Students often refer to Ganita Prakash Class 7 Solutions and Class 7 Maths Chapter 1 Large Numbers Around Us NCERT Solutions Question Answer to verify their answers.
NCERT Class 7 Maths Chapter 1 Large Numbers Around Us Solutions Question Answer
Ganita Prakash Class 7 Chapter 1 Solutions Large Numbers Around Us
NCERT Class 7 Maths Ganita Prakash Chapter 1 Large Numbers Around Us Solutions Question Answer
1.1 A Lakh Varieties!
What if we ate 2 varieties of rice every day? Would we then be able to eat 1 lakh varieties of rice in 100 years? (Page 2)
Solution:
Given, rice varieties eaten per day = 2
Days in a year = 365
∴ Rice varieties eaten in a year = 2 × 365 = 730
Thus, rice varieties eaten in 100 years = 730 × 100 = 73,000 varieties.
Hence, we can eat only 73,000 varieties in 100 years, not 1 lakh.
What if a person ate 3 varieties of rice every day? Will they be able to taste all the lakh varieties in a 100-year lifetime? Find out. (Page 2)
Solution:
Given rice varieties eaten per day = 3
days in a year = 365
∴ Rice varieties eaten in a year = 3 × 365 = 1095
and rice varieties eaten in 100 years = 1095 × 100 = 1,09,500 varieties.
Hence, they can eat all 1 lakh varieties in 100 years.
Estu said, “We know how many days there are in a year — 365, if we ignore leap years. If we live for y years, the number of days in our lifetime will be 365 × y.”
Choose a number for y. How close to one lakh is the number of days in y years, for those of your choice? (Page 3)
Solution:
Let us live for 95 years, y = 95
Number of days in our lifetime = 365 × 95 = 34,675
Now, difference between 1,00,000 and 34,675 = 1,00,000 – 34,675 = 65,325.
Figure It Out (Page 3)
Question 1.
According to the 2011 Census, the population of the town of Chintamani was about 75,000. How much less than one lakh is 75,000?
Solution:
Given population of the town Chintamani in 2011 = 75,000
Now 1,00,000 – 75,000 = 25,000
Hence, 75,000 is 25,000 less than one lakh
Question 2.
The estimated population of Chintamani in 2024 is 106,000. How much more than one lakh is 1,06,000?
Solution:
Given population of Chintamani in 2024 = 1,06,000
Now, 1,06,000 – 1,00,000 = 6,000
Hence, 1,06,000 is 6,000 more than one lakh.
Question 3.
By how much did the population of Chintamani increase from 2011 to 2024?
Solution:
Increase in population of Chintamani from 2011 to 2024 = 1,06,000 – 75,000 = 31,000.
Getting a Feel of Large Numbers (Page 3)
Look at the picture on the right. Somu is 1 metre tall. If each floor is about four times his height, what is the approximate height of the building?
Solution:
Here, Somu’s height = 1 metre
and height of each floor = 4 × 1 = 4 metres
Number of floors in the building = 10
Hence, height of the building = 10 × 4 = 40 metres.
Which is taller – The Statue of Unity or this building? How much taller?
__________ m.
Solution:
Here, the height of the Statue of Unity = 180 m
and the height of Somu’s building = 40 m
Now, 180 – 40 = 140 m
Hence, the Statue of Unity is 140 m taller than Somu’s building.
How much taller is the Kunchikal waterfall than Somu’s building?
__________ m.
Solution:
Here, the height of the Kunchikal waterfall = 450 m
and the height of Somu’s building = 40 m
Now, 450 – 40 = 410 m
Hence, the Kunchikal waterfall is 410 m taller than Somu’s building.
How many floors should Somu’s building have to be as high as the waterfall?
_____________
Solution:
Here, the height of each floor in the building = 4 m
and the height of the waterfall = 450 m
∴ Number of floors required = 450 ÷ 4 = 112.5
Hence, Somu’s building should have at least 113 floors to be as high as the waterfall.
How do you view a lakh — is a lakh big or small? (Page 4)
Solution:
A lakh can seem big or small depending on the context and what it’s being compared to.
For instance, a crowd of 1 lakh people at an event or protest is massive, but in terms of city population, 1 lakh might be considered a small town.
Reading and Writing Numbers
Write each of the numbers given below in words: (Page 4)
(a) 3,00,600
(b) 5,04,085
(c) 27,30,000
(d) 70,53,138
Solution:
(a) 3,00,600 – Three lakh six hundred.
(b) 5,04,085 – Five lakh four thousand eighty-five.
(c) 27,30,000 – Twenty-seven lakh thirty thousand.
(d) 70,53,138 – Seventy lakh fifty three thousand one hundred and thirty-eight.
Write the corresponding number in the Indian place value system for each of the following: (Page 5)
(a) One lakh twenty-three thousand four hundred and fifty-six
(b) Four lakh seven thousand seven hundred and four
(c) Fifty lakhs five thousand and fifty
(d) Ten lakhs two hundred and thirty-five
Solution:
(a) One lakh twenty-three thousand four hundred and fifty-six – 1,23,456.
(b) Four lakh seven thousand seven hundred and four – 4,07,704.
(c) Fifty lakhs five thousand and fifty – 50,05,050.
(d) Ten lakhs two thousand and thirty-five – 10,02,035.
1.2 Land of Tens
Question 1.
The Thoughtful Thousands has only a +1000 button. How many times should it be pressed to show: (Page 5)
(a) Three thousand? 3 times
(b) 10,000? _____________
(c) Fifty-three thousand? _____________
(d) 90,000? _____________
(e) One Lakh? _____________
(f) _____________? 153 times
(g) How many thousands are required to make one lakh?
