Large Numbers Around Us ⇒⇒ Exercise 1

Question 1

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?”

Will they be able to taste all the lakh varieties in a 100-year lifetime?

Solution :

Step 1: Calculate the total number of days in 100 years.

( Number of days in one year = 365 )

Number of days in 100 years =

$$ { 365 × 100 } $$

$$⇒ {36,500} Days $$

Step 2: Calculate the total number of rice varieties eaten in 100 years.

Varieties eaten per day = 2

Total Rice varieties eaten in 100 years =

$$ { 36,500 × 2} $$

$$ {= 73,000} $$

Step 3: Compare the total varieties eaten with 1 lakh varieties.

Target varieties = 1 lakh = 1,00,000

Total Rice varieties = 73,000

Since 73,000 is less than 1,00,000,

They can not eat 1 lakh varieties in 100-year lifetime .

Question 2

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.

Solution :

Step 1: Calculate the total number of days in 100 years.

( Number of days in one year = 365 )

Number of days in 100 years =

$$ { 365 × 100 } $$

$$= {36,500} Days $$

Step 2: Calculate the total number of rice varieties eaten in 100 years.

Varieties eaten per day = 3

Total Rice varieties eaten in 100 years =

$$ { 36,500 × 3} $$

$$ {= 1,09,500} $$

Step 3: Compare the total varieties eaten with 1 lakh varieties.

Target varieties = 1 lakh = 1,00,000

Total Rice varieties = 1,09,500

Since 1,09,500 is greater than 1,00,000,

They can eat 1 lakh varieties in 100-year lifetime .

Question 3

Choose a number for Y.
How close to one lakh is the number of days in y years,
for the y of your choice?

Solution :

Step 1: Let's choose a number for 'y' .

y = 50 years .

( Number of days in one year = 365 )

To get the number of days in y years = 365 × y

Number of days in 50 years =

$$ { 365 × 50 } $$

$$ { 18,250 } $$

Total number of days in 50 years: 18,250 Days

Step 2: Find how close 18,250 days is to one lakh (1,00,000).

$$ { 1,00,000- 18,250 } $$

$$ {= 81,750} $$

For my chosen y = 50 years, the number of days is 18,250 days.

This number is 81,750 days away from one lakh (1,00,000 days).

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 :

Step 1: Identify the two numbers.

Population of Chintamani = 75,000

One lakh = 1,00,000

Step 2: Subtract the smaller number from the larger number to find the difference.

$$ { 1,00,000- 75,000 } $$

$$ {= 25,000} $$

Answer : The population is 25,000 less than one lakh.

Question 2

The estimated population of Chintamani in the year 2024 is 1,06,000.
How much more than one lakh is 1,06,000?

Solution :

Step 1: Identify the two numbers.

Population of Chintamani = 1,06,000

One lakh = 1,00,000

Step 2: Subtract the smaller number from the larger number to find the difference.

$$ {1,06,000- 1,00,000 } $$

$$ {= 6,000} $$

Thus, the population in 2024 is 6,000 more than one lakh.

Question 3

By how much did the population of Chintamani increase from 2011 to 2024?

Solution :

Step 1: Identify the two numbers.

Population of Chintamani in 2011 = 75,000

Population of Chintamani in 2024 = 1,06,000

Step 2: Subtract the smaller number from the larger number to find the difference.

$$ {1,06,000- 75,000 } $$

$$ {= 31,000} $$

The population increased by 31,000 from year 2011 to 2024 .

Question 1

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 :

Given :

Somu's height = 1 metre

Number of floors in the building = 10 floors

Height of a floor = 4 × Somu’s height

$$ { 4 × 1 m} $$

$$ {= 4 m} $$

Height of the building = Height of each floor × Number of floors

$$ { 4 m × 10} $$

$$ { =40 m} $$

Answer: The approximate height of the building is 40 metres.

Question 2

Which is taller — The Statue of Unity or this building?
How much taller? ____________m.

Solution :

Given :

Height of Somu’s building = 40 m.

Height of the Statue of Unity = 180 m.

The Statue of Unity (180 m) is clearly taller than the building (40 m)

To find out how much taller:

Difference = Height of Statue of Unity - Height of the Somu’s building

$$ { 180 - 40 m} $$

$$ {= 140 m} $$

Answer: The Statue of Unity is taller. It is 140 m taller.

Question 3

How much taller is the Kunchikal waterfall than Somu's building? ____________m.

Solution :

Given :

Height of Somu’s building = 40 m.

Height of the Kunchikal waterfall = 450 m.

The Kunchikal waterfall (450 m) is clearly taller than the building (40 m)

To find out how much taller:

Difference = Height of the Kunchikal waterfall - Height of the Somu’s building

$$ { 450 - 40 m} $$

$$ {= 410 m} $$

Answer: The Kunchikal waterfall is taller. It is 410 m taller.

Question 4

How many floors should Somu’s building have to be as high as the waterfall?

Solution :

Given :

Height of each floor in the Somu’s building = 4 m.

Height of the Kunchikal waterfall = 450 m.

Number of floors required

= Height of the Kunchikal waterfall / Height of each floor in the building

$$ {450 \over 4 } $$

$$ {= 112.5 } $$

$$ {= 113 Approx.} $$

So, building would need approximately 113 floors to be as high as the waterfall.

Question 1

How do you view a lakh -- is a lakh big or small ?

Solution :

A lakh (1,00,000) is a term primarily used in the Indian numbering system.

Whether it's perceived as "big" or "small" is entirely dependent on the context of what is being counted or measured.

It’s big for things like the attendance at a small event, 1 lakh people attending an event would be a massive gathering,
for large-scale economic or demographic data, it's a smaller unit.

Question 2

Write each of the numbers given below in words:
(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 thirty-eight.

Question 3

Write the corresponding number in the Indian place value system for each of the following:
(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 hundred and thirty five → – 10,00,235.

Question 1

In the Land of Tens, there are special calculators with special buttons.
The Thoughtful Thousands only has a +1000 button. How many times should it be pressed to show:

Solution :

(a) Three thousand?____________?

$$ {3000 \over 1000 } $$

$$ {= 3 } $$

Ans = Three Times

(b) 10,000?____________?

$$ {10,000 \over 1000 } $$

$$ {= 10 } $$

Ans = Ten Times

(c) Fifty-three thousand?_____________?

$$ {53000 \over 1000 } $$

$$ {= 53 } $$

Ans = Fifty Three Times

(d) 90,000?____________?

$$ {90,000 \over 1000 } $$

$$ {= 90 } $$

Ans = 90 Times

(e) One Lakh?____________?

$$ {1,00,000 \over 1000 } $$

$$ {= 100 } $$

Ans = 100 Times

(f) ____________? 153 times

$$ {153 × 1,000 } $$

$$ {= 1,53,000 } $$

Ans = 1,53,000

(g) How many thousands are required to make one lakh?

$$ {1,00,000 \over 1000 } $$

$$ {= 100 } $$

Ans = 100 thousands.

