Get Class 7 Maths Chapter 1 solutions for Large Numbers Around Us with easy step-by-step explanations, number-system concepts, estimation, multiplication tricks and solved Figure It Out questions.
Students looking for Class 7 Maths Chapter 1 Exercise Solutions can use this page to understand the problems from Large Numbers Around Us, the first chapter of the current NCERT Class 7 Mathematics textbook, Ganita Prakash Part I.
NCERT’s current 2026–27 reprint identifies Large Numbers Around Us as Chapter 1. The chapter introduces students to large quantities expressed through lakhs, crores, millions and billions and develops skills involving place value, number systems, approximation, estimation and mathematical patterns.
Unlike the older NCERT Class 7 Mathematics book, the current Ganita Prakash book does not organise this chapter around a conventional sequence such as “Exercise 1.1, Exercise 1.2”. Instead, questions are presented through several Figure it Out sections and activities. Current solution indexes identify sections around pages 3, 6–7, 9, 14 and 19–21.
This guide gives independently written explanations for the chapter questions while keeping the focus on understanding rather than simply copying answers.
| Details | Information |
|---|---|
| Class | 7 |
| Subject | Mathematics |
| Textbook | Ganita Prakash |
| Part | Part I |
| Chapter | 1 |
| Chapter Name | Large Numbers Around Us |
| Academic reference | 2026–27 reprint |
| Main topics | Large numbers, place value, number systems, estimation and patterns |
The official NCERT textbook explains that Chapter 1 introduces students to lakhs, crores, millions and billions through real-life contexts.
Students work with:
The first set of questions uses the population of a town to help students understand the size of one lakh. The three questions involve subtraction and finding an increase.
A town has a population of 75,000. How much smaller is this than one lakh?
Population:
[
75,000
]
One lakh:
[
1,00,000
]
[
1,00,000-75,000=25,000
]
75,000 is 25,000 less than one lakh.
One lakh is written as:
[
1,00,000
]
To find how much smaller a number is, subtract it from the larger number.
The estimated population of the town in 2024 is 1,06,000. Find how much greater this is than one lakh.
[
1,06,000-1,00,000=6,000
]
1,06,000 is 6,000 more than one lakh.
Do not confuse 1,06,000 with 1,60,000. The position of each digit matters.
Find the increase in population from 75,000 to 1,06,000.
[
1,06,000-75,000
]
[
=31,000
]
The population increased by 31,000.
For an increase:
[
\text{Increase}=\text{New value}-\text{Old value}
]
This section introduces an interesting idea involving a calculator with buttons that represent place values. It helps students understand how a number can be represented in different ways and why place-value notation is useful.
Represent the given numbers in at least two different ways using place-value-style expressions.
One possible representation is:
[
8\times1000+3\times100
]
[
=8,000+300
]
[
=8,300
]
Another representation is:
[
83\times100
]
[
=8,300
]
[
\boxed{8,300}
]
One representation:
[
4\times10,000+6\times100+2\times10+9
]
[
=40,000+600+20+9
]
[
=40,629
]
Another:
[
406\times100+29
]
[
=40,600+29
]
[
=40,629
]
[
\boxed{40,629}
]
Using place values:
[
5\times10,000+6\times1,000+3\times100+5\times10+4
]
[
=50,000+6,000+300+50+4
]
[
=56,354
]
Another form is:
[
563\times100+5\times10+4
]
[
=56,300+50+4
]
[
=56,354
]
One form:
[
6\times10,000+6\times1,000+6\times100+6\times10+6
]
[
=66,666
]
Another:
[
66\times1,000+6\times100+6\times10+6
]
[
=66,666
]
Using place values:
[
3\times1,00,000
+6\times10,000
+7\times1,000
+8\times100
+1\times10
+3
]
[
=3,67,813
]
Another possible grouping is:
[
30\times10,000+60\times1,000+70\times100+81\times10+3
]
[
=3,67,813
]
A number can be represented in more than one mathematically correct way. Place value tells us what each digit contributes to the total.
The next question asks students to find the smallest number of button presses needed for the numbers above.
The efficient method is to use the largest available place-value button first.
| Number | Minimum clicks |
|---|---|
| 8,300 | 11 |
| 40,629 | 21 |
| 56,354 | 23 |
| 66,666 | 30 |
| 3,67,813 | 28 |
For example:
[
8,300=(8\times1,000)+(3\times100)
]
So:
[
8+3=11
]
Therefore, the minimum is 11 clicks.