Solution:
(a) Here, three thousand = 3 times
(b) Here, \(\frac{10,000}{1000}\) = 10 ⇒ 10 times
(c) Here, \(\frac{53,000}{1000}\) = 53 ⇒ 53 times
(d) Here, \(\frac{90,000}{1000}\) = 90 ⇒ 90 times
(e) Here, \(\frac{1,00,000}{1000}\) = 100 ⇒ 100 times
(f) Here, 153 times = 153 × 1000 = 1,53,000
(g) Here, 1,00,000 ÷ 1000 = 100
Hence, 100 thousands make one lakh.
Question 2.
The Tedious Tens has only a +10 button. How many times should it be pressed to show:
(a) Five hundred? _____________
(b) 780? _____________
(c) 1000? _____________
(d) 3700? _____________
(e) 10,000? _____________
(f) One lakh? _____________
(g) _____________? 435 times
Solution:
(a) Here, \(\frac{500}{10}\) = 50 ⇒ 50 times
(b) Here, \(\frac{780}{10}\) = 78 ⇒ 78 times
(c) Here, \(\frac{1000}{10}\) = 100 ⇒ 100 times
(d) Here, \(\frac{3700}{10}\) = 370 ⇒ 370 times
(e) Here, \(\frac{10,000}{10}\) = 1,000 ⇒ 1000 times
(f) Here, \(\frac{1,00,000}{10}\) = 10,000 ⇒ 10000 times
(g) Here, 435 × 10 = 4350 ⇒ 435 times
Question 3.
The Handy Hundreds has only a +100 button. How many times should it be pressed to show:
(a) Four hundred? _____________ times
(b) 3,700? _____________
(c) 10,000? _____________
(d) Fifty-three thousand? _____________
(e) 90,000? _____________
(f) 97,600? _____________
(g) 1,00,000? _____________
(h) _____________? 582 times
(i) How many hundreds are required to make ten thousand?
(j) How many hundreds are required to make one lakh?
(k) Handy Hundreds says, “There are some numbers which Tedious Tens and Thoughtful Thousands can’t show but I can.” Is this statement true? Think and explore.
Solution:
(a) Here, \(\frac{400}{100}\) = 4 ⇒ 4 times
(b) Here, \(\frac{3700}{100}\) = 37 ⇒ 37 times
(c) Here, \(\frac{10,000}{100}\) = 100 ⇒ 100 times
(d) Here, \(\frac{53,000}{100}\) = 530 ⇒ 530 times
(e) Here, \(\frac{90,000}{100}\) = 900 ⇒ 900 times
(f) Here, \(\frac{97,600}{100}\) = 976 ⇒ 976 times
(g) Here, \(\frac{1,00,000}{100}\) = 1,000 ⇒ 1000 times
(h) Here, \(\frac{58,200}{100}\) = 582 ⇒ 582 times
(i) Here, \(\frac{10,000}{100}\) = 100 hundreds
100 hundreds makes ten thousand.
(j) Here, \(\frac{1,00,000}{100}\) = 1000 hundreds
1000 hundreds make one lakh.
(k) Handy Hundreds’ statement is true compared to Thoughtful Thousands because Thoughtful Thousands can’t show multiples of 100 or 10.
However, Handy Hundreds’ statement is false compared to Tedious Tens because Tedious Tens can reach any multiple of 100, although it would take many steps.
Question 4.
Creative Chitti is a different kind of calculator. It has the following buttons:
+1, +10, +100, +1000, +10000, +100000, and +1000000.
It always has multiple ways of doing things. “How so?”, you might ask. To get the number 321, press +10 thirty-two times and +1 once. Will it get 321? Alternatively, it can press +100 two times and +10 twelve times, and +1 once.
Solution:
First method:
Press +10, 32 times ⇒ 32 × 10 = 320
Press +1 once ⇒ 1
Now adding, 320 + 1 = 321
Yes, Chitti reaches 321 this way.
Second method:
Press +100, 2 times ⇒ 2 × 100 = 200
Press +10, 12 times ⇒ 12 × 10 = 120
Press +1 once ⇒ 1
Now adding, 200 + 120 + 1 = 321
Again, Chitti reaches 321.
Question 5.
Two of the many different ways to get 5072 are shown below:
These two ways can be expressed as:
(a) (50 × 100) + (7 × 10) + (2 × 1) = 5072
(b) (3 × 1000) + (20 × 100) + (72 × 1) = 5072
Find a different way to get 5072 and write an expression for the same. (Page 6)
Solution:
Here, (5 × 1000) + (0 × 100) + (7 × 10) + (2 × 1)
= 5000 + 0 + 70 + 2
= 5072
Figure It Out (Pages 6-7)
Question 1.
For each number given below, write expressions for at least two different ways to obtain that number through button clicks. Think like Chitti, be creative.
(a) 8300
(b) 40629
(c) 56354
(d) 66666
(e) 367813
Solution:
(a) 8300:
(i) (8 × 1000) + (3 × 100)
(ii) (83 × 1000)
(b) 40629:
(i) (4 × 10,000) + (6 × 100) + (2 × 10) + (9 × 1)
(ii) (406 × 100) + (2 × 10) + (9 × 1)
(c) 56354:
(i) (56 × 1000) + (35 × 10) + (4 × 1)
(ii) (5635 × 10) +(4 × 1)
(d) 66666:
(i) (6 × 10000) + (66 × 100) + (66 × 1)
(ii) (66 × 1000)+ (666 × 1)
(e) 367813:
(i) (367 × 1000) + (813 × 1)
(ii) (3 × 100000) + (67 × 1000) + (8 × 100) + (13 ×1)
Question 2.