Question 2

In the Land of Tens, there are special calculators with special buttons.
The Tedious Tens only has a +10 button. How many times should it be pressed to show:

Solution :

(a) Five hundred ?______________?

$$ {500 \over 10 } $$

$$ {= 50} $$

Ans = 50 Times

(b) 780 ?____________?

$$ {780 \over 10 } $$

$$ {= 78 } $$

Ans = 78 Times

(c) 1000 ?_____________?

$$ {1000 \over 10 } $$

$$ {= 100 } $$

Ans = 100 Times

(d) 3700 ?____________?

$$ {3,700 \over 10 } $$

$$ {= 370 } $$

Ans = 370 Times

(e) 10,000 ?____________?

$$ {10,000 \over 10 } $$

$$ {= 1,000 } $$

Ans = 1,000 Times

(f) One lakh ?____________?

$$ {1,00,000 \over 10 } $$

$$ {= 10,000 } $$

Ans = 10,000 Times

(g) _____________? 435 times

$$ {435 × 10 } $$

$$ {= 4,350 } $$

Ans = 4,350

Question 3

In the Land of Tens, there are special calculators with special buttons.
The Handy Hundreds only has a +100 button. How many times should it be pressed to show:

Solution :

(a) Four hundred ? ___________ times

$$ {400 \over 100 } $$

$$ {= 4} $$

Ans = 4 Times

(b) 3,700 ? _______________________?

$$ {3,700 \over 100 } $$

$$ {= 37 } $$

Ans = 37 Times

(c) 10,000 ?_____________?

$$ {10,000 \over 100 } $$

$$ {= 100 } $$

Ans = 100 Times

(d) Fifty-three thousand ?____________?

$$ {53,000 \over 100 } $$

$$ {= 530 } $$

Ans = 530 Times

(e) 90,000 ?____________?

$$ {90,000 \over 100 } $$

$$ {= 900 } $$

Ans = 900 Times

(f) 97,600 ?____________?

$$ {97,600 \over 100 } $$

$$ {= 976 } $$

Ans = 976 Times

(g)1,00,000 ?____________?

$$ {1,00,000 \over 100} $$

$$ {= 1,000 } $$

Ans = 1,000 Times

(h) ___________? 582 times

$$ {582 × 100 } $$

$$ {= 58,200} $$

Ans = 58,200

(i) How many hundreds are required to make ten thousand?

$$ {10,000 \over 100} $$

$$ {= 100 } $$

Ans = 100 hundreds

( j ) How many hundreds are required to make one lakh?

$$ {1,00,000 \over 100} $$

$$ {= 1,000 } $$

Ans = 1,000 hundreds

(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.

Ans =

Handy Hundreds' statement, "There are some numbers which Tedious Tens and Thoughtful Thousands can’t show but I can," is partially true, but also partially false.

It is FALSE regarding "Tedious Tens," because any number Handy Hundreds can show (multiples of 100) can also be shown by Tedious Tens (multiples of 10).

It is TRUE regarding "Thoughtful Thousands," because Handy Hundreds can show multiples of 100 that are not multiples of 1000 (e.g., 100, 200, ..., 900, 1500, etc.).

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, it presses +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 :

Method 1 :

Presses +10 thirty-two times:

$$ {32 × 10 }= 320 $$

Presses +1 one time:

$$ {1 × 1 }= 1$$

Total = 320 + 1= 321

Method 2 :

Presses +100 three times:

$$ {3 × 100 }= 300 $$

Presses +10 two times:

$$ {2 × 10 }= 20 $$

Presses +1 one time:

$$ {1 × 1 }= 1$$

Total = 300 + 20 + 1= 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.

Solution :

We need to make 5072. We can combine different numbers of thousands, hundreds, tens, and ones.

(4 × 1000) + (10 × 100) + (7 × 10) + (2 × 1) = 5072

4 is in the thousands place = 4 × 1000 = 4000

10 is in the hundreds place = 10 × 100 = 1000

7 is in the tens place = 7 × 10 = 70

2 is in the ones place = 2 × 1 = 2

Total = 4000+ 1000 + 70 + 2 = 5072

Figure it Out

For each number given below, write expressions for at least two different ways to obtain the number through button clicks. Think like Chitti and be creative:
(a) 8300
(b) 40629
(c) 56354
(d) 66666
(e) 367813

Solution :

We can combine different numbers of thousands, hundreds, tens, and ones.

(a)

( 8 × 1000 )+ ( 3 × 100) = 8300

(83 × 100) = 8300

(b) 40629

(40 × 1000) + (6 × 100) + (2 × 10) + (9 × 1) = 40629

(3 × 10,000) + (10 × 1000) + (5 × 100) + (12 × 10) + (9 × 1) = 40629

(c) 56354

(56 × 1000) + (3 × 100) + (5 × 10) + (4 × 1) = 56354

(5 × 10,000) + (6 × 1000) + (2 × 100) + (15 × 10) + (4 × 1) = 56354

(d) 66666

(6 × 10,000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1) = 66666

(5 × 10,000) + (16 × 1000) + (4 × 100) + (26 × 10) + (6 × 1) = 66666

(e) 367813

(3 × 1,00,000) + (6 × 10,000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1) = 367813

(3 × 1,00,000) + (6 × 10,000) + (6 × 1000) + (15 × 100) + (30 × 10) + ( 13 × 1) = 367813

Question 1

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 :

Largest 3-Digit Number:

To make the largest possible number with a fixed number of presses, we want to maximize the value contributed by each press. This means using the button with the highest value that keeps the total within the 3-digit range (less than 1000).

Presses +100 nine times:

$$ {9 × 100 }= 900 $$

Presses +10 eight times:

$$ {8 × 10 }= 80 $$

Presses +1 thirteen time:

$$ {13 × 1 }= 13$$

Total = 900 + 80 + 13 = 993

Total presses: = 9 + 8 + 13 = 30

Therefore, 993 is the largest 3-digit number made with 30 clicks.

Smallest 3-Digit Number:

To make the smallest possible number (more than 99) with a fixed number of presses, we want to maximize the value contributed by each press, but this time using the lowest value buttons for the majority of presses to keep the number small, and then use higher value buttons just enough to reach the 3-digit threshold (more than 99).

Presses +10 eight times:

$$ {8 × 10 }= 80 $$

Presses +1 twenty two time:

$$ {22 × 1 }= 22 $$

Total = 80 + 22 = 102

Total presses: = 8 + 22 = 30

Therefore, 102 is the smallest 3-digit number made with 30 clicks.

(b) 997 can be made using 25 clicks.

( 9 × 100 )+ ( 9 × 10) + ( 7 × 1) = 997

Total = 900 + 90 + 7 = 997

Total presses: = 9 + 9 + 7 = 25

So, yes, 997 can be made with 25 clicks.