The smallest-click expressions look very similar to the usual place-value expansion of a number.
Consider:
[
56,354
]
Its place-value form is:
[
5\times10,000+
6\times1,000+
3\times100+
5\times10+
4
]
The coefficients are exactly the digits of the number.
Therefore, the minimum number of clicks is related to the sum of the digits when each digit is obtained using its largest suitable place-value button.
For example:
[
5+6+3+5+4=23
]
So the number requires 23 minimum clicks under the given button system.
This section compares the Indian and International number systems. The Indian system uses lakh and crore, while the International system uses million and billion.
Indian system:
Forty lakh fifty thousand six hundred seventy-eight
International form:
[
4,050,678
]
Four million fifty thousand six hundred seventy-eight
Indian system:
Four crore eighty-one lakh twenty-one thousand six hundred twenty
International form:
[
48,121,620
]
Forty-eight million one hundred twenty-one thousand six hundred twenty
Indian system:
Two crore twenty-two thousand two
International form:
[
20,022,002
]
Twenty million twenty-two thousand two
Indian system:
Twenty-four crore sixty-eight lakh thirteen thousand five hundred seventy-nine
International form:
[
246,813,579
]
Two hundred forty-six million eight hundred thirteen thousand five hundred seventy-nine
Indian system:
Thirty-four crore fifty lakh five hundred forty-three
International form:
[
345,000,543
]
Three hundred forty-five million five hundred forty-three
Indian system:
One arab two crore three lakh four thousand fifty
International form:
[
1,020,304,050
]
One billion twenty million three hundred four thousand fifty
Indian grouping:
[
3,2,2,2,\ldots
]
International grouping:
[
3,3,3,3,\ldots
]
[
1,01,01,010
]
First convert to the international value:
[
1,001,001,001
]
Indian notation:
[
\boxed{1,00,10,01,001}
]
[
\boxed{10,20,30,040}
]
International value:
[
9,080,700,600
]
Indian notation:
[
\boxed{9,08,07,00,600}
]
When converting a billion-based number into Indian notation, do not simply copy the International comma positions.
[
30,000<3,00,000
]
Therefore:
[
\boxed{30\text{ thousand}<3\text{ lakhs}}
]
[
500\text{ lakhs}=5\text{ crore}=50\text{ million}
]
Therefore:
[
\boxed{500\text{ lakhs}>5\text{ million}}
]
[
800,000<8,000,000
]
Therefore:
[
\boxed{800\text{ thousand}<8\text{ million}}
]
[
640\text{ crore}=6.4\text{ billion}
]
Since:
[
6.4<60
]
we get:
[
\boxed{640\text{ crore}<60\text{ billion}}
]
This section develops quick multiplication methods. The main idea is to rearrange factors so that products such as 100, 1,000 or 1,00,000 appear.
Rearrange the factors:
[
2\times50\times1768
]
[
=100\times1768
]
[
=\boxed{1,76,800}
]
Since:
[
125=\frac{1000}{8}
]
Therefore:
[
72\times125
=72\times\frac{1000}{8}
]
[
=9\times1000
]
[
=\boxed{9,000}
]
Group convenient factors:
[
125\times8=1,000
]
and
[
40\times25=1,000
]
Therefore:
[
1,000\times1,000
]
[
=\boxed{10,00,000}
]
The trick is not a new formula. It is simply rearranging factors to make easy products.
[
25\times12
=25\times4\times3
]
[
=100\times3
]
[
=\boxed{300}
]
[
25\times240
=25\times4\times60
]
[
=100\times60
]
[
=\boxed{6,000}
]
[
250\times120
=250\times4\times30
]
[
=1,000\times30
]
[
=\boxed{30,000}
]
[
2,500\times12
=2,500\times4\times3
]
[
=10,000\times3
]
[
=\boxed{30,000}
]
We need:
[
___\times___=12,00,00,000
]
One convenient answer is:
[
1,200\times1,00,000
]
Therefore:
[
\boxed{1,200\times1,00,000=12,00,00,000}
]
Do not lose track of zeroes when multiplying numbers such as 250, 2,500 and 1,00,000.
The final section contains 14 mixed reasoning questions. These questions are designed to make students apply place value, number formation, estimation and logical reasoning rather than merely perform routine calculations. Public indexes identify these as the final “Figure it Out” questions of Chapter 1.
Use each digit from 0 to 9 exactly once.