Creative Chitti has some questions for you-
(a) You have to make exactly 30 button presses. What is the largest 3-digit number you can make? What is the smallest 3-digit number you can make?
(b) 997 can be made using 25 clicks. Can you make 997 with a different number of clicks?
Solution:
(a) We have (9 × 100) + (8 × 10) + (13 × 1) = 993
Total clicks = 9 + 8 + 13 = 30
Hence, 993 is the largest 3-digit number made with 30 clicks.
Also, we have (8 × 10) + (22 × 1) = 80 + 22 = 102
Total clicks = 8 + 22 = 30.
Hence, 102 is the smallest 3-digit number made with 30 clicks.
(b) Here, (8 × 100) + (19 × 10) + (7 × 1)
= 800+ 190 + 7
= 997
Total clicks = 8 + 19 + 7 = 34
Hence, 997 can also be obtained in 34 clicks be made by 30 button presses = 30.
Question 3.
Systematic Sippy is a different kind of calculator. It has the following buttons: +1, +10, +100, +1000, +10000, +100000. It wants to be used as minimally as possible.
How can we get the numbers (a) 5072, (b) 8300 using as few button clicks as possible?
Solution:
(a) Here, 5072 = (5 × 1,000) + (7 × 10) + (2 × 1)
Total clicks = 5 + 7 + 2 = 14
Hence, 5072 can be obtained in 14 button clicks.
(b) Here, 8300 = (8 × 1,000) + (3 × 100)
Total clicks = 8 + 3 = 11
Hence, 8300 can be obtained in 11 button clicks.
Question 4.
Is there another way to get 5072 using fewer than 23 button clicks? Write the expression for the same.
Solution:
Hence, (5 × 1,000) + (7 × 10) + (2 × 1)
Total clicks = 5 + 7 + 2 = 14
Hence, 5072 can be obtained in less than 23 clicks i.e., 14 clicks.
Figure It Out (Pages 7-8)
Question 1.
For the numbers in the previous exercise, find out how to get each number by making the smallest number of button clicks, and write the expression
(a) 8300
(b) 40629
(c) 56354
(d) 66666
(e) 367813
Solution:
(a) Here, 8300
(8 × 1,000) + (3 × 100) = 8300
Hence, 8300 can be obtained in 8 + 3 = 11 clicks.
(b) Here, 40629
(4 × 10,000) + (6 × 100) + (2 × 10) + (9 × 1) = 40629
Hence, 40629 can be obtained in 4 + 6 + 2 + 9 = 21 clicks.
(c) Here, 56354
(5 × 10,000) + (6 × 1,000) + (3 × 100) + (5 × 10) + (4 × 1) = 56354
Hence, 56354 can be obtained in 5 + 6 + 3 + 5 + 4 = 23 clicks.
(d) Here, 66666
(6 × 10,000) + (6 × 1,000) + (6 × 100) + (6 × 10) + (6 × 1) = 66666
Hence, 66666 can be obtained in 6 + 6 + 6 + 6 + 6 = 30 clicks.
(e) We have 367813
(3 × 1,00,000) +(6 × 10,000) + (7 × 1,000) + (8 × 100) + (1 × 10) + (3 × 1) = 367813
Hence, 367813 can be obtained in 3 + 6 + 7 + 8 + 1 + 3 = 28 clicks.
Question 2.
Do you see any connection between each number and the corresponding smallest number of button clicks?
Solution:
The smallest number of button clicks = the sum of its digits.
Question 3.
If you notice, the expressions for the least button clicks also give the Indian place value notation of the numbers. Think about why this is so.
Solution:
The Indian place value system helps in identifying and differentiating between numbers easily by grouping them into thousands, lakhs, crores, etc.
1.3 Of Crores and Crores!
How many zeros does a thousand lakh have? (Page 8)
_____________
Solution:
Thousand lakh = 1,000 × 1,00,000 = 10,00,00,000
It has 8 zeros.
How many zeros does a hundred thousand have? (Page 9)
_____________
Solution:
Hundred thousand = 100 × 1,000 = 1,00,000.
It has 5 zeros.
Figure It Out (Page 9)
Question 1.
Read the following numbers in Indian place value notation and write their number names in both the Indian and American systems:
(a) 4050678
(b) 48121620
(c) 20022002
(d) 246813579
(e) 345000543
(f) 1020304050
Solution:
(a) 4050678
Indian system: 40,50,678
Forty lakh fifty thousand six hundred and seventy-eight.
International system: 4,050,678
Four million fifty thousand six hundred and seventy-eight.
(b) 48121620
Indian system: 4,81,21,620
Four crore eighty-one lakh twenty-one thousand six hundred twenty.
International system: 48,121,620
Forty-eight million one hundred twenty-one thousand six hundred twenty.
(c) 20022002
Indian system: 2,00,22,002
Two crore twenty-two thousand and two.
International system: 20,022,002
Twenty million twenty-two thousand and two.
(d) 246813579
Indian system: 24,68,13,579
Twenty-four crore sixty-eight lakh thirteen thousand five hundred and seventy-nine.
International system: 246,813,579
Two hundred forty-six million eight hundred thirteen thousand five hundred seventy-nine.
(e) 345000543
Indian system: 34,50,00,543
Thirty-four crore fifty lakh five hundred and forty-three.
International system: 345,000,543
Three hundred forty-five million five hundred forty-three.
(f) 1020304050
Indian system: 1,02,03,04,050
One arab two crore three lakh four thousand and fifty.