Can you make 997 with a different number of clicks

( 9 × 100 )+ ( 8 × 10) + ( 17 × 1) = 997

Total = 900 + 80 + 17 = 997

Total presses: = 9 + 8 + 17 = 34

So, yes, 997 can also be made with 34 clicks.

Question 2

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?
Is there another way to get 5072 using less than 23 button clicks? Write the expression for the same.

Solution :

(a) Getting 5072 with as few button clicks as possible:

( 5 × 1000 )+ ( 7 × 10) + ( 2 × 1)= 5072

Total presses: = 5 + 7 + 2 = 14 clicks

Yes, 5072 can be obtained in minimum 14 button clicks.

(b) Getting 8300 with as few button clicks as possible:

( 8 × 1000 )+ ( 3 × 100) = 8300

Total presses: = 8 + 3 = 11 clicks

Yes, 8300 can be obtained in minimum 11 button clicks.

Is there another way to get 5072 using less than 23 button clicks? Write the expression for the same.

We already found the minimum number of clicks for 5072 is 14 clicks. Since 14 is less than 23, the answer is Yes, there is a way.

The expression for getting 5072 using the minimum number of clicks (14 clicks) is:

( 5 × 1000 )+ ( 7 × 10) + ( 2 × 1)= 5072

Figure it Out Q. 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) Getting 8300 with as few button clicks as possible:

( 8 × 1000 ) + ( 3 × 100) = 8300

Total presses = 8 + 3 = 11 clicks

(b) Getting 40629 with as few button clicks as possible:

( 4 × 10000 )+ ( 6 × 100) + ( 2 × 10)+ ( 9 × 1) = 40629

Total presses = 4 + 6 + 2 + 9 = 21 clicks

(c) Getting 56354 with as few button clicks as possible:

( 5 × 10000 ) + ( 6 × 1000) + ( 3 × 100) + ( 5 × 10)+ ( 4 × 1) = 56354

Total presses = 5 + 6 + 3 + 5 + 4 = 23 clicks

(d) Getting 66666 with as few button clicks as possible:

( 6 × 10000 ) + ( 6 × 1000) + ( 6 × 100) + ( 6 × 10)+ ( 6 × 1) = 66666

Total presses = 6 + 6 + 6 + 6 + 6 = 30 clicks

(e) Getting 367813 with as few button clicks as possible:

(3 × 1,00,000) + (6 × 10,000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1) = 367813

Total presses = 3 + 6 + 7 + 8 + 1 + 3 = 28 clicks

Figure it Out Q. 2

Do you see any connection between each number and the corresponding smallest number of button clicks?

Solution :

The smallest number of button clicks for each number is the sum of its digits .

Figure it Out Q. 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 expressions for the least button clicks indeed correspond to the Indian (or standard, depending on context) place value notation of the numbers. the standard place value notation is the most efficient way to represent a number using powers of ten.

The Indian place value system helps in identifying and differentiating between numbers easily by grouping them into thousands, lakhs, crores, etc..

Question 1

How many zeros does a thousand lakh have?

Solution :

One thousand: = 1,000 (which has 3 zeros)

One lakh: = 1,00,000 (which has 5 zeros)

Thousand lakh = 1000 × 1,00,000

= 10,00,00,000

This number is also known as 10 crore in the Indian numbering system, or 100 million in the International numbering system.
A thousand lakh has 8 zeros.

Question 2

How many zeros does a hundred thousand have?

Solution :

One hundred: = 100 (which has 2 zeros)

One thousand: = 1,000 (which has 3 zeros)

Hundred thousand = 100 × 1,000

= 1,00,000

This number is also known as 1 lakh in the Indian numbering system.
A hundred thousand has 5 zeros.

Question 1

Read the following numbers in Indian place value notation and write their number names in both the Indian and American systems:

Solution :

(a) 4050678

Reading in Indian Place Value Notation: 40,50,678

Indian Number Name: Forty Lakh Fifty Thousand Six Hundred Seventy-Eight

American (International) Number Name : Four Million Fifty Thousand Six Hundred Seventy-Eight

(b) 48121620

Reading in Indian Place Value Notation: 4,81,21,620

Indian Number Name: Four Crore Eighty-One Lakh Twenty-One Thousand Six Hundred Twenty

American (International) Number Name : Forty-Eight Million One Hundred Twenty-One Thousand Six Hundred Twenty

(c) 20022002

Reading in Indian Place Value Notation: 2,00,22,002

Indian Number Name: Two Crore Twenty-Two Thousand Two

American (International) Number Name : Twenty Million Twenty-Two Thousand Two

(d) 246813579

Reading in Indian Place Value Notation: 24,68,13,579

Indian Number Name: Twenty-Four Crore Sixty-Eight Lakh Thirteen Thousand Five Hundred Seventy-Nine

American (International) Number Name : Two Hundred Forty-Six Million Eight Hundred Thirteen Thousand Five Hundred Seventy-Nine

(e) 345000543

Reading in Indian Place Value Notation: 34,50,00,543

Indian Number Name: Thirty-Four Crore Fifty Lakh Five Hundred Forty-Three

American (International) Number Name : Three Hundred Forty-Five Million Five Hundred Forty-Three

(f) 1020304050

Reading in Indian Place Value Notation: 1,02,03,04,050

Indian Number Name: One Hundred Two Crore Three Lakh Four Thousand Fifty

American (International) Number Name : One Billion Twenty Million Three Hundred Four Thousand Fifty

Question 2

Write the following numbers in Indian place value notation:

Solution :

(a) One crore one lakh one thousand ten

Ans : Reading in Indian Place Value Notation: 1,01,01,010

(b) One billion one million one thousand one

Ans : Reading in Indian Place Value Notation: 1,001,001,001.

(c) Ten crore twenty lakh thirty thousand forty

Ans : Reading in Indian Place Value Notation: 10,20,30,040.

(d) Nine billion eighty million seven hundred thousand six hundred

Ans : Reading in Indian Place Value Notation: 9,080,700,600.

