CLASS 8 · GANITA PRAKASH PART I · CHAPTER 3
Original short and explained answers for all Figure it Out questions and the chapter’s number-system activities, including Egyptian and base-five arithmetic, abacus regrouping, and place-value conversions.
79 worked answers · short steps first, explanations when you need them
Textbook pages 49–54
Question why-count
They needed to track food, livestock, goods exchanged and offerings.
They also tracked time and recurring events such as seasons and moon phases.
Counting helped people keep reliable records of quantities and time.
Question sticks-add
Put the two collections together.
Match or count every stick in the combined collection using the chosen stick sequence.
Combine the sticks; the combined collection represents the sum.
Question sticks-subtract
Start with the collection representing the larger amount.
Remove one stick for each object in the amount being taken away.
The sticks left represent the difference.
Remove matched sticks; the remaining sticks show the difference.
Question sticks-multiply-divide
For multiplication, make the required number of equal groups and combine them.
For division, split the sticks into equal groups; count each group and any leftovers.
Equal groups model multiplication; sharing into equal groups models division and any remainder.
Question letter-number-system
After z, use two-letter strings in order, such as aa, ab, ac,…, az, ba, bb,… .
Continue to three-letter strings after all two-letter strings have been used.
Use strings in a fixed dictionary order, increasing string length when needed. Many consistent extensions are possible.
Question make-own-system
Choose an ordered list of number symbols or names.
State how the list continues and how to decode each representation.
A valid design gives every counting number an unambiguous place in an ordered sequence.
Question Counting with body parts
Touch one body part for each item counted; follow a fixed sequence.
A fixed body-part sequence can serve as a counting record.
Textbook pages 58–60
Question roman-1222
Break 1222 into thousands, hundreds, tens and ones.
Combine the standard Roman symbols for those parts.
1222 = MCCXXII.
Question roman-2999
Break 2999 into thousands, hundreds, tens and ones.
Combine the standard Roman symbols for those parts.
2999 = MMCMXCIX.
Question roman-302
Break 302 into thousands, hundreds, tens and ones.
Combine the standard Roman symbols for those parts.
302 = CCCII.
Question roman-715
Break 715 into thousands, hundreds, tens and ones.
Combine the standard Roman symbols for those parts.
715 = DCCXV.
Question roman-add-a
The values are 232 and 413.
232+413=645=500+100+40+5.
645, written in standard Roman form as DCXLV.
Question roman-add-b
LXXXVII=87 and LXXVIII=78.
87+78=165=100+50+10+5.
165, written CLXV.
Question multiply-v-l
Convert the two symbols to their values: V and L.
Multiply: 250.
250 (which can be regrouped as CCL).
Question multiply-l-d
Convert the two symbols to their values: L and D.
Multiply: 25000.
25000 (which can be regrouped as 25 copies of M).
Question multiply-v-d
Convert the two symbols to their values: V and D.
Multiply: 2500.
2500 (which can be regrouped as MMD).
Question multiply-vii-ix
Convert the two symbols to their values: VII and IX.
Multiply: 63.
63 (which can be regrouped as LXIII).
Question roman-limits
Hindu numerals use ten digits with place value; a digit’s position changes its value.
Zero marks an empty place, and the same written rules work in every place.
Roman numerals use fixed-value symbols and have no zero placeholder.
Place value and zero make Hindu-numeral calculations compact and systematic; Roman calculations require careful regrouping of symbols.
Textbook pages 54–61
Question gumulgal-values
Count one for each urapon and two for each ukasar.
For example, ukasar-urapon is 3; ukasar-ukasar-urapon is 5.
Question gumulgal-arithmetic
The amounts are 9+7, 9−6, 9×4 and 16÷4. In the third expression, 9×4=36.
Convert each result back into a string of ukasar (2) and urapon (1).
Question gumulgal-object-names
Different words can group or classify objects in ways that matter in that community.
Language may reflect customary ways of counting, trading or sharing.
Different sequences may suit different objects or cultural practices; the system can still count accurately if its rules are clear.
Question Gumulgal beyond six
7 = ukasar-ukasar-ukasar-urapon; 8 = ukasar-ukasar-ukasar-ukasar
Seven is three ukasar and one urapon; eight is four ukasar.
Textbook pages 61–70
Question egyptian-10458
Split the number into its non-zero place-value groups.
10458=10,000+400+50+8.
10,000+400+50+8 (draw one Egyptian symbol for each landmark unit in the sum).
Question egyptian-1023
Split the number into its non-zero place-value groups.
1023=1,000+20+3.
1,000+20+3 (draw one Egyptian symbol for each landmark unit in the sum).
Question egyptian-2660
Split the number into its non-zero place-value groups.
2660=2,000+600+60.
2,000+600+60 (draw one Egyptian symbol for each landmark unit in the sum).
Question egyptian-784
Split the number into its non-zero place-value groups.
784=700+80+4.
700+80+4 (draw one Egyptian symbol for each landmark unit in the sum).
Question egyptian-1111
Split the number into its non-zero place-value groups.
1111=1,000+100+10+1.
