NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 1 PDF

NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 1 help students understand the basics of large numbers in a simple and clear way. This chapter covers important concepts like reading and writing large numbers, comparing and ordering them, using commas as per the Indian and International numeral systems, and performing basic operations. These solutions are prepared according to the latest class 7 maths syllabus and are ideal for building a strong foundation in mathematics.

By using these Class 7 Maths Chapter 1 solutions, students can improve their problem-solving skills and get better marks in exams. These NCERT Class 7 Maths Chapter 1 solutions also include step-by-step explanations to make learning easier and more effective.

Class 7 Maths Ganita Prakash Chapter 1 Large Numbers Around Us

Class 7 Maths Chapter 1 Large Numbers Around Us NCERT solutions help students explore the use of large numbers in real-life situations. This chapter introduces learners to reading, writing, and comparing large numbers, including place value, estimation, and operations involving big values. 

These solutions are created to make complex concepts easier to understand through simple explanations and step-by-step methods. With the help of these NCERT solutions, students can strengthen their understanding and improve their accuracy in handling large numbers confidently.

Class 7 Maths Ganita Prakash Chapter 1 Question Answers

Below are the Ganita Prakash Class 7 Chapter 1 Solutions Large Numbers Around Us created to help students understand how to work with big numbers in everyday life.

NCERT In-Text Questions (Pages 1-3)

Q.1. Eshwarappa shared this incident with his daughter, Roxie, and son Estu. Estu was surprised to know that there were about one lakh varieties of rice in this country. He wondered, “One lakh! So far, I have only tasted 3 varieties. If we tried a new variety each day, would we even come close to tasting all the varieties in a lifetime of 100 years?” 

What do you think? Guess.
Solution:
One lakh is represented as 1,00,000 in figures.
If Roxie and Estu try a new variety every day, then they can taste 365 varieties in a year (since there are 365 days in a year).
To taste all 1,00,000 varieties, they would need 1,00,000 + 365 ~ 274 years.

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:

For 3 varieties of rice every day, they would need 1,00,000 ÷ 1095 ~ 91 years.
Here, they would be able to eat all 1 lakh varieties of rice in 100 years.

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:

To get the number of days in y years, we have 365 × y years.
For 1,00,000 days we have 1,00,000 ÷ 365 ~ 273 years.

Thus, we have 365 × y = 365 × 273 ~ 99645 days (closest to 1 lakh)

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:
One lakh = 1,00,000
Now, 1,00,000 – 75,000 = 25,000
Thus, the population of Chintamani in 2011 was 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:
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:
The population increase from 2011 to 2024 = 1,06,000 – 75,000 = 31,000

You may have come across interesting facts like these:

• • The world’s tallest statue is the ‘Statue of Unity’ in Gujarat depicting Sardar Vallabhbhai Patel. Its height is about 180 metres.

• • Kunchikal waterfall in Karnataka is said to drop from a height of about 450 metres.

It is not always easy to get a sense of how big these measurements are. But, we can get a better sense of their size when we compare them with something familiar. Let us see an example.

NCERT In-Text Questions (Page 3)

Q.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:
Somu’s height = 1 m
Height of a floor = 4 × Somu’s height
= 4 × 1 m
= 4 m
Approximate height of the building = Height of 11 floors
= 11 × 4 m
= 44 m

Q.2. Which is taller — The Statue of Unity or this building? How much taller? __________ m
Solution:
The Statue of Unity is 180 m tall, and the building is 44 m tall.
Difference in heights = 180 – 44 = 136 m.
The Statue of Unity is taller than the building.

Q.3. How much taller is the Kunchikal waterfall than Somu’s building? __________ m
Solution:
Kunchikal waterfall is 450 m tall, and the building is 40 m tall.
Difference in heights = 450 – 44 = 406 m
Thus, the Kunchikal waterfall is 406 m taller than the building.

Q.4. How many floors should Somu’s building have to be as high as the waterfall? __________
Solution:
Each floor is 4 m tall. To match the height of the waterfall (450 m), floors needed = 450 ÷ 4 = 112.5.
Thus, building would need approximately 113 floors to be as high as the waterfall.