Question 3

Compare and write ‘<’, ‘>’ or ‘=’:

Solution :

(a) 30 thousand ______ 3 lakhs

30,000 < 3,00,000

(b) 500 lakhs ______ 5 million

5,00,00,000 > 5,000,000

(c) 800 thousand ______ 8 million

8,00,000 < 8,000,000

(d) 640 crore ______ 60 billion

6,40,00,00,000 < 60,000,000,000

Question 1

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 :

(a) Round up

( Rounding up is typically used when you need to ensure you have enough of something, or to err on the side of caution)

1. Purchasing Materials for a function

2. Estimating Time for Tasks

(b) Round down

( Rounding down is often used when you need to be conservative, or when an excess would be problematic or impossible. )

1. Calculating the Number of Groups

2. Calculating Safe Loading Capacity

(c) Either Rounding Up or Rounding Down is Okay

( These situations typically involve estimations where the impact of minor inaccuracies is negligible, or where the choice depends on a secondary preference. )

1. Estimating Crowds

2. Estimating Travel Time (for non-critical appointments)

(d) Exact Numbers Are Needed

( In these situations, any rounding can lead to significant errors, miscalculations, or even safety hazards. Precision is paramount. )

1. Scientific Measurements and Experiments

2. Medical Prescriptions and Dosages

Question 1

Write the five nearest neighbours for these numbers:

Solution :

(a) 3,87,69,957

( "Nearest Neighbours" typically refer to the numbers closest to the given number, both above and below it, in a sequence of integer )

Nearest thousand : 3,87,70,000

Nearest ten thousand : 3,87,70,000

Nearest lakh : 3,88,00,000

Nearest ten lakh : 3,90,00,000

Nearest crore : 4,00,00,000

(b) 29,05,32,481

Nearest thousand : 29,05,32,000

Nearest ten thousand : 29,05,30,000

Nearest lakh : 29,05,00,000

Nearest ten lakh : 29,10,00,000

Nearest crore : 29,00,00,000

Question 2

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 :

The number could be between 4,99,99,501 and 5,00,00,499 as rounding to the nearest thousand, ten thousand, lakh, ten lakh, or crore all yield 5,00,00,000

Question 3

Roxie and Estu are estimating the values of simple expressions.
(1) 4,63,128 + 4,19,682
Roxie: “The sum is near 8,00,000 and is more than 8,00,000.”
Estu: “The sum is near 9,00,000 and is less than 9,00,000.”

Solution :

(a) Are these estimates correct ? Whose estimate is closer to the sum ?

exact sum = 4,63,128 + 4,19,682

= 8,82,810

Roxie's estimate: "The sum is near 8,00,000 and is more than 8,00,000."

The first part ("near 8,00,000") is less accurate, as it's significantly closer to 9,00,000.

The second part ("is more than 8,00,000") is correct.

Estu's estimate: "The sum is near 9,00,000 and is less than 9,00,000."

Both parts of Estu's statement are correct. The sum is indeed near 9,00,000 and it is less than 9,00,000.

Estu's estimate (with a difference of 17,190) is significantly closer to the actual sum of 8,82,810 than Roxie's estimate (with a difference of 82,810)..

(b) Will the sum be greater than 8,50,000 or less than 8,50,000? Why do you think so?

8,82,810 > 8,50,000

So, The exact sum is 8,82,810, which is clearly greater than 8,50,000

The sum 8,82,810 has 8 lakhs as its highest place value, just like 8,50,000. However, looking at the ten thousands place, 82,000 is greater than 50,000, which means the overall value of 8,82,810 is larger than 8,50,000

(c) Will the sum be greater than 8,83,128 or less than 8,83,128? Why do you think so?

The sum we calculated is 8,82,810.
8,82,810 < 8,83,128

So, the exact sum is 8,82,810, which is clearly a smaller number than 8,83,128.

(d) Exact value of 4,63,128 + 4,19,682 = ______________

The sum = 8,82,810.

Question 4

Roxie and Estu are estimating the values of simple expressions.
(1)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”.

Solution :

(a) Are these estimates correct ? Whose estimate is closer to the difference?

exact difference = 14,63,128 – 4,90,020

= 9,73,108

Roxie's estimate: " The difference is near 10,00,000 and is less than 10,00,000."

The first part ("near 10,00,000"), is correct , 9,73,108 is reasonably close to 10,00,000.

The second part ("less than 10,00,000") is correct.

Estu's estimate: "The difference is near 9,00,000 and is more than 9,00,000."

The second part ("is more than 9,00,000") is correct. and the first part ("near 9,00,000") is less accurate

Roxie's estimate (with a difference of 26,892) is significantly closer to the actual difference of 9,73,108 than Estu's estimate (with a difference of 73,108).

(b) Will the difference be greater than 9,50,000 or less than 9,50,000? Why do you think so?

9,73,108 > 9,50,000

So, The exact difference is 9,73,108, which is clearly greater than 9,50,000

(c) Will the difference be greater than 9,63,128 or less than 9,63,128? Why do you think so?

The difference we calculated is 9,73,108.
9,73,108 > 9,63,128

So, the difference will be greater than 9,63,128. The difference exceeds 9,63,128 by 9,980, indicating it is far from the actual difference.

(d) Exact value of 14,63,128 – 4,90,020 = ___________

The difference = 9,73,108

Question

Observe the populations of some Indian cities in the table below

From the information given in the table, answer the following questions by approximation

Solution :

Q 1: What is your general observation about this data? Share it with the class.

Ans: The data shows the vast majority of these cities experienced significant population growth between 2001 and 2011.

Some cities showed phenomenal population increases, almost doubling their size in just 10 years

Kolkata is the only city in this list that shows a slight decrease in population.

Q 2: What is an appropriate title for the above table?

Answer : Population of Major Indian Cities: 2001 and 2011 Census Data

Q 3 : How much is the population of Pune in 2011? Approximately, by how much has it increased compared to 2001?

Answer :The population of Pune in 2011 was 31,15,431.
Approximately, the population of Pune has increased by about
( 31,15,431 − 25,38,473 = 5,76,958 )or nearly 5,77,000 from 2001 to 2011

Q 4: Which city’s population increased the most between 2001 and 2011? 

Answer : Population of Bengaluru experienced the largest population increase between 2001 and 2011. Its population increased by 41,24,644.

Q 5: Are there cities whose population has almost doubled? Which are they?

Ans: Yes, several cities experienced a population growth where their numbers almost doubled between 2001 and 2011.

Bengaluru → 84,25,970 ÷ 43,01,326 (Growth Factor: 1.96)

Vadodara → 35,52,371 ÷ 16,90,000 (Growth Factor: 2.10)

Hyderabad → : 68,09,970 ÷ 36,37,483 (Growth Factor: 1.87)

Surat → 44,67,797 ÷ 24,33,835 (Growth Factor: 1.84)

Q 6 : By what number should we multiply Patna’s population to get a number/population close to that of Mumbai?

Answer :The population of Patna (2011) : 16,84,222
The population of Mumbai (2011): 1,24,42,373

Factor =

$$ {1,24,42,373 \over 16,84,222 } $$

Factor = 7.39

To get a population close to that of Mumbai, you should multiply Patna's population by approximately 7.39.

Question

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?

Solution :

The key here is the relationship between the numbers 5, 2, and 10.
We know that: When you perform (divide by 2) and then (multiply by 10), you are applying successive scale factors:

$$ {10 \over 2 } = 5 $$

We know that 5 × 2 = 10 (multiplication fact) has two division facts:
( 10 ÷ 2 = 5 ) and (10 ÷ 5 = 2 ).

So, the combined effect of dividing by 2 and then multiplying by 10 is exactly the same as directly multiplying by 5.
It's just a different sequence of operations that achieves the same net scaling effect.

Question 1

Find quick ways to calculate these products:

Solution :

Let's find quick ways to calculate these products by using the properties of multiplication, specifically looking for ways to create powers of 10.