International system: 1,020,304,050
One billion, twenty million three hundred four thousand and fifty.
Question 2.
Write the following numbers in Indian place value notation:
(a) One crore one lakh one thousand ten
(b) One billion one million one thousand one
(c) Ten crore twenty lakh thirty thousand forty
(d) Nine billion eighty million seven hundred thousand six hundred
Solution:
(a) 1,01,01,010
(b) 1,00,10,01,001
(c) 10,20,30,040
(d) 9,080,700,600
Question 3.
Compare and write ‘<’, ‘>’ or ‘=’:
(a) 30 thousand __________ 3 lakhs
(b) 500 lakhs __________ 5 million
(c) 800 thousand __________ 8 million
(d) 640 crore __________ 60 billion
Solution:
(a) ∵ 30,000 (<) 30,00,000
∴ 30 thousand (<) 3 Iakhs.
(b) ∵ 5,00,00,000 (>) 5,000,000
∴ 500 lakhs (>) 5 million.
(c) ∵ 8,00,000 (<) 8,000,000
∴ 800 thousand (<) 8 million.
(d) 6,40,00,00,000 (<) 60,000,000,000
∴ 640 crore (<) 60 billion.
1.4 Exact and Approximate Values
Think and share situations where it is appropriate to (a) round up, (b) round down, (c) either rounding up or rounding down is okay, and (d) when exact numbers are needed.
Solution:
(а) Round Up:
Buying food for a group or occasion.
Buying materials so you don’t run short.
(b) Round Down:
The shopkeeper is saying lower prices to attract buyers.
Estimating fuel left to drive more carefully.
Estimating remaining time to a meeting (e.g., saying 10 minutes left when it’s 12 minutes) to create urgency and avoid being late.
(c) Either is Okay:
Casual talks.
Distance estimation between places.
Telling academic results.
(d) Exact Numbers Needed:
Handling money (bank balance, salary, tax).
Doing science experiments or engineering work.
Dialling emergency numbers (like 100 for the police)
Nearest Neighbours (Page 11)
Write the five nearest neighbours for these numbers:
(a) 3,87,69,957
(b) 29,05,32,481
Solution:
(a) Here, 38769,957
Nearest thousands – 3,87,70,000
Nearest ten thousands – 3,87,70,000
Nearest lakhs – 3,88,00,000
Nearest ten lakhs – 3,90,00,000
Nearest crores – 4,00,00,000
(b) Here, 29,05,32,481
Nearest thousands – 29,05,32,000
Nearest ten thousands – 29,05,30,000
Nearest lakhs – 29,05,00,000
Nearest ten lakhs – 29,10,00,000
Nearest crores – 29,00,00,000
I have a number for which all five nearest neighbours are 5,00,00,000. What could the number be? How many such numbers are there?
Solution:
A number between 4,99,99,500 and 5,00,00,499 (inclusive of the first, exclusive of the second) rounds to 5,00,00,000 at all five levels. There are 1000 such numbers.
Roxie and Estu are Estimating the Values of Simple Expressions
1. 4,63,128 + 4,19,682,
Roxie: “The sum is nearly 8,00,000 and is more than 8,00,000.”
Estu: “The sum is nearly 9,00,000 and is less than 9,00,000.”
(a) Are these estimates correct? Whose estimate is closer to the sum?
(b) Will the sum be greater than 8,50,000 or less than 8,50,000? Why do you think so?
(c) Will the sum be greater than 8,83,128 or less than 8,83,128? Why do you think so?
(d) Exact value of 4,63,128 + 4,19,682 = _____________
Solution:
(a) The estimate is correct.
(b) Yes, it would be more than 8,50,000 because if we add the first 2 digits from left, the sum is 87, which is greater than 85.
(c) The sum would be less than 8,83,128.
(d) Exact value of 463128 + 419682 = 8,82,810
2. 14,63,128 – 4,90,020
Roxie: “The difference is nearly 10,00,000 and is less than 10,00,000.”
Estu: “The difference is nearly 9,00,000 and is more than 9,00,000.”
(a) Are these estimates correct? Whose estimate is closer to the difference?
(b) Will the difference be greater than 9,50,000 or less than 9,50,000? Why do you think so?
(c) Will the difference be greater than 9,63,128 or less than 9,63,128? Why do you think so?
(d) Exact value of 14,63,128 – 4,90,020 = _____________
Solution:
(a) Exact difference = 14,63,128 – 4,90,020 = 9,73,108
Estimated difference = 15,00,000 – 5,00,000 = 10,00,000
Roxie’s estimate is closer to the actual difference.
(b) Exact difference = 9,73,108
It is greater than 9,50,000.
Now estimating the numbers 14,63,128 and 4,90,020 to the nearest ten thousands place, we get 14,60,000 and 4,90,000 respectively.
Now, Difference = 14,60,000 – 4,90,000 = 9,70,000
It is more than 9,50,000.
(c) Exact difference is 9,73,108
It is greater than 9,63,128.
Difference = 9,73,108, – 9,63,128 = 9,980
It is far from the actual difference.
(d) Exact value = 4,63,128 – 4,90,020 = 9,73,108
Population of Cities (Pages 12-13)
Observe the populations of some Indian cities in the table below.
From the information given in the table, answer the following questions by approximation:
Question 1.
What is your general observation about this data? Share it with the class.
Solution:
Most cities have seen a significant rise in pollution from 2001 to 2011, with a few cities like Bengaluru, Surat, and Vadodara almost doubling their population. Some cities, like Kolkata, have shown very little growth.
Question 2.