1,000+100+10+1 (draw one Egyptian symbol for each landmark unit in the sum).
Question egyptian-70707
Split the number into its non-zero place-value groups.
70707=70,000+700+7.
70,000+700+7 (draw one Egyptian symbol for each landmark unit in the sum).
Question egyptian-read-symbols
Add the value of each repeated landmark symbol in the first picture.
Repeat for the second picture.
The textbook answers are 276 and 4322, respectively.
Question base5-15
Use the largest available powers of 5 first.
Expanded form: 3×5+0×1.
3×5+0×1; replace each coefficient by that many symbols of the corresponding landmark value.
Question base5-50
Use the largest available powers of 5 first.
Expanded form: 2×25.
2×25; replace each coefficient by that many symbols of the corresponding landmark value.
Question base5-137
Use the largest available powers of 5 first.
Expanded form: 1×125+2×5+2×1.
1×125+2×5+2×1; replace each coefficient by that many symbols of the corresponding landmark value.
Question base5-293
Use the largest available powers of 5 first.
Expanded form: 2×125+1×25+3×5+3×1.
2×125+1×25+3×5+3×1; replace each coefficient by that many symbols of the corresponding landmark value.
Question base5-651
Use the largest available powers of 5 first.
Expanded form: 1×625+1×25+1×1.
1×625+1×25+1×1; replace each coefficient by that many symbols of the corresponding landmark value.
Question base5-zero
Every positive whole number can be made by adding enough 1-symbols.
The system has no symbol for an empty collection.
Zero cannot be represented with the listed positive landmark symbols because there is no zero symbol.
Question base7-landmarks
Start with 7⁰=1; multiply by 7 each time: 1, 7, 49, 343.
Base 7: 1, 7, 49, 343, … . Base n: n⁰, n¹, n², n³, … .
Question base-n-why
Each place is a fixed multiple of the next smaller place.
When a place collects n units, regroup them as one unit in the next place.
The repeated regrouping rule makes addition and multiplication systematic. Roman landmarks do not have one fixed grouping ratio.
Question Egyptian landmark sequence
1, 10, 100, 1,000, 10,000
1, 10, 100, 1,000, 10,000.
Question Base-five representation of 143
143 = 125 + 5 + 5 + 5 + 1 + 1 + 1
One 125-symbol, three 5-symbols, and three 1-symbols.
Question Egyptian addition regrouping
Each next Egyptian landmark is ten times the previous one.
Ten copies of a landmark equal one copy of the next landmark.
Question Egyptian multiplication by ten
10×10ⁿ = 10ⁿ⁺¹
It becomes the next landmark number.
Question Egyptian symbol limit
No; ten matching symbols can be exchanged for one symbol of the next landmark.
No. Regroup ten copies as one next-landmark symbol.
Question Create a base-four system
1₄, 2₄, 3₄, 10₄, 11₄, 12₄, 13₄, 20₄, 21₄, 22₄, 23₄, 30₄, 31₄, 32₄, 33₄, 100₄
The base-four numerals are 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33, 100.
Question Multiply by five in base five
Multiplying by 5 shifts each digit one place left; write 0 in the ones place.
Shift the base-five digits one place left and append 0.
Question Abacus addition
2,907 + 43 = 2,950
Ones: 7+3=10, write 0 and carry 1 ten; tens: 0+4+1=5.
2,950.
Question Figure It Out: Egyptian addition 1(i)
9×1000 + 11×100 + 15×1 = 9,000 + 1,100 + 15 = 10,115
Regroup 11 hundreds as 1 thousand and 1 hundred; regroup 15 ones as 1 ten and 5 ones.
10,115 = 1×10,000 + 1×100 + 1×10 + 5
10,115, represented by one 10,000-symbol, one 100-symbol, one 10-symbol and five 1-symbols.
Question Figure It Out: Egyptian addition 1(ii)
1×1000 + 12×10 + 6×1 = 1,000 + 120 + 6 = 1,126
Regroup 12 tens as 1 hundred and 2 tens.
1,126 = 1×1000 + 1×100 + 2×10 + 6
1,126, represented by one 1,000-symbol, one 100-symbol, two 10-symbols and six 1-symbols.
Question Figure It Out: base-five addition
First numeral: 125 + 2×25 + 5 + 2 = 182
Second numeral: 3×125 + 25 + 2×5 + 2 = 412
182 + 412 = 594 = 4×125 + 3×25 + 3×5 + 4
594, or four 125-symbols, three 25-symbols, three 5-symbols and four 1-symbols.
Question Multiply Egyptian landmarks by 10
10×10=100
100×10=1,000
1,000×10=10,000
10,000×10=100,000
100, 1,000, 10,000 and 100,000.
Question Multiply Egyptian landmarks by 100
10×100=1,000
100×100=10,000
1,000×100=100,000
10,000×100=1,000,000
1,000, 10,000, 100,000 and 1,000,000.
Question Products of Egyptian landmarks
10×10⁵=10⁶
10²×10³=10⁵
10³×10³=10⁶
10⁴×10⁶=10¹⁰
10⁶, 10⁵, 10⁶ and 10¹⁰.