A number is divisible by 5 when its last digit is 0 or 5.
To make the number as large as possible, put the digits in descending order and place 0 at the end:
[
\boxed{9,876,543,210}
]
An even number must end in an even digit.
To minimise a 10-digit number, the first digit must be the smallest non-zero digit, namely 1.
Then place 0 next, while reserving a suitable even digit for the final position.
The smallest arrangement is:
[
\boxed{1,023,456,798}
]
The task is to find a seven-digit number whose English number name contains a large number of letters.
One suitable answer is:
[
\boxed{77,77,777}
]
Its number name is:
Seventy-seven lakh seventy-seven thousand seven hundred seventy-seven.
The important idea is to choose words such as seventy and seven, which contain many letters, repeatedly.
[
\boxed{77,77,777}
]
The question asks for a number with a maximum-length name; the focus is on the reasoning behind the construction rather than memorising one particular number.
We need a 9-digit number such that exchanging any two digits makes the number larger.
For this to happen, the digits must increase from left to right.
Consider:
[
1<2<3<4<5<6<7<8<9
]
Therefore:
[
\boxed{123456789}
]
If any two digits are exchanged, a larger digit moves to the left of a smaller digit, making the number larger.
Using all nine distinct digits, there is only one strictly increasing arrangement:
[
\boxed{123456789}
]
So the answer is:
[
\boxed{1}
]
The given 20-digit number is formed by repeating:
[
12345
]
four times.
Ten digits have to be removed while keeping the remaining digits in their original order.
To make the resulting 10-digit number as large as possible, we should choose the largest possible digits as early as possible.
A suitable maximum subsequence is:
[
\boxed{5534512345}
]
The first digit should be as large as possible. By removing suitable earlier digits, we can bring a 5 forward.
Then we maximise the next available digit, and continue this process from left to right.
[
\boxed{5,534,512,345}
]
The question asks whether we can eventually find two consecutive numbers whose English names have no letter in common.
For example:
The pattern continues.
There is no such pair of consecutive number names under the usual English number naming system.
The activity is mainly designed to encourage students to investigate patterns in number names rather than perform a conventional arithmetic calculation.
Suppose all positive integers are written consecutively:
[
1,2,3,4,\ldots,10,11,12,\ldots
]
The question has three parts.
Numbers 1 to 9 contribute:
[
9\times1=9
]
digits.
Numbers 10 to 99 contain 90 numbers.
Therefore:
[
90\times2=180
]
Total so far:
[
9+180=189
]
Digits still required:
[
1000-189=811
]
Each three-digit number contributes 3 digits.
[
811=270\times3+1
]
So 270 complete three-digit numbers take us from 100 through 369.
The next number is:
[
370
]
The remaining one digit is the first digit of 370.
[
\boxed{\text{The 1000th digit is 3, occurring in 370.}}
]
Digits through 99,999:
[
9+180+2,700+36,000+4,50,000
]
[
=4,88,889
]
Remaining:
[
10,00,000-4,88,889
=5,11,111
]
Each six-digit number contributes 6 digits.
[
5,11,111=85,185\times6+1
]
Thus, after 85,185 complete six-digit numbers, the next number contains the millionth digit.
Starting at 1,00,000:
[
1,00,000+85,185=1,85,185
]
[
\boxed{1,85,185}
]
The millionth digit is the first digit of this number.
The number of occurrences of digit 5 from 1 through 9,999 is:
[
4,000
]
So another:
[
5,000-4,000=1,000
]
occurrences are needed.
Counting through the five-digit numbers, the 5000th occurrence is reached when writing:
[
\boxed{13,495}
]
Thus, the 5000th occurrence of the digit 5 is written while writing 13,495.
The calculator has only two useful buttons:
We express each target using these values.
[
2\times10,000+8\times100
]
[
=20,000+800
]
[
=\boxed{20,800}
]
[
9\times10,000+21\times100
]
[
=90,000+2,100
]
[
=\boxed{92,100}
]
[
12\times10,000+5\times100
]
[
=1,20,000+500
]
[
=\boxed{1,20,500}
]
[
653\times10,000
]
[
=\boxed{65,30,000}
]
[
702\times10,000+57\times100
]
[
=70,20,000+5,700
]
[
=\boxed{70,25,700}
]
The coefficients tell us how many times each button needs to be pressed.