(a) 2 × 1768 × 50

$$ {2 × 50 × 1768 } $$

( We can easily multiply 2 and 50 first, as their product is a power of 10 (100) )

$$ ={100 × 1768 } $$

( Multiply by 100 (add two zeros))

$$ ={176800 } $$

Answer: 2 × 1768 × 50 = 176800

(b) 72 × 125 [Hint: 125 = 1000 ÷ 8]

$$ {72 × 125 } $$

$$ ={72 × {1000 \over 8 }} $$

( Use the hint 125 = 1000 ÷ 8 . This allows us to divide by 8 first, which simplifies the number before multiplying by 1000. )

$$ ={{72 \over 8 } × 1000} $$

( Rearrange to perform division first)

$$ ={9 × 1000} $$

$$ ={9000} $$

Answer: 72 × 125 = 9000

(c) 125 × 40 × 8 × 25

$$ {125 × 40 × 8 × 25 } $$

$$ ={(125 × 8) × (40 × 25) } $$

( Rearrange and group the pairs of numbers that multiply to powers of 10 or easily manageable multiples. )

$$ {1000 × 1000} $$

( Perform multiplication for each pair)

$$ ={ 1,00,000} $$

Answer: 125 × 40 × 8 × 25 = 1,00,000

Question 2

Calculate these products quickly:

Solution :

Let's find quick ways to calculate , The key to quick multiplication with 25 or 250 is to remember their relationship with 100 or 1000.

(a) 25 × 12 = ______

$$ {25 × 12 } $$

( 25 is equivalent to dividing by 4 and then multiplying by 100 )

$$ ={{100 \over 4 } × 12 } $$

( Rearrange to perform division first)

$$ ={100 × {12 \over 4 } } $$

$$ ={100 × 3} $$

( Multiply by 100 (add two zeros))

$$ ={300 } $$

Answer: 25 × 12 = 300

(b) 25 × 240

$$ {25 × 240 } $$

( 25 is equivalent to dividing by 4 and then multiplying by 100 )

$$ ={{100 \over 4 } × 240 } $$

( Rearrange to perform division first)

$$ ={100 × {240 \over 4 } } $$

$$ ={100 × 60} $$

( Multiply by 100 (add two zeros))

$$= {6000 } $$

Answer: 25 × 240 = 6000

(c) 250 × 120

$$ {250 × 120 } $$

( 250 is equivalent to dividing by 4 and then multiplying by 1000 )

$$ = {{1000 \over 4 } × 120 } $$

( Rearrange to perform division first)

$$ = {1000 × {120 \over 4 } } $$

$$= {1000 × 30} $$

( Multiply by 1000 (add three zeros))

$$ {30,000 } $$

Answer: 250 × 120 = 30,000

(d) 2500 × 12

$$ {2500 × 12 } $$

( 2500 is equivalent to dividing by 4 and then multiplying by 10000 )

$$ = {{10000 \over 4 } × 12 } $$

( Rearrange to perform division first)

$$ = {10000 × {12 \over 4 } } $$

$$= {10000 × 3} $$

( Multiply by 10000 (add four zeros))

$$ ={30,000 } $$

Answer: 2500 × 12 = 30,000

(e) ______ × ______ = 120000000

We need to find two numbers that multiply to 120,000,000

We know possible simple solution

12 × 10,000,000 = 120,000,000.

Or ( 120 × 1,000,000 = 120,000,000)

Let's try to involve 250 or 2500.

Since 120,000,000 is a multiple of 12, we can split it.

Let one number be 2500

$$ { ---- × 2500 = 120000000 } $$

$$ = { 120,000,000 \over 2500 } $$

( 2500 is equivalent to dividing by 4 and then multiplying by 10000 )

$$ {= { 120,000,000 \over 10000 } × 4 } $$

$$ = {12000 × 4 } $$

$$ = { 48000} $$

So second number = 48000

Answer: 48000 × 2500 = 120000000

Question

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.

Solution :

(1) Summary of Evaluation and Extension :

11 × 11 = 121

111 × 111 = 12321

1111 × 1111 = 1234321

11111 × 11111 = 123454321

111111 × 111111 = 12345654321

Observed Pattern Rule: the digits in the result ascend from 1 to a peak, and then descend back to 1.
The peak digit is equal to the number of '1's in the original number being multiplied.

(2) Summary of Evaluation and Extension :

66 × 61 = 4026

666 × 661 = 440226

6666 × 6661 = 44402226

66666 × 66661 = 4444022226

Observed Pattern Rule: The number of '4's in the prefix (before the '0') is (N-1).
The number of '2's in the suffix (before the '6') is (N-1).

(3) Summary of Evaluation and Extension :

3 × 5 = 15

33 × 35 = 1155

333 × 335 = 111555

3333 × 3335 = 11115555

Observed Pattern Rule: The pattern is very consistent: If the first factor has 'N' number of '3's (and the second factor has '(N -1)' number of '3's followed by a '5').the product will consist of 'N' repeated '1's followed by 'N' repeated '5's.

(4) Summary of Evaluation and Extension :

101 × 101 = 10201

102 × 102 = 10404

103 × 103 = 10609

104 × 104 = 10816

Observed Pattern Rule: This pattern is a shortcut for squaring numbers close to 100, derived from the algebraic expansion: ( 100+x )² = 100 ² + ( 2 × 100 × x ) + x²
= 10000 + 200x +x ²

Question 4

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?

Solution :

The minimum number of digits in the product of two numbers is the sum of their individual digit counts minus one.
The maximum number of digits is simply the sum of their individual digit counts.

Case 1: when p = ( m + n ) − 1

( here m = number of digits in the first number)

( here n = number of digits in the secoond number)

(A) Single Digit x Single Digit:

2 × 1 = 2

digits = ( 1 + 1 )- 1 = 1

(B) Two Digits x One Digit :

11 × 2 = 22

digits = ( 2 + 1) - 1 = 2

(C) Two Digits x Two Digits :

10 × 30 = 300

digits = ( 2 + 2) - 1 = 3

Case 2: when p = m + n

( here m = number of digits in the first number)

( here n = number of digits in the secoond number)

(A) Single Digit x Single Digit:

4 × 5 = 20

digits = 1 + 1 = 2

(B) Two Digits x One Digit :

15 × 7 = 105

digits = 2 + 1 = 3

(C) Two Digits x Two Digits :

50 × 60 = 3000

digits = 2 + 2 = 4

Question 5

Roxie says that the product of two 2-digit numbers can only be a 3- or a 4-digit number. Is she correct?

Solution :

Case 1: when p = ( m + n ) − 1

( here m = First 2-digit number)

( here n = Second 2-digit number )

Two Digits x Two Digits :

The smallest 2-digit number is 10.

10 × 10 = 100

digits = ( 2 + 2) - 1 = 3

Case 2: when p = m + n

Two Digits x Two Digits :

The largest 2-digit number is 99.