NCERT In-Text Questions (Pages 4-5)

Q.1. 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

Q.2. Write the corresponding number in the Indian place value system for each of the following:
(a) One lakh twenty-three thousand four hundred fifty-six
(b) Four lakh seven thousand seven hundred four
(c) Fifty lakh five thousand fifty
(d) Ten lakh two hundred thirty-five

Solution:
(a) 1,23,456
(b) 4,07,704
(c) 50,05,050
(d) 10,00,235

1.2 Land of Tens

NCERT In-Text Questions (Pages 5-6)

Question 1. The Thoughtful Thousands only has a + 1000 button. How many times should it be pressed to show:

(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:
(b) 10 times
(c) 53 times
(d) 90 times
(e) 100 times
(f) 1,53,000
(g) 100 thousands

Question 2. The Tedious Tens only has 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) 50 times
(b) 78 times
(c) 100 times
(d) 370 times
(e) 1000 times
(f) 10,000 times
(g) 4350

Question 3. The Handy Hundreds only has 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) 4 times
(b) 37 times
(c) 100 times
(d) 530 times
(e) 900 times
(f) 976 times
(g) 1000 times
(h) 58200
(i) 100 hundreds
(j) 1,000 hundreds
(k) Do it yourself.

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:
Yes.

Question 5. Two of the many different ways to get 5072 are 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:

5 × 1000 + 7 × 10 + 2 × 1

Figure it Out (Page 7)

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.

Solution:
(a) 8300
8 × (+1000) = 8000 (8 clicks)
3 × (+100) = 300 (3 clicks)
Total clicks: 8 + 3 = 11
Expression: (8 × 1000) + (3 × 100) = 8300

(b) 40629
4 × (+10000) = 40000 (4 clicks)
6 × (+100) = 600 (6 clicks)
2 × (+10) = 20 (2 clicks)
9 × (+1) = 9 (9 clicks)
Total clicks: 4 + 6 + 2 + 9 = 21
Expression: (4 × 10000) + (6 × 100) + (2 × 10) + (9 × 1) = 40629

(c) 56354
5 × (+10000) = 50000 (5 clicks)
6 × (+1000) = 6000 (6 clicks)
3 × (+100) = 300 (3 clicks)
5 × (+10) = 50 (5 clicks)
4 × (+1) = 4 (4 clicks)
Total clicks: 5 + 6 + 3 + 5 + 4 = 23
Expression: (5 × 10000) + (6 × 1000) + (3 × 100) + (5 × 10) + (4 × 1) = 56354

(d) 66666
6 × (+10000) = 60000 (6 clicks)
6 × (+1000) = 6000 (6 clicks)
6 × (+100) = 600 (6 clicks)
6 × (+10) = 60 (6 clicks)
6 × (+1) = 6 (6 clicks)
Total clicks: 6 + 6 + 6 + 6 + 6 = 30
Expression: (6 × 10000) + (6 × 1000) + (6 × 100) + (6 × 10) + (6 × 1) = 66666

(e) 367813
3 × (+100000) = 300000 (3 clicks)
6 × (+10000) = 60000 (6 clicks)
7 × (+1000) = 7000 (7 clicks)
8 × (+100) = 800 (8 clicks)
1 × (+10)= 10(1 click)
3 × (+1) = 3 (3 dicks)
Total clicks: 3 + 6 + 7 + 8 + 1 + 3 = 28
Expression: (3 × 100000) + (6 × 10000) + (7 × 1000) + (8 × 100) + (1 × 10) + (3 × 1) = 367813

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 depends on the place value of the digit in the number. If the digit in a particular place is small, then less number of clicks are required and if it is big, then more number of clicks are required.

NCERT In-Text Questions (Pages 8-9)

Q.1. How many zeros does a thousand lakh have?