We know:
[
1\text{ lakh}=1,00,000
]
and
[
1\text{ billion}=1,00,00,00,000
]
Therefore:
[
\frac{1,00,00,00,000}{1,00,000}
]
[
=10,000
]
[
\boxed{10,000\text{ lakhs}}
]
Two sets of cards numbered 1 to 9 are available.
The objective is to arrange them to achieve the required result.
To maximise the sum, larger digits should occupy the higher place-value positions.
One maximum arrangement is:
[
9,987,654
]
and
[
87,654
]
Adding:
[
9,987,654+87,654
]
[
=\boxed{10,075,308}
]
One arrangement giving the largest sum is:
[
\boxed{9,987,654+87,654}
]
To make the difference small, the seven-digit number should be as small as possible and the five-digit number as large as possible, subject to the available cards.
A suitable arrangement is:
[
1,122,334
]
and
[
99,887
]
Therefore:
[
1,122,334-99,887
]
[
=\boxed{1,022,447}
]
For a maximum sum, put large digits into high-value positions.
For a minimum difference, make the two resulting numbers as close as the digit restrictions allow.
The available cards are:
[
4,000,;13,000,;300,;70,000,;1,50,000,;20,;5
]
Each card can be used at most once for a particular target.
The aim is to construct a number as close as possible to each target.
One useful construction is:
[
4,000\times(20+5)+13,000
]
[
=4,000\times25+13,000
]
[
=1,00,000+13,000
]
[
=1,13,000
]
Difference:
[
1,13,000-1,10,000=3,000
]
A close value is:
[
\boxed{1,13,000}
]
Use:
[
1,50,000+70,000-(4,000\times5)
]
[
=2,20,000-20,000
]
[
=\boxed{2,00,000}
]
This is exact.
Use:
[
70,000\times5+1,50,000+4,000\times20
]
[
=3,50,000+1,50,000+80,000
]
[
=\boxed{5,80,000}
]
Again, the target is obtained exactly.
One close construction is:
[
4,000\times300+70,000-\frac{1,50,000}{5}
]
[
=12,00,000+70,000-30,000
]
[
=\boxed{12,40,000}
]
Difference:
[
12,45,000-12,40,000=5,000
]
Another equally close construction can also be made.
One construction is:
[
(70,000-4,000)\times20
+1,50,000\times5
+13,000+300
]
[
=66,000\times20+7,50,000+13,000+300
]
[
=13,20,000+7,50,000+13,000+300
]
[
=\boxed{20,83,300}
]
Difference:
[
20,90,800-20,83,300=7,500
]
These questions are not only about arithmetic. They test whether you can choose operations that make a target number or get close to it.
Assume each coin is 1 mm thick.
The height is approximately:
[
180\text{ m}
]
Convert metres to millimetres:
[
1\text{ m}=1,000\text{ mm}
]
Therefore:
[
180\text{ m}=180\times1,000
]
[
=1,80,000\text{ mm}
]
Each coin is 1 mm thick.
Therefore:
[
\frac{1,80,000}{1}=1,80,000
]
[
\boxed{1,80,000\text{ coins}}
]
The calculation is a good example of converting units before solving a problem.
The bird can travel approximately 900–1,000 km per day.
Suppose the journey is about:
[
12,000\text{ km}
]
[
12,000\div1,000=12
]
So it would take about:
[
12\text{ days}
]
[
12,000\div900\approx13.33
]
So it would take about:
[
13.3\text{ days}
]
The trip would take approximately 12 to 13.3 days, depending on the bird’s average daily distance.
The bird travelled approximately:
[
13,560\text{ km}
]
in about:
[
11\text{ days}
]
[
13,560\div11
]
[
\approx1,232.73
]
So approximately:
[
\boxed{1,233\text{ km/day}}
]
There are 24 hours in a day.
[
1,232.73\div24
]
[
\approx51.36
]
Approximately:
[
\boxed{1,233\text{ km/day}}
]
and
[
\boxed{51.4\text{ km/hour}}
]
These are approximate values because the journey duration is given as “about 11 days.”
Somu’s building has 10 floors, each 4 metres high.
Therefore:
[
10\times4=40\text{ m}
]
So the building is:
[
\boxed{40\text{ m}}
]
high.
Lower height:
[
4,500\div40=112.5
]
Upper height:
[
6,000\div40=150
]
Therefore, the eagle’s maximum flight height is about:
[
\boxed{112.5\text{ to }150\text{ times the building height}}
]
[
8,850\div40=221.25
]
Therefore:
[
\boxed{221.25\text{ times}}
]
the height of the building.