99 × 99 = 9801

digits = 2 + 2 = 4

Since the smallest product is a 3-digit number (100) and the largest product is a 4-digit number (9801), any product of two 2-digit numbers will fall within this range, meaning it will either be a 3-digit number or a 4-digit number.

Therefore, Roxie is correct.

Question 6

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?

Solution :

No, we don’t need to do all possible multiplications, check extremes: 10×10 and 99×99

The smallest 2-digit number is 10.

10 × 10 = 100

number of digits in product = 3

The largest 2-digit number is 99.

99 × 99 = 9801

number of digits in product = 4

Since the smallest product is a 3-digit number (100) and the largest product is a 4-digit number (9801), any product of two 2-digit numbers will fall within this range, meaning it will either be a 3-digit number or a 4-digit number.

Question 7

Can multiplying a 3-digit number with another 3-digit number give a 4-digit number?

Solution :

No, multiplying a 3-digit number with another 3-digit number cannot give a 4-digit number. It will always result in either a 5-digit or a 6-digit number.

The smallest 3-digit number is 100.

100 × 100 = 10000

number of digits in product = 5

Since the smallest possible product of two 3-digit numbers is 10,000 (a 5-digit number) .

Therefore, it is impossible for the product of two 3-digit numbers to be a 4-digit number.

Question 8

Can multiplying a 4-digit number with a 2-digit number give a 5-digit number?

Solution :

Yes, multiplying a 4-digit number with a 2-digit number can give a 5-digit number.

Smallest 4-digit number: 1000 and Smallest 2-digit number: 10

1000 × 10 = 10000

number of digits in product = 5

So, the product of a 4-digit number and a 2-digit number can give a 5-digit number .

Question 9

Observe the multiplication statements below. Do you notice any patterns? See if this pattern extends for other numbers as well.

Solution :

Question 1

1250 × 380 = _____ is the number of kīrtanas composed by Purandaradāsa according to legends.

Solution :

1250 × 380 = 475,000

According to legends, 475,000 is the number of kīrtanas composed by Purandaradāsa.

Question 2

2100 × 70,000 _____ is the approximate distance in kilometers, between the Earth and the Sun.

Solution :

2100 × 70,000 = = 1,47,00,00,000 km

147,000,000 km. is the approximate distance in kilometers between the Earth and the Sun.

Question 1

6400 × 62,500 _____ is the average number of litres of water the Amazon river discharges into the Atlantic Ocean every second.

Solution :

$$ { 6400 × 62,500 } $$

$$ = { 40,00,00,000 } $$

40,00,00,000 is the average number of litres of water the Amazon river discharges,

Question 2

13,95,000 ÷ 150 ____ is the distance (in kms) of the longest single-train journey in the world.

Solution :

$$ {13,95,000 \over 150 } = 9300 $$

9300 km. is the is the distance of the longest single-train journey.

Question 1

Adult blue whales can weigh more than 10,50,00,000 ÷ 700 ___ kilograms.

Solution :

$$ {10,50,00,000 \over 700 } $$

$$ = { 1,50,000 } $$

Weight = 1,50,000 kilograms.

Question 2

52,00,00,00,000 ÷ 130 ___ was the weight, in tonnes, of global plastic waste generated in the year 2021.

Solution :

$$ {52,00,00,00,000 \over 130 } = 40,00,00,000$$

Weight = 40,00,00,000 tonnes

Question 1

The RMS Titanic ship carried about 2500 passengers. Can the population of Mumbai fit into 5000 such ships?

Solution :

Total population of Mumbai = 1,24,00,000

The total capacity of 5000 such ships

$$ {5000 × 2500 } $$

$$ = { 12,500,000 } $$

Since 12,500,000 (total ship capacity) is greater than 12,400,000 (Mumbai's population)

the population of Mumbai can indeed fit into 5000 such ships.

Question 2

Inspired by this strange question, Roxie wondered,
“If I could travel 100 kilometers every day, could I reach the Moon in 10 years?”.
(The distance between the Earth and the Moon is 3,84,400 km.)

Solution :

Total distance Roxie could travel in 10 years:

Number of days in a year = 365 days

Number of days in 10 years = 10 years × 365 days = 3650 days

Total distance Roxie could travel = 3650 days ×100 km/day = 365,000 km

Distance to the Moon = 3,84,400 km

Since 365,000 km is less than 3,84,400 km, Roxie would not be able to reach the Moon in 10 years traveling 100 kilometers every day.

Question 3

Find out if you can reach the Sun in a lifetime, if you travel 1000 kilometers every day.
(Distance between the Sun and the Earth from page 16 calculation = 147,000,000 km)

Solution :

Distance between the Earth and the Sun: Approximately 147,000,000 km.

To reach the Sun by traveling 1000 kilometers every day:

Number of days to reach the Sun by traveling 1000 km/day

$$ {147,000,000 \over 1000 } $$

$$ = { 1,47,000 } Days $$

Number of years

( Total Days in one year = 365 )

$$ {1,47,000 \over 365 } $$

$$ = { 402.74 } $$

Total time to reach the sun = 403 years approx.

So, it would take approximately 403 years to reach the Sun at that speed. This is much longer than a typical human lifetime.

Question 4

Make necessary reasonable assumptions and answer the questions below:

Solution :

(a) If a single sheet of paper weighs 5 grams, could you lift one lakh sheets of paper together at the same time?

Weight of one sheet = 5 grams

Weight of one lakh sheets = 100,000 sheets × 5 grams/sheet = 500,000 grams

= 500 kg

Since 500 kg is significantly more than 50-70 kg,
a average human with a 50-70 kg lifting capacity could NOT lift one lakh sheets of paper together at the same time.

500 kg, which is too heavy for any person to lift at once.

(b) If 250 babies are born every minute across the world, will a million babies be born in a day?

Babies born per hour: = 250 babies × 60 minutes = 15,000 babies

Babies born per day:= 15,000  babies × 24 hours = 3,60,000 babies

3,60,000 < 1,000,000.

No, a million babies will not be born in a day if 250 babies are born every minute. Based on that rate, approximately 360,000 babies would be born in a day.

(c) Can you count 1 million coins in a day? Assume you can count 1 coin every second.

Total seconds in a day: = 60 seconds × 60 minutes × 24 hours = 86,400 seconds.

Time taken to count 1 coin = 1 second.

In a single day, we can count 86,400 coins.
86,400 < 1,000,000.

Since 86,400 coins is significantly less than 1,000,000 coins, it is not possible to count 1 million coins in a day at that rate.

Question 1

Using all digits from 0-9 exactly once (the first cannot be 0) to create a 10-digit number, write the:

Solution :

(a) Largest multiple of 5

To make the number as large as possible,

-- For the number to be a multiple of 5, its last digit must be 0 or 5.

-- To make the number largest, we want to put the largest available digits at the beginning.

-- The first digit cannot be 0.