What is an appropriate title for the above table?
Solution:
Population of major Indian cities in 2001 and 2011.
Question 3.
How much was the population of Pune in 2011? Approximately, by how much has it increased compared to 2001?
Solution:
Pune’s population in 2011 = 31,15,431
Pune’s population in 2001 = 25,38,473
Increase in population = 31,15,431 – 25,38,473 = 5,76,958
Approximately, the population increased by about 5.8 lakhs or nearly 6 lakhs.
Question 4.
Which city’s population increased the most between 2001 and 2011?
Solution:
Population increase of some cities:
Bengaluru: 84,25,970 – 43,01,326 = 41,24,644
Hyderabad: 68,09,970 – 36,37,483 = 31,72,487
Ahmedabad: 55,70,585 – 35,20,085 = 20,50,500
Surat: 44,67,797 – 24,33,835 = 20,33,962
Vadodara: 35,52,371 – 16,90,000 = 18,62,371
Bengaluru had the highest population increase: 41,24,644.
Question 5.
Are there cities whose population has almost doubled? Which are they?
Solution:
Yes, several cities nearly doubled their population. They are:
(i) Bengaluru
2001: 43,01,326
2011: 84,25,970
(ii) Hyderabad
2001: 36,37,483
2011: 68,09,970
(iii) Surat
2001: 24,33,835
2011: 44.67,797
(iv) Vadodara
2001: 16,90,000
2011: 35,52,371
Question 6.
By what number should we multiply Patna’s population to get a number/population close to that of Mumbai?
Solution:
Here, Mumbai (2011): 1,24,42,373
Patna (2011): 16,84,222
Multiplier = \(\frac{1,24,42,373}{16,84,222}\) = 7.39
Hence, we should multiply Patna’s population by 7 to get a number close to that of Mumbai.
1.5 Patterns in Products
Using the meaning of multiplication and division, can you explain why multiplying by 5 is the same as dividing by 2 and multiplying by 10? (Page 14)
Solution:
Multiplying by 5 is the same as multiplying by 10 and then dividing by 2 because
\(\frac{10}{2}\) = (10 ÷ 2)/(2 ÷ 2) = 5
Figure It Out (Page 14)
Question 1.
Find quick ways to calculate these products:
(a) 2 × 1768 × 50
(b) 72 × 125 [Hint: 125 = \(\frac{1000}{8}\)]
(c) 125 × 40 × 8 × 25
Solution:
(a) Here, 2 × 1768 × 50
= 2 × 50 × 1768
= 100 × 1768
= 176800
(b) Here, 72 × 125
= 72 × \(\frac{1000}{8}\)
= 9 × 1000
= 9000
(c) Here, 125 × 40 × 8 × 25
= 125 × 8 × 40 × 25
= 1000 × 1000
= 1000000
Question 2.
Calculate these products quickly
(a) 25 × 12 = _____________
(b) 25 × 240 = _____________
(c) 250 × 120 = _____________
(d) 2500 × 12 = _____________
(e) _____________ × _____________ = 120000000
Solution:
(a) Here, 25 × 12 = \(\frac{100}{4}\) × 12
= 100 × 3
= 300
(b) Here, 25 × 240 = \(\frac{100}{4}\) × 240
= 100 × 60
= 6000
(c) Here, 250 × 120 = \(\frac{1000}{4}\) × 120
= 1000 × 30
= 30000
(d) Here, 2500 × 12 = \(\frac{10000}{4}\) × 12
= 10,000 × 3
= 30,000
(e) Here, 25000 × 4800
= \(\frac{1,00,000}{4}\) × 4800
= 1,00,000 × 1200
= 12,00,00,000
How Long is the Product?
In each of the following boxes, the multiplications produce interesting patterns. Evaluate them to find the pattern. Extend the multiplications based on the observed pattern. (Page 14)
Solution:
Observe the number of digits in the two numbers being multiplied and their product in each case. Is there any connection between the numbers being multiplied and the number of digits in their product? (Page 15)
Solution:
Yes, when we multiply
1- digit × 1-digit = 1 or 2 digits
2- digit × 2-digit = 3 or 4 digits
3- digit × 3-digit = 5 or 6 digits
4-digit × 4-digit = 7 or 8 digits
Roxie says that the product of two 2-digit numbers can only be a 3- or a 4-digit number. Is she correct? (Page 15)
Solution:
Yes, she is correct.
Because, product of smallest two 2-digit numbers = 10 × 10 = 100 (3-digit number)
And the product of the largest two 3-digit numbers = 99 × 99 = 9801 (4-digit number).
Should we try all possible multiplications with 2-digit numbers to tell whether Roxie’s claim is true? Or is there a better way to find out? (Page 15)
Solution:
No, we don’t need to do all possible multiplications, check extremes: 10 × 10 and 99 × 99 give the minimum and maximum digits possible.
Can multiplying a 3-digit number by another 3-digit number give a 4-digit number? (Page 15)
Solution:
Product of smallest two 3-digit numbers = 100 × 100 = 10,000 (5-digit number)
Product of largest two 3-digit numbers = 999 × 999 = 9,98,001 (6-digit number)
Hence, a 4-digit number can’t be obtained by multiplying two 3-digit numbers.
Can multiplying a 4-digit number by a 2-digit number give a 5-digit number? (Page 15)
Solution:
Here, product of smallest 4-digit number and a 2-digit number = 1,000 × 10 = 10,000 (4-digit number)
and product of largest 4-digit number and a 2-digit number = 9,999 × 99 = 9,89,901 (5-digit number)
Yes, it is possible.
