply into another power of that base.

Question 14. While adding two Egyptian numerals, a scribe ended up with 13 hundred-symbols, 12 ten-symbols and 7 unit-symbols. Regroup these correctly and state how many of each symbol the final numeral has, and which number it represents.
Answer:
Ten copies of a symbol are replaced by one copy of the next symbol.
12 ten-symbols = 1 hundred-symbol + 2 ten-symbols.
13 hundred-symbols + 1 more = 14 hundred-symbols = 1 thousand-symbol + 4 hundred-symbols.
Final numeral: 1 thousand-symbol, 4 hundred-symbols, 2 ten-symbols and 7 unit-symbols, which is 1427.
In simple words: Carry 10 tens into a hundred and 10 hundreds into a thousand to get 1427.

Teacher's Note:
1. Regroup from the smallest symbol upward.
2. This is just 'carrying' in addition.

Question 15. The Chinese rod numerals used Zong (vertical) and Heng (horizontal) rods alternately for successive places. Explain why they did this, and state what confusion could arise if only Zong rods were used.
Answer:
The rods for one to nine are made of simple strokes, so if every place used the same shape, one could not tell where one place ends and the next begins.
Alternating Zong and Heng rods makes the boundary between neighbouring places visible, and also makes a blank place easier to spot.
With only Zong rods, the five strokes of 41 could equally be read as 32, 23 or even 122.
In simple words: Switching rod direction marks where each place begins, so numbers are not misread.

Teacher's Note:
1. Changing shape plays the role of spacing.
2. Check: 41, 32, 23 and 122 all use five strokes.

Question 16. Distinguish between the use of zero as a placeholder and the use of zero as a number. Name one civilisation that used it only as a placeholder and one Indian mathematician who treated it as a number.
Answer:
As a placeholder, zero is only a mark that fills an empty position so that the other symbols stay at their correct landmark numbers. The later Mesopotamians used it this way (the Mayans also used a shell-like sign).
As a number, 0 stands on its own and can be added, subtracted and multiplied like other numbers (0 + a = a, 0 × a = 0). Brahmagupta codified its arithmetic in 628 CE (Aryabhata also computed with it).
In simple words: A placeholder zero only fills a gap; a number zero can be used in calculations.

Teacher's Note:
1. Either Mesopotamians or Mayans is a correct placeholder example.
2. Brahmagupta is the key name for zero as a number.

Question 17. Show that in a base-n number system no landmark number can be used n or more times in a correct representation. Illustrate your reasoning once for the Egyptian system and once for the Mesopotamian system.
Answer:
Each landmark number is exactly n times the previous one. So n copies of a landmark number make exactly one copy of the next landmark number and must be replaced by it.
Hence each landmark number is used only 0, 1, 2, …, or n − 1 times.
Egyptian (n = 10): 10 ten-symbols make one hundred-symbol, so no symbol appears 10 or more times.
Mesopotamian (n = 60): 60 sixties make 3600, so no position can show 60 or more.
In simple words: Whenever you have n of something, you trade it for one of the next, so you never keep n of anything.

Teacher's Note:
1. This is why base n needs exactly n digits.
2. The same rule is 'carrying' in our addition.

Question 18. Write 45 in base 8, base 5 and base 2. Then state how the number of digits of a numeral changes as the base becomes smaller, and give the reason.
Answer:
Base 8: 45 = 5 × 8 + 5, so 45 is 55.
Base 5: 45 = 1 × 25 + 4 × 5 + 0, so 45 is 140.
Base 2: 45 = 32 + 8 + 4 + 1, so 45 is 101101.
As the base gets smaller the numeral gets longer (2, 3, then 6 digits). Smaller groups hold less, so more groups and more positions are needed for the same quantity.
In simple words: The smaller the base, the more digits you need.

Teacher's Note:
1. Check each answer by expanding it back to 45.
2. Base 2 numerals are long, which is why computers use many bits.

Question 19. In a place value system of some base n, the numeral 132 stands for the number 42 of the Hindu system. Form an equation in n, solve it, and verify your value of n.
Answer:
The place values of 132 are n², n and 1, so 1 × n² + 3 × n + 2 = 42.
n² + 3n − 40 = 0 ⇒ (n + 8)(n − 5) = 0.
The base must be positive, so n = 5. It is also bigger than every digit used (1, 3, 2), as a base must be.
Check: 1 × 25 + 3 × 5 + 2 = 25 + 15 + 2 = 42.
In simple words: Solving n² + 3n + 2 = 42 gives base 5.

Teacher's Note:
1. Reject the negative root; a base is a positive whole number.
2. The base must be larger than every digit in the numeral.

Question 20. Compare how 16 would be written (i) by extending the Gumulgal way of counting in 2s and (ii) in a base-2 place value system that uses ‘ukasar’ for 1 and ‘urapon’ for 0. Which representation is shorter, and what idea makes it shorter?
Answer:
(i) 16 = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2, so the name is ukasar repeated 8 times.
(ii) 16 = 1 × 2⁴, so the numeral is 10000, that is ukasar urapon urapon urapon urapon, 5 symbols.
The place value form is shorter. The Gumulgal way counts groups of one size (2) only, while place value uses the whole chain of landmark numbers 1, 2, 4, 8, 16, … and lets position show which one is counted.
In simple words: Using powers of 2 with place value needs 5 words instead of 8.

Teacher's Note:
1. Note the swap: here ukasar = 1 and urapon = 0.
2. Powers plus position is the big idea of the chapter.