Therefore, the largest multiple of 5
that can be created using all digits from 0-9
exactly once (with the first digit not being 0) is 9,876,543,210

(b) Smallest even number

To make the smallest even number,

-- For the number to be even, its last digit must be 0, 2, 4, 6, or 8..

-- To make the number smallest, we want to put the smallest available digits at the beginning.

-- The first digit cannot be 0.

So, 1,023,456,798 this is the smallest possible number because we filled the positions from left to right with the smallest available digits at each step

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 which has the maximum number of letters.

Solution :

To find a 7-digit number (from 1,000,000 to 9,999,999) whose name, when written out in words, uses the maximum number of letters.

Therefore, the 7-digit number name with the maximum number of letters is 77,77,777
The word name is: Seventy-seven lakh seventy-seven thousand seven hundred seventy-seven.It has 60 letters.

Question 3

Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?

Solution :

A 9-digit number where exchanging any two digits results in a bigger number:

The only such number is 123,456,789.
There is only 1 such number.

Question 4

Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.

Solution :

To make the resulting number as large as possible,
we want to ensure that the leftmost digits are the largest possible.
We iterate through the original number, and for each position in our final 10-digit number, we select the largest available digit from the current remaining part of the original number

Therefore, the number 5534512345 is the largest possible.

Question 5

The words ‘zero’ and ‘one’ share letters ‘e’ and ‘o’.
The words ‘one’ and ‘two’ share a letter ‘o’, and the words ‘two’ and ‘three’ also share a letter ‘t’.
How far do you have to count to find two consecutive numbers which do not share an English letter in common?

Solution :

Number 0: {Z, E, R, O}, and 1: {O, N, E} => Shares O, E

1: {O, N, E} , and 2: {T, W, O} => Shares O

2: {T, W, O}, and 3: {T, H, R, E ,E} => Shares T

3: {T, H, R, E ,E} and 4: {F, O, U, R} => Shares R

4: {F, O, U, R} and 5: {F, I, V, E} => Shares F

5: {F, I, V, E} and 6: {S, I, X} => Shares I

6: {S, I, X} and 7: {S, E, V, E, N} => Shares S

7: {S, E, V, E, N} and 8: {E, I, G, H, T} => Shares E

8: {E, I, G, H, T} and 9: {N, I, N, E} => Shares I, E

9: {N, I, N, E} and 10: {T, E, N} => Shares N, E

It shows that all consecutive numbers have at least one common letter.
So, 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.

Solution :

(a) What would the 1000th digit be? At which number would it occur?

Ans : counting the digits contributed by numbers of different lengths.

1-digit numbers (1 to 9): There are 9 numbers, and each contributes 1 digit = 9 digits.

2-digit numbers (10 to 99):

There are 99− 10 + 1 = 90 numbers. Each contributes 2 digits = 90 × 2 = 180 digits.

Total digits from 1 to 99 = 189 digits

Now we need to find which 3-digit number contains the 1000th digit.

We have already used 189 digits for 1- and 2-digit numbers,

The remaining digits we need to count within the 3-digit numbers are
1000 − 189 = 811 digits.

Since each 3-digit number contributes 3 digits,

Number of 3-digit numbers =

$$ {811 \over 3 } = 270 Numbers.$$

The first 3-digit number is 100

the 270th 3-digit number is 100 + 270 − 1 = 369.

Therefore, the 1000th digit will be the first digit of the next number, which is 370.

The number at which the 1000th digit occurs is 370.

The first digit of 370, which is '3', is the 1000th digit.

(b) What number would contain the millionth digit ?

Ans : counting the digits contributed by numbers of different lengths.

1-digit numbers (1 to 9): There are 9 numbers, and each contributes 1 digit = 9 digits.

2-digit numbers (10 to 99):

There are 99− 10 + 1 = 90 numbers. Each contributes 2 digits = 90 × 2 = 180 digits.

3-digit numbers (100 to 999):

There are 999 − 100 + 1 = 900 numbers. Each contributing 3 digits 900 × 3 = 2700 digit

4-digit numbers (1000 to 9999):

There are 9999 − 1000 + 1 = 9000 numbers. Each contributing 4 digits 9000 × 4 = 36000 digit

5-digit numbers (10000 to 99999):

There are 99999 − 10000 + 1 = 90000 numbers. Each contributing 5 digits 90000 × 5 = 450000 digit

Total digits up to 5-digit numbers: 9 + 180 + 2700 + 36000 + 450000 = 488889 digits.

We are looking for the millionth digit.
We have used 488,889 digits up to the end of 5-digit numbers (i.e., up to 99,999).
This means the millionth digit will be part of a 6-digit number.

Number of remaining digits needed: 1,000,000 − 488,889 = 511,111 digits.

These 511,111 digits will come from 6-digit numbers.

Number of 6-digit numbers contributing to these digits:

$$ {511,111 \over 6 } $$

= 85185 with a remainder of 1.

So, the number of complete 6-digit numbers we've gone through is 85,185.

The number that contains the millionth digit will be the (85185 + 1)th 6-digit number.

Starting from 100,000 (the first 6-digit number), we add 85,185 to find the number that contains the end of the 85,185 full 6-digit numbers:

100,000 + 85,185 − 1 = 185,184

The millionth digit is the very next digit.

The next number is 185,185.

The number that would contain the millionth digit is 185185

The millionth digit is the first digit of 185,185, which is '1'.

(c) When would you have written the digit ‘5’ for the 5000th time ?

Ans : Let's count the occurrences of the digit '5':.

Numbers from 1 to 99:

Units place: 5, 15, 25, ..., 95 (10 occurrences)

Tens place: 50, 51, ..., 59 (10 occurrences)

Total '5's: 20 times.

Numbers from 100 to 999 (Three-digit numbers) :

Units place: 105, 115, ….., 995 (90 occurrences)

Tens place: 150-159, 250-259, ……, 950-959 (90 occurrences)

Hundreds place: 500-599(100 occurrences)

Total '5's: 90 + 90 + 100 = 280 times.

Numbers from 1000 to 9999 (Four-digit numbers) :

Units place: 1005, 1015, …, 9995 (900 occurrences)

Tens place: 1050-1059, 1150-1159, …, 9950-9959 (900 occurrences)

Hundreds place: 1500-1599,2500-2599,…, 9500-9599 (900 occurrences)

Thousands place : 5000-5999 (1000 occurrences)

Total '5's: 900 + 900 + 900 + 1000 = 3700 times.

From 1 to 9999, the digit '5' has appeared = 20 + 280 + 3700 = 4000 times

This means the 5000th '5' will occur in a 5-digit number (10000 and above).

We need an additional 5000−4000 = 1000 occurrences of the digit '5'.

Let's count incrementally from 10000.

Count '5's up to 10999:

Units place: 10005,10015,...,10995 (100 times)

Tens place: 1050-1059, 1150-1159, …, 9950-9959 (900 occurrences)

Hundreds place: 10050−10059,...,10950−10959 (100 times)

Thousands place : 10500−10599 (100 times)

Thousands: No '5'

Total in ( 10000−10999 ) : 100 + 100 + 100 = 300 times.