Question Products of landmarks in any base
5ᵃ×5ᵇ=5ᵃ⁺ᵇ
nᵃ×nᵇ=nᵃ⁺ᵇ
Yes. The product is another landmark number.
Question Egyptian multiplication by ten: examples
522×10=(500+20+2)×10=5,000+200+20=5,220
1,010×10=(1,000+10)×10=10,000+100=10,100
5,220 and 10,100.
Question Multiplication rule in Egyptian symbols
Each 1-symbol becomes a 10-symbol; each 10-symbol becomes a 100-symbol; continue the same shift for every symbol.
Replace every symbol with the symbol for the next higher landmark value.
Question Multiplying Egyptian numerals
Write each number as a sum of powers of 10.
Use the distributive property to multiply each pair of landmark values.
Regroup equal powers of 10 in the partial products.
Expand into landmark values, multiply the parts, and regroup the result.
Question A number times an Egyptian ten-symbol
For example, 352=3×100+5×10+2.
352×10=3×1000+5×100+2×10=3,520.
Every symbol moves up one landmark place; 352 becomes 3,520.
Question Refine an invented number system
Choose a base and a fixed set of digit symbols.
Give each position a power-of-the-base value.
Use a zero symbol for an empty place.
Use a positional base system with a zero placeholder.
Question Abacus addition: combine counters
Thousands: 2+0=2; hundreds: 9+0=9; tens: 0+4=4; ones: 7+3=10.
Exchange ten ones for one ten. The tens become 5 and the ones become 0.
The abacus shows 2 thousands, 9 hundreds, 5 tens and 0 ones: 2,950.
Question Abacus addition: regrouping
Ten counters in one place have the same value as one counter in the next place.
Move ten counters off that line and add one counter to the line on its left.
Regroup ten counters as one counter in the next higher place.
Textbook pages 71–77
Question mesopotamian-63
Use 60²=3600 and 60 as place values.
63=1×60+3.
1×60+3.
Question mesopotamian-132
Use 60²=3600 and 60 as place values.
132=2×60+12.
2×60+12.
Question mesopotamian-200
Use 60²=3600 and 60 as place values.
200=3×60+20.
3×60+20.
Question mesopotamian-60
Use 60²=3600 and 60 as place values.
60=1×60+0.
1×60+0.
Question mesopotamian-3605
Use 60²=3600 and 60 as place values.
3605=1×3600+5.
1×3600+5.
Question mayan-77
Choose coefficients from 0 to 19 at each place.
Expanded form: 3×20+17.
3×20+17.
Question mayan-100
Choose coefficients from 0 to 19 at each place.
Expanded form: 5×20+0.
5×20+0.
Question mayan-361
Choose coefficients from 0 to 19 at each place.
Expanded form: 1×360+0×20+1.
1×360+0×20+1.
Question mayan-721
Choose coefficients from 0 to 19 at each place.
Expanded form: 2×360+0×20+1.
2×360+0×20+1.
Question mesopotamian-zero
A blank position can be mistaken for no place at all.
The same cluster of signs might then be read at different place values.
Without a clear zero placeholder or spacing convention, a numeral can be ambiguous.
Question Mesopotamian 3600 ambiguity
60 = 1×60; 3600 = 1×60². Without a place marker, the same symbol can be read in different positions.
The position is unclear, so the numeral can be ambiguous.
Question Mesopotamian place coefficients
Sixty units of one place regroup as one unit of the next place.
Each coefficient is from 0 to 59.
Question Mayan landmark values
1, 20, 360, 7,200
1, 20, 360, 7,200.
Textbook pages 78–80
Question rod-numerals
Alternating direction distinguishes neighbouring place positions.
With no space marker, a row of identical orientations could be grouped in multiple ways.
Alternation helps identify place boundaries; an unspaced all-vertical 41 could be misread as other groupings, such as 23, 32 or 122, depending on the grouping rule.
Question binary-compare
Repeatedly divide by 2 or count with place values 1,2,4,8.
The representations are 1, 10, 11, 100, 101, 110, 111, 1000.
1=1₂, 2=10₂, 3=11₂, 4=100₂, 5=101₂, 6=110₂, 7=111₂, 8=1000₂.
Question uses-of-zero
Money, dates, measurement, addresses and digital communication use written numbers.
Computers and many technical systems rely on place value and zero.
They are used in trade, science, navigation, communication and everyday records; without zero and place value, many modern calculations and technologies would be much harder.
Question twenty-five-base-8
Express 25 as a sum of powers of 8.
25=3×8+1.
31₈.
Question twenty-five-base-5
Express 25 as a sum of powers of 5.
25=1×25+0×5+0.
100₅.
Question twenty-five-base-2
Express 25 as a sum of powers of 2.
25=16+8+1.
11001₂.
Question Base two using Gumulgal words
1, 10, 11, 100, 101, 110, 111, 1000 (base 2)
1=1₂, 2=10₂, 3=11₂, 4=100₂, 5=101₂, 6=110₂, 7=111₂, 8=1000₂.