Observe the multiplication statements below. Do you notice any patterns? See if this pattern extends to other numbers as well. (Page 15)
Solution:
1-digit × 1-digit = 1 or 2 digits
2-digit × 1-digit = 2 or 3 digits
2- digit × 2-digit = 3 or 4 digits
3- digit × 3-digit = 5 or 6 digits
4-digit × 4-digit = 7 or 8 digits
5- digit × 5-digit = 9-10 digits (Min = 5 + 5 – 1, Max = 5 + 5)
8 digit × 3-digit = 10-11 digits (Min = 8 + 3 – 1, Max = 8 + 3)
12 digit × 13-digit = 24-25 digits (Min = 12 + 13 – 1, Max = 12 + 13)
Fascinating Facts about Large Numbers
1250 × 380 __________ is the number of Kirtanas composed by Purandaradasa according to legends. (Page 16)
Solution:
Kirtanas = 1250 × 380
= (125 × 10) × (38 × 10)
= 125 × 38 × 10 × 10
= 4,750 × 100
= 4,75,000
2100 × 70,000 __________ is the approximate distance in kilometers between the Earth and the Sun. (Page 16)
Solution:
Distance = 2100 × 70,000
= (21 × 1,000) × (7 × 10,000)
= 21 × 7 × 1,000 × 10,000
= 147 × 1,00,00,000
= 1,47,00,00,000 km
6400 × 62,500 __________ is the average number of litres of water the Amazon River discharges into the Atlantic Ocean every second. (Page 17)
Solution:
Required Average = 6,400 × 62,500
= (64 × 100) × (625 × 100)
= 64 × 625 × 100 × 100
= 40,000 × 10,000
= 40,00,00,000 litres.
13,95,000 ÷ 150 __________ is the distance (in kms) of the longest single-train journey in the world. (Page 17)
Solution:
Required Distance = \(\frac{13,95,000}{150}\)
= \(\frac{1395}{15} \times \frac{1000}{10}\)
= 93 × 100
= 9300 km
Adult blue whales can weigh more than 10,50,00,000 ÷ 700 __________ kilograms. (Page 18)
Solution:
Here, weight = \(\frac{1,05,00,00,000}{700}\)
= \(\frac{105}{7} \times \frac{1,00,00,000}{100}\)
= 1,50,000 kg
52,00,00,00,000 ÷ 130 _____________ was the weight, in tonnes, of global plastic waste generated in the year 2021. (Page 18)
Solution:
Here, weight (in tonnes) = \(\frac{52,00,00,00,000}{130}\) = 40,00,00,000
1.6 Did You Ever Wonder….?
The RMS Titanic carried about 2500 passengers. Can the population of Mumbai fit into 5000 such ships? (Page 19)
Solution:
Here, passengers in one ship = 2500
and passengers in 5000 ships = 5000 × 2500 = 1,25,00,000.
Also, population of Mumbai = 1,24,00,000
Hence, the population of Mumbai can easily fit into 5000 such ships.
Find out if you can reach the Sun in a lifetime, if you travel 1000 kilometers every day.
(You had written down the distance between the Earth and the Sun in a previous exercise.)
Solution:
Distance between the Sun and the Earth = 14,70,00,000 km
Distance travelled in one day = 1000 km
Distance travelled in one year (365 days) = 365 × 1000 = 3,65,000 km
Time (in years) to reach the Sun = \(\frac{14,70,00,000}{3,65,000}\) = 403 years
Since a man can’t live up to 403 years.
Hence, you can’t reach the Sun in a lifetime.
Make necessary reasonable assumptions and answer the questions below: (Page 19)
(a) If a single sheet of paper weighs 5 grams, could you lift one lakh sheets of paper together at the same time?
(b) If 250 babies are born every minute across the world, will a million babies be born in a day?
(c) Can you count 1 million coins in a day? Assume you can count 1 coin every second.
Solution:
(a) Here, the weight of a single sheet = 5 grams
Weight of 1 lakh sheets = 1,00,000 × 5
= 5,00,000 grams
= \(\frac{5,00,000}{1000}\)
= 500 kg
Since a typical person can’t lift 500 kg.
Hence, you couldn’t lift one lakh sheets of paper.
(b) Here, Babies born in a minute = 250 babies
Babies born in an hour = 250 × 60 = 15,000 babies
Babies born in a day = 15,000 × 24 = 3,60,000 babies
Now, 3,60,000 is less than 1 million (10 lakh).
Hence, a million babies will not be bom in a year.
(c) Here, coins counted per sec = 1 coin
Coins counted per minute = 60 coins
Coins counted per hour = 60 × 60 = 3,600 coins
Coins counted per day = 3,600 × 24 = 86,400 coins
Since 84,000 is much less than 1 million (10 lakh).
Hence, you can’t count 1 million coins in a day.
Figure It Out (Pages 19-21)
Question 1.
Using all digits from 0-9 exactly once (the first digit cannot be 0) to create a 10-digit number, write the
(a) Largest multiple of 5
(b) Smallest even number
Solution:
(a) Arranging digits in decreasing order = 9, 8, 7, 6, 5, 4, 3, 2, 1, 0
All multiples of 5 can end only in 5 or 0.
Hence, the largest multiple of 5 = 9876543210.
(b) Arranging digits in ascending order = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Since, number can’t start with zero.
So, the smallest number = 1023456789
Swapping the last two digits to make the above number even.
Hence, the smallest even number = 1023456798
Question 2.
The number 10,30,285 in words is Ten lakhs thirty thousand two hundred eighty five, which has 43 letters. Give a 7-digit number name that has the maximum number of letters.