Cumulative '5's: 4000+300=4300. (We need 700 more)

Count '5's up to 11999:

Same as above (300 '5's).

Cumulative '5's: 4300 + 300 = 4600. (We need 400 more)

Count '5's up to 12999:

Same as above (300 '5's).

Cumulative '5's: 4600 + 300 = 4900.(We need 100 more)

We have 4900 '5's up to 12999. We need 100 more

We need the 100th '5' starting from 13000

Count '5's from 13000 onwards:

Units place: 13005,13015,...,13995 (100 times)

The last '5' in the units place within 13000s is 13995. This accounts for 100 '5's.

So, 4900 + 100 = 5000.

Therefore, the 5000th occurrence of the digit '5' is in the 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:

Solution :

(a) 20,800

= (2 × 10,000) + (8 × 100)

Total clicks = 2 + 8 = 10 clicks.

(b) 92,100

= (9 × 10,000) + (21 × 100)

Total clicks = 9 + 21 = 30 clicks.

(c) 1,20,500

= (12 × 10,000) + (5 × 100)

Total clicks = 12 + 5 = 17 clicks.

(d) 65,30,000

= (653 × 10,000)

Total clicks = 653 clicks.

(e) 70,25,700

= (702 × 10,000) + (57 × 100)

Total clicks = 702 + 57 = 759 clicks.

Question 8

How many lakhs make a billion?

Solution :

1 lakh = 1,00,000

1 billion = 1,000,000,000

Number of lakhs making a billion =

$$ {1,000,000,000 \over 1,00,000 } $$

$$ = { 10,000. } $$

So, 10,000 lakhs make a billion.

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

Solution :

(a) largest possible sum

To get the largest possible sum, we want both numbers to be as large as possible. This means we should place the largest available digits in the most significant (leftmost) positions.

For the 7-digit number, place 9 in the millions place, then 8 in the hundred thousands, and so on.

For the 5-digit number, place 9 in the ten thousands place, then 8 in the thousands, and so on.

Largest Possible Sum = 9,876,543 + 98,765 = 9,975,308

(b): Smallest Possible Difference

To get the smallest possible difference, we want the two numbers to be as close to each other in value as possible.

This is achieved by making the smaller number as large as possible and the larger number as small as possible, or more generally, by making the most significant digits of both numbers very close.

For the 7-digit number : We want it to be as small as possible. So, the leading digit should be 1, followed by 2, 3, 4, 5, 6, 7.

For the 5-digit number : We want it to be as large as possible to reduce the gap . So, the leading digit should be 9, followed by 8, 7, 6, 5.

Smallest Possible Difference = 1,234,567 − 98,765 = 1,135,802

Question 10

You are given some number cards: 4000, 13000, 300, 70000, 150000, 20, 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 for making a particular number.

Solution :

(a) 1,10,000: Closest I could make is 4000 × (20 + 5) + 13000 = 1,13,000

Closest Calculation: 70,000 + 13,000 + 20,000 + 6,000 = 1,09,000.

(b) 2,00,000

Closest Calculation: 1,50,000 + ( 70,000 − ( 4000 × 5)).

1,50,000 + 50,000 = 2,00,000

(c) 5,80,000

Closest Calculation: ( 70,000 × 5 )+ 1,50,000 + (4,000 × 20)

= 3,50,000 + 1,50,000 + 80,000

= 5,00,000 + 80,000 = 5,80,000

(d) 12,45,000

Closest Calculation: (1,50,000+70,000+13,000)×5+(4,000×20).

=(2,33,000 × 5) + 80,000

= 11,65,000 + 80,000 = 1245000

(e) 20,90,800

Closest Calculation: ( 1,50,000 − 70,000 ) × ( 20 + 5) + 4,000 + 13,000 + 300.

= 80,000 × 25 + 4,000 + 13,000 + 300

= 2000000 + 4000 + 13000 + 300 = 2017300

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 :

Statue of Unity = 180 metres

Since 1 meter=1000 millimeters, we convert the height:

$$ = {180 × 1000 } $$

Height of Statue of Unity in millimeters = 180,000 mm

The thickness of each coin is given as 1 mm

Number of coins = Total height in mm/Thickness of one coin in mm

$$ {180,000 \over 1 } $$

$$ = { 180,000 Coins. } $$

Therefore, 180,000 coins would need to be stacked to match the height of the Statue of Unity.

Question 12

Grey-headed albatrosses have a roughly 7-feet wide 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 :

We'll calculate the duration for both the lower and upper bounds of their daily travel:

Using the lower end of daily travel (900 km/day):

Number of days = Total distance / Daily distance

$$ {12,000 \over 900 } $$

Number of days = 13.33 days

Using the higher end of daily travel (1000 km/day):

$$ {12,000 \over 1000 } $$

Number of days = 12 days

Therefore, it would approximately take between 12 to 13.3 days for an albatross to complete a 12,000 km journey, such as crossing 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 :

Total distance covered by the bar-tailed godwit = 13,560 km

Total duration of the flight = 11 days

Find the approximate distance covered every day:

Number of days = Total distance / Daily distance

$$ = {13,560 \over 11 } $$

Distance per day = 1232.73 km

Find the approximate distance covered every hour:

Distance per hour = Total distance / Total number of hours

$$ = {13,560 \over { 11 × 24 }} $$

$$ = {13,560 \over 264 } $$

Distance per hour = 51.36 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 :

Height of Somu's building = 40 m

1. Comparing Bald Eagles' Flight Height to Somu's Building

Bald eagles are known to fly as high as 4500 m to 6000 m.

For the lower range 4500 m

Eagle's lower flight height / Somu's building height

$$ = {4500 \over 40 } = 112.5 $$

Bald eagles can fly approximately 112.5 times higher than Somu's building.

For the upper range 6000 m

Eagle's upper flight height / Somu's building height

$$ = {6000 \over 40 } = 150 $$

Bald eagles can fly approximately 150 times higher than Somu's building.

2. Comparing Mount Everest's Height to Somu's Building

Mount Everest is approximately 8850 m high.

Mount Everest's height / Somu's building height

$$ = {8850 \over 40 } = 221.25 $$

Mount Everest's height 221.25 times higher than Somu's building.

3. Comparing Aeroplanes' Flight Height to Somu's Building

Aeroplanes can fly as high as 10,000 m to 12,800 m

For the lower range 10,000 m

Aeroplane's lower flight height / Somu's building height

$$ = {10,000 \over 40 } = 250 $$

Aeroplanes can fly approximately 250 times higher than Somu's building.

For the upper range 12,800 m

Aeroplane's upper flight height / Somu's building height

$$ = {12,800 \over 40 } = 320 $$

Aeroplanes can fly approximately 320 times higher than Somu's building.