Solution:
1,000 lakh = 10,00,00,000 (8 zeros)

How many zeros does a hundred thousand have?
Solution:
100 thousand = 1,00,000 (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) Indian System
40.50.678 → 40 lakh 50 thousand and 678
Forty lakh fifty thousand six hundred seventy-eight

American System
4.050.678 → 4 million 50 thousand and 678
Four million fifty thousand six hundred seventy-eight

(b) Indian System
4.81.21.620 → 4 crore 81 lakh 21 thousand and 620
Four crore eighty-one lakh twenty-one thousand six hundred twenty

American System
48,121,620 → 48 million 121 thousand and 620
Forty-eight million one hundred twenty-one thousand six hundred twenty

(c) Indian System
2,0, 22,002 → 2 crore 22 thousand 2
Two crore twenty-two thousand two

American System:
20,022,002 → 20 million 22 thousand 2
Twenty million twenty-two thousand two

(d) Indian System
24,68,13,579 → 24 crore 68 lakh 13 thousand 579
Twenty-four crore sixty-eight lakh thirteen thousand five hundred seventy-nine

American System
246.813.579 → 246 million 813 thousand 579
Two hundred forty-six million eight hundred thirteen thousand five hundred seventy-nine

(e) Indian System:
34,50,00,543 → 34 crore 50 lakh 543
Thirty-four crore fifty lakh five hundred forty-three

American System:
345,000,543 → 345 million 543
Three hundred forty-five million five hundred forty-three

(f) Indian System:
1,020,304,050 → 1 arab 2 crore 3 lakh 4 thousand 50
One arab two crore three lakh four thousand fifty

American System:
1,020,304,050 → 1 billion 20 million 304 thousand 50
One billion twenty million three hundred four thousand 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) 1,001,001,001
(c) 10,20,30,040
(d) 9,080,700,600

Question 3. Compare and write ‘>’, ‘<’ or ‘=’:
(a) 30 thousand __________ 3 lakh
(b) 500 lakh __________ 5 million
(c) 800 thousand __________ 8 million
(d) 640 crore __________ 60 billion

Solution:
(a) 30 thousand < 3 lakh
(b) 500 lakh > 5 million
(c) 800 thousand < 8 million
(d) 640 crore < 60 billion

NCERT In-Text Questions (Pages 11-12)

Question 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) By adding 4,63,128 and 4,19,682, we get 8,82,810 and the estimated sum is 5,00,000 + 4,00,000 = 9,00,000
The exact sum is 8,82,810, which is closer to 9,00,000.
Thus, Estu’s estimate is correct and closer to the actual sum.

(b) The exact sum is 8,82,810, which is clearly greater than 8,50,000.
If we estimate the two numbers (4,63,128 and 4,19,682) to the nearest ten thousands, we get 4,60,000 and 4,20,000.
By adding them, the results will be closer to 8,80,000, which is well above 8,50,000.

(c) The exact sum is 8,82,810, which is less than 8,83,128.
The sum falls short of 8,83,128 by only 318, making it closer to the actual sum.

(d) Exact value of 4,63,128 + 4,19,682 = 8,82,810

Question 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) The difference is: 14,63,128 – 4,90,020 = 9,73,108
The estimated difference = 15,00,000 – 5,00,000 = 10,00,000
But the numbers are rounded off to the highest value.
So, the difference should be less than 10,00,000.
Thus, Roxie’s estimate is closer to the actual difference.

(b) The exact difference is 9,73,108, which is clearly greater than 9,50,000.
If we estimate 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.
By finding the difference, we get 9,70,000, which is more than 9,50,000.

(c) The exact difference is 9,73,108, which is 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 = 9,73,108

NCERT In-Text Questions (Page 19)

Q.1. The RMS Titanic carried about 2500 passengers. Can the population of Mumbai fit into 5000 such ships?

Solution:
Total population of Mumbai = 1,24,00,000
Capacity of 5000 ships = 5000 × 2500 = 1,25,00,000
Yes, Mumbai’s population can fit easily into 5000 ships, as the capacity of the ship is more than the total population of Mumbai.

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.)

Q.2. How far would she have travelled in a year?

Solution:
Distance travelled by Roxie in a year = 100 km/day × 365 days = 36500 km

Q.3. How far would she have travelled in 10 years?

Solution:
Distance travelled by Roxie in 10 years = 100 km/day × 365 days × 10
= 36500 km × 10
= 365000 km

Is it not easier to perform these calculations in stages? You can use this method for all large calculations.
Solution:
Yes, it is easier to perform these calculations in stages.