Solution:
Using 7 (seven) and 8 (eight) to make such a number since both numbers contain 5 letters when written in words.
77,77,777 = Seventy-seven lakh seventy-seven thousand seven hundred seventy-seven.
Letters in 77,77,777 = 60.
88,88,888 = Eighty-eight lakh eighty-eight thousand eight hundred eighty-eight.
Letters in 88,88,888 = 57.
Hence, 77,77,777 has the maximum number of letters in its number name.
Question 3.
Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?
Solution:
Number must be the smallest possible = 123456789
Any swap, for example, 213456789, is larger.
Question 4.
Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.
Solution:
1. Consider the First Digit: We have 20 digits and need to strike out 10, leaving us with 10 digits. To maximize the resulting number, we want the leftmost digit to be as large as possible. Looking at the original number 12345123451234512345, the first few digits are small. We can strike out the initial 12345 to potentially get a larger digit in the first position.
2. Look at the next available digits: After striking out the first five digits, we have 123451234512345. The first digit is now ‘1’. Let’s see if we can do better. If we instead only strike out the first four digits (1234), the remaining number starts with ‘ 5 ’. This is better than ‘ 1 ’.
3. Continue the process: Now our number starts with ‘5 ’. We’ve struck out 4 digits so far and need to strike out 6 more. The remaining number is 5123451234512345. We want the next digit to be as large as possible. We can strike out the ‘1234’ following the ‘5’, leaving us with 551234512345. We’ve now struck out 4 + 4 = 8 digits.
4. Final Steps: We need to strike out two more digits from 551234512345. To keep the number as large as possible, we should strike out the ‘1’ and the ‘2’. Therefore, by striking out the digits 1234, then 1234, then 1, then 2, the remaining number is 5,53,45,12,345.
Question 5.
The words ‘zero’ and ‘one’ share letters ‘e’ and ‘o’. The words ‘one’ and ‘two’ share a letter ‘0’, and the words ‘two’ and ‘three’ also share a letter ‘t’. How far do you have to count to find two consecutive numbers that do not share an English letter in common?
Solution:
Here;
words one & two share o
words two & three share t, e
words three & four share r
words four & five share f
words five & six share i
words six & seven share s
words seven & eight share e
words eight & nine share e, i, n
nine & ten share n, e
It shows that all consecutive numbers have at least one common letter.
Hence, there is no such pair of consecutive numbers that do not share an English letter in common.
Question 6.
Suppose you write down all the numbers 1, 2, 3, 4, ….., 9, 10, 11,….. The tenth digit you write is ‘1’ and the eleventh digit is ‘0’, as part of the number 10.
(a) What would the 1000th digit be? At which number would it occur?
(b) What number would contain the millionth digit?
(c) When would you have written the digit ‘5’ for the 5000th time?
Solution:
(a) Here, digits from 1 to 9 = 9 digits
Digits from 10 to 99 = 90 numbers × 2 = 180 digits
Total digits from 1 to 99 = 180 + 9 = 189 digits
Remaining digits to reach 1000th digit = 1000 – 189 = 811
Numbers with 3-digit numbers = \(\frac{811}{3}\) = 270
full numbers + 1 digit left over.
The first 3-digit number is 100
270th 3 digit number is 100 + 270 – 1 = 369
The next number is 370.
Hence, the first digit of 370, which is 3, is the 1000th digit.
(b) 1. 1-digit (1-9) numbers = 9 digits
2. 2-digit (10-99) numbers 90 × 2 = 180 digits
Total = 9 + 180 = 189
3. 3-digit (100 – 999): 900 × 3 = 2700
Total = 189 + 2700 = 2889
4. 4-digit (1000 – 9999): 9000 × 4 = 36,000
Total = 2889 + 36000 = 38,889
5. 5-digit (10000 – 99999): 90000 × 5 = 4,50,000
Total = 4,50,000 + 38,889 = 4,88,889
6. Now we’re in 6-digit numbers
Needed = 1,000,000 – 488,889 = 511,111 digits left
Each 6-digit number = 6 digits
So, \(\frac{511,111}{6}\) = 85,185 full numbers, remainder 1
So, the millionth digit is the 1st digit of the 85,186th 6-digit number.
7. First 6-digit number = 100,000 + 85186 – 1 = 1,85,185
The millionth digit occurs in the number = 1,85,185
(c) 1. 1-digit (1-9): ‘5’ comes once
total = 1
2. 2-digit (10-99): ‘5’ comes 19 times
total = 1 + 19 = 20
3. 3-digit (100-999): ‘5’ appears 280 times
total = 280 + 20 = 300
4. 4-digit (100-9999): 5 appears 3700 times
Total = 300 + 3700 = 4000
5. Numbers starting from 10000 onward.
For the 5000th number, we require 5000 – 4000 = 1000 more numbers that lie in 10001-10999.
6. Among 10000-10999, one digit 5 appears in 100 numbers (e.g., 10005, 10015,000).
The digit 5 appears in 100 numbers (e.g., 10500-10599).
Total = 4000 + 300 = 4300
In 11000-11999
5 at unit place = 100
5 at tens place = 100
5 at a hundred place = 100
Total = 4300 + 300 = 4600
In 12000-12999
4600 + 300 = 4900
In 13000-13999
Unit = 100
Total = 5000
Final number = 13995.
Question 7.
A calculator has only ‘+10,000’ and ‘+100’ buttons. Write an expression describing the number of button clicks to be made for the following numbers:
(a) 20,800
(b) 92,100
(c) 1,20,500
(d) 65,30,000
(e) 70,25,700
Solution:
(a) Here, 20,800 = (2 × 10,000) + (8 × 100)
Total clicks = 2 + 8 = 10 clicks.