At 10,000 m: [ 10,000\40=250 ]
At 12,800 m: [ 12,800\div40=320 ]
Therefore:
[
\boxed{250\text{ to }320\text{ times}}
]
the height of the building.
The comparison is based on ratios of heights, rather than the everyday use of the phrase “times bigger.”
The Indian system uses:
For example:
[
5,43,21,678
]
is read as:
Five crore forty-three lakh twenty-one thousand six hundred seventy-eight.
The International system uses:
For example:
[
54,321,678
]
is read as:
Fifty-four million three hundred twenty-one thousand six hundred seventy-eight.
[
1\text{ lakh}=1,00,000
]
[
1\text{ crore}=1,00,00,000
]
[
1\text{ million}=10,00,000
]
[
1\text{ billion}=1,00,00,00,000
]
Therefore:
[
1\text{ billion}=10,000\text{ lakhs}
]
Do not write Indian and International comma patterns interchangeably.
Indian:
[
1,23,45,678
]
International:
[
12,345,678
]
Remember:
[
1\text{ million}=10\text{ lakhs}
]
In:
[
5,43,21,678
]
the digit 5 is in the crore place, while the digit 4 is in the ten-lakh place.
When rounding, look at the digit immediately to the right of the place being rounded.
Large-number multiplication often produces several zeroes. Write the calculation carefully instead of relying entirely on mental arithmetic.
A useful preparation sequence is:
Learn the difference between Indian and International notation.
Convert figures into words and words into figures.
Convert quantities to the same system before deciding which is larger.
Round numbers correctly and judge whether an answer is reasonable.
Look for factors that can produce 100, 1,000 or 10,000.
Do not skip the puzzle-style questions. They are useful for developing mathematical thinking.
A district has a population of 4,86,000 people. After five years, its population becomes 5,32,000.
Find the increase in population and express the result in lakhs.
To find an increase:
[
\text{Increase}=\text{New population}-\text{Old population}
]
[
5,32,000-4,86,000
]
[
=46,000
]
Now:
[
46,000=0.46\text{ lakh}
]
The population increased by:
[
\boxed{46,000}
]
or
[
\boxed{0.46\text{ lakh}}
]
Calculate quickly:
[
125\times32\times8
]
Look for factors that make 1,000.
Group 125 and 8:
[
125\times8=1,000
]
Therefore:
[
125\times32\times8
]
[
=1,000\times32
]
[
=\boxed{32,000}
]
[
\boxed{32,000}
]
The current NCERT Class 7 Mathematics Chapter 1 in Ganita Prakash Part I is called Large Numbers Around Us.
The current Ganita Prakash Chapter 1 is organised through Figure it Out sections and activities rather than the older-style Exercise 1.1 format. The chapter’s questions occur at several points, including pages 3, 6–7, 9, 14 and 19–21.
The chapter covers large numbers, Indian and International number systems, place value, exact and approximate values, rounding, estimation and patterns in multiplication.
[
1\text{ billion}=10,000\text{ lakhs}
]
First understand the Indian and International number systems. Then practise number names, comparisons, estimation, rounding and multiplication patterns. Finally, solve the reasoning-based questions without looking at the solutions.
No. Use solutions to understand the method. After reading an explanation, close the page and solve a similar question independently.
Estimation helps students judge the approximate size of an answer and identify unreasonable calculations without performing lengthy exact calculations.
Since exact SyllabusStudy.in URLs were not supplied, the following are recommended anchor-text targets only:
A strong internal-link structure would be:
Class 7 Maths → Chapter 1 → Chapter 1 Solutions → Chapter 1 Notes → Important Questions → MCQs → Practice Test
This gives students a natural learning path instead of creating many nearly identical keyword pages.
Class 7 Maths Chapter 1 – Large Numbers Around Us builds an important foundation for working confidently with large quantities. Students learn how the Indian and International number systems represent numbers, how to compare lakhs, crores, millions and billions, and how estimation can simplify real-world calculations.
The chapter also goes beyond routine arithmetic. The number-card puzzles, digit-counting problems and multiplication activities encourage students to look for patterns and explain their reasoning.
For the best results, students should first attempt each problem themselves, then use the step-by-step solution to identify mistakes, and finally solve a similar problem independently.
NCERT’s current Ganita Prakash textbook specifically aims to encourage mathematical thinking, exploration, problem-solving and connections with everyday situations.
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