Q.4. Find out if you can reach the Sun in a lifetime, if you travel 1000 kilometres every day.
(You had written down the distance between the Earth and the Sun in a previous exercise)

Solution:
Distance between the Earth and the Sun = 2100 × 70,000 = 147000000 km
Distance travelled by you = 1000 km/day
Time required = 147000000 km ÷ 1000 km/day
= 147000 days
= 147000 days ÷ 365 days/year
~ 403 years
Not possible, as an average human being has a life expectancy of less than 100 years.

Make necessary reasonable assumptions and answer the questions below:
(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 bom every minute across the world, will a million babies be bom in a day?
(c) Can you count 1 million coins in a day? Assume you can count 1 coin every second.
Solution:
(a) Weight of a single sheet = 5 grams.
Weight of 1 lakh sheets = 1,00,000 × 5 = 5,00,000 g = 500 kg, which is too heavy for any person to lift at once, as it exceeds normal human lifting capacity.
Thus, we couldn’t lift 1 lakh sheets together.

(b) Number of babies born every minute = 250
Total babies born in a day = 250 × 1440 = 3,60,000 babies
[Number of minutes in a day = 1440 minutes]
Thus, a million babies (1,000,000) will not be bom in a single day, as the daily count is 3,60,000 babies.

(c) Time taken to count 1 coin = 1 second.
In a single day, we can count 86,400 coins.
[Total seconds in a day = 24 × 60 × 60 = 86,400 seconds]
Thus, we cannot count 1 million coins in a day at the rate of 1 coin per second, since it would take approximately 1,000,000 ÷ 86,400 ~ 12 days to complete the task.

Figure it Out (Pages 19-21)

Question 1. Using all digits from 0-9 exactly once (the first cannot be 0) to create a 10-digit number, write the
(a) Largest multiple of 5
(b) Smallest even number
Solution:
(a) To form the largest multiple of 5, the last digit must be 0 or 5.
Starting with the biggest digits in decreasing order, the largest multiple of 5 is 9876543210 (10 digits).

(b) To form the smallest even number, the last digit must be even (0, 2, 4, 6, 8), and the digits should be written in ascending order.
The even number is 1023456798 (10 digits).

Question 2. The number 10,30,285 in words is Ten lakhs thirty thousand two hundred eight five, which has 43 letters. Give a 7-digit number that has the maximum number of letters.

Solution:
77,77,777 (Seventy-seven lakhs seventy-seven thousand seven hundred seventy-seven).
This has 61 letters, making it one of the longest 7-digit numbers.

Question 3. Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist

Solution:
To ensure swapping any two digits increases the value, the digits must increase from left to right. So the arrangement would be: 123456789. There is only 1 number that satisfies the given condition.

Question 4. Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.

Solution:
The given number is 12345123451234512345.
After removing the 10 smallest digits from the left, the number we get is 5534512345.

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 7’. How far do you have to count to find two consecutive numbers that do not share an English letter in common?

Solution:
The problem involves finding two consecutive numbers whose English names share no common letters.
Here, zero and one share “e” and “o”.
one (1) and two (2) share “o”.
two (2) and three (3) share “t”.
……………………………..
Nineteen and twenty share: ‘t’, ‘e’, ‘n’
…………….. and so on.
Therefore, there are no consecutive numbers that do not share a 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:
Numbers 1-9 contribute 9 digits (1 digit each).
Numbers 10-99 contribute 90 × 2 = 180 digits (2 digits each).
Numbers 100-999 contribute 900 × 3 = 2700 digits (3 digits each).

(a) To find the 1000th digit:
Digits so far: 9 + 180 = 189.
So, the 1000th digit will lie in the 3-digit numbers range.
Remaining digits: 1000 – 189 = 811.
Number of 3-digit numbers to reach 811 digits:
811 ÷ 3 = 270, with 1 remaining number.
Thus, first we need to write the first 270 3-digit numbers starting from 100.
So, the 270th 3-digit number = 100 + 270 – 1 = 369.
The next number is 370.
Thus, the 1000th digit is the 1st digit of 370, which is 3.