(b) Here, 92,100 = (9 × 10,000) + (21 × 100)
Total clicks = 9 + 21 = 30 clicks.
(c) Here, 1,20,000 = (12 × 10,000)
Total clicks = 12 clicks.
(d) Here, 65,30,000 = (653 × 10,000)
Total clicks = 653 clicks.
(e) Here, 70,25,700 = (702 × 10,000) + (57 × 100)
Total clicks = 759 clicks.
Question 8.
How many lakhs make a billion?
Solution:
We know, 1 lakh = 1,00,000
1 billion = 1,000,000,000
Number of lakhs making a billion = \(\frac{1,000,000,000}{1,00,000}\) = 10,000
Question 9.
You are given two sets of number cards numbered from 1-9. Place a number card in each box below to get the (a) largest possible sum, (b) smallest possible difference of the two resulting numbers.
Solution:
(a) To get the largest possible sum, both the 7-digit and 5-digit numbers need to be the largest.
7-digit number = 98,76,543
5-digit number = 98,765
Longest possible sum = 98,76,543 + 98,765 = 99,75,308.
(b) To get the smallest possible difference, the 7-digit number needs to be the smallest, and the 5-digit number needs to be the largest.
7-digit number = 12,34,567
5-digit number = 98,765
Smallest possible difference = 12,34,567 – 98,765 = 11,35,802
Question 10.
You are given some number cards: 4000, 13000, 300, 70000, 150000, 20, and 5. Using the cards, get as close as you can to the numbers below using any operation you want. Each card can be used only once to make a particular number.
(a) 1,10,000: Closest I could make is 4000 × (20 + 5) + 13000 = 1,13,000
(b) 2,00,000
(c) 5,80,000
(d) 12,45,000
(e) 20,90,800
Solution:
(a) 1,10,000
Closest estimate = 70,000 + (4,000 × 5) + 13,000
= 70,000 + 20,000 + 13,000
= 1,03,000
(b) 2,00,000
Closest estimate = 1,50,000 + 70,000 – 4,000 × 5
= 1,50,000 + 70,000 – 20,000
= 2,00,000
(c) 5,80,000
Closest estimate = 70,000 × 5 + 1,50,000 + 4,000 × 20
= 3,50,000 + 1,50,000 + 80,000
= 5,80,000
(d) 12,45,000
Closest estimate = 70,000 × 20 – 1,50,000 – 4,000 – 300 × 5
= 14,00,000 – 1,50,000 – 4,000 – 1,500
= 12,44,500
It is very close to 12,45,000
(e) 20,90,800
Closest estimate = 13,000 × 300 – 70,000 (20 + 5) – 1,50,000 + 4,000
= 39,00,000 – 17,50,000
= 20,04,000.
Question 11.
Find out how many coins should be stacked to match the height of the Statue of Unity. Assume each coin is 1 mm thick.
Solution:
Here, the height of each coin = 1 mm
Height of Statue of Unity = 180 m = 180 × 100 × 10 mm = 1,80,000 mm.
Number of coins to be stacked = (1,80,000 mm)/(1 mm) = 1,80,000 coins.
Question 12.
Grey-headed albatrosses have a roughly 7-foot wingspan. They are known to migrate across several oceans. Albatrosses can cover about 900 – 1000 km in a day. One of the longest single trips recorded is about 12,000 km. How many days would such a trip take to cross the Pacific Ocean approximately?
Solution:
Here, total trip = 12,000 km
Distance covered in a day = 900 – 1000 km
Estimated Number of days if it flies 900 km/day = \(\frac{12,000}{900}\) = 13.3 days
Estimated Number of days if it Hies 1000 km/day = \(\frac{12,000}{1,000}\) = 12 days
Hence, it would take approximately 12 to 14 days for a grey-headed albatross to complete a 12,000 km trip across the Pacific Ocean.
Question 13.
A bar-tailed godwit holds the record for the longest recorded non-stop flight. It travelled 13,560 km from Alaska to Australia without stopping. Its journey started on 13 October 2022 and continued for about 11 days. Find out the approximate distance it covered every day. Find out the approximate distance it covered every hour.
Solution:
Here, total distance = 13,560 km
Duration = 11 days Each day = 24 hours
Duration in hours =11 × 24 = 264 hours
Hence, distance covered per day = (13,560 km)/(11 days) = 1232.73 km or 1233 km.
Hence, distance covered per hour = (13,560 km)/(264 hours) = 51.36 km or 51 km.
Question 14.
Bald eagles are known to fly as high as 4500 – 6000 m above the ground level. Mount Everest is about 8850 m high. Aeroplanes can fly as high as 10,000 – 12,800 m. How many times bigger are these heights compared to Somu’s building?
Solution:
Here, the height of Somu’s building = 40 m
(i) Now, Bald eagles’ flight height = 4500 – 6000 m
Lower estimate = \(\frac{4500}{40}\) = 112.5
Upper estimate = \(\frac{6000}{40}\) = 150
∴ Bald eagles fly about 112 to 150 times higher than Somu’s building.
(ii) Also, Mount Everest = 8850 m
\(\frac{8850}{40}\) = 221.25
Mount Everest is about 221 times taller than Somu’s building.
(iii) Now, Aeroplanes flight height = 10,000 – 12,000 m
Lower estimate = \(\frac{10,000}{40}\) = 250
Upper estimate = \(\frac{12,000}{40}\) = 320
Hence, airplanes fly about 250 to 320 times higher than Somu’s building.