(b) Using the same logic:
Numbers 1-9: 9 digits
Numbers 10-99: 180 digits
Numbers 100-999: 2700 digits
Numbers 1000-9999: 9000 × 4 = 36,000 digits
Numbers 10,000-99,999: 90,000 × 5 = 4,50,000 digits
Numbers 1,00,000-9,99,999: 9,00,000 × 6 = 54,00,000 digits

To reach the millionth digit:
Upto 5-digit numbers: 9 + 180 + 2700 + 36,000 + 4,50,000 = 4,88,889
Remaining digits in the 6-digit range: 1,000,000 (or (10,00,000) – 4,88,889 = 5,11,111
The number of 6-digit numbers required: 5,11,111 ÷ 6 = 85,185, with 1 remaining number.
So, the 85,185th 6-digit number is 85,185 + 1,00,000 – 1 = 1,85,184.
The millionth digit occurs in the number 185184 + 1 = 1,85,185.

(c) Single-digit numbers (1-9): 1 (only 5)
Two-digit numbers (10-99)

• (15, 25, 35,…, 95), totaling 9 occurrences.

• 50, 51, 52, …, 59, totaling 10 occurrences.

Thus, 19 occurrences of the digit 5 in the range 10-99.
Total occurrences so far: 1 + 19 = 20

Three-digit numbers (100-999)
(i) Units position: Numbers like 105, 115, ….., 995 contribute 10 occurrences per 100 numbers. Across 900 numbers, there are 90 occurrences.

(ii) Tens position: Numbers like 150-159, 250-259, ……, 950-959 also contribute 10 occurrences per 100 numbers, and 90 occurrences in all.

(iii) Hundreds position: Numbers like 500-599 contribute 100 occurrences in this range.
Thus, 90 (units) + 90 (tens) + 100 (hundreds) = 280 occurrences
Total occurrences so far: 20 + 280 = 300

Four-digit numbers (1000-9999)
Now it gets more intense! Here, 5 appears in four positions (units, tens, hundreds, thousands):
(i) Units position: Every 10 numbers, e.g., 1005, 1015, …, 9995 = 900 occurrences total.

(ii) Tens position: 1050-1059, 1150-1159, …, 9950-9959. That’s 900 occurrences total.

(iii) Hundreds position: 1500-1599,2500-2599,…, 9500-9599 = 900 occurrences total.

(iv) Thousands position: 5000-5999 = 1000 occurrences
Adding these up: 900 (units) + 900 (tens) + 900 (hundreds) + 1000 (thousands) = 3700 occurrences
Total occurrences so far: 300 + 3700 = 4000
Numbers starting from 10000 onward
For the 5000th number, we require 5000 – 4000 = 1000 more numbers that lie in 10001-10999.

(v) Among 10000-10999, one digit 5 appears in 100 numbers (e.g., 10005, 10015,….., 10995).
The digit 5 appears in 100 numbers (e.g., 10050-10059, …, 10950-10959).
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) 20,800 = 2 × 10,000 + 8 × 100
Number of clicks = 2 + 8 = 10 clicks

(b) 92,100 = 9 × 10,000 + 21 × 100
Number of clicks = 9 + 21 = 30 clicks

(c) 1,20,500 = 12 × 10,000 + 5 × 100
Number of clicks = 12 + 5 = 17 clicks

(d) 65,30,000 = 653 × 10,000 + 0 × 100
Number of clicks = 653 + 0 = 653 clicks

(e) 70,25,700 = 702 × 10,000 + 57 × 100
Number of 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
So, 1,000,000,000 ÷ 1,00,000 = 10,000.
Thus, 10,000 lakhs make a billion.

NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 1 PDF Download

Students looking for easy access to the solutions can get the class 7 math chapter 1 pdf from the link given below. This PDF includes answers to all the questions from Chapter 1, Large Numbers Around Us. 

Download the PDF and strengthen your understanding of large numbers with step-by-step solutions.

Benefits of Using NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 1 

Here are the Benefits of Using NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 1 

• Follows the latest NCERT guidelines ensuring no important topic is missed.

• Helps students understand the method and logic behind each answer.

• Useful for quick revision before exams with concise and accurate answers.

• Reduces mistakes in calculations by guiding through proper methods.

• Strengthens understanding of large numbers, place value, estimation, etc.

• Regular practice with these solutions improves speed and accuracy during exams.